Anurag Mishra Mechanics 1 with www.puucho.com
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
't§l
·' !II Published by:
SHRI BALAJI PUBLICATIONS
(EDUCATIONAL PUBLISHERS & DISTRIBUTORS)
6, Gulshan Vihar, Gali No. 1,
Opp. Mahalaxmi Enclave,
Jansath Road, Muzaffarnagar (U.P.)
Phone: 0131-2660440 (0), 2600503 (R)
website : www.shribalajibooks.com
email : sbjpub@gmail.com
!II First edition
2009
!II Fourth edition
!II Fifth edition
!II Reprint
2012
2013
2014
!II © All rights reserved with author
!II Price : {
428.00
!II Typeset by :
Sun Creation
Muzaffarnagar
!II _printed at :
Dayal Offset Printers
Meerut (U.P.)
!II All the rights reserved .. No part of this publication
may be reproduced, stored in a retrieval system
or transmitted, in any form or by any means,
electronic, mechanical, photocopying, recording
or otherwise, without the prior permission of the
author and publisher. Any violation/breach shall
be taken into legal action.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
,
·
Preface
I have been involved in teaching Physics for last 16 years. This book is an opportunity to
present my experiences. During my interaction with UT-JEE aspirants. I realised that most
feared topic is mechanics. Some of the reasons put forward by students behind this thought
were:
@
No spontaneous thoughts.appear after reading a problem. Mind goes blank. Can not
. proceed in a problem.
@
How to ~roceed in a problem? Which law is applicable; that is a given problem will
involve conservation of energy or momentum or both.
@
If some one says solve the problem in non-inertial reference frame, horrible thoughts
appear inmind.
@
Total confusion about CM frame.
@
Proper understanding of constraints.
@
S)lort cut approach in relative motion.
@
No single book available that gives large no. of solved examples with elaboration of
concepts in asolµtion.
This book will help the students in building analytical and quantitative skills,
addressing keyl,Jlisconceptions and developing.confidence in problem solving.
I sincerely wish that this book will fulfill all the aspirations of the readers. Although
utmost
full care has been taken to make the book free from error but some errors.
.
ina_dvertently may creep-in. Author and Publisher shall be highly obliged if suggestions.
regarding improvement and errors are pointed out by readers. I am indebted Neeraj Ji for
providing me an opportunity to write a book of this magnitude.
-
.
I am indebted to my father Sh. Bhavesh Mishra; my mother Smt. Priyamvada Mishra, my
wife Manjari, my sister Parul, my little kids Vrishank and Ira for giving their valuable time
which I utilized during the writing of this book and people of Morada
bad, who supported.
.
·me throughout my career.
In the last, !also pay my sincere thanks to all the esteemed members ofM/s. ShriBalaji
Publications in bringing out this book in"the present form.
Anurag Mishra.
www.puucho.com
r
Anurag Mishra Mechanics 1 with www.puucho.com
Our Other Publications
for
(Engineering Entrance Examination€)
Er. Anurag Mishra
Er. Anurag Mishra
Mechanics
!•JEE
• Simple Harmonic Motion
• Wave Motion
Electricity&
Magnetism
forJEE
• Solid and Fluids
• Gravitation
Vol. II
• Electrostatics
, •, Electric Current
j•
yapacitors
I
l•
1•
The Magnetic Field
/
and A.G. Circuits
'
Electromagnetic Induction
'I-----~
Er. Anurag Mishra
)feat&
Thermodynamics
/•JEE
I'
.
1 •
Temperature, Heat &
tl)e. equation of State.
Heat Transfer
Optics
• Thermodynamics
.I ,
'
.
: : ·,-:' \ ~
www.puucho.com
. foJ EE
.
Geometrical Optics'
l"
Wave Opti~
Anurag Mishra Mechanics 1 with www.puucho.com
How to face the challenge ?
Following are some doubts which arise in the mind of almost all the students
but may face them by taking some care.
1.
I can not solve numerical because my concepts are not clear. In fact numerical
solvingitselfis an exercise t~ learn concepts.
·z.
1 /can
not study because I am in depression, I fell into it because I was not
'
studying! Depression is escape mechanism of people afraid of facing failures.
Failure is integral partoflearning.
3.
I understand everything in class but can not solve on my own.WRITING work is
· vital. It is a multiple activity, initially idea comes in mind then we put into
language to express it, we are focussed in hand eye coordination, eyes create
visual impression on brain which isrecorded there. WRITING WORKS ARE
EMBOSSED ON BRAIN LIKE CARVINGS OF AJANTACAVES.
4.
In exams my brain goes blank, but I can crack them at home. Home attempt is
your second attempt! you are contemplating about it while home back. You
do not behave differently in exam you replicate your instincts. Once a fast
bowler was bowling no balls. His coach placed a-stump on crease, in fear of
injury he got it right. CONCEPTUALIZATION, WRITING EQUATION, SOLVING,
THEN PROBLEM GETS TO CONCLUSION!
5.
I am an average student. It is a rationalization used by people afraid of hard
work. In their reference frame Newton's first law applies "if I have a
misconception I will continue with it unless pushed by an external agent
even I will surround him in my web of misconception yielding zero
resultant:' AVERAGE IS NOT DUE TO CAPACITY LACUANEBUT DUE TO LACK
OF DETERMINATION TO SHED INERTNESS.
6.
A famous cliche "/ do not have luck in my favour' PRINCIPLE OF CAUSALITY:
CAUSE OF AN EVENT OCCURS IN TIME BEFORE OCCURRENCE OF THAT
EVENT i.e., cause occurs first then event occurs. SHINING OF LUCK IS NOT AN
iNSTANTANEOUS EVENT IT IS PRECEDED BY RELENTLESS HARD WORK.
Sow a seed ofaspiration in mind, water it with passion, dedication it will bear
fruit, luckcan give you sweeter fruit.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
t.
Do not take study as a burden actually )ts a skill like singing and dancing. It
has to be honed by proper devotion and dedication.
·
2.
Withou't strong sen_se of achievement you can't excel. Before entering the
cmppeti.tive field strong counselling by parents is must. Majority do not
· know what for they are here. No strategic planning, they behave like a tail
: ende_r batting in frontofSteyn's bouncers.
3. - 'science is not a subje,ct based on well laid dowh procedures or· based on
learning sonie facts, it ipvolves very intuitive and exploratory approach.
Unless their is desire and passion to learn you can not discover new ideas. It
requires p'atience and hard work, whose fruits may be tangible later on.
4.
Some students realize very late that they are studying for acquiring skills and
, .honingthem. Their is a feeling that they can ride at the back of instructor and
· , achieve ~xcellence. Study comes as tqrtu·rous exercise enforced on them and
their is some mechanism that can take this burden-of them. -
5.
Science is not about gaining good marks, up toXth by reading key points good
marks are achieved but beyond that only those survive who have genuine
interest in learning and exp to ring. Selfstudy habit is must.
6. ·
IF YOU WANT TO GAIN LEAD START EARLY. Majority of successful students
try to finish .major portion-elementary part of syllabus before they enter
Coaching Institute. Due to this their maturity level as comparecl to others is
· more tliey get ample time to adj4st with th~ fast pace. They are less·
traumatized by the scientific matter handed over. For those who enter fresh
must be counselled to not get bullied by ·early starters but work harder
initially within first two months initial edge is neutralized.·
7.
Once a sl:l,ldent lags behind due to scime forced or unforced errors his mind
begins to play rationalization remarks like I am an average student, my mind
is not sharp enough, I have low IQ etc: These words are mechanisms _used to
· a\Toid hard work. These words a,re relative terms a person who has .early start
may be intelligentr~lative t9y6u.' .
.
.
,
'
' :i.
0
lntelligence means _cu~ulative i-~sult of h_ard _work of previous years, that
hard work has eventuallY. led to a developm~nt of instinct tci crack 'things easily.
_., ·
/
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
I
CONTENTS
--
(j UNIT AND DIMENSIONS
(j I. DESCRIPTION OF MOTION
Subject of kinematics (19), Vector notation (20), Displacement
I-
vector (20),Parallelogram law of vector addition (21).,
Component ofvector(23), Unitvector(24), Expressing a vector
in unit vector notati_on (25), Position vector (27), Rectangular
resolution of a vector in three dimensions (28), Vector
multiplication (31), The scalar product of two vectors (32),
The vector product of two vectors (35), Rectilinear motion of
a particle (37), Calculus supplementary (41), Rectilinear
motion (44), Instantaneous velocity ( 45), Integration ( 48),
Interpretation of graphs (53), Average velocity & Average
speed (58), Two-dimensional motion with constant acceleration (69), Projection on an inclined plane
(82), Relative motion (87), Application of advanced concept of relative motion (88), Equation of
motion for relative motion (98), Projection of a particle in an accelerated elevator (101), Projection of
a ball in horizontally moving trolley (101 ), Closest distance of approach between two moving bodies
(102), Problems: Level-1 (106), Level-2(113), Level-3 (117),Answers(123), Solutions (124).
Q
2. FORCE ANALYSIS
The concept of force (138), Reference frame (139), Ideal string
~~~~~p=--=~-=::;~
(143), ideal pulley (143), Contact force (143), Concept of
external and internal force (144), Pulley system (145), Tension
in a hanging rope (145), Constrained motion (148), Pulley
constraint (154), Normal constraint (155), Elastic force of
spring (158), Parallel combination (159), Friction (165), The
laws of sliding friction (165), Direction of kinetic Friction
(171), FBD when arm is in_ deceleration (185), Circular
motion (186), Angular velodty vector (187), Concept of ,
pseudo force (192), Non-inertial reference frame (193),
Whirling rope (195), Lift Force on an airplane (196), Non-uniform circular motion on horizontal plane
{196), Problems: Level-1 (210), Level-2 (221),Level-3 (230),Answers (238), Solutions (240).
1
I
\_
I
______________ ---------------- www.puucho.com
--------··~
'
Anurag Mishra Mechanics 1 with www.puucho.com
--·
,·
-CJ 3. WORK AND ENERGY
Work done (265), Unit of work (266), Conservative and non-conservative
force (268), Concept of potential energy (269), Classical work energy theorem
(270), Conservation of mechanical energy (271), Work done by friction
(271), Work done by spring force (271), Work depends on the frame of
reference (272), Work due to internal force (friction) (272); Work energy
theorem in a non-inertial reference frame (273), How to apply
coriservation of energy_ equation (273), Vertical circ_ular motion (283),
Power (293), lntemai energy so!-lrces & work (296), Problems Level-1
(299), Level-2 (306), Level-3 (310),Answers (316), Solutions (318).
CJ 4. IMPULSE AND MOMENTUM
Impulse (328), Conservation of momentum (328), Conservation
momentum for a two particle system (329), Relative velocity and the
conservation of momentum (330), Recoil disintegration, explosions (335),
Impulsive force (336), Centripetal acceleration revisite~ (338), Centre of
mass (340), Position of COM of two particles (340), Centre of gravity (341 ),
Motion of the centre of mass (34,1), Kinetic energy of a system of particles
(342), Most important concept (343), Finding the centre of mass by
integration (353), Collisions (361), Models for elastic & inelastic
collisions (362), Oblique impact (365), The velocity of the centre of
mass for collisions (370), Elastic collisions in the CM reference frame (371), Inelastic collisions in CM
reference frame (372), System of variable mass; Rocket propulsion (380), Problems Level-1 (383),
Level-2 (395), Level-3 (399), Answers (406), Solutions (408).
·
(J 5. RIGID BODY MOTION
What is rigid body(421), General rigid body motion (421 ), Rotation about centre
of mass (422), Kinematics of fixed axis rotation (422), Vector representation of
rotational quantities (425), .Torque (427), Newton's second law for rotation
(429), Rotational kinetic energy and moment of inertia (430), Rotational
kinetic energy of a collection of particles (431), Perpendicular axis theorem
(437), Dynamics of a rigid body(441), Angular momentum (447), The ladder
(450), Work done due to torque (457), Angular
,, momentum of a projectile
(462), Angular momentum of an i[]verted conical pendulum (462), Angular '=--~='-~=-~=~-impulse0angular momentum theorem (464), Two bodies rotatory system
(466), Kinematics ofrigidbody rotation (475), Total kinetic energy of body (484), Dynamics of rigid body in
plane motion (486), Torque on the rotating skew rod (506), Problem: Level-1 (511); Level-2 (519), Level,3
(525), Answers (531), Solutions (533), Exercisg_,_advanced problems (544), Comprehension based
pro blems (553), Assertion and reason type problems (562).
www.puucho.com
________________
.,..,.,,-/
I
,.I
Anurag Mishra Mechanics 1 with www.puucho.com
i
UNIT AND DIMENSIONS .
Physics is that branch of science in which we observe,
measure and describe natural phenomena related to matter
and energy. Like all the science, physics is ultimately based
on observation. To ~ssemble the relevant observations into a
coherent picture by, constructing a logical framework is
called theory. Theory enables the physicist to account for
past observations and to decide how new ones should be
made. Nearly all physical observation are quantitative; they
require measurement.
(a) Magnitude of Physical Quantity=
Numerical value x Unit
th us for a given physical quantity when the unit will
change, numerical value will also change, e.g. density of
water =lg- cc- 1 = 10 3 kg- m-3 and not lkg- m-3 .
Every measurement is a comparison of a
quantity with a standard quantity that is, an
agreed upon quantity of the same kind. To measure a
Criteria for Standards
The choice of the standard is arbitrary. However, several
criteria must be met if a standard is to be as useful as
possible.
1. Stability : The standard should not vary with time.
If this criterion is satisfied, measurements made at different
time, using the same standard, can be meaningfully
compared.
2. Reproducibility: The standard should be accurately
reproducible so that copies, ideally identical with the
standard itself can be used elsewhere. If this criterion is
satisfied, measurements made at different places can be
compared.
3. Acceptability : The standard should be universally
accepted so as to eliminate clumsy and possibly inaccurate
comparisons among measurement made with separate
standard.
4. Accessibility : As nearly as possible, the standard
should be readily accessible to everyone ·vho needs to use it.
5. Precision : It should be possible to measure the
standard itself with a precision at least as great as the
precision with which any comparable measurement can be
made.
length for example, you adopt as your standard a convenient
measuring rod, whose length you use as the unit of length.
You count the number of times that the rod fits into the
length to be measured. This number given the length in
terms of the chosen unit.
Physical Quantities
The quantities by means of which describe the laws of
physics are called Physical Quantities. A physical quantity is
complete specified if it has ~---e>(A) Numerical value only ratio e.g.,
refractive index, dielectric constant, etc.
or
Magnitude only Scalar e.g.,
mass, charge,current etc.
or
Magnitude and direction Vector e.g.,
displacement, torque, etc.
In expressing the magnitude of a physical quantity we
choose a unit and then find physical quantity how many times
that unit is contained in the given physical quantity, i.e.
(b) Larger the unit smaller will be the magnitude and
vice-versa, e.g., 1 kg= 1000 gm then as 1000 is greater than
1, gm is smaller unit than kg of mass.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
6. Security : The standard should be as safe as
possible from and preferably' immune from all possible·
causes of damage. · . . ·
When a' standard meets these criteria as nearly as.
possible, it can. be·taken as the-primary standard. The 1
primary standard Ca.I\ 'the!J be,yse_d to produce secondary·
standards'calibrateo in terms of the primary, and so on.
· A set of base units, toget:li,(!r with the ~es required to
express·all other units fn tenns·of.them, constitutes a system
of units. The systein,in general use throughout the world is
called _the Interil.ati(!nal System, of Units. The, short
form SI (from the French System International) is used in
'
all languages.
Systems of Units'. ·. ,.
A complete ·set ~f tmits; both.fundamental anc\ derived
for all kinds of phys.ical quantities, is called ·a system of
units. There are· several systems of units· which liave been·
employed for describing measurements. A few common
' ..
systems are give'n ):,elo.;,. : '
A. CGS system
. _
,
The systeni is' also <;ailed· Gaussian system' of.units. In it
length, mass and time have lieen taken as, the fundamental
quantities, and corresponding _fundamental units are ·
centimeter{cm), gram (g) and second (s) respectively. The
unit of force in _this system is dyne while that of work- or
energy is erg. ,
B. FPS system ·
It uses foot, poupd ;md second for the length, n:iass, and
time measurements respectively.' In this system force is a
·unit poundal.
·
·'
derived quantity
c. MKS system
In this system the lf?ngth, mass .and time have ]:,een
taken as the fundamental.quantities, and the corresponding
fundamental miitsaietiie-nietre, kilogtani and' second.,'Fhe
units ofl all ~ther mechanica\ 'quantities like force, work
power, etc., are derived in terms of t:Jiese fundamental uni~.·
For example, ·the unit of force is" thitt force which will
produce and acceleration of .1 m/ s 2 in a body of mass 1 kg
.
'
'
and is called newton. Tlie unit of work or energy is joule,
while, of power is watt: ·
Table:1 Units of Some Physical,
Quantities in Different Systems ·
D. International system of units [SI units] , ,
In 1971 the International Bureau of weight and
measures held its meeting and deci,ded a system of units
which is known.as the international system of units._.If is
abbreviated ·as SI units from .the French name Le 'system·
International d Units and is the extended MKS system
applied to the whole of physics. Now-a-days, most _of the
engineers and physicists use this system of units. Table 2.
gives the fundamental quantities and their SI units.
Besid~s ·-the above seven fundamental units, two
supplementary units are also defined viz, radian (rad) for
plane angle and steradian (sr) for solid angle. '
The SI Unit of Angle ('l"he Radian)
A circular arc is divided into 360 degrees. The degrees
are. subdivided into 60 minutes of arc, ·each of· which
contains 60 secon4s. Angle is measured as' the ratio of arc
length and the radius of the scale (Fig. 1). The measured
angle does not depends on the radiu~ of the scale, sinqi'the
length of the arc is proportional to its radius, In figure the
shaded portion of the cir~le is a fracti~n f of the whole cirde,
where
'
ItI
1-
witJ:i
Force · ; / Y '' dyn~ _':' >~~o.i4N
Workoren~;gy
e~·,._\
[PO~er t. ~,.~ •1 X .,.~rgts'~' .
.::> .
. ';~
.
:".J&tUe/~.J.
·;_..·.w~~-~·~.: ~
. ---- ---- ,'
'
'
.
- - - - - - -
.,
·,
',
--·_"_··f_lg_:~_..,,___ ,;
f = angl!!(degrees) ~
.,
,
.
L
)
.j-
' '
j
,.'
' :,, .
aidength = arc length
circumference 21t x radius
·) ' 360° iirclength ' · . . '
so
ang1e Cd egrees =
---''- ·
21t ; radius ·
. The number (360°/21t) has no physical significance. We
may define a unit of angle by cutting ·a circle into any
number of pieces, we obtain a· particularly conveniel)t unit,
the SI unit of angle or radius (abbreviated rad), by choosing
21t pieces.
An angle ofl radian corresponi:ls to an arc length equal ·
to the radius of the circle.
·
.
arclength S
Angle(radians) = - - " . radius
R
;. _ poiipd;,y ,;"\
ft:-powidal
.· ~'·,:-~
L
I
1
••
ft~P~~~dal/s-"'~
www.puucho.com
360°
Anurag Mishra Mechanics 1 with www.puucho.com
-
-
' UNiT AUD DIMcllSIONS
Concept : How to express an angle of 1° in tenns of
radians.
The arc lertgth c~rresponding to 1' is 1/360 of the
circumference.
Thus S
=
(2rrr)
360
~ = ~ = 1.745 x10-2 rad
rt z
: \ Fundamental
quantity
\
3 , we mqy use 1° z
_!_ rad.
60
Fundamental and Derived Units
Normally each physical quantity requires a unit or
standard for its specification, so its appears that there must
be as many units as there are physical quantities. However, it
is not so, It has been found that if in mechanics we choose
arbitrary unit of ariy three physical quantities, we can
express the units of all other physical quantities in
mechanics in terms of these. Arbitrarily the physical
quantities mass, length and time are chosen for this purpose.
So any unit of mass, length and time in mechanics is called a
fundamental, absolute or base unit. Other units, which can
be expressed in terms of fundamental units, are called
derived units. For example, light-year or km, kg/m are
derived units as these are derived from units of time, mass
and length respectively.
Note:
(i) Only four additional fundamental quantities temperature
current, luminous intensity and amount of substance are
needed to deal all other branches of physics.
(ii) Apart from fundamental and derived units we also
sometimes come across practical units.
(a) These may be of fundamental or derived quantities e.g.,
light-year is a practical (fundamental) unit of distance while
horse-power is a practical (derived) unit of power.
(b)These may or may not belong to a system but can be
expressed in any system of units, e.g.
1 mile= 1.6 km= 1.6 x 103 m =1.6x10 5 cm.
I
Name of unit \symboi \Represen,
.tatlon ,
metre
m
L
2. Mass
kilogram
kg
M
3. Time
second
s
T
A
A
kelvin
K
e (or) K
candela
cd
cd
mole
mo!
Mo!
1. Length
R 108°
The degree is a small fraction of a radian_ For rough
calculation, with
Table-2
I
1° (measured in radians)
=
5. Fullstops are not written after the abbreviations and
units, e.g. 1 litre = 1000 cc (and not c.c.) emf, amu, etc.
4. Electric current
ampere
5. Temperature
6. Luminous intensity
7. Amount of substance
,·-
STANDARDS OF LENGTH, MASS AND TIME
The Unit of Length
Length is the measure of intervals in space. The SJ unit
of length is the meter (s-ymbols m). The name is derived
from the Greek word metron meaning 'measure'. The meter
is now defined to be the distance the light travels, through
.
1.
vacuum, 1n - - - - - s .
299,792,458
Three imponant considerations underlie this definition.
First, the speed of light is now defined to be precisely
299 792 458 meter per second. Should more precise
measurements be made of the speed of light, the effect
would be to change the length of the meter slightly. Second,
length and time can now be measured with comparable
precision. Third, and most important of all, the
speed of light in vacuum is precisely the same for
all observers. This is a fundamental in this theory,
and so strong is the confidence placed in it by
scientists, that the definition of the meter can be
soundly based on the constancy and universality
of the speed of light in vacuum.
We also have some other practical units which are
frequentiy used for small and large lengths. They are :
(a) 1 fermi = 1 fm = 10-13 m
(b) 1 X-ray unit = lXU = 10-13 m
1. Even if a unit is named after a person the unit is not
written with capital initial letter. Thus we write newton (not
(c) 1 angstrom = 1A ~ 10-10 m
Newton) for unit of force.
(d) 1 micron = 1 µm = 10-6m
2. For a unit named after a person the symbol is a capital
letter. Symbols of other units are not written in capital
(e) 1 astronomical unit = 1 AU= 1.49 x 1011 m
letters. For example, N for Newton (and not n) while m for
[Average distance between sun and eanh i.e., radius of
metre (not M).
earth's orbit]
3. The symbols or units are not expressed in plural form.
(t) 1 light-year = 1 ly = 9.46 x 101s m
Thus we write 50 m or 7 erg and not 50 ms or 7 ergs.
[Distance that light travel in 1 year in vacuum]
4. Not more than one solidus is unused. For example, 1
poise should be written as 1 poise = 1 g/s cm or lg s- 1 cm- 1
and not 1 g/s/cm.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
14
-
(g) 1 pars~c =~pc = 3.08 x 10 m = 3.26 light-year
[The distance at which a star subtends an
angle of parallax of 1 sec at an arc of 1 AU]
The Unit of Mass·
Mass is a basic property of matter. The standard kg is
the mass of a platinum - iridium cylinder stored in a special
vault in the International Bureau of standards in severs,
France, th~ accuracy.of this standard is 1 part in 10 8 parts.
To measure mass of atoms or molecules we use the unit
'atomic mass unit' abbreviated as amu (or now u). At
present atomic mass unit is defined as (1/12) th the mass of
an· atom of carbon-12 isotope. Now as the mass of a
carbon-12 atom is ,
12
12
-----=
g
Avogadro's no. 6.02 x 10 23
so 1 amu (or u)
12
= 2_ X ( ,
) = 1.67 X 10-24 g = 1.67 X 10-27 kg
23
12
6.02 X 10
6
_ --- __-- __ . _______ MECHANics.il
The numerical value in this definition of the
second was so chosen as to make the new
standard compatible with the old one. The new
standard is, however, about 1000 times as precise as the old
one. Just as fluctuations in a human pulse rate can be
measured by comparing the pulse with the swing of a
pendulum, fluctuations in the earth's rotation rate (which
determines the length of the day) can be measured in terms
of the period of the microwaves produced in an atomic
clock.
As a result of the redefinition, the day is no longer
exactly 86400s long. This is awkward for astronomers and
others who continue to use the 'mean solar second' , defined
1
as - -- day. To keep the two systems compatible, an extra
86400
'leap second' is added to the mean solar day every few years
as needed, by international agreement.
Atomic standards have advantages other than
precision over arbitrarily constructed standards.
Because all atoms of a given kind are indentical,
The Unit of Time
there is no need to construct and maintain a
We measure a time interval by comparing it with a unit
standard in a central laboratory. We need not
of the same kind a unit of time. The unit of time must be
worry about the possible destruction of the
defined in terms of some physical system that behaves in a
standard and we need not transport secondary
repetitive way. We can use the time interval between
standards to it for checking.
repetitions, called the period, to define the time unit.
Every good physical measurement shares a number of
common features. Suppose, for example, that you mr·asure
When it had become possible to measure and
a sailboat's length as 10 meter infact it is a compari: n of
periods of atomic phenomena far more precisely
the boat's length with that of another object, the metet :ick.
than the periods of larger systems, and to use
Every physical measurement is a comparison
such phenomena in establishing standards. A
of two similar physical quantities.
standard based on a atomic phenomenon is called
Second, we accept the meter stick as a valid device for
an ato~c standard. The atomic standard of time
measuring
the boat's length. At the factory, the meter stick
relies on the fact that an atom emits a specific
was
marked
by a machine, itself adjusted by comparison
kind of electromagnetic 'light' wave when the
'
with
a
standard
length.
arrangements of its electrons undergo a specific
To be valid, a measuring device must be
cha,:ig~ ·called an atomic transition. Like all repetitive
compared
against a widely accepted standard.
waves, electromagnetic waves are periodic. Since 1967, the
Also,
the
procedure must be stable ·so that we know how
second has been defined in terms of a particular atomic
to
compare
different
measurements made at different times.
transition, in which an outer electron of a cesium - 133 atom
Accuracy describes how much a measurement might
'flips' its orientation relative to the atom' nucleus. This flip
differ
from another measurement made with greater care.
causes the atom to emit a wave that has a very sharply
For
good
measurement requirement is adequate precision.
defined period. The device used to measure the period is
The
precision of a measurement is the
called an atomic clock. The clock contains electronic
smallest
amount of the measured quantity that
components that both stimulate and detect the repeated
can reliably be distinguished.
flipping in the many cesium atom within the clock. The
Greater precision requires a more carefully
second is defined as the time required for 9, 192, 631, 770
manufactured device.
periods of the microwaves that stimulate these transitions.
The process of comparing a particular
Like all SI units, the name second has an international
measuring instrument against a standard is called
standard symbol, which is s.
_calibration.
'
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
IUNIT AND DIMENSIONS
· Table 3, Some SI Prefixes and
Multiplication Factors
4.
I
I
F~ai:- \ Prefix, '\Symbol \ ~r~c-, \ Prefix \ Symhe1
<t1on.
_
. ,,
lion
·
I
I
10·'
I
'
10-2
I
I
deci
.centi
1
d
10'
deca
da
,,
C
102
hecto
h
I
kilo
k
I
10-J
milli
m
103
10"'"
micro
µ
10•
mega
M
II
I
10-9
nano
n
109
giga
G
I
I
!
l
10012
pico
p
10''
tera
T
I
f
10'5
peta
p
a
1018
exa
E
I
'''
'
10-JS
10-18
femto
atto
'
Dimensions of a Physical Quantity
If any derived unit depends upon the r th power of the
fundamental unit, it is said to be of r dimensions in that
fundamental unit.
The unit oflength is represented by [L], the unit of mass
by [Ml, unit of time by [Tl, the unit of current by [A], the
unit of temperature by [Kl and that of intensity of
illumination by [CJ. In mechanics the various quantities
depend only on the units of length, mass and time. As an
example
The area of a square of side L·= Lx L = L2
and volume of a cube of side L =.Lx Lx L = L3
Thus, the area and volume are said to be of 2 and 3
dimensions in length respectively. The unit of area which is
the product of two lengths is represented as [L x L] or [L2 ].
Similarly, for volume we can write [L3,].
Distancetravel.led
.
The speed or ve1oc1ty = - - - . - - Time
= [L'J = [L'r'J
IT']
Since area does not depend upon mass and time the
dimensions of M and T are zero. The area is thus completely
represented as [M 0 L2 T0 J.
Dimensional Equation (Formula) of Some Physical
Quantities
1. Area = Length X Length = L1 X L1 = L2 = [M 0 L2 T0 J
2. Volume = Length x Length x Length
= L1
X
L1
X
L1 = L3 = [M 0 L3 :r 0 J
3. Velocity= Distance= [L'J = [MoL'r'J
Time
IT'J
Thus dimensions of unit of velocity are O in mass, 1 in
length and -1 in time.
.
Velocity [i}T,_1 ]
Acceleratmn = - - ~ = =-~~
Time .. [T1 ]
[M 0 L1 r
2
]
Thus dimensions of unit of acceleration are O in mass,
1 in length and -2 in time. ·
5. Force = Mass x Acceleration
= [M1 ][L1r 2 J = [M1 L1T-2 ]
Thus dimensions of unit of force are 1 in mass, 1 in
length and -2 in time.
6. Work (energy ) = Force x Distance
= [M1 L1r 2 ] [L1 ] = [M1 L2 r 2 ]
Thus dimensions of unit of work or energy are 1 in
mass, 2 in length and -2 in time.
7. Power = Work= [M'L'T-'] = [M1 L2 r
Time
IT']
3]
Thus dimensions of unit of power are 1 in mass, 2 in
length and -3 in time.
8. Momentum = Mass x Velocity = [M 1L1 r 1 ]
Thus dimensions of unit of momentum are 1 in mass, ·1
in length and -1 in time. ,
9. Impulse= Force x Time,= [M1 L1 T-2 ]IT1 ]
= [M1 L1 r 1 J
Thus dimensions of unit of impulse are 1 in mass, 1 in
length and -1 in time.
P
10. Pressure,
1 1
2
Force =---~=
[M L r ] [M'L-'T-'J
=-2
·
.
Area
[L ]
Thus dimensions of unit of pressure are l,in mass, -1
in length and -2 in time.
11. Kinetic energy =
.!2·Mass x (Velocity) 2
= [M1 ][L1 r
1 2
]
= [M1 L2 r
2
]
These are the same as those of work.
12. Potential energy = Force x Distance
= [M1 L1 r 2 ][L1 ] = [M1 L2 r 2 ]
These are also the same as those of work.
13. Couple or Torque = Force x Length of arm
= [M1 L1 r 2 ][L1 ] = [M1 L2 T-2 ]
Thus dimensions of unit of couple, or torque are 1 in
mass, 2 in length and -2 in time. ·
.
14. Ang1e = -Arc
- - = -[L'J = No d"1mens1ons
Radius [L1 ]
15. Angular velocity = ~gle = IT- 1 ]
Time
Thus dimensions of unit of angular velocity are -1 in
time
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
15--·
1 ,.,, ._,
,
.:·'
, · An ·
.
gu 1ar acce1eratton
16•
·
.
•"'
'
=
1
= rr- l = rr-•1
= Couple
x Time
= [M1L2T-2]['I"1] = [M'L-2T-']
The dimensions of unit of stress are 1 in mass, -1 in
length and -2 in time.
.
Change in length
19. Stram=--~-~~ [L: l = No dimensions
Original length
[L]
. 8 or o ther T.
. ratios
.
ngonometr1c
20. sm
= -[L']
[L'J
Mass = [M'J = [M1L-3]
Volume [L3]
......~••",.L•'
h = Jj__ = [~'1•;.-2l.= [M1L2F'J,
,: [ r
)
,
_-,
Force '
27. Force constant = - - - - Displacement .
, '
M
[ 1L1F 2_]
= =-------"-" [M1r•1
[L']
28. Surface tension : It is defined as the force per unit
length in the surface of a liquid.
: ·.,.,
1
1
.
Force
[M L T-2]
:. Surface tensmn = - - - - -1- - [M1T-2]
,,
Length.
[L ]
1
29. Temperature = [8 ]
Now-a-days it is taken as fundamental quantity in SI
units and is expressed in kelvin (8).
30. Heat = Energy = [M1r:'r2]
2
]
Energy
31. Specific heat = - - - - - - ' ~ - Mass x Temperature .
2]
·
·
. -
= [ ML r . = [L2T-20-'J
[M1 ][81 ]
.
32. Latent heat ,
= Heat
energy = [M'L2T-2J
Mass
[M1J
Pressure X Volume
33. Gas constant , R = - - - - - - - . ·
Moles x Temperature
[M1L-1T-2frL3]
= '[mol1][8]
[L•r•J
= [M1L2T-20-1mol-1 ]
24. Gravitational constant : The force of attraction
between two masses m1 and m 2 lying a distance r
apart is give~ by F =· G m; m2
r•
where G is the gravitational constant. Thus
Fr 2
G=-m1m2.
2
'"'-r-r".;":
,.,, ,.. JJ"'''-···"MECHANICS-1:j
--~-_c---·---~·..,,,__,
I 2
= [M:] = No dimensions
[Ml
1
23. · Frequency; v =
[T-1]
Timeperiod
11
,,.~-, -;:, •
It has the same dimensions as those of work i.e.,
The, dimensions of unit of density are 1 in mass and ~3
in length.
"fi
.
Mass of body
·
.
22 • Spee, c grav,ty =
,
Mass of equal volume of water
or
·~-.
, V
[M1L2 r
= No dimensions
=
• ,-
Time interval
The dimensions pf unit of angular impulse are 1 in
mass, 2 in length and -1 in· time.
2
_ Force _ [M1L1r J _ ·[M11_,T_21 .
18. ,Stress - - - - ~ - 2- ~ .
· Area . [L ]
21. · Density
.
,:";:i
•. ',
Change in angular v~locity'"".--'-'-'----
. ['I"l]
17. Angular impulse
,,
,,::
2
[G] = [M L r ][L ] = [M""'L3r2]
[M1 ][M1 ]
34. Boltzmann's constant,
Heat energy· · · [M1L2r 2]
k -=
Temperature
[81]
·
[M1L2T-28-1]
These are the same as those of the gas constant ·R
35. Coefficient of thermal conductivity : The total
quantity of heat Q flowing through an .area A of a slab
of thickness d in time t, when the two opposite faces
are at temperatures 8 2 and 8 1 is given by
•
Q=KA.(0 2 .:.0,)t ·
d
25. Young's modulus of elasticity: It is defined as
the ratio of the stress to the.longitudinal strain. Thus
y =Stress= F/A = F.L
Strain
1/L
A. I
[M1L1T-2][L']
or
[Y] = =--___..;=--=[M1 1:1·r2 J
2
1
where K is the co-efficient of thermal conductivity.
Thus,
[L ][L ]
26. Planck's constant : According to Planck's law, the
energy E in a wave of frequency v is given by
E=hv
':"her~ h is the Planck's constant. Hence
Applications of Dimensional Equations
To change from one system of units to the
other : If the measure of a quantity in units u1 is n 1 and
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
that in terms of units u 2 is n 2 , then as the quantity measured
is the same in both ·the systems, we .have
n1u1
= n2u2
6[
,
Suppose a given quantity q has dimensions fl, b and c in
mass [Ml, length [L] and time [Tl respectively, then the
dimensional equation for this quantity is [M"LbT'].
If the fundamental units are M1 , L1 and T1 of the first
system in which the numerical value is n1 , then the quantity
q = n1 [MfL~T{l
[E~l~t~1
~
I
Solution. Dimensions of Y = [M1 L-1 T
2]
=n,[::]'[~:r'[;:r
2
Now,
n2
= 20_x 1011[~]'_ [lcm]-'[lsec]-2
·
1kg .- lm
1sec
1
= 2ox1011 x --x 100.
1000
= 20 X 10 10 Nm-2
n 1 [MfLiT{] = n,[M~L\T~]
~2
3
--,
Find by dimensional ll!ethod the value of.Yin SI units when
Y=20x10 11 dmecm-2 •
_ · _ . ·_. _ _ _ _ _ 1
Similarly, for the second system for which the
fundamental .units are M2,L 2 and T2 and the numerical
value is n 2; we have
.
. '·
a b C
'
q = n,[M2J;2 T2l
Hence,.
1]1[ ]-'[1]-2•
n 2 =1.013x10 - - -11000 100
1
2
6
3
n2 = 1.013 X10 x 10- x 10 = 1.013 x 10 5 Nm-•
=n,[::r[~:r[;:]'
1- -= --
. . . .,-:..::::~~
~~_g~m;,~
4 ~ '
/dsin,i,ilime~ions find the value of'g' in'MKS system. TheJ
\Value_iri;cgs syst~98{). _ " •:,'. ' C
:
'
Scil!,ltic:m: Her~ n1 = 980,. n 2 = ?
L1 = 1cm, i. 2 = lm = 100cm
.
T1 = 1sec, T2 = 1sec
Dirirensions .of acceleration are L1 T-2.
"'"."'a": ~--~r::m:r[;J
0
1
'
= 98o[M
M2
2
J [__!__]
[I:]100 1
1
= 9.8
Hence the value ofg in MKS system is 9.8 m sec-•.
b~~~~.t~~;f21)a>
C
. . .
------.
of. light, acce/eration due. to gia_vity· and normal
atmospheric pressure are taken asf ~~~'fun~amental units,
what will 'be t1ie units of. mass, length 'and time ·?-·Given
velocity of.light 3 X 10 8 ms-1, fl!'Celeratfon
to gravity j
10 ms-2 and normal p.ressure -·1Q_5 Nm~2 .:
_ _J
'if velocity
uue.
Solution.
Velocity= [L1T 1 ] = 3 x 10 8 ms-1
•.. (1)
Acceleration= [L1T 2] = l0ms-2
... (2)
Pressure= [M1 L-1 T 2] = 10 5 Nm~2
... (3)
Dividing, eqn. (1) by (2), we get
3 108
T1 = x
= 3x 107 sec
... (4)
10
Multiplying, eqn. (1) and (4), we get
11 = 3 X10 8 X 3 X107 = 9 X10 15 m
From eqn. (3), we get
,----...
iConvert a pressure of.76 cm of. me~cury into.Nm - 2 • Density of.:J
imercl!!)'. is 13.q_g,111.'._g:. __·___- - ~ - - - - - - - - - - ' -
M'=
~
Solution. Pressure P = hpg
= -76Xl3.6X980
= 1.013 x 106 dyne/cm 2
Now
n1 = 1.013 x 10 6 n 2 =?
M1 = lg
1 1 ·= 1cm
T1 = 1sec
Dimensions of pressure =
M2 = 1kg = 1000 g
L2 = lm = 100cm
T2 = 1sec
[M1L-1T 2]
105
[L-1r21
.
]
F
= 10 5 x 9x.1015 x 9x 1014
,.
=8lxl0 34 kg
~
-------------~
(E~F!l'7d
5
ck the accuracy of. the equati~ii · ..
'
1' (ii
1
'·,
n= 21,Vrif
·
·)
!where l is.the length of.the string, m. iis mqss p~r unit length,
~he stretching force and n the frequencyyf.yibration. ____ '
www.puucho.com
Fj
Anurag Mishra Mechanics 1 with www.puucho.com
\8
MECHANICS-f 1
Solution. Dimensions of left hand side n =
.! = [ r1 ]
T
or
t=kff
I
[11 ]
M 1L-1
The constant k can be found out experimentally. It
comes out to be equal to 21t.
2
2
Dimensions of right hand side= __!_[M'L'T-
]
= [L-1 ][11 T-1 ] = [ r 1 ]
t
AB the dimensions of left and right hand sides are equal,
the relation is correct.
~~Btm,l?J~J
1 · -··- ·-
··-·-
6 ~-
-~-- --
-.-----------1
-·-
:check the a~c~racy of the relation
.
----r--· --··;
s=ut + -2 at
2
where s is)
.
w/
;the distance travelled, by the body th. unifonn acceleration a:
,in time. (and having ,initial velocity u. . . ___ · . · _ .. · __ J ·
Solution.
1
Dimensions of left hand side = [1
]
Dimensions of right hand side ut = [L1 r 1][T1 ] = [11 ]
_!,at 2 = [L1 r 2][T 2] = [L1]
2
AB the dimensions of each term on the right hand side
are equal to the dimensions of the term on the left hand
side, the relation is correct.
l2E~~~·Gl>
---- ·-·--,
''Deduce the relation for the time period ofa simple pendulum.J
Solution. The time period 'of a simple pendulum can
possibly depend upon :
(i) The length I of the pendulum,
(ii) The acceleration due to gravity g,
(iii) The mass of the bob m and
(iv) The angular amplitude 0.
Since the circular measure of an angle has no
dimensions, let the time period t be proportional to a th
power of 1, b th power of g and c th power of m, then
t = kl"gbm'
... (1)
where k is the dimensionless constant of
proportionality.
Substituting the dimensions of the various quantities,
we have the dimensional equations of both sides.
M0 L0 T1 = [L"(L1 r 2 )bM'] = [La+bT-2bM']
According to principle of homogeneity by comparing
the dimensions on either side, we have
a+ b = 0, c = 0 and -2b = 1
1
or
= k11/2g-1/2mo
27tff
~~~~~~
r;;:riv; dimensionally the relatio~
I. ...... -·---------·---
!_;
2 • ·--- --- ·--:
S = ut +
- - - - - _;;/____ ··- ___,. ___,
Solution. The distance s travelled by the particle
depends upon its initial velocity u, acceleration! and time t.
Let
s =· ku" fbt'
Substituting the di~ensions of various quantities, we
have
[L1 ] = [L1 r 1]"[L1 T-2 ]b [T1 ]' = [La+b][T-a-2b+,J
Comparing the dimensions of similar quantities, we
have
a+b = l; -a-2b+_c = 0
Since there are three unknown quantities a, b and c,
two equations are not sufficient to find their valµes. To solve
it the problem can be split into two parts :
(1) When the particle has.no acceleration. In such a case
s = k1uatc
[11 ] = [L11"1 J"[T1 ]' = [L"TC-a+cl]
or
a=l,. -a+c=O
or
a=c=l
Hence,
s = k1ut
(2) When the particle has no initial velocity. In such
a case
s = k 2 Jbt'
1
or
[L ] = [L1 r 2 Jb[T 1 J' = [LbTC-2 b+cJ]
or
Hence,
b = l, -2b + C = 0
c=2b=2
When the body has both an initial velocity as well as
acceleration, the equation of its motion must contain both
the parts.
s = k 1ut + k 2 ft 2
The value of k1 comes out to be equal to unity and k 2
1
equal to-.
2
Hence,
1
b=--anda=+2
2
Substituting values in eqn. (1), we have
t
=
www.puucho.com
1 2
S=Ut+-ft
2
Anurag Mishra Mechanics 1 with www.puucho.com
: UNIT AND DIMENSIONS
Substituting the dimensions of all physical quantities.
[ML2 T-2 ]
[h] ;
If a composite physical quantity in tenns of moment of inertia
I, force F, velocity v, work W and length L is defined as,
Q ; (IFv 2 I WL3 ),
find_the dimen_sions _of Q and identify i(.
Solution. As, [I]; [ML2J,
[FJ=[MLr2J
and [W]; [ML2T-2]
[ML2][MLr2J[Lr1 ] 2
[Q]; - - ~ - - - [MT-2 ]
[ML2r 2] [L] 3
[v]; [Lr 1 ]
As [MT-2] are dimensions of surface tension, force
constant or surface energy, i.e., energy per unit area, the
physical quantity may be any one of these.
Note:
From this problem it is evident that if dimensions are given,
the physical quantity may or may not be unique.
TO FIND DIMENSIONS OF PHYSICAL
CONSTANTS OR COEFFICIENTS
Write any formula or equation incorporating the given
physical constant and then substitute the dimensional
formulae of all other quantities to find the dimensions of the
required constant or coefficient,
or
-
From Newton's law of
From the relation between
gravitation, we have
G and 'g' we have
F
G;
or
GM
g;F?
; Gm1m2
T
Fr
or
m1m2
gR2
G;M
'(ii) There are also physical constants and coefficients which
;are dimensionless. For example, mechanical equivalent of
'heaJ J.
CONVERSION OF UNITS
This is based on the fact that for a given physical
quantity. Numerical value x Unit = Constant
So when the unit changes, numerical value will also
change.
A. The Newton into Dyne
The newton is the SI unit of force and has dimensional
formula [MLT-2] ;
1 newton= 1 kg m/s 2
so
or
[G]
= [M-1 L3 r 2 J
So its SI unit is m 3/kg-s 2 or Nm 2 / kg 2
B. Planck's constant h
Method-I
Method-II
According to Planck :
de-Broglie :
E ;hv
,. ; _!:_
mv
or
E
h;v
1kg; 10 3g and lm; 10 2 cm
but
lN; (103g)(l02cm)
10s gem; lOsdyne
s2
s2
B. Gravitational Constant G from CGS to MKS
System
The value of G in CGS system is 6.67 x 10-8 CGS units
while its dimensional formula is [M- 1 L3T-2];
so
G;6.67x10-8 cm 3/g-s 2
but
Substituting the dimensions of all physical quantities.
[MLr2] [L2]
[Lr2J[L2]
[MJ[M]
=
[M]
[GJ ;
-
Concept (i) From examples (A) and (B) it is clear that
'if a physical quantity is given, its dimensions are unique.
so
Method-II
_c
[r'J
[h] ; [ML2 T- 1 J
So SI unit or Planck's constant is kg-m 2/s which can also
be written as (kg - m 2/ s2 ) x s. But as kg- m 2/s 2 is joule, so
unit of h is joule x sec, i.e., J-s.
A. Gravitational Constant G
Method-I
c....__
so
1cm; 10-2 m and lg= 10-3 kg
G; 6.67 x 10-8
2
(l0" m)3
(10-3 kg) (s 2 )
; 6.67 x 10-11 m 3 /kg - s 2
G ; 6.67 x 10-11 MKS units (or SI units)
i.e.
C. Density from a given System to a New System
Suppose we have a new system of units in which unit of
length is 5 cm and unit of mass 20 g, i.e.,
5 cm= 1 La (say) and 20 g = 1 Ma (say)
then density of a substance, which is (say) 8 glee, in this
new system will be,
_g_; [lMa / 20] SO Ma
8
8
cm 3
[lLa / 5] 3
La 3
i.e., in this new system the value of density will be 50
units.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
[10
. ·~=-c":C=:-c--==-=MECHANICS-II ·
-TO
CHECK THEDINIENSIONAL····---'~'--A-.-E-in_s_t-ei-n=M_a_s_s_-E_n_e-rg~y~R'e~l-a-ti-on-~-,"'---~~
:=::-===c:~
·.
•..
CORRECTNESS OF A GIVEN PHYSICAL
RELATION
'Principle of homogeneity' states that the
dimensions of each term on both sides of an equation must
be the same. Mass can be added to mass· to give mass and
not to length or time.
If the dimensions of each term on both sides are same,
the equation is dimensionally 'correct, otherwise not, A
dimensionally correct equation m·ay ·or may not be
physically correct.
A. Check the Correctness of the Formula
2
F = mv /r 2
Dimensionally,
[MLT-21= [Ml [Lr 1 l 2/[Ll 2
i.e.,
= [Mr2J,
As in the above equation dimensions of both· sides are
not same ; this formula is not correct dimensionally, so can
never be correct physically.
B. Check the Correctness of the Formula
S=Ut-(½)at
2
.
Dimensionally,
.[Ll = [Lr1 l[Tl-[Lr2l[T 2l
i.e.,
[Ll = [Ll - [Ll
As in the above equation dimensions of each term on
both sides a_re same, so . this .equation. is dimensionally
correct.
.
However, from equation of motion· we know that
s = ut + (1/2)at 2 .
.
'
So the given equation is phys/~ally wr_ong thoU:gh it is
correct dimensionally.
..
C. Check the Correctness of the Formula
T = 21t.,fI/mgL
Dimensionally,
[T]
I
,,
If it is known that when mass is convened into energy.
Let the energy produced depend on the mass (m) and speed
of light (c), and the function to be product of power
functions of m and c, i.e.
E =Kmxcy
Where K is a dimensionless constant. If the above
relation is dimensionally correct.
[ML2 r 2 1 = [M]"[Lr 1 ]Y
or
[ML2 r
2
l = [MxI!rYl
Equating the exponents of similar quantities on both
sides of the equation
x=landy=2
Thus the required physical ,relation becomes
E =Kmc 2
The value of dimensionless constant is found unity
through experiments
E = mc 2
B. Stokes' Law
When a small sphere moves at low speed though a fluid,
the viscous force F, opposing the motion, is found
experimentally to depend on the radius r, the velocity of the
sphere v and the viscosity Tl of the fluid.
·
If the function is product of power functions of 11; r and
v,
F = KTtxryvz
... (1)
where K is dimensionless constant. If the given relation
is dimensionally correct.
[MLT-~] = [ML-1 r 1 ]"[L]Y[Lr1 ]z
or
Equating the exponents of similar quantities on both
sides of the equation.
x=I;-x+y+z =1 and -x-z =-2
Solving these for x, y and z, we get x = y = z = 1
So eqn. (1) becomes
[ML2l
= [Tl
[Ml[Lr2 l[L]
As in the above equation the dimensions of both sides
are same, the given formula is dimensionally correct. It may
or may not be physically correct. However,.from the theory
of physically pendulum we know that T = 21t.jI I mgL. So
the given formula is both dimensionally and physically
correct.
AS A MATHEMATICAL TOOL TO
DERIVE NEW RELATIONS
F=Kwv
On experimental grounds, K = 6Jt;
so
F = mtTtrv
C. Planck's Length
=·
Construct a new physical quantity hi~ng-dimensions of
length in terms of universal constants G, c and h
If the function is the product of power functions of G, c
and h,
... (1)
The principle of homogeneity of dimensional analysis
provides us with a powerful tool to discover new laws
relating different physical quantities.
Following examples will illustrate the method :
where K is a dimensionless constant proportionality.
If the above relation is dimensionally correct,
[LJ.=[M-113 2]"[Lr1 JY[ML2 r 1 Y
i.e.,
[Ll = [M"x+zL3>c+y+2zT-2x-y-zl
www.puucho.com
-r-
Anurag Mishra Mechanics 1 with www.puucho.com
·-~!U_Nf_TAN_D_Df_M_EN_SI_ON_S_ _ _ _ _ _ _ _ _ _ _ _ _ _-'--_
Equating the exponents of sl~ilar quantities on both
Sides Of the equation, ' . ,
'i I
,
-x+z =·o, 3x·+ y + 2z·=·l 'and -ix·-y-z = 0
Solving ·these for x, y and z, we get
1
·, .:.3
1
and Z=x=2; Y =2
0
2
So eqn .. (1) becomes,
QL
=KG112c-,12h v2
If the constant K is assumed, to be unity
. QL =~Gh/c 3,'
_;__-'----'·-:,.,.~,:~·c,;..c\"----~----·''·i_j]
if dimensions· are g!ven. ·For example,._if.)he di111ensional
formula of a physical quantity is'[ML2T-2l, it may be"work or
'
.
energy or torque.
- ,., . , ,
.
(2) Numerical constant [Kl having no dimensions such
as (1/2), 1 or 21t, etc., cannot be expressed by,.the niethods
.,.of dimensions.
·
·
(3) The method of dimensions cannot used. to derive
relations involving produci:,of physical qua~tities.,'it-cannot
be used to' derive relations other )hari product of power
functions. For example,
·
' .
,, ,
s = ut + (1/2) at. 2 , Qr·... y =.asincot '
Cannot be derived by USlng- thi~, th~Ory.: , ' , I ,
.
0
•
L.!=~"A-t;ll\RJ~,.@>
fij ::e~city,. f~;c~ ~d- time are tizl~n to be fundamental
tquantities
find dimensional formula for (a) Mass, and
·
~(l,)_Efl~r~--_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _.,
Solution Let the quantity be Q, then
· Q = f (v,F,T)
Assuming that the function is the product of power
functions of v, F and T,
·
Q =K vxpyyz
.... (1)
where K is a dimensionless constant of proportionality.
The above equation dimensionally becomes.
[Q] = [LT- 1 Y[MLT-2]Y[T]'
i.e.
.. .. (2)
We can check the dimensional correctness of these
relations.
, .
,
(4) Dimensioni,J analysis is 'not ~seful for· deriving
formula for a physical quantity that depends·on more than 3
physical quantities as then ,there will be less number of
equations than the unknowns. However,. still we can check
correctness of the given equation dimensionally.
For example, T =21t~I I mgL cannot ·b~ derived _by
theory of dimensions but its dimensional correctness can be
checked.
(5) Even if a physical quantity depends on 3 physical
quantities, out of which two have same dimensions, the
formula cannot be deriv~d liy theory of dimensions, e.g.'
· formula for the frequency of a tuning fork f =.(d/L2 )v
cannot be derived by· dimensional analysis.
'
Now
i.e., [Ql
Q = Mass
So eqn. (2) becomes
[Ml = MyLx+yT-x- 2y+zl
(a)
= [Ml
its dimensional correctness requires.
y = 1,x+y = Oand-x-2y +z = 0
which on solving yields
x=-1,y=landz=l
Substituting it in eqn. (1), we get
Q=Kv-1Ff,
(b)
'
PHYSICAL QUANTITIES FROM ., .
. HEAT AN!;> TH~,R~ODYNAMICS
Q = Energy i.e. [Q] = [ML2T-2l
So eqn. (2) becomes
[ML2T-2l = [MYLx+yrx-2y+zl
Which· is the light of principle of homogeneity yields
y = l;x+y = 2and-x-2y +z = -2
which on solving yields
1. Temperature: It is _a 'fundamental quantity with
dimensions [0l and unit kelvin '[Kl.
2. Heat: By definition," it is energy transferred due to
energy transferred, so its dimensions' ate [ML2T-2l arid SI
unit joule (J). ~racticai unit of heat .is calorie (cal} and 1
calorie = 4.18 joule.
· .
· .
3. Coefficient of lin~ar-Expansion a
'
So
Le.
Q=KvFf
Limitations of Theory of Dimensions
(1) If dimensions ate given, many physical quantities
have same dimensions. Physical quantity may not be unique
Af.
4. Specific heat c
As
X=y=z=l
So eqn. (1) becomes
'
It is defined as a= - ' .
L,i0
Le.,
'[al =[B-1 l . '
So its unit is (C 0 i-1 or K~J'
Q= mc,iB;
Q'
c=--mi\0·. -:
. [ML2 F 2 l
[cl= c......,~~
'
"[M][0l
[cl= [L2 r 20-1 l
So its SI unit -will be . J/kg:~. "".hile· practical unit
i.e.
cal/g-co.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
12
[p +
5. Latent Heat L
By definition, Q = mL ,
[L] = [ML2T-2] / [M]
i.e.,
i.e.,
or
. [L] = [L2r2]
[ML2 r
2
As this equation is dimensionally correct, each term on
either side will have same dimensions i.e.,
[a/V] = [PV]
or
[a] = [ML-1r2J [L3] [L3] = [MLsr2]
and
or
/T]
[P x b]
[K]=--~2
11 _~
[L ][0/L]
[Kl= [MLr30-11
Its SI unit is W/m-k while practical unit is caVs-cm-C'.
7, Mechanical equivalent of heat J
According to Ist law of thermodynamics work and heat
are related as
l~rt.
is es_ timat·e·d· ;h.at p-;;;;;_.in.~te ~~c-h ~~--_i_~i~~rth.: -~_;_e_"iv;;:;j
about 2 c°;lorie of heat energy from the sun. This constaizr ~
1
called solar constant S. Fxpress solar constant in SI units._;_!
Solution. Given that
S = 2cal/ cm 2 -min
W=JH
But as 1 cal= 4.18J, 1 cm= 10-2 m'and 1 min= 60 s
w
or
J=-
S=
H
= [ML2 r 2 ] '
2
or
i.e.,
[J]
- [J]
·
[ML
2
r
or
]
S=l.4k-W/m 2
~~-----[i21"--,-;---.
3
r----------·· -- ---------------·
I
iflnd_@IJ)~nsions_of a.,. where P~= pressure, t
So
[R]
[ML-1r 2][L3]
=- - - ~ -
[mo!] [0]
or
[Rl = [ML2r 20-1mol-1J
So its SI unit is J/mol-K while practical unit is
cal/mol-K. It is a universal constant with value 8.31 J/mol-K
or 2 cal-K.
10. van der Waals constants a and b
According to gas equation, for one mole of a real gas
__J
[a.t 2 ]
[a.]T2
= [M0 L0T 0 ]
= [MOLOTO]
[a.J = r2
= [ML2r 20-11
So its SI unit is J/K and its practical value
1.38 x 10-23 J / K
9. Gas constant R
According to gas equation, for perfect gas,
PV=µRT
= lime.~_
Solution. Exponential and trigonometric function are
dinlensionless.
2
i.e.
~
Ip= P0 e-cot2J
E =-kt
2 2
[kl= [ML r ]
[0]
2x4.18
-1. 4 x 10 3_J_
c10-2 mJ2(60s)
m 2s
r
.
.d:E,xa•:"u.;.,i;e:
12.
t.-·--··--~:\ii.~-:..=.:.~.-J
= [M 0 L0 T 0 ]
i.e., J has no dimensions. Its practical unit is J/cal
and has value 4.18 J/cal.
8. Boltzmann constant k
According to kinetic theory of gases, energy of a gas
molecule is given by
-Le·
= [PV]
= [L3 ]
[bl= [V]
.
i.e.,
= RT
a
ab
PV+--Pb--=RT
V
V2
or
So its SI unit will be J/kg while practical unit caVg.
6. Coefficient of thermal conductivity K
According to law of thermal conductivity, Heat
transferred per second
dQ =KA dB
dt
dx
Va2 ] (V - b)
f-=Exam,,.,.il,e
- -- --"-""' ___ ;! l j~
~
_____ ___ ------,.
2
J
',
a
.
a-ct
The dimension of b in the equatwn P_= _b_x_ whe-re
P = pr:_es_sur_e,_~ =_g~p_lacement aml L.=.. time _ _ __
Solution. [Pl
=[
b:]-[ c::]
By principle of Homogeneity,
[P] =
[b:] = [ c::]
[i]i=[ML-1T-2]
[i]=[Mr2]
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
13
(uNIT AND _DIMENSIONS
Method-II
r=-
.
.
----- -
= p•pb
V = kP"pb:
V
---,
The position of a particle at time t, is given by the equation, I
x(t) = v; (1-e.-ac ), where v 0 is~ constant and
_dimensions ofv 0 • & ct are respectively.
ct>
t
[1r1] = [ML"1T-2]"[M1-3
0. 'The\
1
a=-
_. __________ ;
Solution: [v 0 ] = [x][cx] = [M 0 11 r 1 J
and [ct][t] = M01°T 0 , [ct]= [M0 1°r11
=>
kJ;_:>S.QnilRl&J 17
E,E~o:~f~,t~J.ii];;>'
i
'When a solid sphere moves through a liquiq, the liquid
opposes the motion with a force E The magnitude of F depends
10n the coefficient ofviscosityT] of the liquid, the radius r ofthe
lsphere and the speed v of the sphere. Assuming that E is
!proportional to different powers of these quantities, guess a
iformulafor F using the method of dimensions.
Solution: Suppose the formula is F = kri"rbv'
~-;;-ng,; mod~lus oi s;eel is 19 XJ010N/m
,clyne/cm
Equating the exponents of M, 1 and T from both sides,
•
Solving these, a = l, b = l and c = 1
Thus, the formula for Fis F = /crirv.
Thus,
[Y] =
Express i~-ij
Force
(distance) 2
[Fl = [M1F2] = [ML1r2]
or,
JE
2
:r\~:r
1
2
: ; ~ =(\:g)(1
so,
or,
b~E?<943ll'l~"I~,__~
1
= lO00x-xl = 10
100
1 N/m 2 = 10 dyne/cm 2
19x1010 N/m 2 =·19xl011 dyne/cm 2, -
1-,E~x:~mpJg.J~
.,;i;,,
__
imensional formula for viscosity of fluids is,
.
T\
= [M'i-ly-:l J
Dividing eq. (1) by (2),
[Pp-1] = [12r2]
=>
Thus, 1 poiseuille = 10 poise
[P] = [ML-1r2]
... (1)
[p] ': [M1-3]
... (2)
]
Fmd how many poise (CGS _unit of viscosity) is equal to 1
lp_oiseuill~(SI unit of.viscosity)]__ · ___________
Solution: T\ = [M1 1-1 r 1 1
1 CGS units= g cm-1s-1
1 SI units_= kg m-1s-1
= 1000 g·(loo ~mi-1 s-1
. = 10g cm-1 s-1
Solution: Method-I
=>
•
Here dyne is the CGS unit of force.
This suggest that it has dimensions of
· -a-c = -2
[1 r1] = [P112p-112]
2
Solution: The unit of Young's modulus is N/m 2.
a=l
-a+b+c = l
Cc!LP-112r>.-1/2
2
[12]
2
N/m is in SI units,
,
So,
1 N/m 2 = (1 kg) (1 m)-1 (ls)-2
and
1 dyne/cm 2 = (1 g) (1 cm)-1 (ls)-2
= [M"1-•+b+cT-a-c]
tc)_P-11~P.112
f~
[12]
Then, [M1r2] = [M1-1r 1]"[1]b[½]'
/If P is-~he pressure o_if a gas andp. is its d.!dimension ofvelocity.
(a) pl/2p-112
(b) pl/2pl/2
1
b =-2.'
2
. [v] = [P1'.2P-112]
[v] = [P112p-1121
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
\ji
s
'j
i.
V
-~~----h-:-·~
~
~")
1""'
In two systems of tmits,"the relation betw~en velocity,
. by
acceleration ,': ,:, iµid ''. ,force : is ' given
2
'
'v·e
'· ·
,
, F
•
,
.
v 2 =-1-·-,a:i=·a1E-r,F2 =_!.; where e and 't are
't_
I,.~
.
.E't
, .•,'
1
constants then.find in this new·,system:
m2 .
. . . ', (b.) L2,
( a) .
.
.
.
. ',
'
ml
' '
'
.
'
'
, [Ans. (a). ·}
E -r
'
'. . ~I
~ ; (9)
;1 ·
3
'.
't .
•· . .
.
a
.
.
nRT - - . ·
·
2. In the formula P =- - e ·Rn,;, find the dimensions of
.
,: V-b
.
' . ·
a and b where p ' = pressure, n = no: of moles,
T "" temperature, -V = volume'. and R = universal gas.
constant. . , . ·
·
[Ans. (a),.;. [ML5 T-2 inol~1 ], (b) = [L3 Il
3. The loss of.pressure when a fluid_ flows through a pipe
is'!liven by P-;= -kp"l Vbd'µ .where d and l are diameter
arid"length\,ftlie pipe respectively, p, d and µ are the
ma~s, ·density arid coefficient of viscosity of the fluid, V
is the mean ·velocity of flow through the pipe and k is a
huinerka! ·constant. Find the ~alues .of a, .b and c.
[Ans a·=ib'"h'=
.
M°gh·, ·~.;. i·
. ' c:'='-21.
•
•
i__ ,
,.,
nxYT - - ·
.
.
.
.
4. P =--. e .nxi;; where n is number of moles, P is
Vo ·,
. .
. .
,, .
.
pr~ssui-e, Tis te~perature, Vo i; vol~~e, ,M is mass, g
~epresent~ ~ccel~ration iliie to g~avity and h is height.
····Finddiinehsioii'ofxandvalueofy., _·. · ·
· ,,
[Ans. [Mi.2 r 2 K-1 mo1-rJ; y = 1l ·
5. The uni~· CU) ~fvel~city,:;ic~eleratio~ ruid force in two
systems are related as under : .
2
(a) U~
Cc)
'
= ~Uv
p ', .
u~ = [_!_];F
. ap.
. (b)
.
u: = ,.
'cct')
.
.
I-'·-;,'
.
' i:_. •., , '' '~
·.' ; ' ..
u~ = (p\ )uPJ
6. Specific heat of hydrogen at constant pressure, ·
C P = 29 joule _kelvin-1 mo1-1 •
(a) Find dimensions of C P.
(b) Unit of length is changed 'to 50 cm, unit•of ti.me is
changed to 2 sec, unit of temperature is changed to
2K. keeping units of mass· and amount of substance
same. Find the value of specific heat of hydrogen in,
.
new system of units:
[Ans. (a) [ML2 T-2 K-1 mol- 1],(b)928]
m_oniy~o;;';"Altefu';;tiC'ilifoI@~
1. E, m, L, G denote energy, mass, angular momentum &
·gravitation constant _respectively. The dimensions of
EL2
""'"s'2 will be that of :
mG
(a) angle
(b) l~ngth
(c) mass
(d)' time
,
2. The dimensional formula for which of the following
pair is not the same ?
. ''
(a) impulse and momentum '
(b) torque and work
(c) stress and pressure
(d) momentum and angular momentum
3. If the speed of light (c), acceleration due to gravity•(g)
and pressure (p) are taken as fundamental units, the
dimensions of gravitational constant (G) are : ·
(a) [c2g3p2]
(b) [cog2p-']
(c) [c2g2p-2]
(d) [cog p-3]
never be a meaningful quantity?
(a) PQ - R
(b) PQIR
,
'All th~ pnme<hymboii,
b~long to one system and
· unprim¢'d' pnes CU) b.elong to the other systems ..a and
p are· climensiohless ~onstants: How:m9mentum units
of the' !WO systems. are 'related,? , . . : . .
.
.
[Ans.
4. Which of the following combinations of three
dimensionally different physical quantities P, Q, R can
(aP)U•.
.
.
,,. ,~· '
',;-r::~:t
·s.
(c) (P-Q)/R
(d) (PR-Q 2 )/QR
In a view unit· system, I unit of time is equal to i'o
second, I unit of mass is 5 kg and I unit of length is 20
m. In the new SVStem of units_ ,, nnit nf Pm>ro-v is equal
to:
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
\UNIT AND DIMENSIONS ·
{a) 20 joule
(b)
20
(d) 16 joule
(c) 4 joule
.
.
(c) Dimensional formula 9f k)s [MLT2]
Jc_ joule
a.
.
(d) Dimensional formula of
a-t 2
6. The d1mens1ons of - m the equanon, P = - - where
b
bx
P is pressure, x is distance and t is time, are:
(b) [MT-2]
(a) [M2LT-s]
3
(c) [LT ]'
(d) [ML3T 1]
7. The time dependence of a physical quantity p is given
.
2
by p = p 0 /-a< J where cxis constant and tis time. The
l
is [T]
13. If P and Q have different non-zero dimensions, which
of the following operations is possible?
(a) P +Q
(b) PQ
(c)P-Q
(d)l-~
Q
2
14. In the formula X = 3¥Z , X and Z have dimensions of
capacitance and magnetic induction respectively. What
are the dimensions of Y in MKS system?
(b) [M-3L-2T4Q4]
(a) [M-3L-tT3Q4]
(c) [M-2L-2T4Q4]
(d) [M-3L-2T4Q']
2
15. A cube has a side of length 1.2 x 10- m. Calculate its
volume:
(a) l.7xl0-6m 3
Cb) l.73xl0-6 m 3
(c) l.70xl0-6 m 3
(d) l.732xl0- 6 m 3
constant a:
(a) is dimensionless
Cb) has dimensions [r-2]
(c) has dimensions [r 2]
(d) has dimensions of p
8. If area (A), velocity (v) and density (p) are base units,
then the dimensional formula of force can be
represented as:
'• 16. Pressure depends on distance as, P =!:exp(--az}
~
k8
(b) [Av 2p]
(a) [Avp]
where ex,~ are constants, zis distance, k is Boltzmann's
(d) [A 2vp]
(c) "[Avp 2 ]
constant and 8 is temperature. The dimensions of ~
9. Two forces P and Q ad at a point and have resultant R.
are:
2
2
(a) [MoLoTo]
(b) [M-lL-IT-1]
If Q is replaced by (R - P ) acting in the direction
Q
(c) [M 0 L2T 0 ]
(d) [M-1L-.1T 2]
opposite to that of Q, the resultant :
17. A 'wire of length I =6 ± 0.06 cm and radius
(a) remains same
Cb) becomes half
r = 0.5 ± 0.005 cm and mass m = 0.3 ± 0.003 g.
(c) becomes twice
(d) none of these
Maximum percentage error in density is :
10. If instead of mass, length and time as fundamental
(a) 4%
·Cb) 2%
quantities, we choose velocity, acceleration and force
(c) ·lo/o
(d) 6.8%
as fundamental quantities express their dimensions by
18. Which of the following sets have different dimensions?
v, a and F respectively, then the dimensions of Young's
(a). Pressure, Young's modulus, stress
. modulus will be expressed .as :
Cb) Emf, potential difference, elec)ric potential
(a) [Fa 2 v-4]
Cb) ·[F 2 v-1 a]
(c) Heat, work done, energy
(c) [Fa 2 v-1 ] . . ·
(d) [Fav-2 ]
(d) Dipole moment, electric flux, electric field
19. Which of the pair have same dimensions ?
·
11. Which of the following statements is correct about
{a) Force and strain
conversion of units, for example 1 m = 100 cm?
(a) Conversion of units have identical dimensions on
Cb) Force and stress
(c) Angular velocity and frequency
each side of the equal sign but not the same units.
(d) Energy·and strain
Cb) Conversion of units have identical dimensions on
20. Tlie physical quantities.not having same dimensions are:
each side of the, _equal sign but not the ss1me .units.
(c) If a larger unit is used then numerical value of
(a) torque and work
(b) momentum and Planc~s constant
physical quantity is large.
(c) stress·and Young's modulus
(d) Due to ·conversion of units physical quantity to be
(d) speed and (µ 0 e 0 )-1/ 2 ; ·
measured will change.
12. If the speed v of a particle of mass m as function of
21. The dimension of coefficient of viscosity is : .
(a) [ML-1 r 11
Cb) [MLT2l
time t is given by v = ro A sin[(
~ere A has
2
0
(c) [ML T ]
(d) [MLT1 ]
22. A particle is moving eastwards with a velocity of 5
dimension of length.
rn/s. In 10 sec, the velocity changes to 5 rn/s
(a) The argument of trigonometric function must be a
· northwards. The average.acceleration in this time is:
dimensionless quantity.' ,
(a) zero
Cb) Dimensionai'formula of ro is [LT-1]
l}] .
-----.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
MECHAf/1~
\16
}z ms(c) }z ms-
2
(b)
2
(d)
.
5. The pairs of physical quantities that have the same
towards north-west
dimensions in (are):
(a) Reynolds number and coefficient of friction
(b) Curie and frequency of a light wave
(c) Latent heat and gravita,tional potential
(d) Planck's constant and torque
towards north-east
.! ms-2 towards north
2
23. Out of the following the only pair that does not have
identical dimensions is:
(a) angular momentum and Planck's constant
(b) moment of inertia and moment of a force
(c) work and torque
(d) impulse and momentum
24. Which of the following units denotes the dimensions
ML2 /Q 2 , where Q denotes the electric charge?
(a) weber (Wb)
(b) Wb/m 2
(c) henry (H)
(d) H/m 2
25. The dimension of magnetic field in M, L, T and C
(Coulomb) is given as :
(a) [MLr'c'J
Cb) [MT 2 C-2 J
(c) [Mr'c'J
(d) [MT-2 C- 1 J
m:F~~i~:.~· "~
A student forgot Newton's formula for speed of sound
but he knows there were speed (v), pressure (p) and
density (d) in the formula. He then start using
dimensional analysis method to find the actual
relation.
V = kpxdy
Where k is a dimensionless constant. On the basis of
above passage answer the following questions:
1. The value of x is :
. Cb) I
(a) 1
2
1
(c) -2
mM~re.~~!!!2~1ternative:,:~"c;;~~
2. The value of y is :
(P + ; )cv2
b)
= nRT where P
is the pressure, Vis
the volume , Tis the absolute temperature, R is the molar
gas constant and a, b are arbitrary constants. Which of
the following have the same dimensions as those of PV?
(a) nRT
(b) a/V
(d) ab/ V2
(b)
(a) 1
1. Which of the following dimensions are correctly
matched? (8 = temperature)
(a) Angular momentum{M1 L2 T-1 ]
(b) Torque{M1 L2 r 2 J
(c) Stefan's constant{M1 r 3 e-4J
(d) Planck's constant{M1 L2 T-2 ]
2. The gas equation for n moles of a real gas is
(d) 2
.!
2
1
(c) -2
(d) 2
3. If the density will increase the speed of sound will:
(b) decrease
(a) increase
(c) unchanged
(d) none of these
=M=~!!,i~g Typ~];obi~~~
1. Match the column :
I
(a) Energy density
(Energy per unit volume)
1
(p). dyne/ cm2
(c) Pb
3. The dimensions of the quantities in one (or more) of
I (b)
Force constant of a spring
(q) kg-m/s
the following pairs are the same. Identify the pair(s) :
(a) Torque and work
(b) Angular momentum and work
(c) Energy and Young's modulus
· (d) Light-year and wavelength
4. The dimensions of length are expressed as Gx cY h •,
where G, c and h are the universal gravitational
constant, speed of light and Planck's constant
respectively, then:
I (c)
Pressure
(r)
erg/cm2
(s)
pascal
(a) X = (1/2),y = (1/2)
(b) X = (1/2), Z = (1/2)
(c) y = (-3/2),z = (1/2)
(d) y = (1/2), z = (3/2)
I
J
/
'
2. Suppose two students are trying to make a new
measurement system so that they can use it like a code
measurement system and others do not understand it.
Instead of taking 1 kg, 1 m and 1 sec, as basic unit
they took unit of mass as a kg, the unit of length as f3 m
and unit of times as y second. They called power in
new system as ACME then match the two columns.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
UNIT AND DIMENSIONS
I
\
Column-I
17
\
\
Column-II :
(a)
IN in new system
(p)
o;-lp-2y2
(b)
lJ in new system
(q)
o;-lp-ly2
(c)
(d)
1 pascal (SI unit of pressute) in (r}
new system
o:-'p y2
o; ACME in watt
cx'p2y-J
:j
I
\
Column-I
\
\
Column-II
(p) [ML2 r 2l
i
:
(b) Torque
(q) [ML2r1l
'
:
(c) Inductance
(r)
(d) Latent heat
(s) [ML'Q-2l
(el Capacitance
(t)
(f)
(u) rer2l
Resistivity
[ML3 r 1Q-2l
Column-I
ii...
v
C ln(Dx)
For above equation to
dimensionallv correct
\
\
.2
(b) Pressure = P + -1 pv
+ gX
(q) [Al =[M 0 L1 r
(c) X=At+
(r)
2
:
'
'
Column-II
(p) [Al =[M 1 L1 r'l,
[Bl= [M 0 L0 r'J,.
be
[CJ =[M 0 L0 r 1J
V
B ln(Ct)
1
I
1,
[BJ = [M L T-1],
[CJ =[M 0 L0 r 1J,
[A] =[M 1 L1 r 2),
0
0
[Bl= [M 0 L0 r'J,
[CJ =[M"1 L0 T 1],
1
(a) Angular momentum
[M-1L-2T'Q2l
\
(a) F=Asin(Bt)+
'
3. Match the physical quantities given in Column I with
dimensions expressed in terms of mass (M), length
(L), time (T), and charge (Q) given in column II.
I
4. Match the following:
:
(s)
1
(s)
Dimensionally
incorrect
(Where F = force, P = pressure, p = density, t = time, ,
= velocity, a = acceleration, X = displacement)
v
'
'
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
.
- ------.-- --.----···-·--·•,--.-----·---·--------···T----~----.
'-1!1 --
AN9WER9 ------ --
1.
(a)
2.
11
(a)
12. (a)
21.
(a)
22.
1. (a, b, c)
(d)
(b)
(b)
(a)
4.
(c)
13. (b)
14.
(b)
15. (a)
(b)
24.
(c)
25.
3.
-- .
23.
2. (a, b, c, d)
3.
(a, c)
5.
(1,)
2. (c)
-- -
.
. (b)
7.
(b)
8.
(b)
16._
(c)
17.·
(a)
18.
(d)
(a, d)
6.
(b, c)
9. .
19.
(c)
4. (a, b) 5.
3. (b)
=~l~~:,j'~f~~op1emj~
1. (a)-p, s; (b)-r ; (c)- p, s
2. (a)-q ; (b)- p ; (c)- r ; (d) s
3. (a) -q ; (b) -p ; (c) -s (d) -u; (e) -r; (f) -t
4. (a) -r ; (b) -s ; (c) -q
www.puucho.com
~
- MECHANICS-I '
---
6.
=.!!!s~age:~ =:~
1.
. .
.
··-- ----------·-·· - - __ J
7.-
(a, b, c)
(a)
10.
(a)
(c)
20.
(b)
Anurag Mishra Mechanics 1 with www.puucho.com
-·,
\\
DESCRIPTION OF MOTION
Mechanics is the branch of physics for studying the
motion of bodies, i.e., the change in their position in space
and time.
The position of a body in space can be specified only
relative to some other body or bodies. Therefore, when we
speak about motion, we mean relative motion, i.e., the
motion of a body relative to another body which is
conditionally assumed to be fixed. If we mentally attach a
coordinate system to the body taken as fixed and called the
reference body, this system, together with the chosen
method of measuring time;forms the reference system.
Normally, the Cartesian coordinate system, is used.
SUBJECT OF KINEMATICS
Kinematics is the part of theoretical mechanics in which
the mechanical motion of particles and rigid bodies is studied
without regard to the acting forces.
1. In order to describe the movement of an object we
must specify its position relative to observer. One of
the most convenient coordinate system is
Cartesian coordinate system. It consists of
three mutually perpendicular axes designated as
x-axis and y-axis and z-axis. Location of any point is
specified by three coordinates x,y, z as shown in Fig.
1.1.
y
y
y(t) ••
•••• •• t _ __
y(t)
I
·~
x(t)
o ......: :'.
z<t> ......... ::Y
(a)
X
Flg.1.1
0
X
x(t)
(b)
2. Position of a particle in space is determined relative
to some fixed
point. Position of
a particle is not
absolute;
it
depends on the
position
of
AJ
observer.
Fig. 1.2
Consider a train
moving
with
velocity 10 m/s. A block kept inside the train is
visualized by two observers one on the ground and
the other inside the train. For an observer in the
train the block is at rest and for the ground observer
the block moves with the train (Fig. 1.2). In the
language of physics a technical term frame of
reference is used to describe a coordinate system
and position of observer. We say that in reference
frame of ground the block is in motion and in
reference frame of train the block is at rest.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
,20.
-
. ---- -- - -- MECHANICS-fl
'"··----
---- --------- ----
- - - · - - - - - - - - - - - - _J
------ -----·
magnitude and also follow laws of vector algebra are
called vector quantities.
Concepts. 1. In order to define the position of a particle
in space, it ls tzecessa,y to have a fixed body or a system
o/
co-ordinate axes attached to it which i.s called a reference
system.
2. By a reference sjstem ls meant an absolutely rigid boay,
or a co-ordinate system invariably attached to it with respect to·
wh_ich a given motion i.s considered.
'
C
The motion of a given body ls revealed only by
comparison with a reference system.
In some cases a moving reference system which executes
motion with respect to the basic reference system ls considered•
3. Sometimes
polar
coordinates
are
used to specify the
position
of
a
partide. In it spatialposition is denoted
II
>,
by length r from
origin and. angle e is
generally measured
from ,p6sitive x-axis
(Fig. i.3).
and
x = rcos0'.'
y
] ]·
r=~X2+y2
and.
A'
(c)
(b)
(a)
Fig.1.5
in kinematics.
3. In nature, no fixed bodies exi.st and consequently there
can be no fixed reference systems. A fixed reference system i.s
usually assumed to be a system of co-ordinate axes attached to
·
the earth.
4"'
A
A
!
The displacement from a point A to a point B is a
vector quantity. Its magnitude is the straight line distance
from A to B; its direction is that of an arrow that points from
A to B. Points B and C are equidistant from point A but the
two displacements are different because they have different
directions. Two displacements (vectors) are equal if they
have same length and same direction (Fig. 1.5).
VECTOR NOTATION
A vector quantity is represented by a bold later with an
arrow above it or a bar above it.
e
~::::==::=;~x
x=·rcose
...,
A or A
:
...,
The magnitude of vector is represented as IA I and it is
...,
Flg.1.3
= rsin0
tan0 = ~
y
X
4. Trajectory of a particle
y
denotes the actual path
followed· by it. Path
length s(t) is defined as
the distance travelled
along a trajectory in
time t. It is measured
'--------•x,
from the_ starting point
Flg.1.4
of the motion at t = 0.
Path length is the total
laistance covered; it can only increase with time as a
particle moves, hences is·a]ways a positive quantity
(Fig. 1.4).
5. We always express results of our measurement in
terms of a number, e.g., room temperature is _25°C.
The value 25 is called the magnitude of the quantity.
Some quantities do not have direction associated
with them; such quantities are called scalar
quantities. Motion is a quantity that involves
direction as well as magnitude. We say a car is
movin~ with velocity 10 km/h eastward. Such
quantities which have direction as well as
referred as modulus of A.
Geometrically a vector is shown _by an arrow dr& wn to
an appropriate scale. The direction of arrow represents
direction of vector and length of arrow represents
magnitude of vector.
Displacement Vector
Displacement vector represents change in position of a
moving object. A car starts from Kota and travels north-east
to reach Delhi. Its displacement vector will be represented
by an arrow joining starting point Kota to terminal point
Delhi.
--- ---- - l
. _ _ ___ Fig._ 1;6 _______ ..•
www.puucho.com
'
Anurag Mishra Mechanics 1 with www.puucho.com
, -DESCillPTION OF MOTION
:__________
- ·-----
21,
_, _, _, 7
Displacement vector is straight line segment from initial
point Kota to terminal point Delhi.
ii'
2:::·
Significance of Resultant Displacement
A team of hikers begins from A travel to B, C and D the
successiVe destinations. Net displacement, resultant
--;
displacement, total displacement mean same thing.
Resultant displacement vector is straight line segment from
initial point A to final point D.
A
--;
--;
Draw-A first and then B
--;
~---~A
o'.i··:..~--~
.
r.__,
.,
.
A
--;
,._1
--;
--;
Draw B first and thenA
Fig.1.9
·A" ...
<\s' ;_,;,.i"···-·····
S2
Parallelogram Law of Vector Addition
_,
_,
-..J ..Actual path followed
Fig.1.7
--)
SResultant
--)
--)
--)
=S1 + S2 =+ S3
The above expression is_ symbolic representation of
vector summation. It_ just expresses that net displacement
can be obtained by vector summation of individual
displacements, but how summation is to be carried we still
have to learn.
Consider two vectors P and Q, draw these vectors tail to
tail such that they represent the two adjacent sides of a
parallelogram, the resultant of these vectors is represented
by the diagonal of the parallelogram passing through that
point.
.............................. C
.,'
--;
---~/4
R
.: ''
--;
Geometric_ ~eprese!1!ation_ of Sul!! of Vectors
p
--;
s,
Fig.1.10
Diagonal AC completely represents resultant vector
from triangle AEC
AC 2 ~AE 2 +CE 2
--;
s,
~ Tail of second
/
'\
Head of first
vector
Fig.1.8
vector
Substituting
and
We get,
Each vector in a sum is to be drawn with its tail at head
of preceding vector. Resultant vector has tail at the tail of
first vector and head at head of final displacement vector.
_, . _,
Any of the vectors A and B can be drawn first.
=(AB+BE) 2 +CE 2
=AB 2 +BE 2 + 2(AB)(BF'
BE 2 +CE 2 =BC 2
BE =BC cusr
= AB 2 +BC 2 + 2ABBC cos8
AC
R
and
Note: -
L.'£ 2
=~P 2 +Q 2 + 2PQcos8
Qsin8
tanct=-~-P+Qcos8
... (1)
... (2)
------------------
Alv. ys substitute magnitude of P and Qin above equations
(1 J and (2).
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
'22
MECHANICS-I
Case I:
...,
If
(1) Vector addition can be represented in two ways
...,
IPl=IQI
...,
IRI= 2Pcos8/2
and
R
+
->
v,
a=B/2
aLJ~
~
->
v,
->
p
vectors to be added
are placed tai~ to tip.
Fig.1.11
Fig.1.15
Significance of Vector Addition
Illustration 1. Consider a
Fa 100N
block that is pulled by two boys
simultaneously. Each boy exerts
force of 100 N. We can easily
~Fa100N
guess that block will move at 45°
Fig. 1.12
angle. From rule of vector
addition we can see that resultant force is
t ··"
FR
··~·.·········...-
V2
(2)
->
a
v,
:
~
->
v,
Correct
representatio'n
Fig.1.16
=..JlQ0 2 + 1002
= 10Q-J2 N
Vectors to be added are placed tail to tail such that
they represent consecutive sides of a parallelogram.
Diagonal of parallelogram represents resultant.
Net force is ·vector sum of all the forces acting on the
object.
1/
....v,
Illustration 2. A boat moves with velocity Y8 in still
water [Fig. 1.13 (a)]. Velocity measured by observer on
+
v;/"'(~·-··/;
~
....
=
v,
ground will be y8 . But if water flows at a certain rate it will
make the boat moye f~st':r ~s!<'~e~ ~:11e11d~ng_ on whether
t
.Fig.1_._17 _
Incorrect
representation
_. ---· ··- _ _
Vector Addition is Commutative
It is not important in which order vector are added
-+
--+
--+
--+
A+B=B+A
....Va
-,
V
->
V
(b)
(a)
Flg.1.18
Fig.1.13
'.
'----~----··-----
...,
v
8
...,
Draw A first then draw B
the boat moves along the stream or opposite to the stream.
The velocity measured by an observer on the shore is the
vector sum of the velocity of the boat
___ J
I------ --
->
A
and the current
velocity ,I c ·
->
...,
...,
V=Ve+Vc
Illustration 3. Similarly if an
airplane moves in a wind its res.ultant
velocity will be a combination of its
own velocity and velocity. of wind.
--+
V resultant
--+
···- --_,--·--·-- I
->
f-VwJnd
!
, Vairplane
--+
' !
V resultant I
...,
_F_!g. ~-1_4 __
_,
->
Draw B first then draw A
Resultant obtained is same in both cases.
'
, --+
=V airplane~ V wind
i
Fig, 1.19
l
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
DESCRIPTION OF MOTION
'
....
~-:
-
' . ""ii]
----·--·
·,,_
1
p (X, y).'
v,!L'•
Qi
v,
-->
A
Fig. 1.20
I".. •
v,
'i
t
t0.
• t ., ,_v, •
\lL+·'v;
·c;i
-
A vector sum is independent of the order in which the
vectors are added, Le., it obeys the associative law (Fig.
1.21).
it.,
?
.,.
,..'il"
.
,.//
C,
I
1
'1
)(
)'~
r"
•
~ rU<.
J(
+'
rca:
...'! ._
'o
~•••• •••
i
! .•
....
i-.. ··:'
•. ·
--+
'B
:
Fig. 1.22 (b) ·
A
Component expresses effective value of a
vector in_ a particular direction. Consider a ball
Fig. 1.21
--+
--+
--+
(A+B)+C
--+
--+
v
--+
= A+(B+C)
Component of Vector
Resolution of vector means separating a vector into sum
--+
of two or more vectors. We can write A as the sum of two
--+
--+
--t
--+
--+
--+
--+
vectors Ax andAy:A=Ax+Ay,whereAx andAy are
projections of A on x- and y-axis. [Fig. 1.22 (a)]
1
y
l
"' i ' .
C
-~A,
.'.<('!..!"
.
I
•" ~"" """".
A
8
:
::
+--A,--.
X
moving with velocity in
y
. north-east direction [Fig.
1.22 (b)]. After a certain
P (x, y)
time interval the ball will
be at position P relative
;;II
to the origin [Fig. 1.22
>,
(c)]. It displacement
0
along east (x-axis) takes
7~::====;;r-+x
X = Yx I
place with x-component
of velocity whereas
Fig.1.22 (c)
displacement
along
north takes place with
y-component of velocity.
Figure shows a block being pulled by a force on a
horizontal table
i"
A.,_ =Aco~e
_____ Fig.1.22_(a) __ ·
.
y
_i
4
Ax =Acose;
Ay =Asin0
Magnitude of A in terms of components is given by
--+ '
--+
F
f/':..___...........
;rr·····--+
IAl=~A;+A;
Fy
Direction of vector A in terms of its components is given
by
.
:•'
'
X
F,
Fig. 1.22 (d)
Fx = x-component of force vector
Fy = y-component of force vector
When block is displaced it travels in plane its position
can be described by x and y-coordinates. Cause of motion
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
r,-·-··
•
______________________ ~--·------ --
,24 --------- _-- ----------~
along x-axis is x-component of force, similarly cause of
motion along y,axis is y-component of force.
illustration 4. A boat
-- -- - --- is tied to river bank with
ropes as shown in figure.
45°
Force exerted by ropes on
37°
boat is shown in figure. What
....
,
N
N-W
N-E
w
---->
IF21•
ate x and y-components of F1
MECHANICS-I
---
-
30N [
E
....
and F2 vectors?
Fig. 1.23 (a)
....
Force F1 can be written as
.... .... ....
F1 = F1x+ F1y
.....
S-W
.-
(a)
....F1
[F1xl= F1 cos45°
....
= 40cos45°
=2WZN
N
'
:
':
45"
-) :
30° North
-----4'~~±.!:Fi,:
[F1y[ = F1 sin45°
= 40sin45°
=2WZN
of east
30°
w
E
60°
(b)
y
.....' '
....
....
F2,
[F2xl= F2 cos37°
....
S-E
s
....
= 30cos37°
=24N
F2,
X
37°
60° South of
west
/
s
Shore line
(c)
(b)
....
Fig. 1.25
F2
(c)
[F2y[= F2 sin37°
Fig.1.23· __
j
= 30sin37°
= 18N
Note that y-component of F2 is negative.
illustration 5. A box of mass m is placed on a
smooth frictionless incline. i.,yeiglit of an object acts in
vertical direction. Consider x and y-axis parallel and
perpendicular to incline. What are x and y-components of
weight W of box?
!~::·· J.i(,' .
:~
Wease WWsin8(r·
______ c,
v .L
v 11
v .L
v 11
UNIT VECTOR
....
Consider a vector A, shown in figur':
A vector has magnitude and direction. If we divide
....
vector A by its magnitude, we are left with a
quantity that has unit magnitude and its
....
direction is same as that of vector A, this left
over quantity is a unit vector represented by
A
I1
Thus
_ Fi!J. 1_02~
Wx = W sin0 = mg sin0
WY = -W cos0 = -mg cos0
* Weight of object= mass x acceleration due to gravity
=mg
illustration 6. A man rows a boat with a speed 10
m/s along N-E direction. The shore line is 15° south of east.
. What are components of the velocity vector along and
perpendicular to shore?
y
Fig. 1.26
....
A
A=...,
A
....
[A[
....
Also vector A can be expressed in terms of unit vector A
as
....
....
A
A=[A[A
....
Suppose a force of magnitude 10 N acts in direction of A
....
....
vector, this force F can be expressed as F = (lOA)N.
www.puucho.com
.,
= component perpendicular to shore
= component parallel to shore
= v sin 60° = l0sin 60° = 15-J2 m/s
=vcos60°= 10cos60°= Sm/s
Anurag Mishra Mechanics 1 with www.puucho.com
DESCRIPTION OF MOTION
25
Concept: 1. Unit vector is dimensionless vector it can
be multiplied to any quantity without changing it dimensions.
2. Sole purpose of unit vectors is to represent a direction in
space.
_,
- a=axi+ayj ... (2)
_,
ay = ayj
and
Thus
Iitxl= acosa.
Also
A unit vector is a dimensionless vector with unit
magnitude. It is a mathematical device to convert a scalar
quantity into a vector. They are used to specify a particular
direction in space. In Cartesian coordinate system there are
three unit vectors
k whose directions are parallel to
coordinate axes x-, y- and z-axes respectively (Fig. 1.27).
layl= asina
Finally we have
--+
l ],
k
y
A
A
or
component of vector along y-axis.
A
j
J
-;
y
A
A,i
...../
Concept:
Unit vector in direction ofa
_,
A
X
X
j
a
a=lal
acosai+ asina]
=
a
a= cosai + sinaj
Fig. 1.27
_,
We can define any vector A through a combination of
unit vectors
-
-
-
_,
A =Axi+Ayj+A,k
-
Similarly (i)
vector A,i has magnitude Ax and direction positive
x-axis. Similarly, we can define Ay] and A, k.
b = -b cosai + b sin a]
.6 = - cosai + sin a]
Expressing a Vector in Unit Vector Notation
' f /4,l
Figure shows a vector in x-y plane.
_i_:_:.;
- Fig.1.28
a
Redraw vector with its tail at origin.
lz_.,
Fig, 1.30
(ii)
, Fig, 1.29 (a)_
--+
';
':
C=-Cxl-CyJ
= -ccosai- csinaJ
C=-cosai-sina;j
y
Concept: A vector can be displaced parallel to itself,,
anywhere in space without changing its value. It is referred as·
translation of vector.
_,
_,
_,
a= ax+ay
We can write
... (3)
or
component of vector along x-axis
.
f
_,
ay = a sma = y-component o vector a
,•'
~
A
':
We can express vector aby any equation (1), (2) or (3).
1'-~--,.:"
A
•
_,
ax = a cos a= x-component of vector a
__....... A,k
k
';
a= acosa1+ asmaJ
z
z
Fig. 1.29 (b)
... (1)
www.puucho.com
.
Fig.1:31 _
Anurag Mishra Mechanics 1 with www.puucho.com
26
(iii)
....
d
.
= dx i-
d = (cos45°)i-(sin45°)j
dy]
= dcosai-dsina]
=
d=cosai-sinaj
i-]
J'2
N
~·
s
+......... .
: · __ Fl~,
Fig. 1.32
illustration 7. Write unit vector in direction of N-E,
N-W, S-W andS-E.
'
.
:.
,
~:]
l
.
l
s '
Fig, 1,13
''.
j
'
a= (cos45°) i +(sin45° )j
: ' y._···:
~.
3
/.'
....... •. )···.
I -4j
', c
I
x_'l
Fig. 1,3l
!
,
,
I
.' . I
--·- ---" o--- -.•-·'-'
~ = (lOm/s)(-ii+~J)
i+]
b = -(cos45° )i + (sin45° )j
-i+j
= Fz
:~ .,]
.
= (-8i+6])rn/s
illustration 9. A hill is inclined at 0 with horizontal.
Write a unit vector in direction parallel and perpendicular to
hill, in standard x-y-coordinate system.
Let the required vectors be a and b respectively.
.
..
----·-- ..........
_
.....
''}
flop~J'\' ..· I
r· . . :··---~Lx!
'
s.
_Fig: 1:Jj,_
I
c = -(cos45° )i- (sin45° )j
- Fz
d
SE,
· '\
Illustration 8. A boat is moving in
direction -4i + 3] with ,a speed of 10
rn/s. Write velocity vector of boat in unit
vector notation.
Direction of motion of boat is along
unit vector
->
-4i+3j
V=-,,====
.J42 +32
4, 3 •
=--1+-j
5
5
Thus velocity vector of. boat is
= Fz
- -i-j
,
1~~
.
) '' -~-t=-c_:~~-~f~-~1n_~j
!
,*;
0.
'
" .
ir ·
1.
W.
'.
·sw
45°
1
' , ~ ·
ti='sin81+cos8j
8 .
.
E·
C
i.
Fig."1.38
s
Fig. 1:35
a= - cos0i + sine]
b = sin8i+cosej
www.puucho.com
!
!
'!
Anurag Mishra Mechanics 1 with www.puucho.com
-~ ·---·- ·--------·---27
.
DESCRIPTION OF MOTION
....
.1:r -
Concept: Vector A is indc;·endent of choice of cuordinatc
axes, although components ofvecrordepend on the choice ofco·
ordinate axes (Fig. 1.39).
l:2 _ 1!1
- final position vector - initial position vector
At time t 1 , the particle is at point P1 , and 'its position
rl = X1 i + )' 1J. Similarly at time
. .
= x i + y j and displacement of
vector can be expressed as
....
t 2 position vector is r2
....
particle Li r
= (x 2 -
.
2
2
x 1 ) i + lv 2
-
.
y 1 )j.
Average Velocity
Speed is a scalar quantity which describes rate of
motion, but velocity is a vector that gives the direction of
motion as well as rate.
A body's average velocity is defined as its displacement
divided by time M. Velocity vector is parallel to the
Fig. 1.39
....
displacement S.
y
Path of particle
t + t>t
'
Magnitude of a vector is independent of the choice of,'
coordinate axes, hence it is a scalar; whereas vector
1component depends on the arbitrary choice of a coordinate!
axis; it cannot be a scalar; we call it simply a vector:
component.
1. If a vector is zero then all its components are
individually zero
A,i+Ayj +A,k. = Q
then
A X =A y =A Z =0
If two vectors are equal, then their components along
2.
---.--------. --. -.-,:
,Ix
y
1~
L . . . . .J"'
v,
-1----------'--X
(x + t>x)
X
Fig.1.41
Velocity components vxor vyequal to the change in
corresponding coordinate Lix or Liy divided by Lit
Lix
Liy
V =- V =x
At' y
At
Position Vector of a Point with Given Coordinates
Consider a point' P' with Cartesian coordinates (x,y, z)
relative to the origin 'O' then the position vector of' P' is
given by
the rectangular axes are also equal i.e., if
A,i + Ayj + Azk. "B,i + Byj + B,k.
(Ax -Bx )i+ (Ay -By)j + (A. +B. )k = 0
Ax -Bx= 0, Ay -By= 0, Az -B,
:t>y
Displace~ent
. .
_,
=0
OP = r = xi+ yj + zk
• l
(x, y, z)
y
p
POSITION VECTOR
The position vector
of a particle is a vector
drawn from origin to the
position of particle (Fig.
1.40). For a particle at
the point P(x, y) position
vector is
....
.
Y
.
displacement
vector · Li 1
is
the
difference in position vector.
r
"'--------------x
Flg.1.42
_,
If vector OP makes angle o:, pand y respectively with x, y
r=xi+yj
The
_,
Path of a particle
0
Fig. 1.40
and z coordinate axes then components of vector are
www.puucho.com
rx
ry
rz
= r cos ex
= rcosp
= rcosy
Anurag Mishra Mechanics 1 with www.puucho.com
[38________ ---- -- _ ·.:.,---'------ . - - - - - - ' - - - - ' - - - - - - - - · : - '-·_,_MECH~IC~:U
Angles a, p and y are referred _as direction cosines.
thus
cos 2 a+ cos 2 P+ cos 2 y = 1
r
The direction cosines of are,
'X
y
cosa =-,cos!}= and
r
r
Z
cosy=-,
r
where,
r =-1-r!"" ~ x 2 + y~ + z 2 position vector of A w.r.t. B defined
-->
as, rA/B
-->
-->
= rA- rB
Position
vector
of
point
A(x,y,z)
with
w.r.t.
B(x 2 ,y 2 ,z 2 ) is given by,
Where ·cosa, cos!}, cosy are known as direction cosines,
these are the cosines of the angles chat the vector makes
with the ·x, y and z-axes respectively.
How to Obtain a Unit Vector in a Given Direction ?
If we are given two points in space, we want to define a
mrit vector along a· line which begins at the point
'A'(x1 ,y 1 ,z 1 ) and passes through the point'B'(i2,y 2,z 2).
First we find position vector of point ':B' w.r.t. point' A'.
Position vector of A is given by
, 1A/B = (X1 - X2)i(y1 + Y2)J+ (z1 -z2)lc
. Rectangular Resolution
Dimensions ·
.
-->
of a Vector in
Three
Where rx, ry and rz are the magnitudes of components
·
!)long x, y and z-axes respectively.
By the geometry of Fig. 1.43,
r = l,2 + ,2 + 1'.2
,._
rB = X2i + Y 2J + Z 2lc
,--+
--+
--+
rB/A =
(X2
-X1)i+(y2 -y1)J+Cz2-z1)lc
rB/A = rB-rA
Now unit vector in the direction of tl:iis position vector is
given by,
·.1
-,=======.====~~==~=
i1)
•
rBJA.
(X2·-X1)i+(y2-Y1)J+Cz2-Z1)lc
rB/A = - - =
2
2
2
ri!B/A I
z
y
X
•
Similarly, position vector .ofB "is
-->
Suppose the vector r is to be resolved into three
inutually perpendicular component vectors along the
directions of x-axis, y-axis and z-axis.
In accordance with polygon law of addition of vectors,
Le.,
r = rxi+_ryj +rzlc
.Y
•
rA =x1i+yd+Z1K'
~(X2 -
X1)
+ (y2 - ,Y1) + Cz2 .-
Sinlilarly unit vector in the opposite direction of this
position vector is given by,
-->
rB/A
--+
--+
r A/B = ----::;-- = - rB/ A
lrB/A I
r
Note that vector A/B is opposite to vector 1B/A.
Shortest Distance Between Two Points
If the rectangular Cartesian coordinates of two poims' A'
'
X
Flg.1.43 ·
.
C •• '
l Jand 1c- respectively then
rx = r cosa, ·ry = r cos!}, rz = r cosy
So chat
r
r
cosa = ..l =
x
.r
~r2 + r2 + r.-2
X
y
z
.
r
.
· sinrilarly cosp =
Y ·
J 2
2
2
' -vrx +ry +rz
and-
1'.
COS"(=
.
---+
--+
__,
-->
AB= rB/A
if a, p and y are the angles which r makes the direction
of
and' B' with position vector rA/0 and rB/O relative to the
origin 'O' be (x1 ,y 1 ,z1 ) and (x2 ,y 2 ,z 2) respectively, then
--+
--+
= rB;o- rA/o
Where
and
Therefore AB= rB/A
= (X2 -
i + (y 2 -
Zt) le
Therefore, the shortest distance between the points
(x1 ,y1 ,z1 ) and (x 2 ,y 2 ,z 2 ) is
IABI = ~r(_X_2---X-1-)2~+-(y_2___Y_1_)2~+-(z_2___Z1_)_2
z
+ r2
+ r.2
"Jfr2
X
y
Z
www.puucho.com
X1)
y 1) j + (z 2
-
Anurag Mishra Mechanics 1 with www.puucho.com
i DESCRIPTION OF MOTION
-----!
.
29
or
Also
-- - --- ---- - -- --•
j
'A bird ,moves with velociry 20 m/ s in a direction making an;
:angle of 60° with the eastern line and 60° with verticalj
lupward~present the velocity vector in rectan!(Ular form. __ _j
Solution: Let eastern line be taken as x-axis, northern
as y-axis and vertical upward as z-axis. Let the velocity v
makes angle ex, pand y with x, y and z-axes respectively, then
ex= 60°, y 60°. We have cos 2 a+ cos 2 ~ + cos 2 y = 1.
or
cos 2 60°+ cos 2 ~ + cos 2 60° =1
=
or
cos~=
->
1
.fi.
A
A
,.
· v = vcosexi+vcos~j +vcosyK
=l0i+1o.fi.J+10k
LEx:a!ltn'""'te
-127......._
~
--·---:S::I:!::.
---------- ----- ---·· -- -----
or
given
--..:::i~
,--
;Two vectors, both equal in magnitude, hav.e their resultant!!
!equal in magnitude of the either vector. Find the angle
'./!etw_g_en the vectors.
·--·-- _
_ __._!
Solution: Let 0 is the angle between the vectors
A 2 =A 2 +A 2 + 2AAcos0
1
which gives cos0 =- -
r-
(Q +P)(Q-P) =144
P+Q =18
18(Q-P)=144
or
Q-P= 8
Now from equation (ii) and (iii), we get
P=5N
Q=l3N
""'
],> ·
---1
!The sum of the. magnitudes of two forces acting at a point isl
i 18N and the magnitude of their resultant is 12N. If thel
:resultant makes an angle of 90° with the force of smallerj
bzm1itucle, what are the mwmit!!..d§..9f.l&.lWO ft,mc.,cc,,,escc?_ __,
Solution: Method-I:
Let P < Q and 0 is the angle between them.
tan 900 = Qsin0
P+Qcose
----------,
:;;;;-;:;r-;:,-_- -
-
,____J
,around and walks 5.0 m back towards the classroom. He
stops 15.0 mfrom the door. Tota1time of motion is 25.0 sec.
What is his average speed and average velociry?
-~- 5~-:~: ::?a:-2~:-~
'---- -----'
--- Fl~--1_E_.4____
I
l._~_~.:,., >· I
_,_I
_______ j I
Solution: According to definition of average velocity,
/!,x
I
Vav
i't \\···-......Q
... (iii)
!origin, walks 20 m down the corridor, then stops, turns
!~
-.s- ----------------··
Q
... (i)
... (ii)
--·- ·---------·-···-----· --------·--·-1
IA student starts from his physics classroom considered to bel
0 =120°
•••••••••••••••••••••
->
r
e-.·f:: ,=2$1!eJI$\~
g"'-4~i>.,_
i
2
1
LExa~""-'e
=
~.:..,:.~-1:'!:~M~ 3
->
resµltant of P and R is equal to Q.
p2.+R2 =Q2
or
Q2-P2 =R2
=122 =144
=20[.! i +-2... j +
2
.fi. 2
or
P +Q =18
given
or
Q=(l8-P)
From equation(i),
Qcos0 =-P
Substituting Q =(18 - P) and Q cos0 =P in equation
(ii), we get
p 2 + (18-P) 2 + 2P(-P) =144
P 2 + (324+ P 2 - 36P).:. 2P 2 =144
or
P=5N
Which gives
Q= 18-P = 13N
and
Method-II: It is clear from· the figure that the
->
.!k]
... (i)
... (ii)
... (iii)
P+Qcos0=0
(·:tan90°=oo)
2
2
R =P +Q 2 +2PQcos0
=-=
Xf - X,
t.t
t1 -t,
. - 15 ·0 - O.O 0.600 m/s
25.0-0.0
total distance travelled
Average spee d = ---.- - . - - - tune mterval
= 20.0+5.0 l.00 m/s
25.0
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
r-,30
-
------------
---- - -- - --~---
-+
-+
...,
"""7-+
-+-+-+
"""?-+
,subtractions: (a) A+B, (b) A+B+<:; (c) A-B, (d)
!-+
-+
-+
-)
,.
"
-+
It.
"
"
"
Consider two vectors A =2i -3 j + 5 k and B=-1-2 j + 7k .
Consider three vectors A, Band C as shown in Fig. lE.5.
:Perform graphically the following vector additions and,
'
--)
C=-(A+B)=-A-B
or
..., ...,
-+
MECHANICS-I I
-+
-+-+-+
We have to find a vector C such that A+ B + C = 0. Clearly
-+
-+
-+
"
,.
I\
C=-(A+ B)which is -i + 5j-12k.
-+
'A+B-C.
- -
- 1
l{i
--),._,._,...
-)
'Vector A= 3i+5j -2k and vector B
-+
-+
-+
A.A
= -3j+ 6k
Find a,_
-+
,vector- C such that 2 A + 7 B + 4 C = 0.
• I
Solution: Let
and
Flg.1E.5
...,
...,
-+
Solution:
-+
-+
--+
-+
-+
In case (d) A+B-C=(A+B)+(-C),
...,
we have to reverse the direction of vector C and add it to the
-+
-+
-+
-+
resultant of A and B, Le., A + B .
~
"~·; ~-·:
l'c,
.......
to
+
tai
+
t<
(a)
~'
......
A-B
(c)
(b)
...,
...,
...,
Vector 2A + 7 B + 4C is zero if each of its components is
zero,
i.e.,
2Ax+7Bx+4Cx=0,
2Ay+7By+4Cy=0
and
2Az + 7B. + 4C, = 0
Thus we have three independent equations to determine
Cx, Cy and c •..
On substituting numerical values, we have
2x3+7x(0)+4Cx =0 or
Cx =-1.5
2 x 5 + 7 x (-3) + 4C y = 0 or Cy = 275
2 x (-2) + 7 x (6) + 4Cz = 0 or
c. = - 9.5
-+
A
.A
A
Thus vector C = - l.5i + 2.75j - 9.5k.
1'0/
i
l~~c;im~~~~- 177>-
fCJ
tci,
+
T..;
J
1\vo unit vectors i and are directed along x-axis and y-axis'
respectively. What is the magnitude and direction of the
(d)
vectors
Fig.1E.5
Remark:-------------------
i_+ J and i-) ?
d
y
............
h
In the figure shown, -C =A+ B
j
:
h
i
PR
(a)
...,
The x-, y- and z-components of vector 2A+ 7 B+ 4C are
respectively
x-component = 2Ax + 7Bx + 4C x
y-component = 2Ay + 7By + 4C Y
z-component = 2A. + 7B. + 4C.
X
a
~-'-:--,-+--x
h
-j
<.,~
.
·······~~··
goo
,j
i .
I
Flg.1E.7
(b)
Solution:
www.puucho.com
From parallelogram law of vector addition,
Anurag Mishra Mechanics 1 with www.puucho.com
DESCRIPTION,OF MOTION
·---~~---------------(b) The magnitude of average velocity vector is
1i + j I,;, J1 i 12 + I j 12 + 2I111 j I cos 90'
I 'if av I =Jc2.50) 2 + (2.50) 2
\i\=IJl~l
where
=3.54 m/s
Average speed is measured by the length of the path
travelled
1
.
--- (21t X 25.0)
Average speed =_,4_ _ __
\i+J\=-J2wrlt
tan a
=- ' -\j\sin0
=-'------
\i \+ \ J\cos 0
=
or
.
1-sin90' =l
1 + 1 · cos 90°
= 3.93 m/s
a= 45°
Similarly
We have· to reverse the vector j and add it tp
Ii
-11 =J1 i \
=J1
2
i.
2
+ I j \ + 21 11111 cos 90°
+ l 2 + 0 =-J2 writ
IJ I sin (-90°)
tan a=
v~~:rs .'.'.JI
= _1
vector A is the vector sum
of
the
vectors
...
->
;A girl is jogging along a circular path of radius. 25. 0 m. In 1
110.0 second she jogs a. quarter ofa circle starting from point1
P. (a) Compute her displacement and average velocity, and
(b) Compute the magnirude of the runner's average velocity
and her <_1verage speed.
------y
1
'
'"""'--
vector A by a scalar a
changes all the cartesian
components ' by the same
factor:
->
•
•
a A= a(Axi + Ayj + A,k)
I
Fig.1E.8
(a) Her initial and final position vectors
ri
,...--+
=(25.0) i ;
r1
F-
->
->
->
~r=r1 -ri
=(25.0)j - (25.0) i
The average velocity is
_,
Vav
,...
C-25.0i + 25.oj l
= -----~10.0
:•
:A".t
, yl
:
:
:
:
:
: AI :
A
:
xi
1)·:::::·:......... l.-··
: ~- _
:
.':!!I.:. 1.,_4~J'!L____ j
1 =(F)+Fyj +F,k) + (fxi+ fyj + f,k)
+ fx)i+(Fy + fy)i+(F, + f,)k
->
f =(F)+Fyj +F,k)-(Jxf+ fyj + f,k)
=(Fx
=(25.0) j
The displacement is given by
:-)
, A
You can see ):hat the components of vector sum (F+ f)
are the scalar sums of the respective components of the
individual vectors. Similarly,
are
--+
:,
->
xJ
'--------..... -~-
Solution:
/"~,(~
/~
~--------~------- ' -Azk :
2. Vectors in cartesian form may be added or subtracted
provided they are of same type, representing same kind of
physical quantities such as displacement add to
displacements, forces add to forces. For example
= (Fx
1
.I
=aA)+aAyj +aA,k
F+
P
j
AxtAy],A,k.
The multiplication of a
--------·---------------- ----------~
ri
---- YA
--- - -- --- ---- --,
I
->
I 1I+ I j I cos C-90°)
a= -45°
or
VECTOR MULTIPLICATION
1. The components of a vector are 1
Components
scalar quantities. The multiplication of
l-) - ~ - "
the unit vectors, by the cartesian ,A=:A,_:i +:A,:j +:~:kl
1
components leads to a vector sum of
1
\ __
.
I
three mutually perpendicular vectors.
,_ _'29· 1.44 (a)___J
In the figure shown the
i
i-J=i+(-j)
2
10.0
- fx)i+ (Fy - fy)j +(F, - f,)k
->
->
Hence the components of the vectors F- f are the
scalar differences of the respective scalar component of the
individual vectors.
3. In vector addition and subtraction the vectors
involved. must be of the same type, While in vector
multiplication there are two distinct ways; each of these
ways has its own set of rules notations and applications in
physics.
·
=(-2.50! + 2.50j) \11/s
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
l32
MECHANICS:!'.] .
THE SCALAR PRODUCT OF TWO VECTORS
Multiply a vector by a scalar produces another vector.
The scalar product is different from this multiplication. The
scalar product is a way to multiply two vectors to yield a
,..
-+
-+
scalar result.
product of any two A and B vectors is
..., Scalar
...,
(iv) When the angle 0 between vectors is acute
(0 < 90°' cos0 > 0) the scalar product is positive; if the angle
is obtuse (0 > 90°, cos0 < O) the scalar product is negative.
The scalar product of two vectors that are perpendicular
.(orthogonal) is zero.
written as A · B and defin~d as
..., ...,
A·B=ABcos0
...,
...,
where A and B are magnitudes of the vectors A and B,
and 0 is the angle between them when they are drawn tail to
t.ail.
(i) The
angle
between
two vectors always is taken to be
the smaller angle between the
vectors when they are ·drawn from
a common point. In Fig. 1.45 shown
...,
90s 8 > O;·so scalar
prOdvqti~," Po~itiVe ,
})'\,,,
J.,f
L
...,
angle between vector A and B is 0.
With this convention 0 is always less
than or equal
to 180°.
_
f'lg, 1.45
';"s 90°
(ii) In order to geometrically interpret the scalar
...,
~
Az~~
O; So scalar product' is
...,
product we draw A and.B drawn with their tails together.We
.
.
-+
-+
drop a perpendicular from the dp of B to line containing A.
...,
The quantity B cos0 is called projection of B or component of
...,
...,
...,
...,
perpendicular to A then the shadow of vector B on the line
...,
...,
So we have
component of B on line of A.
j-i= 0
k·i=O
,
...,
-.,.
perpendicular,"to'A , , ' - •
The li~e
-,
cont~~~1ngA.
, .·.
f ....
· ' af'·-'... /2'. -Th~projec~n
~
~
ofBalongAis
·<B cos 8 B cos ·, •
~~
(Note that,B cos a
can be·< 0 if-8 > 90~}
8
·v, ,"
A·A = (Axf+Ayj+A:k)·(~xf+A_,j+A,k)
= AxAx (i-i) +AxAy(l · 1) +AxA, (i· k)
+AyAx cj · i) + AyAy cj · 1) + AyA, cj · k)
:
.
: The hne
~
: containing A
- -i-A,Ax (k -iJ + A,Ay (k- j) + A,A, (k · k)
,_
or
A2
We can also take projection the other way around.
..., ...,
-+ -+
-+ -+
(iii) The scalar product is commutative.
The scalar product is also distributive, Le.,
-+-+
---t
A, (B1 + B 2
-+-+
-+-+
r= A·
B1 + A: B2
= A2 +A2y +A2
X
Z
A= '\J1A X2 +Ay2 +AZ2
A· B = A (B cos0) = (A cos0)B
A·B=B-A
_,
taking the scalar product of A with itself
Flg.1.46
---~.~--
i-J=0. i-k=O
J-1=1 J-k=O
k-1=0 k•k=l
(vi) We can find magnitude
of any vector A by
_...,
Acr:Je ~-;;;,
,t
__J
i- j = (1) (1) cos 90° = O._
also
i-i= 1
Light
I
•
(v) The scalar product of a cartesian unit vector
with itself is unity. For example
i- i = (1) (l)cosO = 1
containing A has length equal to the projection of B or
...,
, - ,
Flg.1.47 ,.
B on i:he line containing A. Imagine light shining
'.
cos 0,< O;,SO scalar"
,. eroduct is negative
(vii) When two vectors are expressed in cartesian form
the scalar product becomes
-+ -)
A
• ',.
A
A
A-B = (Axi+Ayj +A,k):(Bxi+Byj +B,k)
= AxBx + AyBy + A,B,
(viii) Angle between two vectors can be calculated
with the help of dot product:
www.puucho.com
o,
Anurag Mishra Mechanics 1 with www.puucho.com
r
DESCRIPTION OF r~OTION - ...
---+---+
k1!,~~~)~l~:19f~
A·B
cos0=-AB
---+---+
Where
+ AyBy + A,B,
A· B = AxBx
~!:!~h~i;jr_onent of
A= ~A; +A; +A;, B = ~BX2 +By2 +B z2
and
---+
(ix) Work done :
Solution: Let b
---+---+
-:i': 2i + 3] along·~;~; ~irectio~:j
•
•
= (i + j)
---+
W = F- s =Fscos0
---+
The component of a along b
(x) Angle between the vectors:
---+---+
We have,A- B = AB cos0
acos0b=[-;~b},
---+---+
A-B
cose = - -
(2i + 3J) · (i + Jl (i + Jl
)1 2 + 12 . )1 2 + 12
AB
= -'-~====,,....=-'· ,=;;===;;=
A 1 B1 + A 2B2 + A3B3
=,=;;===§===§c-'i=;;====§===;;=
~Af +A~ +A~~Bf +B~ +B~
2x1+3xl (i+Jl
=--~./2.
-/2
5 • •
= -(i+ j)
(xi) Component or projection of one vector .
along other vector :
i----· --------+··-------·-
!
B
/
2
L1;:¥~,~:t~~
I--+_
•
•
---+
•
•
:If A= 3i + 4j and B = 7i +24j, find a vector having the
~
i
~me
'L
Solution: The required vector is= BA
;:
B=)7 2 +24 2 =25
Fig.1.48
·-------· -
---+
--------+
---+
(a) component of vector A along vector B
and
A=A=
A
A cos0B = AB case B
•
•
3i+4j
) 32 + 42
= .!.(3i+ 4J)
B
5
-[A·BJ.
-
~
"
mamftude as B and paralle!Jo A - - - - - - ~
•
1 • •
BA= 25 x-(3i+ 4j)
-- B
5
B
....
= 15i+20J
---+
(b) Component of vector B along vector A
s
r----·-
IO LiJ_cos e
C
..
.-----
-;;·------·------
.
!Under a force (10i - 3 j + 6 k) newton a. body of mass 5 kg!
/moves from position (6i +5]-3k) m to position!
•
•
I
1
j(IO i - 2j + 7 k)m Dedu_c_e the work don!__ ___
~
....A
Flg.1.49_ _ __
Solution: As displacement
i.e.,
---+
---+
---+
s = r 2 - r1
s = (lOi- 2j +7k)- (6i + -j- 3k)
= (4i-7J + 10k) m
Bcos0A = ABcose A
A
=[A~iJA
So
i.e.,
w= i.1 = c101-3j + 6kl-c41-7J + 101ci
---+
W = (40+21+ 60)
= 121joule
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
g· _es "h-.. -,.;l'·~
--·-1.~ - -l··~
~·
f'F~am,,i!.ia
12fo-·
.
~ parti~ie tli~vd.in th~ ·x-~~la~~'µha~i ~he action of &force ~
such th~qlie value of its lin~~r,n;t_~1~iitum p at any time tis
~- -~
.
·~-" .
-~
p';,:",_ '2.cost and.Py = 2 ~int. Wha£is the angle 0 between F
i··_
~
,!/'\;·,_.','.
,,.,·
.
.
'
7
l~~i; at:~
-._,,~·,::,:·_",·i_·
",_
.
~~ t?
given
.
,,
..
'
.
'/
··
..., ...,
~.
'
.
,
...,
...,
...,
Now as,.
.
...,
--+ --+
..
• ( , 'I'
'
•
..
'
~
.. . . •
.
-
•
A
rY
A
=.!.= xi+_yj
r
r
i, .= i~os0+ }sine.
•
•
-+
a=-Psin0
. cos0
Since 10 is a vector of unit magnitude :
a2+P2=1
Therefore, p = + cos0
and
a= -siri0·
Thus,
le = -lsin8 + j cos0
We have taken p = + cos0 and not the other solution,
p = - cos0 because we define a system i,, i0 vectors in the
directions of increasing r and 0.
Or,
: ~~::~[t·;]c:::~-~2[co~t)](:s:::~1= 0
0
Fp2x2
. --+ .
"~~:,::;;;r:-:-1,--,,..,
~€!fjl~J\~J~
13 . ~
b~~:tijlfl'fl~~ 15 ~
[v1-·. two , ;~~-;e;.~ .vectors A' :;:;~-:~
-t
~olution: Here v =
31 + 2j + 3k.
Let 0 ·be the angle
between the line and the velocity-of the particle. Then
3-2+3
4 ,
..
. cos0 = -T22 -./3 ./66.
~
-+
-t·
.-:.,, '., ="22-=-.
~ut
'.,.
., ' '
"i-}+k·
-t-t
. '
,
-t-t
(A+B) · (A+B)
-t
-t
-+
-+ -t '-+ -t
-t-t
-t-t
=(A-B)·(A-B)
-t -t
-t
-+
-t -t
-t ~
-t -t
orA-A+A·B+B·A+B·B=A·A-B-A-A-B +B·B
... (1)
-t-t-t-)-t-t
2
-t-t•.
Substituting _these values in equation no. (1)
.
.
,, /i
Vect~r comp~nent
-t·
(A+B) =; (~-B)
Taking self product of both sides
We know that A·B= B-A; A·A =A and B·B= B 2
· · '
: .-0..,;=. -~
-t
Solution: Given
-t
-If'
·.. .-. ·c. ' . '·.·.
-./3 .
Vector component =c./3 d is the unit·vector along the
obey
(. .
[A+B = A:-B, the angle,betw~enthem_is :.
1(g}_l2Q."...:._.,_·_(b}_90° .· . (f).-60° • '(cJ)~0~0- - ~
. Compon~nt of velocity along this line=lvlcos0
.. •.
.- ' · .. · 4
4
line, :
·.··· . • .
H,
i.e., 0 .= 90° i.e., F and p are orthogonal,
lli
.•
lie of unit !71Cig7Jitude normal to _the ,vectori;a.nd)yi~_(h'tli~
·
'
"
;x,yp)ane.' __ · · __
Flg.1E.14 ', '
where a and. p are coefficients
to be determined. ··
·,
Using the definition of scalar product; we have ·
i,.i 0 = (icos0 + j siri0)(ia + JP)
Now as ·· F'.p = FxPx +FyPy
0
'T,'~-"'l'.~"1,..--::t
x = rcos0, y = tsinB
}low; Let_ le= la+ jp,.' ·
IFI= ~[(-2sint) 2 + (2cost) 2 = 2
.! .
I··
Or,
F = i(-2sillt) +j (2cost)
with
" .
\
dt '
,..., d •. . .. •
F·= dt[i(2cost)'+j(2sint)]
So,
, .---·
I~ (xf + y J),maFirig an an~l-~. 0 with;th,e_jcccms. F/~1:~,teff~r
,ir · of unit rr/agqitu!fe in the qirectl9n of} vector _r and·'! 'vector
ir
IPI= ~[(2cost) 2 + (2sint) 2 ,;, 2
--+
.-'>
dp
_. F=-·I,
'
,--+
= i (2cost) + j (2sint)
'
,
Solution: By definition
Solution: As p=ipx+JPy
.
r-
(A point P)ie-s i!l rl1e x-y plan~. Its po~itiq/l ca~:be ip~iifi,,ed cy
litf -·~· y ~tgoi;,dinates. orby 1;·i11,HaJ1y direc1,i~/~~ptok
=. j3 (f - 'j + ~) .
-:·:'•-i
--~l · .
-t-t
-t-t
. A 2 +2A·B+B 2 ':'A 2 -2A·B+B 2
-t -t
or
-t
4A·B=0
...,
...,
or
-+
C·
.,
4IAIIBlcos0=0
'
Because A and B·are non-zero vectors hence.-,
"
'
·~;;~'
www.puucho.com
1:r
Anurag Mishra Mechanics 1 with www.puucho.com
....
IA!;,, 0
....
(iii) If~ is the angle between vector (2i + 3]) and (I+ j)
and !Bl¢ 0
cos0 = 0 = cos90°
0 = 90°
Angle between two. vectors is 90° hence alternative (b)
is correct.
,.
.....
then by the dot product
ry·
~
A
--·---------- .. - ---"·-·:7
Jand Jare unit. vectors along x,axi;; and y-axi;; respectfve(y._l
IWhat is the magnitude and direction of the vector i + j and:
iY
IfI - ]? What are ·_th_e. magnitudes
of c.omponents of a vectoj
• I
:; = 2f + 3] along the --directio~
of i + j and i :- ]?
---- --------. --------~J:
r;·----
Solution: Use the relation;
IY
....
'h
....
A
(i + i)
~-~----
la+hi=.Ja 2 +b 2 +2abcos0
bsin0
and a= tan· 1
a+bcos0
(i) Angle between i and
is 90°
I(2i + 3])/ cosp I(i + ]JI =(2i + 3J) · (i + j)
.i
li+]I
2+Sj.i+3
5
j
Fig. 1 E.16
--=--- = -
(a)
2
../z
y'
_ 2i-i+ 3J-i-2i-J-3j. j
-
../z
acos(90°+1i)'= -~
.
v2
THE VECTOR PRODUCT OF. TWO VECTORS
(i) The vector product or cross product of two
vectors yields another vector. The vector product of two
~
~
~
~
vectors A and B is written symbolically as Ax B. The
magnitude of the vector product is defined to be
.... .... ....
ICl=IAx Bl= ABsin0
'
' .
where 0 is the smaller of the angles between the two·
....
vectors, The direction of vector C .is defined to be
....
....
perpendicular both A and B. Keep the two vectors until the
tails of the two vectors coincide. The two vectors then define
a plane as shown in Fig. 1.50. The direction of the vector
product is perpendicular to this plane,
7
,_-J
J) ·(i- j)
../z
1. sin 90°
l+l.cos90°
, IJI sin(-90°)
tana =-.~~.---Iii +Ul cos(-900)
-l-sin90°
-1
tan a.' - - - - =
'l+lcos90° l+O
r = -1 = tan(-45°)
-J~~
J) is
.
=.Ji 2 + 12 + O =../z units
~-
Magnitude of component of a along the direction of
(i+
=-1-=ll+O
tana = 1 =}a= 45°
Ciil .-. Ii- j I= J~Ii-12-+~I
J-12 -+2-1,-.Il~Jl-co-s-90-0
j
../z
....
.
umts
. / (2i + 3J) Icos(90°+~) = (Zi + 3
Iii sin0
••
lil+Ulcos0
A
2i-i+ 2i-]+3J-i+ 3j. j
I (2i + 3j) Icosp
=.J1 2 +1 2 +0
Ii+ Ji= ../z units
. -1
D
a-b = abcos0.
(2i + 3])- (i + ]J =I (2i + 3])/1 (i+ ]JI cos~
. _,_,
Ji+ Ji=~/ i./ +1]1 +2i.l. ~os90°
a=tan
A
,_ !:!~: _1E.16Jc)_. j
........
Since these.are unit vectors therefore /i/=IJ I= 1
2
A
(i -
....
(ii) The direction of ve~tor C can be determined by
: = 90° {
-J
1
!
_ _ Ftg.1E;1~(b) ___ J
a vector product right hand rule. Curl fingers of your
right hand and imagine swinging them from ·the directilln of
first vector in the vector product to the direction of the
second vector as shown in Fig. 1.50. The extended thumb of
your right then indicates the appropriate direction of the
vector product.
·
/
www.puucho.com
I
., -
Anurag Mishra Mechanics 1 with www.puucho.com
r-'·
1·. 1i~·,._;:._:·-=-~---~-~--"'--'-,,-~-~-"--'-------'-'.h
r~----..
-
)·
A
A
A
A
'
Ax B = (Axi+Ayj +A,k) x (Bxi+Byj+B,k)
'!
'
I
Ii
--+
--+
~
= (AyB, -A.By)l+(A.Bx -AxBz)J
+(AxBy -AyBx)k
In determinant form we can write it as
j
k
-i
-, -,
"l
l
: -. ,· :· ____ ____ ,
Flg.1.~0 ___ _
AxB= Ax
Bx
j
Ay
!1.
By
B,
-,
(iii) The angle between any vector A and itself is 0',
hence the magnitude of the vector product with itself is zero.
-, _,
, IAxAI= Msin0'= 0
rhe vector product of any vector with itself is zero.
(iv) The magnitude of the vector product is the· area
of the parallelogram formed by two vectors, also, the
magnitude of the vector product is twice the area of the
tri~ngle formed by connecting the tips of the vectors.
(v) If the vector product of the two vectors is zero
and neither vector is of zero magnitude then the sine of
angle between the two vectors must be zero i.e., the two
·vecto;s_ ~re·either parallel or antiparallel.
·
,' 1
--+
. --+
If
AxB=0
-,
:...
_,.,·'
B
~
0, then
-,
--+
'. . --+
--+
--+
--+
::--::7
.....
-,
_, _,
-,
Solution: Resultant vector= a+ b + c
= 1+2] + 3k:-l+2j +k+ 31+ j
= 31 + sj +4k
.
_,
Unit vector"tl. by the property is given by = ~
_,
IAI
Unit vector of resultant 31 + sj + 4k
--+
•
--+
=
--+
';'
_,
·
jixJI= Clf (l) sin 90'= 1
From right·hand rule the direction of vector product is
perpendicular to both i and J, i.e., parallel to k. Similarly
JxJ=O
_,
-,
.
.
,._
. .
,._
+3kx 2j +ix k+ 2jxk+ 3kxk
, = 0+2k-3] + 2k+ 0-61+-j + 2i+ 0
= '-4l-41+4k = --4(i+ j-k)
itself is zer-o.
Jxi=-k
s./2
-,
= '-ix i-2j x i-3kx i.+ ix 2j + 2j x 2f
--+
(viii) ix i = 0, as vect_or product of any vector with
ix'J=k
(3i + sj + 4k)
r = axb = (i+ 2j + 3K) x (-i+2j +K)
Ax(B,+ B,) = Ax B,+AXB2
'ixi=O
~(3)2 + (5)2 + (4)2
=
Vector perpendicular _to vector a and bis cross product
,--+.
--+
3i + sj + 4k
_,
AxB=-BxA
i.e~, ·
. ..
'
The order of the terms in a vector product is important.
(vii) The vector product obeys distributive law as long
as the order of the terms is preserved.
--+
--+
ft= 3l+sj+4k
J3i+ sj + 4kl
Although Ax-Band Bx A have.the same magnitude in
accordance with right- hand· rule their directions are
opposite . .- ,- · .:
•
,,.
-,
, implies
sin9 = 0
Thus 9 is either 0' or iS0'.
--+
.,,.,
un_it vector_ r w_hicli is nonnal to both a and I>. What is thej4
--+
~
[nc!ingtio[l_oi_r and c?
. ·· '
,
·
Ax B = ABsin9 =·o
--+
--+
'
_,
and -
and
---+·
Given a =i+2j +3k, b;= -i +.2j tk and c = 3i + j.
Find the argle of t.,_;ultant :with ~:aii,s: Also find a unitv_'iicto~
in the direction of the resultant of these vectors. Also.find a
ixk=-J
--+ --+
--+
--+
r-c=lrllclcos9
-4(i + j- k) · (3i + J) = l--4Ci + J-klll (3i + JJI cos9
-4(i- 3i+ j · 3i-k- 3i+ i- j + j ·J-k· j)
= ~42[1 2 + 1 2 + (-1) 2] ~(3) 2 + 12 cos9
=}
]xk=i
kxi=j
kx]=-i
kxk=O
(ix) The vector product of two vectors expressed in
cartesian form ,as
'
www.puucho.com
-4(3 + 0-0+ 0+ 1- 0) = 4,,13.JTh cos9
' -4
cos9=--
../3o
1
e~cos-
(~)
Anurag Mishra Mechanics 1 with www.puucho.com
[~sc111PJl~N o(r.ijT101(__ -_-___________
~>i~mi:ili Gal>
,- -
X
.
,If~= 2i -
3] + 6k and b = 6i + 3] - 2k,
:
--+
find the angle,
Xo -••••••••••••••••••
--+
,between vector a and b. Also find unit vector perpendicular to
->
->
poth a _and b.
Solution:
->->
a- b
(i)
lo
=ab case =a, b, + a2b2 + a3b3
Fig.1.51
where
=~(2) 2 + (-3) 2 + (6) 2 and
b =~(6) 2 + (3) 2 + (-2) 2
Concept: 1. Note that in this manner we obtain the
coordinate x of the point on the given moments but not the
distance travelled.
and b =7
7.7 case= 2.6+ (-3)(3) + 6(-2)
49cose =-9
-1 -9
e=cos 49
a=7
or
or
(ii) We know
->
2. The distance travelled can be found from the
coordinate x only in the case when the particle moves in one
direction.
->
ax b represents a vector which is
->
->
perpendicular to both a and b.
i j
->
axb= 2 -3
->
Now
6
3
k
6
-2
=-12i+ 40] + 24k
->
r--------ax bl= ~(-12) 2 + (40) 2 + (24) 2 = 4v'145
->
1
->
->
•
axb
.
-Unit vector n : : : _, ->
-12i+ 40] + 24k
-3i+ 10] + 6k
4v'145
"'145
n=---=~--
In the graph of x-t show the coordinate of point cannot
be greater than x 0 although after time t 0 the distance s(t)
travelled by the point exceeds x 0 while the coordinate x
becomes less than x 0 •
Concept: If a car moves constant(y in one direction the
distance traveled ls equal to the coordinate describing the
motion but when the direction of motion changes to the
oppo.site direction the distance travelled still in creases while
the coordinate decreases.
Velocity of a Particle in Rectilinear Motion
lax bl
•
The distance x can be measured by taking snapshots of
the moving particle at definite moments with a fixed
cemera.
RECTILINEAR MOTION OF A PARTICLE
When a particle is in a rectilinear motion it moves along
a straight line, its distance from a fixed point on the line
increases or decreases with time. In such cases we associate
a reference frame with that _straight line and consider fixed
reference point as origin.
In order to completely determine the law of motion of
particle the coordinate x of the point with respect to origin
and as function of time must be known.
To plot the graph of the dependence of the coordinate x
on time t we choose a certain length scale and put values of
the coordinate x on the axis of ordinates and time t on axis
of abscissas.
The velocity of a point is a physical quantity
determining rate of change of the coordinate with time. The
magnitude of the average velocity is equal to the ratio of the
distance travelled by the point to the time taken. If particle is
at x 1 at time t 1 and at a point x 2 at time t 2 then its average
velocity is
Concepts: 1. Average velocity depends on the time
interval for which it ls computed.
' average velocity ls the same
2. If in a given motion, the
for any time interval the motion has the constant velocity
and ls said to be uniform.
3. In case of uniform motion which starts from the origin
there ls no difference between the value of the coordinate and
:that of the path travelled.
4. If a particle travels unequal distances in equal time
intervals its motion ls seid to be non-uniform. In a
non-uniform motion the average ve_locit.)r ls no longer a
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
~[3::_:s=··==·=·=:0='>·::::·:~::;1~::L$::>:'.:;====:=:'·::7!7::;·,.1::::1_;::·-~;:::===::::::::::~::;:::;:;{li(\r,~:·=~--'~~r·;t""';-~~~~-·~-~·---"--·-::=~·-:,;~:._~}>
5
constantqii~nt.ityanddependso~ti~eintervalforwhit:li tt-;;J
• d
.
;~
.
.
l .
'
co.mpu..te : F,·._P___ r,_a. non-um,_,orm. mo.ti.on av_e.r.age ve oc1ty·c-·annot1·
describe. the:V9riations of motion of.the body. Such '.a IJ!(?tion
can be described adequately by' instantaneous velocity. . ..
v(t) = Lim v·
M--+ 0
av.
-/ :~~·.'
'ME($~1,CS-:I'
= Lim t,x = dx(t)
. dt
M--+ 0 --6.t
i.e., the instantaneous velocity is the derivative w.r.t.
time of the position function.
+In the graph of xversus t the slope at each point (at
each instant of time) is equal to the value of
instantaneous velocity v at that point. Note that slope
at t is zero, i.e., instantaneous velocity is .zero.
~--~-,-- -~= .... - ... - - ~ - - - - ··---~
1. Average velocity:
Fig. 1.52 (a) shows the
positions of a car at times t 1 and t 2 •
Average velocity is defined as the ratio of displacement
t,x to time interval .M.
2
X
.j
(a)
F>O~itlQn_ yersu~_ume,:_g;:apti,.
:.: ...'........ ~:.' _, ... 7i:~>~2
+Fig.1.52 (b) depicts average velocity graphically. Join
initial point P1 and final point P2 by a straight line.
Slope of this li;,e is t,x. Hence, the average velocity is
.
t,,~
.
the slope of the straight line connecting points (t 1 , x 1 )
and (t 2 , x 2 ).
2. Instantaneous velocity: If we decrease time
interval M, for very small time interval, the line P1P2 will be
tangent to the curve at point t 1 • The slope of this line is
defined as the instantaneous velocity at time t 1 •
Insta;,taneous velocity is velocity at a single instant of.time.
Mathematically, it is defined as
I ... X
r
''·'·
i
lt.
J''"
• •
-~~
·S
,£1 ·...
:,c-,.-;~-.
..
.•
,,i~ . . '.
R2
A_tri
•
'
:
+The speed of an object is the magnitude of its
velocity.
_,
d r · ..
Speed= I_V I= I dt I
Since sp~ed is the magnitude of a vector, it is .a scalar
quantity that is never negative.
3. Acceleration·: Acceleration of an object signifies
how rapidly the object's velocity is changing, both in
magnitude and direction, whether the object is speeding up
·
·
or slowing down.
..:.'·.:jJ
V2 - V1
Average acceleration =--"-~
t 2 - t1
·..
;', .. I•
'.
I[,_.
·~~·~·F~ig~1.5~.:_~~
I
. a
Instantaneous acce1eration
www.puucho.com
=
,iv
M
L;~ ,iv
t..,-,q M
llll
-
=-dv
dt
Anurag Mishra Mechanics 1 with www.puucho.com
--as;;i
+ In velocity versus time
graph (Fig. 1.55), the slope
of line P1P2 gives average
acceleration and the slope
of tangent at point P gives
instantaneous acceleration.
+ In accelerated motion v aJ?,,d
a are in the same direction.
I
I
I
ds2
l ds
L,'____F1~g._1_.s_1_ _ __,
Similarly for distance
Jds =total distance
Jas= as,+ as 2 + dss ;t-, .. +as.
=> total path length or distance
From Fig. 1.58. Displacement vector can be expressed as
·.,!1
Fig.1.55
_,
_,
4. Equations describing motion with constant
acceleration:
vector sum of d x and d y vectors
=v 0 + at
2
x(t) = x 0 + v~t + (1/2) at
ds=dx+dy
· v(t)
v
2
=v~ + 2a(x -
x
=x 0 + (l/2)(v 0 +v)t
_,
... (1)
_,
_,
... (2)
... (3)
x0)
... (4)
v(t)--, velocity at time t
v O --, velocity at time t =0 (initial velocity)
x(t) --, position at time t
x 0 --; position at time t =0
a ~ acceleration
+Thus displacement in time interval tis x(t) - x 0 •
+For constant acceleration, the velocity varies linearly
with time hence _the average velocity is the mean
value of-the initial and final velocities
Vav. = (1/2)(v 0 + v)
This relation is valid only for constant acceleration
motion.
Total Displacement and To~I Distance
Consider a particle that moves along path represented
by arrows. Entire path can be divided in very small
displacement segments.
--+-
Path of
particle. ""'. -
as =Id-; I= -Jcdx) 2 + Cdy) 2
J-Jcdx) 2 + (dy) 2 =total distance or total path~length
Graphical Representation of Motion in One Direction
If maximum power of xis 1 and maximum power of y is
1 graph is straight line.
·
• y=mx+cmx
X
I
~:--i
I
'
---->
vector AB represents net displacement
--+
'~~~·1,; :
y2 =4ax
0
x2 = -4ay
-- ----'
Flg:1.60
:
y2 =-4ax -
'... . - .
·--~------~-'---'
If maximum power of x and y is 2 graph may be circle,
ellipse, two straight line ~tc.
Flg:1.56
·--+
[,
-~~~
-
--+
.
V
'
i ·;· "'-""'.._.~·l.:.n.•
-~
-,1
. _...___ ___
.. .,,~~-~L----------'--~Flg.:1'.59·
x2 =4ay
--+
.
:iz····,;,v.
- i x ~ x " •'; -~, ,/'' -··_ '·_'''x
L----~-'-'- - - - ~ - ~ -
--+
sR
Ex,
. ',
()/.. r) ...
,,..--,._-..'::::,.:_~~- si
.'
X
,,
_,
i~S1
$~,
y-=:mx-c
If maximum power of xis 2 and 'maximum power of y is
1 cir vice-versa graph is parabola.
•.
Jd s =net displacement
-,..;
dS3
Fig.' 1.1!8
--+
= Jds=ds1 +ds 2 +ds 3 + ... +dsri
s-t
curve
If we puts on y-axis and t on x-axis for every value oft
we have a value of s.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
(b) Uniform acceleration:
We have a particle moving
with uniform acceleration a
and initial velocity u. Its
displacement s at any time t
can be represented as
1 2
s =ut+-at
1. The average velocity from time t 1 to t 2 will be
v av = s2 - s, = slope of ll~e joining p 1 and' P2
' . t2 -t,
For a particle moving along a straight line when We
plot a graph of s versm t, v av. is the slope of the
straight line that connects two particular points on
the s(t) curve: one is the point that corresponds to
s2 and t 2 , and the other is the point that
corresponds to s1 and t 1 • Like displacement, v av. has
both magnitude and direction (it ,is another vector
quantity). Its magnitude is the magnitude of the
line's.slope. A positive Vav. (and slope) tells us that;
the line slants upward to the. right; a negative v av. · .
(and slope), that the line slants downward to, the
right.
2. Instan~ani,ous velocity:
According to definition
.
P2
s ------,·--·-.. t,i
$ ,,2
"' '
v= 11m-
t;
M--)o .1t
In ·curve, if M ~ 0 the i s, ., ...
point p 2 comes very close ,t,"b , :·
to point p 1 .•
t-
Notg;)------~---~::;::==~:;:;:;
ts,
If velocity is· uniform slope of
curve
must
remain
unchanged.
Curve
with
uniform slope is straight line.
If velocity is 1 ms-1
=> s = vt => s = t
tane = 1
r-·· -,-.. -
-., ..--- -----··· .......... - ·. ·""'" ..•. -,
a!
1kcydiststa,;t;1ngfrom a point Atravels 200 ~ due' north'to
!point B, at constant speed of 5 ms"'' He. rests at B for, 30i
:sec?ndf q~d ~he,:, :travelsc300_ :m <ll!lc south a point~ ~~
1cons.tant speed p{J O ms 1 • F,1nd~qvera1<e velocity._ . , , ... , i I
Solution: ...,..., ... ~---· ____" ..
w
7
B200
•'
100
A
C-100., _20_ ••.• _( ... 0·1
r. ·
'Fig.1E.11!_,__
: _ __
1.·
From A toB
x=200; v=S; t=40sec
FromB to C
Displacement = 300 m, time taken = 30 sec
Net displacement = -100 m
Total time = 100 sec
,
l
-1
. · = - 100
Average ve1oczty
- - = - ms ..
100
,
v-t curve
Cases:
(a) Uniforin velocity:
Wir-=~-.'>!G-~-,~~ii;J 19 ~
~1
'~---''-~;..;_\~j
The instantaneous velocity can be found
by determining the .slope of the tangent to
the displacement time graph atthat instant.
Velocity at point p1 or time t1 is v.
v = tan a
2
Curve is parabola.
Velocity at t 1 is tan 8.
t
J'
. ,S .
21-,,~~.--
By using dependence of v on t we
can plot a v-t graph.
* Slope of·v-t curve at any point
represents acceleration at that
instant.
tan 8 = acceleration at time t 1
Area underv-t graph andt-axis.
As we know dx = vdt
and f.vdt = x = Area ~nder v-t graph.
~--.....-n--...~
t
.
1-
www.puucho.com
" 't,
j
l
2
~"~-~·-~Fig0 J_.67 _ _:___~
al
Anurag Mishra Mechanics 1 with www.puucho.com
[P.escRrPrioti"o(~.-o_ti_ori_·-----'----~-·-~---~~,_.::·~-------~--__ __···_··--_-_-- -·-·-- .. ·.i1J
Thus, area under curve will represent displacement
in that time period.
Alternative:
= 10-2t
V
No\11_:~.-------------------
Area (1) = .! x lOx 5 = 25
2
(1) Area above /-axis +ve displacement.
(2) Area below I-axis is -ve displacement.
Area (2) = .! x 3 x 6 = 9
2
Displacement= 25- 9 = 16 m
Distance = 25 + 8 = 34 m
Objective:
Thus,
1. Total displacement will be sum of areas with
appropriate signs.
. 1. Using graph, distance can be calculated directly.
2. Total displacement will be sum of areas with
appropriate signs.
2. Total distance will be sum of areas without sign.
Cases:
(1) For uniform velocity:
acceleration = 0
slope= 0
!!~_!I
I
i
1
Ivlu
;I_
(slope is -ve) i.e., 0 > 90°
Note: 0 is always with +ive
x-axis.
[kl..i:=2&9me;'fec-.:....~~
. .
. --r::-1----
L
r:·----- -. -.. . ---- --------- ·---- . - --- -A particle moves .in a straight line with constant velocity of 5 '.
rr acceleration l !ms
· for 2 seconds. It then moves With a constant accele~ation'
(2) For uniform straight line curve:
tan 0 = acceleration
For increasing velocity:
tan 0 = acceleration
For decreasing velocity:
1
Flg.1.68 (a)tj
3. Total distance will be sum of areas without sign.
4. To plot straight line using equation of motion.
it
lv
'!
I,
8
t-+ii
Fig. 1.68 (b)
of-2 ms·2 for 8 seconds. Draw velocity-time graph for 10:
!seconds of motion and find.
'
t!l) _Ejrzql 11_ijgs:_ity_ ___(/J) Qisp_lgc_e_me_nt__ _(c)_ 1htal_dfstance '
1
Solution: Area (1) = 5 x 2 = 10
,r·
!
decreasing
l
1:......:
~ __ Fig.1.68(c)
5
I'
a
·1
I
s-.! x 2 x s x s
V
--- 2--_-·p-:--7
I10
2
i
_
·A particle is travelling in a straight line. It has a. initial/
;velocity of 10 ms-1 • When it is subjected to an acceleration of:
-2 ms-2 for 8 seconds. Find displacement and distance]
:tn:1Ver~<il1J {l_s~fol!!fs~-- ___ _ _ _
__ ____
:
1 :' 2
'
6
8 10
4
3
j-11+-----'>I
-----------·---- --- - - -- -- ------,
Solution: s = 10 x
-- V- ---
I
,_ _ ---- _F_i!!:..~~-2_1__ --------'
Area (2) = .!_x 5 x 2.5 = 6.25
2
Area (3) = _ _! x (11) x 5.5 = -30.25
2
Displacement= -14m
Distance = 46.25 m
r - - - - · . . - - - -- _,:, _________ - - -- - - --·
L._ -~AlCULUS SUPPLEMENTARY
Differentiation
Derivative of a Constant Function :
~(c)
1-6
dx
I
Fig.1E~D____ j
Displacement= 16 m
Displacement = s1 = 10 x 5 Now u
.! x 2 x 25 = 25 m
2
= 0; a= 2; t = 3
1
s2 =--x2x9=-9
2
=0
The Power Rule :
If n is a positive integer, then
- d ( X ") ::::;nx n-1
dx
The Power Rule (General Version) :
If n is any real number, then
www.puucho.com
-d
dx
( X ") ::::;nx n-1
Anurag Mishra Mechanics 1 with www.puucho.com
Solution: (a) since f(x) = X-2 , we use the power
rule with n =-2 :
) _- -dX
( -2) _ -X
z -2-1
f '( X
dx
.
= -zx-3 =
_2
x3
dy = ~',/[x2 = ~ (x2/3)
dx dx
dx
2 (2/3)-1 ==-x
2 -1/3
=-x
3
· 3
The Constant Multiple Rule
If c is a constant andf is a differentiable function, then
(b)
d
d
4
d· 4
3
3
e.g., (a) -(3x
) = 3-(x ) = 3(4x ) = l2x
dx
dx
Thus dy/dx= 0 if x = 0 or x 2 - 3 = 0, that is, x = ±-Ji
So the given curve has horizontal tangents when x =0, ../3,
and-.J3. The corresponding points are (0,4), (../3,- 5) and
(-../3,-5). (~ee Fig. lE.23)
The Product Rule :
If f and g are both differentiable, then
! [f(x)g(x)] = g (x) ! [f(x)] + f(x) ! [g(x)J
Solution:
(bl ~(-x) = ~[(-l)x] = (-l)~(x) = -1(1) = -1
dx
dx
dx
The Sum Rule :
If f and g are both differentiable, then
.
d
d
d
-[f(x) + g(x)] =-· f(x) +-g(x)
dx
·dx
dx
The Difference Rule :
If f and g are both differentiable, then
By the product rule, we have
d
. d
d
f (x) = -(xex) = x-(ex) + ex -(x)
dx
dx
dx
= xex + ex. l = (x + l)ex
l&~~~lru\BJ~~l25j~
'Differentiate the function f(t) 7 ./t(l-t). ·_ ·. ·_ .•/
![f(x)-g(x)]= !t(x)- !g(x)
~
(x 8 -12x 5 dx
4x 4 + 10x3 - 6x + 5)
.
. .
= ~(x 8 )-12~(x 5 )-4~(x 4 ) + 10~x 3 - 6~(x)
dx
dx
dx
dx
dx
+
!
(5)
=Bx -12(5x ) - 4(4x 3) + 10(3x 2 ) - 6(1)- 0
=Bx 7 - 60x 4 -16x 3 + 30x 2 - 6
7
•.• _ Flg.fE.23 _ .. ··-
1 E•Yd"'? · I e '! 24 t>,c..
l'.s:+-~3.lf:iiiJ!.t':.\~~-:.~
d
dx [cf(x)] = c dx f(x)
e.g.,
("3,-5)
4
Solution: Using the product rule, we have
r:: d
d
f (t) ="'' -(1-t) + (1-t)--./t
dt
dt
=-./t(-1) + (1- t) .!r-112
2
=-.Jt + 1-t = 1-3t
2.Jt
2.Jt
We can also proceed directly without using the product
rule.
f(t) = .Jt -t.Jt = t!/2 -t.3/2
f (t)
~~~·p~~
!Find thep~~nts on the curve
'la~nt.line_js horizontaL ·.
i. = x 4
=.!t-1/2 2
-
6x 2 + 4 wher;-thei
......J
•.•·
.:7
Solution: Horizontal tangents occur where the
·derivative is zero. We have
~t!/2
2
-----··-----···]
--·······-·-·---·-.
':". .ftg(x) where g(4) =>2 and g' ( 4) =3,find']' (4}'.
i1J f(x)
Solution: Applying the product rule, we get. .
dy =~(x;)-6~(x 2 )+~(4)
dx dx
dx
dx
www.puucho.com
f (x) = ~[.ftg(x)] = .ft~[g(x)] + g(x)~[.ft]
dx
dx
dx
Anurag Mishra Mechanics 1 with www.puucho.com
. = ./xg' (x) +g(x).! x- 112
'
2
= -fig' (x) + g(~
.:;
:Differentiare-~X = x 2 sinx
2'1X
'
' (4)
.
Solution: Using the product rule we have
dy_ = x 2 .!._ (sin x) + sinx.!._ (x 2)
dx
dx
'
dx·. '
2
= x cosx+ 2xsinx
2
f(4l = "4g'(4J+L = 2-3 + - = 6.s
..
2"4
2-2
·
The Quotient Rule:
If f and g are differentiable, then
.
d
d
.!._. [f(x)] = g(xla;:[f(x)]- f(xla;:[g(x)]
dx g(x)
y=
e.g., Let
d
'
y=
.
~' ' '
'2
.
',,
u.
sinx.
.
1 we
smg tan x = - and quonent
rue,
cosx1
have
d
2
~(tanx) = j__(sinx)
dx
dx cosx '
3:
(x +6)-(x +x-2)-(x +x-2)-(x +6)
. dx
dx
d,(" )··.
d()
cosxs1nx - s1nxcosx
= _ __,,dx=-·_ _ _ _.,,dx,,___
(xs + 6)2
3
(x +6)(2x+ll-:-(x 2 +x-2)(3x 2)
=
(xs + 6)2
=
1
='tan x
.
S oIut10n:
+x~2.
'
.·, ·I'
x 3 +6 ·· ·,
:'
X
then
3
!Differentiate y
·[g(x)]2
,
·2 ~
;
- - - - :..:::_~ ~ !
(2x 4 + x 3 + 12x + 6)- (3x4 + 3x 3
-
'cos 2 x
cosx. cosx-s'inx(-sinx)
coS 2 x
2
cos x+sin 2 x
1
2
=
=--=sec
x
2
2
=
6x 2)
cos x · ,.
cos · x
d
2
dx (tan x) = sec x
"
Derivatives of Trigonometric Functions
!Find an . ~qll~dpn 'oA th~ ta'ns-~nt" Une to the ·curve!
. "
I
"'ex /(I+,:X~~a(the·poiht '(1,e/2),, ·
'
---·--'l
IY
Solution: According to the quotient rule, we have
(l+x 2 )~(ex)-ex .!._(l+x 2)
dy
dx
dx
dx=
(l+x2)2
- (l+x2)2
,-'
•
'•
C
dx
·: '
"
- (l+x2)2
dyldx -o
x=I
This means that the tangent line at (1, e/ 2) is horizontal
and its equation is y = e / 2
b~~~.Jii
Differentiation Formulas
[;nd F (x) ~{1,'(x) =
I
I
"
~(e"J=e'.·
"'
II
(j+g)'=f'4:g'-·
(f-g)'e. f"-g'
l
'
<
(cf]'= cf.'
(Jg)'= h'fgf,
·>-
.
.,'' - (x") =·n,x':',, ·..
dx
/"
d
'\
(f)'
dx
)• .
.,
= gf:'C.:.~'
\
• -., ·-2- .•
g'.
.. '
'
'
'"
',•,
di,
.
.,
~
,
'.r ,-' -~-:,~~· '
-;·
,.,d . - ' ' ·. : '
··-(sec·xJ a:a secxtanx
'dx
·,.
·-~ -.
,; '
d'' ' .·
.
· ,~{cotx) = -cosec2 x
,:dx
'
'
The Chain Rule :
If y = j(u) and u = g(x) are both differentiable
functions, then
dy dy du
-=-dx du dx
so the slope of the tangent line at (1, e/2) is
d'
' .'
-(c)=O
dx
' '
d'-
~(~ose<:x) = 7 cose~x,cotx
',
d ' ·, . ·- ' :
- (ta11x) a= sec2 x
, _ (l+x 2 )ex -ex(2x) _ ex(l-x) 2
-
'd
-(smx) = cosx
dx i
d .. .
dx (cos~) = -sin x
30
~
'
E¾1) ::·:·~~-;- ?"'T\, :.i]
Solution: If we let u = x 2 +land y
'
.
www.puucho.com
=·.Ju, then
F (x) = dy du = -2_ (2x)
du dx 2./u
· . 1
X
= ~==(2x) = ~==
2.Jx 2 +l
.Jx2 +1
Anurag Mishra Mechanics 1 with www.puucho.com
:44
-~- ~- :.~ -_~_______MECHANICS-lj
Concept: In using the chain rule start from the outside
to the inside. In chain rule we differentiate the outer function·
f [at the innerfunction g (x)] and then we multiply by the,
.derivative of the inner function.
i
2
1. If y = sin(x ), then the outer function is the sine
function and the inner function is squaring function. So the
chain rule gives.
dy = ~ sin
dx dx '---,---'
(x 2 )
outer
function
=
evalilated
at inner
function
cos
(x 2 ) • 2x
'---,---' ' - - - v - - ' '-v---'
derivative evaluated
of outer
at inner
function
function
derivative
at inner
function
Solution: Firstrewritef:f(x) = (x 2 + x+ 1)-113 • Thus
f (x)
outer
function
derivative
of outer
function
~--~
evaluated
at inner
function
derivative
inner
function
= 2sinxcosx
In general, if y = sin u, where u is a differentiable
function of x, then, by the chain rule,
du
-dy = -dy- du
- = cosudx du dx.
dx
d
(
'
)
du
Thus
- smu = cosudx
dx
2. Ify = [g(x)]", then we can writey = g(x) = u" where
u = g(x). By using the chain rule and then the power rule, we·
,get
dy dy du
n-1 du
[ ( )] n-1 , ( )
dx = du dx = nu
dx = n g x
g x
3.
then
If 11 is any real number and u = g(x) is differentiable, i
-d (u ") =nu n-1 -du
dx
dx
dx
3
= _ _!(x
3
;-L
t,._
.
.. ·. . .
2
+ X + l)-413(2x+ l)
r:::::l -:-
,l;;,>fJ::!.!\'.QJ21.~ , 33 I_'"'.>
~
- ~--'""·-----. -·
,,
·Find the derivative of the function
t-2
g(t)= ( 2t + 1 )
= 2xcosx 2 i
Similarly, _note that sin 2 x = (sin x) 2
dy
d
2
- = - (sinx)
= 2
(sin x) . cosx
dx d x ~ - - ~
...._,_.,
'------,--'
= _.!(x 2 + x+ 1)-413 ~(x 2 + x+ l)
9
.. ,'
Solution: Combining the _power rule, chain rule and
quotient rule, we get
g'(t) =
=
9(;t-}lr ! (;t-}lJ
9( t - 2
2t + 1
J (2t + 1)1- 2(t - 2) = 45 (t - 2)
8
8
(2t + 1) 10
(2t + 1) 2
LE~,~~Rl~ 1341;>
Differentiate y
= e'inx.
Solution: Here the inner function is g(x) = sinx and
the outer function is the exponential function f(x) = ex. So,
by the chain rule.
dy
d
.
. d
.
- =-(esmx) = esmx -(sinx) = esmx cosx
dx dx
dx
Note:·-------------------
We can use the chain rule to differentiate an exponential
function with any base a > o
8 x =(elnay =e(lna)x
the chain rule gives
r:.,E-~fl-~Rf~ _: ~1 )>
!!.... (a')=!!_ (el lna)x) =e(lna)x !!__(Ina )x
dx
:Differentiate y =_ (x~ -1) 100 .
Solution: Taking u = g(x) = x 3 -1
we have
dy
dx
and
n = 100,
= ~ (x3 -1)100 =100(x3 -1)99 ~ (x3 dx
Find f (x)
,
if f(x) =
dx
In a =ax In a
because In a is a constant. So we have the formula
!!....(a')=a' Ina
dx
1)
dx
= 1OO(x 3 -1) 99 3x 2 = 300x 2 (x 3 -1) 99
lJ;:~~~-~gJ~:
dx
=e(fna)x_
RECTILINEAR MOTION
G'~.L>
1
~x 2 +x+l
(MOTION ALONG A LINE)
We will assume that a point representing some object is
allowed to move in either direction along a coordinate line.
This is called rectilinear motion. The coordinate line might
be an x-axis, a y-axis or an axis that is inclined at some
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
45
LD~S~RIPTION OF MOTl~N__
angle. We will denote the coordinate line as the s-axis. We
will assume that units are chosen for measuring distance
and time and that we begin observing the particle at time t
=O. As the particle moves along the s-axis, its coordinate is
some function of the elapsed time t, says= s(t). We calls (t)
the position function of the particle, and we call the graph of
s versus t the position versus time curve.
For example, in figure the rabbit is moving in the positive
direction between times t = 0 and t = 4 and is moving in the
negative direction between times t = 4 and t = 7.
2. There is a distinction between the terms speed and
velocity. Speed describes how fast an object is moving without
regard to direction, whereas velocity describes how fast it is
moving and in what direction. Mathematically, we define the
instantaneous speed of a particle to be the absolute value of its
instantaneous velocity; that is,
instantaneous ]
[ speed at time t
Particle
is on the
=[v (t) [= Idsl
dt
For example, if nvo particles on the same coordinate line
have velocities v = 5 m/s and v = -5 m/s, respectively, then
the particles are moving in opposite directions but they both
have a speed of [v [= 5 m/s.
positive side
of the ori in
Particles is on th
egative side o
the origin
L E;:x~_t:r\p}~
Flg.1.69
Fig. 1.69 shows a position versus time curve for a
particle in rectilinear motion. We can tell from the graph
that the coordinate of the particle at time t = - s0 and we
can tell from the sign of s when the particle is on the
negative or the positive side of the origin as it moves along
the coordinate line.
INSTANTANEOUS VELOCITY
L.~5_i:-->
Let s(t) = t 3 - 6t 2 be the position function of a particle
moving along an s-axis, where s is in meters and t is in
seconds. Find the instantaneous acceleration a (t) and show
the graph of acceleration versus time.
v (t)
Solution: The instantaneous velocity of the particle is
= 3t 2 -12t, so the instantaneous acceleration is
dv
a(t) = - = 6t -12
dt
a
40
The instantaneous velocity of a particle at any time can
be interpreted as the slope of the position versus time curve
of the particle at that time. The slope of this curve is also
given by the derivative of the position function for the
6
8
particle.
,
Concept: 1. The sign of the velocity tells us which way
the particle is moving a positive velocity means thats position
of particle w.r.t. origin is increasing with time, so the particle
·is moving in the positive direction ; a negative velocity means
that s is increasing with time, so the particle is moving in the
positive direction; a negative velocity means that s is
decreasing with time, so the particle is moving in the negative
direction (Fig. 1.70).
s(t)
s(t) increasing
v(t) = s' (t) "° 0
s(t)
s(t) decreasing
v(t) = s' (t) < O
(b)
(a)
Fig.1,70
-40
Acceleration versus time
Fig. 1E,35
and the acceleration versus time curve is the line shown
in Fig.lE.35. Note that in this example the acceleration has
units of m/ s 2 , since v is in meters per second (m/s) and time
is in seconds (s).
Concepts: 1. A particle in rectilinear motion is
speeding up when its instantaneous speed is increasing and is
slowing down when its instantaneous speed is decreasing. An
object that is speeding up is said to be "accelerating" and an
object that is slowing down is said to be "decelerating", thus,
one might expect that a particle in rectilinear motion will be
speeding up when its instantaneous acceleration is positive
and slowing down when it is negative.
2. This is true for a particle moving in the positive
direction and it is not true for a particle moving in the
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
f45 ..· ..... ··- ··-· L--------·
~
·-····-· -- ----, -·- - - ...
;negative direction-a particle with negative velocity is speeding
•up when its acceleration is negative and slowing down When:
its acceleration is positive.
i
'
'
i " This is be.cause a positive acceleration implies an:
'increasing velocity and increasing a negative velocity:
idecreases its .absolute value : similarly, a negativei
;acceleration implies a decreasing velocity and decreasing a_l
;negative velocity increases its absolute value.
'
3. Interpreting the sign ·of acceleration
A:
'particle in rectilinear motion is speeding up when its velocity
;and acceleration have the same sign and slowing down when
·
,
,they have opposite ,signs.
,
4. From the velocity versus time curve and the!
!acceleration versus time curve for a particle with position
functions s(t)=t 2 -6t 2•
I
I
.
'
•
Over the time interval O < t < 2 the velocity and'
!acceleration are. negative, so the panicle is speeding up. This·
iis consistent with the speed versus time. curve, since the· speed!
is increasing over this time interval. Over the time interval!
1
i2 <t < 4 the 'velocity is negative a:nd the acceleration isl
•positive, so the particle is slowing down. This is also consistent:
;with spied versu,s time. curve, since'.the, speed is decreasing overl
!this time interval. Finally, on the time interval t > 4 the!
!velocity and acceleration are positiVe, so the particle isl
:speeding up, which again is constant with the speed versus
ltime curve. .
.
·'
--·--·---·---·-·-----·~-
-
--
-·----·--·-·-·-·-·----
-~---·-·-·
,-
.
.
: MECIIANics:,-1
- ~---------··~-,----- ··-········--· ·1
·,,_)
- -· .. ·-·-. .
Concepts regarding position versus time: curve;
The position versus time curve contains all of the j
·significant information about t/le position and 'velocity of a 1
'particle in rectilinear motion.
I
: !· If s(t) > 0, the particle is on the positive side qfthei
;s-axzs.
_
_
_
1
·
2. Ifs (t) < Q the particle is on the negative side of thei
s-axis
·
··
;
3. The slope of the curve at any time is equal to the 1
:instantaneous velocity at that, time.
· . /·
I
4. Where the curve has positive slope, the velocity is 1
:positive and the particle is moving in the positive direction.
5. Where the curve has negative slope, the velocity is
negative
and the particle is moving in the negative direction. 11
1
I 6. Where the slope of the curve is zeta, the velocity is ;zero, I
:and the particle is momentarily .stopped.
· '
:
7. Information about the acceleration of a particle in 1
:rectilinear motion can also be d~duced from the position!
'versus time curve by examining its concavity. Observe that ilie
:position versus time curve will be concave up on· intervctl.s
;where s' (t) > Q and it will be concave down on intervals
1where s" (t) < U But we know from ( 4) that s' Ct) is the
1instanta11eous acceleration, so that on intervals where "the
,position versus time curve is concave up the particle has a,
,positive acceleration, and on inte_rvals 'where it is. concave'
!down the particle has_ a negative. a~celeration. -- , __ · _• '
j
l
J
_._,d _ _,
Summarizes Our Observations About the Position versus Time Curve
i Position versus time curve
kl
lli!
_______ , M,!
I
I
'
.
lo . ,
-- ,,---
·~·
'
.
•
_·
'
'
I
I
l
I
I
tI
'
"
·- ··- .
-
!
!
, ... : -_to:
* Curve has positive slope
* Curve is concave down
*s(t 0 )>0
* Curve has negative slope
* Curve is concave down
*s(t 0 ) < 0
* Curve has negative slope
* Curve is concave up
'
'
6;
,_
>0
,
,'
~
*s(t)
- ti.
. .. !o __ i
--~ -
.
Characteristics of the curve at t
.. ,
=! 0
Behaviour ofthe particle at time t
= t,
* Particle is on the positive side of the origin.
* Particle is moving in the positive direction.
* Velocity is decreasing.
* Particle is slowing down.
* Particle is on the positive side ol the origin.
* Particle is moving in the negative direction.
* Velocity is decreasing.
* Particle is speeding up
* Particle is on the negative side of the origin.
* Particle is moving in the negative direction.
* Velocity is increasing.
* Particle is slowing down.
I
.
I
lI
*s(t 0 ) > 0
* Curve has zero slope
* Curve is concave down
* Particle is on the positive side of the origin.
* Particle is moving stopped.
* Velocity is decreasing.
-~
www.puucho.com
I
'
Anurag Mishra Mechanics 1 with www.puucho.com
DESCRIPTIONOF tAOT!ON
described schematically by the curved line in Fig.I. 71 (c). At
time t = 0 the particle is at s (OJ = 3 moving right with
velocity v (O) = 60, but slowing down with acceleration
a (OJ = -42. The particle continues moving right until time
t = 2, when it stops at s (2J = 55, reverses direction, and
begins to speed up with an acceleration of a (2J = -18. At
time t = 7 I 2 the particle begins to slow down, but
continues moving left until time t = 5, when it stops at
s (SJ = 28, reverses direction again, and begins to speed up
with acceleration a (SJ = 18. The particle then continues
moving right their after with increasing speed.
Conceptual Examples: Suppose that the position
function of a particle moving on a coordinate line is given by
s(t) = 2t 3 - 2lt 2 + 60t + 3. Analyze the motion of the
particle for 1 ;;,, 0 .
Solution: The velocity and acceleration at time t
are v (t) = s' (t) = 6t 2 - 42t + 60 = 6(t - 2)(t - 5)
a(t) = v' (t) = l2t - 42 = 12(t -7/2)
At each instant we can determine the direction of
motion from the sign if v (t) and whether the particle is
speeding up or slowing down from the signs of v aad a(t J
together Fig.1.71 (a) and (b)]. The motion of the particle is
V
5
0
2
O++++++O- ________ O+++++++++ s;gn of v(t) = 6(1 - 2) (t - 5)
.Pos_itive
direction
Negative
direction
Positive
60
Direction of motio_n
40
direction
20
-Analysis of the partiCles direction
(a)
a
7
0
2
2
5
t
0++++++0- ________ O+++++++++ sign ofv(t) = 6(t- 2) (t- 5)
Positive
direction
Positive
direction
Negative
direction
40
20
- - - - - - - - - -O++++t++++++++++++ sing of a(l) = 12 (t- 7/2)
sloWlng
speeding
slowing
speeding
dOwn
up
·down
up
Change in speed
1
-20
Analysis of the particle's
(b)
t =5
t=0
..
r___,,...1=....1,_2~
'--
:,'
-------~'t=2
03
28
55
(c)
Fig.1.71
.
~.:,
...
www.puucho.com
2
4
5
6
7
Anurag Mishra Mechanics 1 with www.puucho.com
f4a -"
is!
Concept: The curved line is above Fig. 1. 71 (c)
'descriptive only. The actual part of particle is back and forth·
'on the coordinate line.
· '
-
·- "--~
·--.-' -1
'The position of a particle is given by the equation.
s = f(t) = t 3 - 6t 2 + 9t
:Where tis measured in seconds ands in meters.
(a) Find the velocit,y at time t
(b) What is the velocit,y after 2s ? after 4s ?
'(c) When is the particle at rest ?
!
:(d) When is the particle moving forward (that is, in thej
positive direction) ?
[
'(e) Find the total distance traveled by the particle during the:
first five seco_nds
_ __ __ __ __ _ __ _ _
i
- . -·- MECHANICS-I I
-·-·-- ·-~·-,,~-- --·-· -'
lf(S)- f(3) l=l 20- DI= 20m
The total distance is 4 + 4 + 20 = 28 m
r--
1
INTEGRATION
1
Indefinite Integrals
Fundamental theorem of calculus establish connections
between antiderivatives and definite integrals, if f is
continuous, then (f(t) dt is an antiderivative off and
J:f(x)dx can be found by evaluating F (b)-J (a), where Fis
an antiderivative off
J f(x) dx is traditionally used for an antiderivative off
and is called an indefinite integral. Thus.
f(x) dx = F(x) means
F' (x) = f(x)
For example, we can write
f
2
Solution: (a) The velocity function is the derivative of
the position function, that is ,
s = f(t) = t 3 - 6t 2 + 9t
v(t) = ds = 3t 2 -12t+9
dt
(b) The velocity after 2 s means the instantaneous
velocity when t =2_. that is,
dsl = 3(2) 2 -12(2) + 9 = -3 m/s
dt t=2
The velocity after 4s is
2
V (4) = 3(4) -12(4) + 9 = 9 m/S
(c) The particle is at rest when v (t) = 0, that is,
3t 2 -12t + 9 = 3(t 2 -4t + 3) = 3(t -l)(t - 3) = 0
v(2) =
and this is true when t = l or t = 3. Thus, the particle is
at rest after 1 s and after 3 s.
(d) The particle moves in the positive direction
when v (t) > 0, that is,
3t 2 - 12t + 9 = 3(t -1) (t - 3) > 0
J x dx = x; + C because
x; + C) = x 2
So we can treat an indefinite integral as representing an
entire family of functions (one antiderivative for each value
of the constant C ) .
A definite integral J:f(x)dx is a number, whereas an
indefinite integral Jf(x) dx is a function (or family of
functions).
Any integral formula can be verified by differentiating
the function on the right side and obtaining the integrand.
For instance
d
2
J sec xdx = tanx+C because dx (tanx+C) = sec 2 x
!
[F(t)] = f(t)
f f(t)dt = F (t)_ + C
and
are equivalent statements,
TABLE
[)eriyative
·-,F~nnula
This inequality is true when both factors are positive
Equivalent
Integration formula
~[x 3 ] =3x 2
(t > 3) or when both factors are negative (t < 1). Thus, the
dx
particle moves in the positive direction in the time intervals
t < l and t > 3. It moves backward (in the negative
direction) when 1 <t<3.
(e) Because of what we learned in parts (d), we need to
calculate the distance traveled during the time intervals [O,
1], [1, 3] and [3, SJ separately.
I 2"1dx=.fx+C
.E.+ixJ=:!r
dx.
2vx
Jsec
i_ [tan t] =sec 2 t
dt
~[u312]
dx
The distance traveled in the first second is
lf(l)- J(O) I=14- DI= 4m
From t = l to t = 3 the distance traveled is
I f(3)- f(l) l=IO- 41= 4m
From t = 3 to t = 5 the distance traveled is
!(
2
tdt =tant+C
=i!u112
2
For simplicity; the dx is sometimes absorbed into the
integrand, for example,
Jldx can be written as Jdx
J_!__
x2
www.puucho.com
can be written as
J dx
x2
Anurag Mishra Mechanics 1 with www.puucho.com
'DESCRIPTloN oFMoriori -- --- - . ·- _____ ·- --- --- - ....... -- - - . -..
49'
[_
Integration Formulas
Integration is guesswork - given the derivative f of a
function F, one tries to guess what the function F is.
Jx 2 dx=
x3
r=2
+c
3
x4
Jx 3 dx=-+C
4
1
r=3
_5
1
x-5+!
J-dx=
Jx dx=--+C=--+C
x5
-5 + 1
4x 4
f Jxdx= f x
112
!+1
x2
dx = --+C
r=-5
I
5
= 2(3) = 6
3
3
4
1
2
TABLE
1
!
--
~ cos x
d
2
5. dx [tan x] =sec x
2
(c)
The
graph
of
y = -J1- x 2
is the upper
Jcosxdx=sinx+C
semicircle of radius l, centered
at the origin, so the region is
the
right quarter circle
extending from x = 0 to x = l
(Fig. lE.38 (c)], Thus,
2
' 7. -[secx]
d
==·sec.x tan x
; dx
y
-+----'---..x
xdx;:::: tan x+C
= cosec x cot x
fcosec xdx =- cot x + C
2
Jsecx tan x dx = secx+C
Jcosec x cot x dx-- -cosec x + C
dx
JI vlC--x- ,dx (area of quarter-circle)= 1
2
-it
4
0
Evaluate
2
2
(a) J(x-l)dx
(b) Jcx-l)dx
0
I
.
!Sketch the region whose area is represented by the definite
'
.
•integral and evaluate the integral using an appropriate
:fonnula form geometry.
4
(a)J2dx
_1
2
Fig, 1E.38 (c)
I
'
1 2 3 4
Jx'dx =-+C
(r;t -1)
r+l
_Jsec
cot x] = cosec2x
i' B. -d[-cosec x]
-1
2
Jsinxdx=-cosx+C
I
-! [-
-2
J(x + 2) dx = (area of tapezoid) = -1 (1 + 4) (3) = -15
xr+l
d
4. dx[-cosx]=sinx
16.
y=x+2
-1
[x'•'] =x'(r,e-1)
[sin x]
Fig. 1 E.38 (a)
Fig. 1 E.38 (b)
dx r+l
_3.
1
-+-<~~-+-+-.. x
2 3 4 5
-+--+~-~~-+--•X
Integration formula
f dx=x+C
I2.d
y=2
3
r=Differentiation
Formula
3
2
y
2
r·
4
(b) The graph of the
integrand
is
the
line
y = x+ 2, so the region is a
trapezoid whose base extends
from x = -l to x = 2 [Fig.
lE.38 (b)]. Thus,
2 312
2
= -x
+C = -(Jx)3 +C
~+l
y
4
J2dx = (area of rewu1gle)
I
2
(b) J(x+2)dx
-1
(c)
J,,/1 - x 2 dx
0
Solution: (a) The graph of ,Le integrand is the
horizontal line y = 2, so the region is a rectangle of height 2
extending over the interval from 1 to 4 [Fig. lE.38 (a)].
Thus,
0
Solution.
The
graph of y = x - l is
shown in figure and
we leave it for you to
verify that the shaded
triangular
regions
1
both have area - .
y
y=x-1
2
Over the interval [0,2]
the · net signed is
Fig.1E.39
www.puucho.com
1t
(1 2 ) = -
4
Anurag Mishra Mechanics 1 with www.puucho.com
1
A 1 -A 2 =
-
1
2 2
= 0, and over the interval [0,1] the net
f
i
l
1
foI (xl)dx = -.
2
and
0
Fundamental Theorem of calculus
!ff is continuous on [a, b] and Fis any antiderivative off
on [a, b], then
b
.
f J(x)dx =F(b)-F(a)
Cexctmfut~>r40·.1:";;.>
'y~-=-~i(',.r'~"tF:::::::;.~;:{;
-·-
-
-
-~
;·.
••
and
[O, it]•
••
' •-'"
.
fc:osec 2 xdx = -cotx+C
2
Jsec xdx=·tanx+C
\f secx tan xdx = secx + C f cosec x cot xdx .= -cosecx + C l
l · 1 .- " ·
I l dx . x+.C ,;I
l J·-.-dx'= tan- x+C · '·.r:---z
·1
=S!Il
x +1 ., .·' •.
I
2
I
'
.•
. • •
-
..
•.
•
"•
--~-
.
-1
.
-v.1-x·
. .
__
•
I(a) Find th~ a,rea undfr the CU[".e y = cos
;[o, -rr/2]
·f co.s:xdx = sin.x+C
'
fsi~xdx'.=,-cosx +C
'I
The most general antiderivative on a given interval is
obtained by adding a constant to a particular antiderivative.
Thus we write
a
,-
X
f cixdx=_':__+c
Ina
+C
I
2
f (x-l)dx= 0
JeX.dx = ex
I
signed area is -A 2 = _ _!_Thus,
2
'
X
>
~
··--~·.
.----
over th~ int~rval)
·
" - · · - - · .••
. ••
. ·I
f_!_dx=-.!+c
2
x
'
.
,
y
- - _-~:.
l.c:E•xa:1m1r:1fl e
1· -- - -- :_3. , . '
41__ j
I. ~
-~
·J
!b~·..='r--::;:.;;J-i,:r~--£:~;;\
J
iEvaluate Cic 3
' I
X
-:,--~--. ' .;---7 .. _
-
6x}dx
I_. __ -- -o_· .. , ...
Solution: We have
I
J3(x
-,1
I
'!
3
3
4
2]
6x)dx = ~ - 6~
-
4
0
·}•
• l
Fig.1E.40
i - - - ,.,
----~
·-
.........
-·· .
(1 4
, a!
= 4.3 -3-3
.,,,,,-~.
= -81 -
Solution.Ca) Since cos x;:: 0 over the interval [O, Jt/2] ,
the area A under the curve is
2
0
2) - (14·0 4-3-02)
27 - 0 + 0 = -6. 75
4
n/2
A= Jcosxdx=[sinx]~/ 2 =sin2:-sin0=1
o
2
(b) The given integral can be interpreted as the signed
area between the graph of y = cos x and the interval [O, it].
The graph in figure suggests that over the interval [O, it] the
portion of area above the x-axis is the same as the portion of
area below the x-axis, so we conjecture that the signed area
is zero ; this implies that the value of the integral is zero.
This is confirmed by the computations.
i
iFind
{
f:2(,zx,'
3
"-;-
0 .
L--- -~ ..... J.
X
+l
-~-···
. _.,:: ·
., "
· ·'
4
'J
1
ri+l, ._,
= .!x 4
2
= .!c2J 2
jU(x)'+ g(x)] ,ix•, : '
2
'
.
., ·.'.1 .. ' ..
·n;t-1
-·
-
1
x]
-3x 2 + 3tan-1 x]~
3(2) 2 + 3tan-1 2-0
=-4+3tan-1 2
This is the exact value of the integral.
\.aJJ(x)~'.,-Jg,(~)dx
···cc·
. .n:t-1J . ' .. J:-cdx = lnlx!·f:C
.. ••:, X,,. .
., ,
x .. ·
X "dx_ =_-·-_}-::.+
2
J(2x -6x+---i-)dx= 2x -6x +3tano
X +l
4
2
fkdx=}:x+C
!
. "" -- ,_,
-
3
Table of indefinite _integrals
CIE-x~mele.;
.;;,-::--· -= ~~~..::::.,.:·
.
www.puucho.com
"'1-:: .;_
'
'i
'
--·-··-· .J
Solution: The Fundamental theorem gives
0
f c f(x) d. cff(x) dx
- ~--
~
I
•
---
·rcos xdx = sin x]" = sin 1t - sin O = 0
0
- -- -
• ). ·
6x + + dx
l~
43
-~-:V
2
O
Anurag Mishra Mechanics 1 with www.puucho.com
-. -----·~~.'-.'-'-'"';,._----"'---,.._--- ~,~-' __ . _ _ ____ _;.·,_, ~--'·':'"'""'~- ______·_s1~.,I
,,
DESCRIPTION OF MOTION
Solution: First we need to write the integrand in a
simpler form by carrying out the divi~ion
9
2
2r:
9
J2t +t i"t-1dt = Jc2+t112 _r-2)dr
1
t
1
9
r 312 c 1 ·
3 312
=2t+----. =2t+-t
~
-1 ]
2
2
.
I
+-]
1
t
9
1
l] (
3 3/2 +9 = [ 2·9 +2(9)
4. If the rate of growth of a population is dn/dt, then
2
' dn
J-dt =n(t 2 )-n(t 1 )
,, dt
is the net change in population during the time period
from t I to t 2 • (The population increases when births
happen and decreases when deaths occur. The net
change takes into account both births and deaths)
5. If C (x) is the cost of producing x units of a commodity,
then the marginal cost is the derivative C' (x). So
3 3/2 +l
1)
2-1 +2·1
<2
JC' (x)dx = C (x 2 )-C(x1 )
,,
·
1
2
4
= 18+ 18+- -2---1 = 329
3
9
Applications
Fundamental theorem of calculus says that if f
continuous on [a, b], then
is
is the increase in cost when production is increased
froin x 1 and units to .'C 2 units.
6. If an object moves along a straight line with position
functions (t), then its velocity is v (t)- s'(t). So
,,
Jv(t)dl = s(t 2 )-s(t1 )
b
JJ(x)dx = F(b)-F(a)
a
where Fis any antiderivative ofJ. This means that F' = f,
so the equation can be rewritten as. We know that F (x)
represents the rate of change ofy = F (x) with respect to x
and F (b)-F (a) is the change iny when x changes from a to
b.
.
The Net change Theorem
The integral of a rate of change is the net change :
b
JF' (x)dx =F(b)-F(a)
.
is the net change of position of displ~cement, of the
particle during the time period from t 1 10 t 2 . It is always
true. In order to calculate the distance trnveled during the
time interval, we have to consider the: intervals when
v (t) > 0 (the particle moves to the right) and also the
intervals when v (t) < 0 (the particle moves to the left) note
that during motion particle may reverse its direction of
motion. In both cases the distance is computed by
integrating lv(t) I, the speed. Therefore
,,
Jv (t) Idt = total distance traveled·
a
Here are few applications.
1. If V (t) is the volume of water in a reservoir at time t,
then its derivative V' (t) is the rate at which water
flows into the reservoir at time t . So
,,
,,fV'(t)dt = V(t
2
~
.
.
·is the change in the concentration ofC from time t 1 to
t2,
3. If the mass of a rod measured from the left end to a
pointx ism (x), then the linear density is p(x) = m' (x)
So
Fig. 1. 72 shows how both displacement and distance
traveled can be interpreted in terms of areas under a velocity
curve.
Jp(x)dx = m(b)- mra)
.. f'ig._!,72
,,
displacement= Jv(t) dt = A 1 -A 2 + A 3
<1
,,
distance= Jiv(t)jdt =A 1 +A 2 +A 3
,,
,,
The acceleration of the object is a (t) = v' (t), so
,,
<1
.
)-V(t1)
is the change in the amount of water in the reservoir
between time t 1 and time t 2 •
2. If•[C] (t) is the concentration of the product of a
· chemical reaction at time t, then the rate of reaction is
the derivative d[C]/dt. So
2
' d[C]
f - d t =[C](t 2 )-[C](t1 )
~
'1
J a(t)dt = v(t 2 )-v(t 1)
'1
is the change in velocity.from time t 1 to time t 2
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
·:cc· _.· : ·: · · .· ~ -~·:, -- ·-- · ~-
1s2
7
L ~-:
'>"~
M~CHA~icBJ
displacement is positive if the final position of the particle is
to the right of its initial position, negative ifit is to the left of
.
._-"".t:,~~'-·=
" ' -· "
- --·,_ - ,___ .,._ ·-·- -~,-;
[..4. partfrlemov~ ii}lmi a l_ine so piat its veloi:ity at time tis:
~(t)~t?cfel~~(nieqsured in mii~rs persecoryd). ' . :
tiJ/cEi'},~JIJ1i'dr&~1a~.einent of S~~ Rarticle during ihe :time;
k,,,-.. penod:!.·=,;,t :;;,.4; '·· .
·, ·
·
·
.
~.:.Einci.nu(iJJ.stance
travel§§Audrti
this
ti1jie
pe.rioiL:.
.
.- . ·.
Solution:· '(a) By equatio~, the displacement is
. ·
.·. , 4
s·(4)__;_·s(l)',;; Jv (t) dt
-.-.-·)·:·:·.>. . :.
= f (t 2 -t ·1_
3
4
2
9
2
1
This'171eans.that the particle moved 4.5 m toward the
'
left. .· . :._."......
'.
.
(b) Nii'i1'·that ..v(t)=t 2 -t-6=(t-3)(t+2) and so
V (t) ,,; 0 op)!i,e''i~terval [!, 3] and V (t) :2 0 on [3; 4]. Thus,
the distah~e 'traveled is .
.
..
' •.4· ••
..
.
3 '
4
1. - .
3
.
3
4
=.f (-t 2 +t + 6)dt +f (t 2 -t -
: t3 -~2 ... ·J3 1. [t3 -t2
[
6)dt
.]4 3.=-=10.l?m
61
·
· ~-+·-+6t · + -+-+6t
3 ·. :2 . ' \ . 3 . 2.
3
6
--1~'!'::1.1
.
l•(lo)
·~
"----~_i,;.:;,,,,r,,...,__,; ·• ,'---'
. ·*1-·~----
. ,s(t,J ; '
:.~(t,)
-·--
. ···.-· ,,
[Find ~,;~V'j;_i.,~itiPl1.i ':fi!ncticm oj>a: particle that moves with
fveuici~ v[t'.fi
along a,coqrdinate line, assuming .that
lthe partiqle'li,g:s cgiirdinate s=4 at timet = Q
·
•
rtli:ct
i=c...--~""'i-¾~_~:~-::-~:--..------,_,,-,_ -- ---·
.
. ,
. -- ..... , ... :··:--r,---·
.
. ·--::·~
is . th_r;re/~tionship betwee:1/Jize
;displacement of a pa,rti.cl~ and it~~ distance it ti:'ayf/Sif·'i/1e
'particle moyes iri tlie negative direction without reyefsitig the, ·
:direction ofmotion?
· · , 1· ',' :. ·, . • ·'. ;/ Integr:p.ting the velocity fu~ctidti:of a particle ov¢rd,time
!interval yields the displizcement'of a:particle.s .over tha't/tiine
;interval. Whereas to find the,fofal d~tance traveled:0by't~'e
,particles over. the time ,inierv'?il (th~ ·_distance'., traveled'i~, tlie
'positive direction plus ihe /listimce' traveled the, 1'1egativ~1
,direction); we must integrate the absolute valii~ oftheyelqcityl
junction; that is, we must-integrate the speed
.
· ,, ...
.
.
'
,
' '[total distance·trav~l~d]'; '... ,
I
during ti~e inte1:yi :,;, J;;iv(t) ldt_
[to,t1J
·
. ·,--·
1= 4 when t "" 0, it follows that
, 4~ s(O) = .!.sinO+C = c·
Thus,
.
1
·1
Conc,ept: Wh~t
"··-
.', -~:.:· :S{t) = fv(t)dt = f cosittdt = ¾sin1tt +C · r~l$;~iAfurii:l~J46~
.
}\o! ·.
Distance Jraveled in Rectilinear motion
In general, the displacement of a particle is not the same
as the distance trayeled by the particle .. For example, a
particle t4at travels 200 min the positive-ditection and then
200 m in 'the negative' direction travels a distance of400 m'
but has a displacement or zero, since it returns to its.starting
position. The only ciise in which the displacement and the
distance traveled are the )ame oc_curs when the particle
moves in the positive direction without reversing the
direction of its motion.
Solutjon: the position function is
Sin~e. · '•
·, . '
ih
~-.
~71"'~,;...~-~
•:c'.~~b,f;e:i 45 ·r,~
--'--...:
:-+--'------'+-+-~ -1-----'---'---~-'-;__,l...'--
1
flv(t).Jdt = fr-v.(t)]dt+ fv(t)dt
1_'; 1 '.·"
. ..
-~1
, - ·l
'•·- .., ·--~-~='""-·--···--------Flg.·1,,Z3
-. .. ' __
• -.,.. •.p:.
·•
,.. -~~
6) dt
,• · =~[_c_£.:__6t] =
.
: --P.o.sitiv~:isp;ace~e~t - -~-, Ne~ative : : : : : ; "
4
1_
.. •.·. , , .
its initial posi#,on, and zero if it coincides with the _initial
position (see Fig. 1.73) .
, ·
. • ,· '·
It
s(t) = - sin1tt + 4
It
Displacement in. Rectilinear Motion
Suppose thats (t) and V (t) aje the position ~nd velocity
functions of a particle moving on a coordinate line. Since
v (t)is the rate of change of s (t) with respect to t, integrating
v (t)°<'ver an interval [t 0 ,t1 ] will produce the change in the
value ,f s(t) as t increases from t 0 to t 1 : that is,
J:>(t)dt ,..·J;;s•(t)d_t =S(t1)-s(to)S
The expression's(t 1Y- s (t 0 ) in this formula is called the
displacement or change in position of the particle over the
time interval [t 0-, t,J. For a particle moving horizontally, the
-----~~
;;;o~~-~~
~o.t/i~t'
:~upp~;e th~~ a. p~rticl~
I~traight line
j~
,velocity attime tis v (t)= (t 2 ":' 2't) nVs. ·
·
· ••· \
(a) Find the displacement of the .particle during·thl ,~ime,
interva!0s;ts3.
· · ·, .
. .
· ·, _·
I
,Cb) ~ind the distance u:avel~d by.~he, particle duri% Ihe
111terval. Q ,,; t ,,; 3, _ .. 0 •• _ .::. ". ·,,, ... _ ..__ •• . ·• • ·:..;J
. )
Solution: (a) The displacement is
tpne1
'
JlvCtl ldt = Ji Ct
0
0
2
-
it) Icit=
•
[.c -r
2
3
3
.=
]
o
0
Thus, the particle is at the same position at time t = 3 as
at t = 0
written,
as
(b) · The
velocity
can
be
v(t)=t 2 -2t=t(t-2), ·from· which we • see that ·
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
[DESCRIPTION OF MOTION
v(t) s Ofor Ost s-2 _and v(t)
distance traveled is
~
Ofor 2 st s
. - - ·3. Thus, the
!?3 I
- - - -- ·- ---------,-----'-"-----'---'
' .
---~--J.:7~
p,E~~~H~ , ~
d,;;~_~;rv~_
J31v(t)[dt
=J2-v(t)dt+
s:v(t)dt ,
0
0
_
= J02-(t 2-2t)dt+f"2 (t 2 -2t)dt .
For each of the _veloc_,·ty .Ve;~;;·
in Fig. 1E.4S,~nil'
:the total distance traveled by, the paracle over the time
interval Os t s 4.
·
----~ .
=-[t:-t2J: +[t:-t2I =i+i=~m
Analyzing the Velocity versus Time Curve
A valuable information can be .obtained from the
velocity versus
time
curve. The integral v dt
V
can
be
interpreted .
geometrically as the net
area between the graph of
v (t) and the interval
[t 0 ,t1 ],
and
it . can
interpreted physicany as
Fig.1.74
the displacement of the
particle over this interval.
For a particle in rectilinear motion, the net signed area
between the velocity versus time· curve and an interval
[t 0 ,t1 ] on the t-axis represents the displacement of the
particle over that time interval (Fig. 1.74).
.
-
,
u (mis)
.
,--~
~g~p.L~-,i~
V -- ----
: C
6
-=--1''-.---i'--r-+---''c--'f,
0
2
3
4
time (sec)
o:'
5
-1
''
'
-2 ---------------------------- '
._._____
Flg.1E.48
- - -· --
-
Solution: In an three pans of figure' the -total area
·between the curve and the interval [O, 4] is 2, so the.particle
travels a distance of 2 units during the time period in an
three cases, even though the displacement is different in
each case.
l~..i9i.p;~~,J~>
.th4
'Find. th; total area b,"tween the curve. y = 1 - ~ 2 and
over the interval [O, 2/ (!':ig, ·_lE:.~9~ _ . _
·
x-axis
,
1
V
I.
.
.'
l
J
I'
!
le'
-2
I
-,3
Fig.1E.49 _
I
, ;
---''---,f--.'1,-~- t
-1 -1
'
V
1- : 1
i
I
B
!Fig. lE.47 shows three velocity versus time curves for a,
!particle in rectilinear motion along a horizontal line. In each
case, find the displacement of the particles over the time
interval Ost .s 4, and explain what it tells you about the 1
motion of the particle.
I . -,,-
I
-,...----,
2
J
r~i=-:;.
.. - -.. --- ------------ -- .
r
·---=-='- ·____,_:,_,_::•___
Solution: The area A is given by
2
A=
1
2
Jl1-x 2 ldx= JC1-x 2 )dx+ J-_(l-x 2 )dx"
0
0
1
=[x-X:-J:-[x-X:r =1-(-34 )=2
(c)
-
Solution: In part (a) of figure the net ·signed area
under the curve .is 2, so the particle is 2 units to the right of
its starting point at the end of the time period. In pan (b) the
net signed area under the curve is - 2 units, so the particle is
2 units to the left of its starting point at the end of the time
period. In pan (c) the net signed area under the curve is ·o,
so the particle is back at its starting point at the end of the
time period.
We- can also interpret geometrically the total distance
traveled by a _particle in rectilinear motion by calculating net
area.
Finding Distance Traveled from tile Velocity versus
Time Curve
. ,
. .
For a partide_ in rectilinear ni~tion,. the total area
between the velocity .versus time. curve and an interval
[t 0 ,t1 ] on the t-axis represents the distance traveled by the
p~rticle over that· time interval.
INTERPRETATION OF GRAPHS
(i) Given the s-t Graph, Construct the 11-;t Graph:
The velocity at any instant is determined by mea~uring the
slope of the s-t graph, i.e.,
·
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
-- u--- -
I
s
u -
- ---
~
0-dtt=O
So
s·2.
s,
,Q
t1 .
ti..
53
..
u,
t3
• 1,
. Fig. 1.75 (a)
Fig.1,76 (a)
ds
} a
:.............=v
dt
Slope of s-t graph ~ velocity ·
For example, measurement of the slopes v 0 , v 1 , v 2 , v 3 at
the intermediate points (0, OJ, (t 1 ,s 1 ), (t 2 , s 2 ), (t 3 , s 3 ) on the
s-t graph. Fig. 1.75 (a), gives the corresponding points on the
v-t graph shown in Fig. 1.75 (b).
a,
a,
a0 = o
10
83
t,
''
t,
\3
t,
Flg.1,76 (b)
Concept: Since differentiation reduces a polynomial of1
:degree ~ to that of degree ,n -1, then if the s-t graph is'
.parabolic (a second-degree curve), the v-t graph will be a
'straight l1ne (a first-degree curve), and the a-t graph will be a
~ons!_<1n_£ a.r a_ horiz_ol!!_al line (<)_ zero-degr.ee ~urve):
u,
.. _ _ i
. --- ----r:::7""--.
\·-,:p~.'l@:mJ?.J,~;J
50 ~v
r--~
Fig 1.75 (b)
(ii) Given the v-t Graph, Construct the a-t
Graph: The acceleration at any instant is determined by
measuring the slope of the v-t graph, i.e.,
bi;cl; ~~v~ - ~l~~ ~ straight road s~~h -th~~-i~s1
iposition' i5 described by the graph shown in Fig. lE.50 (a).
!construct the v-t and a-t graphs for Ost $ 30s.
l --~
- . . . . ... . .
I 500s (mis)
· .du
~;a
dt
.. _ ~!_ope pf_v-t graph_": ~ccelera_tion_ !
For example, measurement of the slopes a0 , a1 , a 2 , a 3 at
the intermediate points (O,O),(t 1 ,v 1 ), (t 2 ,v 2 ), (t 3 ,v 3 ) on
the v-t graph. Fig. 1.76 (a), yields the corresponding points
on the a-t graph shown in Fig, 1.76 (b).
I
'
'
I 100
S
I
'I
= t2
' I
'I
'
10
30 t(s)
Fig. 1E.50 (a)
www.puucho.com
A
.
I
I
•=201-100
'I
.
!
Anurag Mishra Mechanics 1 with www.puucho.com
DESCRIPTION OF MOTION
55
Solution:
Concepts: v-t Graph: Since v = ds/dt, the v -t graph
can be determined by differentiating the equations defining
the s-t graph, Fig. IE.SO (a).
/',.v=fadt.
change in velocity = area under a-t graph
First calculate the particle's initial velocity v O and then
u
We have
ds
dt
ds
lOs<t,;; 30s;
S = 20t - 100, V = = 20
dt
Values of vis the slope of the s-t graph at a given instant.
For example, at t = 20 s, the slope of the s-t graph is
determined from the straight line from 10s to 30s, i.e.,
0 5 t < 10 s;
V
= - = 2t
u(m/s)
u = 2t
20 r,1--,,-------i
"---.,>10~----3-'-0--t (s)
Fig. 1 E,50 (b)
1',.s
v=-
t = 20s;
1',.t
= 500-100
30-10
= 20m/s
Concept: a-t Graph.
Since a= dv/dt, the a-t graph
can be determined by differentiating the equations defining,
the lines of the v-t graph.
u,
Uo
(a)
(b)
Fig. 1.77
add to this small increments of area (1',. v) determined from
the a-t graph. In this manner, successive points,
v 1 = v 0 + I',. v, etc., for the v-t graph are determined, Fig.
1.77 (b). Notice that areas lying above the t-axis correspond
to an increase in v ("positive" area), whereas those lying
below the axis indicate a decrease in v ("negative" area).
Concepts: If the a-t graph is linear (a first-degree
curve), integration will yield a v-t graph that is parabolic (a
second-degree cur-ve), etc.
(iv) Given the v-t Graph, Construct the s-t
Graph. When the v-t graph is given, Fig. 1.78 (a), it is
possible to determine the s-t graph using v = ds/dt
ils
= Jvdt
s
u
a (m/s2)
s,
.1.s:=~1udt
2>----~
t,
t,
(a)
(b)
Fig. 1.78
'-----'-10_ _ _ _ _3_0_ 1 (s)
Fig. 1 E.50 (c)
This yields
displacement = area under v-t graph
We calculate the particle's initial position s0 and add
(algebraically) to this small area increments ils determined
from the v-t graph, Fig. 1.78 (b).
l!;§?<~iit~J_jiJ>
dv
a=-=2
dt
dv
'An experimental car in Fig. lE.51 (a) starts from rest and
lO<t 5 30s;
a=-=0
V = 20
dt
·travels along a straight track such that it accelerates at a
·constant rate for 10 s and then decelerates at a constant rate.
The result are plotted in Fig. lE.55 (c). Show that
,Draw the v-t and s-t graphs and determine the time t' needed
a = 2 m/ s 2 when t = 5 s by measuring the slop of the v-t
'_to_ stop the car. How far has the car travelled?
graph.
(iii) Given the a-t Graph, Construct the v-t
Graph. If the a-t graph is given, Fig. 1.77 (a), the v-t
graph may be constructed using a= dv/dt.
www.puucho.com
05t <10s;
V
= 2t
Anurag Mishra Mechanics 1 with www.puucho.com
- ---
-
;
Solution: Concept: v-t ·Grnph: Since dv =,adt,:
the v-t graph is detennined by integrating 'the straight~line,
',s_egments of t/ie a-t graph,
·
.
.
-
,
~
-
When t = lOs, s = 5(10) 2 = 500 m. Using this initial
condition,
l0ss;t s;60s;
' s (m)
~-
a (m/s2)
3000
s"= 5t2
10
500
A1
t'
10
-2
I
A2
;=-t?+ 1201-600
t (s)
'""'---1.--------'---t (~) i
1
10
60
Fig. 11:.St (a)
Fig. 1E.51
= 0 when t = 0, we have
< 10s·' a= 10·' o
Jv dv = f'o
lOdt·
' v = lOt
I'
Using the initial condition v
0 -< t
500
V
= 101
100
V
Second method : Ti}e s-t graph is shown in Fig.
lE.51 (c). The triangular area under the v-t-graph would
yield the displacement I!. s = s .'.. 0 from t = 0 to t' = 60 s.
I
Hence,
= -21 + 120
i(__...L_ _ _ _ _ __:,,._
10
I!. s = .!:. (60) (100) ~ 3000 m
2
.
.
I (s)'
t' = 60
(v) Given the a-s Graph, Construct the v-s
Graph. For given a-s graph for the particle points on the v-s
graph can be determined by using v dv = ads. Integrating
this equation between the limits v = v 0 at s = s0 , and v = v 1
ands= s1 ,we have,
1 2 -v 2 ) = <1 ads
-(v
Fig.1E.51 (b)
10s<t ~t 1 ;
a==-2;
f1:o dv = f:o -2dt,
V
= -2t + 120
The time taken to stop the car can be obtained by
substituting. v = 0.
t' = 60 s
Second method : The area under the a-t graph is
equal to the change in the car's velocity. Thus net area under
curve gives I!. v = 0 = A1 + A2 . Fig. lE.51 (a). Thus
0 = 10m/s2 (10s) +(-2m/s 2 )(t'-10s) = 0
2
1
0
V
ia
Sin'ce ds =v dt, integrating the
equations of the v-t graph yields the corresponding equations,
of the s-t graph. Using the initial condition s = 0 when t ;, 0, i
we have
, ds. = I, lOt dt,
O
'•
V1
.I
I
··1
1
Concept: s-t Graph:
O
J
= area und~r a-s graph
t'= 60 s
0s;t s;lQs;
= f'10 (-2t+l20)dt,
s = -t 2 + 120t-600
Position at the moment car. stops is obtained.
Whent'= 60s
s = -(60) 2 + 120(60)-600= 3000 m
, , v· (m/s)
'
ds
(<;t ______ _J!
s- 500 = -t 2 + 120t-[ -{10) 2 +120(10)]
When t = lOs, v = 10(10) = l00m/s. Using this as the
initial condition for the next time period, we obtain velocity
as function of time.
I
= -2t+l20;
V
I
J
s,
(a)
_Flg,~_.7y . __ , .... ___ -
= l0t;
s = 5t 2
v
www.puucho.com
i
'(Ii)
__
J
Anurag Mishra Mechanics 1 with www.puucho.com
J
._ I I
'
'
•
'DESOIIP'tiON
•
.
" OF,MOTION
.
a
Area under the a-s gra;h,
initial value of v 0
1
2
V1 = (2fs ads+v~)1/ ,
'
,,_ i
fsos, ads, is determined and the
s0 = 0 is
at
known,
5
motion-= . u=5.
.__ _·..;F:..,11!. l E,52
(vi) Given the v-s Graph, Construct the a-s ·
Graph- If the v-s graph is known, the acceleration a at any
position s can be determined using ads = v dv, written as
dv)
·
a=v ( ds
- ; - - - - - - --.: ---=--7
acceleration = velocid times slope v-s graph
)
.
V
distance = 2 x Us + 228 =
l •.
1
L
1
Solution: Jal= dv; kv 2 = dv; kdt = dv
dt
dt
v2
On multiplying both sides of eq. by 'v'
ds
dv
dv
-·kdt=v-;
kds= -;lnv=ks+Cats=O;
dt
v2
V
f
'I
' .
Thus, at any point (s, y)in Fig. 1.SO(a), the slope dv/ds of
the v-s graph is measured Then since v and dv/ ds are known,
the value of a can be cak:ulated, Fig.1.80 (b).
~~~riiJi~
4J 2t ·
1 )dt
f dv = f (4;--2t
Findl
Solution: a=
0
·
C=lnu; ln(~)=ks
v
= ueks
·
jGiv:~ ~ '." "'cos.t; at t =0; u =·o; x=.1
IPosttwn at t = 1t
~l!!d d~J.JJ.!!£.e:.Ji:om O to 21t,
Solution: a= - cost
Jdv
2
= 4t - t + 5 = -t + 4t + 5
- = -(rt - 4t - 5) = -(t - 5)(t + 1)
t
0
0
f dx = f (4t -t
2
+ l) dt
=}
X
= 2t
2
3
=-sint
"'
•
f dx = f-sint dt
t3
1
+ St
[In forward direction]
0
0 to 2it, v will change sign. velocity upto it :will be
3
3
x 5 = x = 2(5) 2 C5l + 25
0
V
X
- -
= -f costdt
0
After 5 sec, velocit: will be negative
X
t
V
2
V
f
.-
[:::,ea:a~eu;J;til:~i~~~velocity. at -t = 0, u = 5; .
S
,
If a particle acce_1er_ates with a .= °icv_ 2 ana initial ;~iacity = u
then fim:1 velocity aftgsdisp)acement.
·
I
(bl
_;e c~xi use
ij~xam~4e.;;;1 53 ~
~-~___.___--'--- S,
r-- ---i
,
i--- s-i
-----ca>- - - ,E!!!c..1:~J___ - . -
+ 228 m
Concept: If a is gi_·ven asfun_ction of-s or v
~~~ation vdv = ads
, ., .
· _
/_
'-t..___-_-S___J---"·-'------i1 s
3
1 ·-
a =v (du(ds)
!
200
Objective
Equation of motion not applied if acceleration is
variable.
,I
t
''
then
so
V
u•O
negative and
100
r::11::
:!1:11:::t dtl = 2+ 2 = 4.
=-m.
3
and
X12
t = 12 sec
·
• (12) 3
.
= X = 2(12)'.:.. - - + 5(12) = -228 m
3
kEx~IB:l;e.cj 55 ~
~here
porti<1, """" o/O"K ;;~ """'"'""'
~
t is in second. If the particle. is initially at the ong{n. and ·
0 a ;{, -
w.
I
[
it moves alimg positive x-axis with v 0 =2 m/s,jind the nature
of motion oftlte particle.
.
· -: -~
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
158 ---------------------dv
.
Average Velocity and Average Speed
Solution: G1vena=dt'
-dv = 6(t-l)
We have
If a body is moving along a straight line and its velocity
as function of time is known.
V = f(t)
total displacement
then average velocity
total time
t, V dt
t=O t=t v
--<:>--------o-+-'-o- x,
Vo
dt
s
We have
2
Jvo"dv = Jort6(t-l)dt = 3t -6t
Substituting v 0
Fig. 1E.55 (a)
= Ito
= 2 mis, we have
cJx = 3t
J'" dt
2
- 6t + 2
dt
Again integrating both sides, we have
=
V
s:
= J~(3t 2 -
dx
= t3
X
-
, 'o
Concept: Average speed may be greater than or equal to'
,the magnitude of average velocity. If a particle turns during
:motion then speed and magnitude of average velocity' will be,
'different.
6t + 2)dt
3t 2 + 2t = t(t -1) (t - 2).
!
Put v =0 and find the different instants of time when!
velocity' is zero, let they be t 1 ,t 2,t 3 , ... tn. Split the limits of,
integrationfor those instants where velocity becomes zero and;
;lie between t 0 and tn and take modulus for every individual'
1
integratioTL Reversal of motion is possible at those moments:
,on(y when velocity' becomes zero and that is why we have
'splitted the (imit of time.
Following points are noted from the above
example.
Putting x = 0 in x = t (t - l)(t - 2) we have t = 0, 1
and 2. That means, the particle crosses the origin twice at
t = 1 and t = 2. After t = 2, x is positive. Hence its
displacement points in positive x-direction and goes on
increasing its magnitu~e.
This gives total distance.
t=O
J;~ v dt j-t;I J:,2 v dt J+ ... + JJ::, v dt J
total distance =J
1=1
1
t,=1- ,/3 ,
1
12=1 + ,/3 '------{}-t-=2_.....,_
average speed =
Js:~
Flg.1E.55_(b)
Since
t 1 = 1-
v
= 3t
2
-
6t
+2,
putting v
= 0,
we
have
~ andt 2 = (1+ ~} Hence the particle remains
(i) O<t<(1- ~)
x> O;v> O;a< 0
···--·-
• - · - - • •• ,---l
LaJ~:~,q~~,,t,~ -I
x > O; v < O; a < 0
2
t
'
x/t graph
A p;rticl; moves. ~long a stra~ht ll~e a;d its ~eloci~·d;pencb;I
on time as v = 6t-3t 2 where•,]_• is in m/sec and 't' is in,
,second. Find average velocity' and. iVerage speed for first four,
rseconds. __ _ _
_ _ _ _ \ . __ .
_ _ _j
f--'',--,..-,~-..,....-t
v-t graph
·Vo
J:vdt
'
a-t graph
x<0;v<0;a>0
Flg.1E.55 (c)
(iv) 1 + ~ < t < 2
x<O;v>0;a>0
x> 0·v> 0·a> 0
=
J: dt
'5J:tdt-3J:t 2 dt
\
4
48- 64 I
=- - =-4m/sec
+1 ... ,........ -----I
=
6>·,
Solution: Average velocity'~ Total displacement
'
Total time
+vo
(iii)l<t<(l+ ~)
(v) 2 < t <
56
dt
I
j'
Vav
00(1- ~)<t<l
2
vdt J+IJ~ vdt J+ ... +JJ::, vdtl
Jto'"
f
stationary at t = t 1 and t 2 •
. l(
The displacement, velocity
and acceleration of the
particle in different time
: 1
intervals are given in the 1'
-Xo .t-:-43-:-·• •
1
following table and shown I
'
'
•1+V
i i
,/3
in the following graphs.
total distar.ce
_
total tlmi
v:
For average speed, put
0 antetermine roots oft for
which reversal of motion can take ~ace.
0= 6t-3t 2 I
I
=> t = 0 and t = 2 sec.
www.puucho.com
\I
Anurag Mishra Mechanics 1 with www.puucho.com
J.
59!
DESCRIPTION OF MOTION
Thus,
~X9..!Jn.~l.ej~
average speed= IJ: v dt l+I J; vdt I
J; dt
= I6J:t dt -3 s:t
2
dt
1+16 s: t dt -3 s: t
2
dt
I
4
\ 4[+1- 20[ = 24 = 6m/sec
4
4
· IA particl~ moves along a straigh.t line, x. At time t = Q ;;I
!position is at x = 0. The velocity, v, of the object changes
!function of time t, as indicated in the Fig. lE.58; t is in
\seconds, V in m/sec and X in meters.
(a) What is x at t = 3 sec ?
1
2
(b) What is the instantaneous acceleration (in m/sec ) at,
t = 2·sec? t-- ________ ....._, _____
a:~!
Average Velocity by Integration
If velocity is a function of timev = f(t)
then
f
7
+3
I
I
\
I
+2
\ v (m/sec)
J vdt
-Jis the time average of velocity
dt
+1
,·
3:
If velocity is a function of positionv = f(s)
Jvds
1
then-J- is the space average of velocity i.e.,
ds
.
-1
I
...;i
-------·-·----------------~
'
!Velocity vector of a particle is given as
I
--+
\
V=4ti+3j
,.,
,.,
!At t = 0 position of the particle is given as :i'0 = 2j:
IFind the position vector of the particle at t = 2 sec and the
b,ergge acceleratioru,f the p_article for t = 0 to t = 2 sec.
--+
Solution:
. ,..
A
v=4ti+3j
...
=>
--+,
r- r 0
t
,..,
t
,..
,.._
r = 2j+ Bi+6j
= Bi+ Bj
For average acceleration initial velocity at t = 0
->
1:,Vi
(c) What is the average velocity·(i~-m/sec) between t = 0.andj
\
t=3sec?
·
rd) tw::~~s! t;e average speed (in ~s~:__~etween t = 1 an~
Solution: (a) x at t = 3 sec
= Area (0- l)sec+Area(l - 2)sec-Area(2- 3) sec
"
1
1
=3xl+-x3xl--x3x1=3m
2
2
(b) Acceleration is constant fort = 1 sec tot = 3 sec
A
=3j
v, -v'
t f -ti
2
Att = 2sec.
,..
I
I
t; = 1 sec, t1 = 3sec, vi= +3 m/s, v 1 = -3m/s
-3-3
/ s2
a=--=-3m
,..
= J0 4t idt+ JO 3dt j
= 2t 2 i+ 3tj
--+
Fig.1E.58
Acceleration at t = 2 sec : a =
--+ J'->
_ dr=
vdt
J~
0
ro
--+
I
-2
averaged over position.
b~~~!'J¥))'1L~~
I
. . displacement
3m
(c) Average velocity=
.
= - - = 1 m/s
tune
3sec
Distance
(d) Average speed = ='--.- time
Area with magnitude only
=
time
=
Ar(l-2)+Ar(2- 3)
!_x3xl+!_x3xl
2
2
3
3
1 m/s
Final velocity at t = 2 sec.
_,
A
A
v 1 = Bi+3j
_,
--+
_,
VJ-Vi
a=~-t
Bi
2
= 41
Given that x = 120-15t-6t 2 +t 3 (t > 01 find the time
when the velocity is zero. Find the displacement at this
instan."'t·~-------
www.puucho.com
\·
Anurag Mishra Mechanics 1 with www.puucho.com
Solution:
X
= 120-15t - 6t 2 +t 3
Solution:
(a) Distance travelled is given by the area under
velocity-time curve.
Therefore distance = Iarea of A I + Iarea of BI
+ Iarea of Cl+ Iarea of DI
(All areas are considered +ve irrespective of the nature .
of the quadrant in which they lie).
dx
v=-=-15-12t+3t~ =0
dt
Solving, we get · t = 5, - 1 sec. As t > 0, t = 5 sec.
putting t = 5 in the expression of x, we get
x=20m
~!il:m,i-ii,l,e~f7ol~
~ ,_::
CE v & : ~ ~
-r- - - -
,
,___ ... - --- ~- - . - ~.----·
.•,,
(The figure shows the (V,t) graph for the train .accelerating'
1
ifrom resi up to a maximum speed of v
. and then
1
decelerating to a speed of l Oms·- . The acceleration and
deceleration. have , the same magnitude which is equal to
z ., .
.
- . • ,
0:5'·.
. m s. •
./ -~--_ · .. ~=-.-"'--
ms-
I
v (ms·')
I
V
i
·'
,
~
1
:
i
Fig. 1E.60 (a)
·
..
IA birdfl.iesfor 4s with a velocily v = (t -
·-~J ~:;l!;!e::~e;e~~;i;:e~f::i~
§h6'0h<it the~distance travelled)s_02V!::J,OOtr,;etre. _
v 2 -u 2
Solution:.
For the motion between 0
andB
Distance (s 1
1
~:~~e,!~J62lb>,
t (s)
t
0
1
=-x2x2+2x2+-xlx2--xlx2
2
2
2
=6m
10
o
.,
=~x2x2+2x2+~xlx2+~xlx2
2
2
2
=Sm
fb) Displacement= area of A+ area of B + area of C
+ area of D
(Proper signs of areas are considered according to the
nature of the quadrant in which they lie.)
2) m/s in a straigh~i
Ca.lc~late the displace~ent,arutj
v2 -u2
= 2as
s=--2a
Solution : The displacement is given by
4
-·----·--v,(rps·1)
4
s=J 0 vdt=J 0 (t-2)dt
.
', •V
·v 2 -o 2
) = -.- - = v
=1t:-2t["=o
2x0.5
', For the motion between A
. and B
102 -V2
,Distance (s 2 ) = - - - 2x(-0.5)
The velocity of the bird become zero at;
0 = t - 2 .=> t = 2 s
10
JJ: vdt HJ: vdt I
'
=Js:ct-2)dt J+Js: (t-2)dt I
Distance s =
= v 2 -100
Total distance = s1 + s2 = 2V 2
1.00
-
=1t:-2ti: + [t:-2.ti:
particle
2
2
=1 :-2x21+1(~~2x4)-( :-2x2)/
The velocily-time graph for a
travelling along a
straight line is.shown in the Fig. 1E:61. Find
-·-
. '
=2+2=4m
Graphical Method
The velocity-time and
speed-time graphs of motion of the bird are as follows:
. ' v (mis)
2 ···········~--~
A:
0
-1
I
1,
-2
1.
2
B
C
3
4
6
5
o:'
time (sec)
''
'
---------------------------- '
Fig. 1 E.61
·-------~~-------(a)·distance travel(edfrom zero to 6sec.
_ __,__,
'@Jlispjactimentinft<Laboyetime'interval. _ _ _www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
I-L_ DES(ilPTION
OF MOTKlN -••• __
velocity (mis)
_,
[a[=l
(d)
V
_,
= t-2
ar
(e)
_,
= (a-v)v
2
=
+
__ .. -- ·- --- ----------- 61 !
([j.
B
(i+t j+k)]J (i+t j+k)
·~t 2 +2
~t 2 +2
0 + - - - - 7 ! " ' - - - - t - - - - time (s)
A
4
2
_,
(
t
)·
ar = ~t2+2 v
or
t (i+tj+ k)
2
(t + 2)
-2
Variation of velocity with time
(a)
speed (mis)
2
----------------------------
A
+ B
+
O + - - - ¥ ' - - - - ' - - - - - t i m e (s)
2
4
Variation of speed with time
A particle moves in xy plane with a velocity given by,
_,
Flg.1E.62
From the graph,
displacements =I area Al-I area Bl= 0
and
distances =[AreaA[+[AreaB[
1
1
=-x2x2+-x2x2=4m
2
i,_€,i~qi:p_p_l)~ ;~-~
A
A
v = (St - 2) i+ 2j. If it passes through the point (14, 4) at'
t = 2 sec, then give equation of the path.
(b)
= 8t - 2
Jdx = JC8t-2)dt
vx
Solution:
x=4t 2 -2t+c
At
t = 2;
2
X
= 14 =} 14 = 4 (4) - 2 (2) + C
x=4t 2 -2t+2;vy=2
2
the velocity, (b) the speed, (c) the acceleration, (d) the·
'magnitude of the acceleration, (e) the magnitude of the
-component of acceleration along velocity (called tangential'
acceleration), (f) the magnitude of the component of
acceleration perpendicular to velocity (called normal
acceleration).
(a)
(b)
"'
t2,._
_,
dr
O
"
r =ti+-j+tk
2
_,
~
A
V=-=HtJ+k
dt
speed l;l=~t 2 +2
=2
y = 2t+c
Att=2;y=4
=;
·time tis; =ti+-1:.t 2 j+tk. Find as a function of time (a)'
--)
C
f dy = f 2dt
;:.>
A particle moves in such a way that its position vector at any[
Solution:
=}
Equation
=;
c=O
y = 2t
x=(2t) 2 -2t+2
x=y 2 -y+2
=;x+y-y 2 =2
First, an auto starts from rest and accelerates uniformly for 15
s acquiring a velocity of 30m/s. Secondly, the auto then moves:
at this constant velocity of 30 mfs for the next 15 s, after,
'which thirdly, the auto decelerates uniformly by braking at
1.5 m/s2 until it stops.
(a) Sketch the velocity-time graph.
·(b) Sketch the acceleration-time graph for auto.
,(c) Sketch displacem_eT)t-time graph for guto. _
(c)
www.puucho.com
''
Anurag Mishra Mechanics 1 with www.puucho.com
.\
162
MECHA~·~-·
[2v 112 ]~ 0 = -at
Solution:
(a)
r:·;;~~---··
-- ----·-7
ell
'
§.
"
•
·!
2[
'
:
~
.'
~
~
t=
--
!
,50
J
t (sec)'
.
!
a
(d) Velocity at any time t is
dv
ft
- = - adt
f v 0 -v1/2
o
i
------------......1
Fig.1E.65 (a)
r----oo:i
IA 2
'
V
,-
I
[2_vlf2 ]~0
I
I
2[v1;2
I!
C
' :8
ij
o
15
2
''
30
O, ', .15
-v¼ l = -at
I
I
30
I
= -at
-v¼2] = -at
50t(sec)I
v=(Fo-a;r
, ~-1.5
'
I
'
i _____ Fig._1E.65(b) _
V
..S_'.J
(c) Displacement between t (0-15) = !. x 30 x 15
\I
975 · - - ·-· . - •
'E
I:;:-
Iii
E
1
•
675
I
'
.
rl
~250
15
2 v'f2
3 a
X=--0-
i
30
50 t (sec) j
I
~
Displacement between t (30 - SO) = !_ x 30 x 20
.come to rest at infinity.
•
a= -a.Jv
dv
112,
-=-av
dt
fo
v0
dv
v1/2
o/1
j
= 0 and is moving with a velocity given by (-
J
3·f + 2 j) .
2v3/2
.
j
...,
A
A
Solution: a=Scosti-3sintj
Ji;= Jscostdti-J3sintdtj
f v, dv, =f' Scostdt
-3
0
v,=Ssint-3
0
by the particle is - - •
a
.
... ,
,.
2 v3/2
. ., .
0
by the particle is - - , .
3 a
the correct opJion: _ _ __ · ________ _
Solution :(a)
A
, ·
·.
'(!,J_Jhe_p~/JjQJJ~l'_e_c_to_r_pfJlz~:p_aJ:tisk.at..ti11i.e_(t 2'......0)_.___ _
I
I
l_ Mark
1
IFind
.
1
( a) the velocity .at time t and·
~ parric!; ~i~i_a v;l~u;, ~ =v O at_ t = 0 is decele~ated ~~
the rate Ia I= a ,hi,'where a is a positiye constant.
. - 2.,Fo.
(a) The particle comes to rest at t = -·- - '
bJ The particle will
'(c) The distance travelled.
I(d) The distance travelled
A
time t
2
=300m
1·e?~ 66.....J~-L1z,x~mti
h.D~ """
--·)'· -~'.(?~~ll
particle . ti-av.els
· so that . its acceleration
is .giv.~
--,
'
a = 5 cost i- 3 sint j. If the particle is located at (-3, 2) at
IA
=450m
,•
t=-a
I
L _ .... _ Fig. 1E.65 (c) ···- _ _ !
Displacement between t (15 - 30) = 30 x 15
I
2.,Fo
at
"
~-·
2
2
at 3 at Fo
X=Vot+-12
2
---·7
--
dt
f~ dx= f~ (Fo-a:rdt
2
=225m
r::....
cJx = (Fo - at)2
=
dx = (Ssint-3)
dt
f-3x cix=J'0 (Ssint-3)dt ',·
x+3=5-Scost-3t =}_X=2-Scost-3t
Similarly,
dvy = -J'3sintdt
J"'
2
vy-2=3(cost-l)
vy = 3 cost -1
=J' -adt
o
www.puucho.com
··~ 0
Anurag Mishra Mechanics 1 with www.puucho.com
i- DESCRIPTION
OF MOTION- - - - ------ - - J: dy = J~ (3cost-l)dt
63
.
projection_ Generally initial direction of motion is
considered to be positive. Since g is almost constant,
equations of motion with constant acceleration can be used.
(a) Body dropped from a height:
x 0 = 0, v 0 = 0, a=+g,
V =gt
Equations of motion are
y-2=3sint-t
y=2+3sint-t
""7
A
A
v = (Ssint -3)i+(3cost -l)j
Thus,
""7
and
-
A
h =_!gt2
A
2
s = (2- 5 cost - 3t) i+(2 + 3sint -t)j
-
-
'>
v
r--~
l=.~Pl'.l'\P,~=- 1 68
(i) A particle is moving in three dimensions.Its position vector
is given by
~
A
A
= 2gh
:-··,
A
r = 6i+ (3+ 4t)j-(3+ 2t -t 2 )k
1Distance are in meters, and the time) t, in seconds.
( a) What is the velocity vector at t = + 3 ?
(b) What is the speed (in m/sec) at t = + 3 ?
(c) What is the acceleration vector and what is its magnitude
(in m/sec 2) att = +3?
(ii) Now the particle is moving only along the z-axis, and its
position is given by; (t 2 - 2t - 3) k at what time does the
''
''
'''
'
''
''
'''
:'
,o'
¼
(a)
(c)
(b)
Fig.1.81
(b) Body projected upwards:
x0
-particle stand still? _
Solution:
2
= 0,
v0
= u,
a
=-
g
Equations of motion are
v = u - gt,
2
2
2
h =ut-½gt ; v =u -2gh
"i! = 6i+(3+4tJJ-C3+2t-t 2Jk
-+
""7
(i)
(a)
V
dr
= - = 4j-(2-2t)k
dt
A
""7
At t = 3,
V
A
= 4j+4k (m/s)
x0
if2 m/s
A
h
dt
I
""7
A
-+
V
dr
= (2t - 2) k
dt
=-
v0
= u,
= ut - -1 gt 2
2
x0
a= -g
= 0,
v 0 =u,
... (1)
... (2)
v=u+gt
v2 = "2 + 2gh
... (3)
h = ut + .! gt 2
2
a=+g
___ (l)
... (2)
... (3)
In both the cases if we want to calculate time taken to
reach the ground we should solve quadratic eqn, (3) for
time t
In first case, eqn. (3) will be
-h =ut _ _! gt 2
2
r = (t 2 -2t-3)k
""7
= 0,
v=u-gt
v 2 = u 2 - 2gh
dV
a= - = 2k (constant)
~
A/At t = 3 also -+
a= 2k
m s2 +
, I al= 2m s
(ii)
(1) Upward projection (2) Downward projection
A
(b) speed =l~I= ~42 + 4 2 =
(c)
(c) Body projected from a height:
A
A
2
or
gt 2 -2ut-2h = 0
-+
2
stand still means v = 0 ~ t = 1 sec
or
Motion Under Gravity:
Experiments show that when air resistance is neglected,
all bodies near the earth's surface fall with the same
constant acceleration, denoted by g. We call this the
acceleration due to gravity and for practical purposes its
magnitude is 9 .8 m/s 2.
In describing motion of an object projected near earth's
surface, we use the y-coordinate with origin at point of
t
- Bgh
= -2u-±-~4u
'---=-
2g
In second case, eqn. (3) will be
l
2
+ h =ut+-gt
2
or
www.puucho.com
gt 2 + 2ut - 21! = 0
Anurag Mishra Mechanics 1 with www.puucho.com
.
j!i4._._ _ _ _ _ _ _ _ __
-2u
or
--
.
. - - --- MECHANICS-I I
---···-·----- ----------------------. -----:'.2..l
...
± ~4u 2 + Sgh
distance (s)
s
t=------2g
2h
In both the cases we will neglect the negative root oft.
&152&g.m;p~ 69 ~
h ................... .
;~;e;!
[A·b;dy ;rojec;;d vertical~-up;;r-:i;-;;;,; ;he ;;;-oj-;,
ireaches the. ground in time t 1 • If it is projected vertically:
' downwards from the same position with the same velocity, it\
reaches the ground in time t 2 . If a body is released from rest•
and from the same position, then what will be the time (t),
~required by the bod·po reach the ground? .. _. ____ . .. _ I
Solution:
.r-·· - ..···- -· -- -...... ·
II
t
0
·{2hlg
o·"""''---1---L--1,
2-,J2h/g
-,J2h/g
2-.j2h/g
1·
Ql<'----'---~---
-.j2h/g
'
... (i)
"1
:t i
For B,
1 2
-h = -Ut2 --gt2
2
... (ii)
= o-.!:.gt 2
t: I
'
'.lI
... (iii)
2
...
(v)
speed
0
A
Fore;
-h
C '
a
!
:t
:
91--------
: -.J2gh ....... ..
'
:
'
Multiplying equation (i) by
__ ~ig. 1_E_,6_9__ .
t 2 and (ii) by t 1 and adding
togethei; we get
1
-h(t 2 +t 1 ) = --gt 1t 2 (t 2 +t 1 )
'
I.
'
0
t
0
Fig, 1E.10·
• .!
2
-h
1
= --gtit2
... (iv)
2
·r
Equating equations (iii) and (iv), we get
1 2
1
-2gt = -2gtit2
~
!
b~<fy-Jai,; fr~~·~~m; he~h; and·
•
-;;;;,-:.u· b~ck to
initi;li
.
(v) -time (t), speed (v), time (t) and;
•
I
!accelera~~n_ (~) :~Tl!e _(~2!5fEPhs jj_~_t~e !'!O!ion_<Jf_ tJ,e /Jody. ,
Solution:
J¥
.'
'>-'Jo
-)
'
'
0
~
~
1
!Cv)-time (t), speed (v) time (t) ana acceleration (a), - timej
[(t) _graph_sfqr: the mptio_n of t_h~ body._
,
·position. Draw displacement(;) -time (t), distance (s) -time\
;Ct) and velocity
~-
!displacement (s)-time (t), distance,(s) -time (t) and velocity!
,.;:t = ,tlrl2
~J;:x~me1•~rm1>
[.4.
--~---·-··---
;A body thrown up and return back to its initial position. Draw:,·
Let body falls from height h. It takes time
to strikes the ground, its velocity just before strike is
i
Solution: Let the body is thrown with initial velocity
u, it takes time t!. to reach the highest position. It goes to a
2 g
'
height h = !:.... Neglecting air resistance we have following
graphs:
2g
Di.stance (s)
....
u2f--',----~
•s
g
u2
.J2gh. Neglecting time of collision,' we have following
graphs:
2gl---7"'
· 2u
___g -
www.puucho.com
I
o,'----'u'----2.1.u_,. t.
.. --
g
g
Anurag Mishra Mechanics 1 with www.puucho.com
rDESCRIPTION OF rionoN
i...,.
-
--
, _ _ ,,_ - · - -
--
-,
-·-·---
-----·-·
-----· _______ 6aj
~-
phase corresponds to vertical lines on the velocity versus
time graph.
speed (u)
V
u
-------r::7
73 ~
L~'==~~,mpll~a:'.:1
' -- - -
g
-
o/
·A boy throws a ball vertically upward with an initial speed
'15.0 m/s. The ball was released when it was at 2.00 m above,
'ground. The boy catches it at the same point as the point of
.
.
'
pro1ectzon.
(a) What is maximum height reached by the ball?
(b) lfo'!!! lQIJg is_ t!Je f/all in the air?
Of----u""'"-~2u~-
g
-u --------------------
-,
a
Solution: (a) The ball will continue
to move upward as long as it has velocity. At
the maximum height v 1 = 0. We choose
point of projection as origin and upward
direction to be positive.
From equation,v; = u; + 2ayy
01-----+-----+u
2g
2u
g
--g
Fig. 1E.71
V2 =U2 - rym,
y
y
"&I'
At maximum height,
A ping pong ball is dropped from a height H and bounces I
three times before it is caught. Sketch graphs of its position, ·
velocity and acceleration as functions of time. Take upwardi
direction 11§ positive._ _
_____ ;
Ymax. - 2g
(15) 2
= 2x 9.8
Solution: Position versus time graph is parabolic.
Velocity versus time graph is a straight line.
Is
le
lo
le
Fig.1E.73
u2
y
11.5m
hmm. =Ymax. +2= 11.5+2= 13.5m
(b) When the ball returns to starting point, y
tF
From equation,
0 = Uyt
~~-~~~---~-1(s)
or
'
= 0.
1
2
y = u y t + -2 ay t
1 gt 2
- -
2
2uy
t=--
g
= 2(15.0) = 3.06 s
9.8
Remark:------------------(i)
~-+11--+---lll---+--clll--l-l(s)
1
~
"'
(ii)
-4
-81---lll--+--llf-__,_ _-+-_
-12~~-~~----~Fig.1E.72
At each collision the velocity of ball changes from
negative to positive. When the ball is in contact with the
floor the velocity changes substantially during a short time
interval which shows that the acceleration is very large. This
Students can choose either of the directions upward or
downward as positive. But one should apply this· sign
convention to all the vectors throughout the problem.
If upward is positive,
a= -9.8 m/s 2
If downward is positive, a=+ 9.8 m/s 2
Draw a coordinate axis on your diagram and assign
origin. In the problem, identify the special condition
regarding the object, e.g., maximum height reached
implies v, = D at the topmost point, or if the body returns
to origin then y =0.
Use letters x, x 0 , v, v 0 or t for unk ·owns in equations.
Look for equations which involve unknown quantities.
One of them might provide you a solution. If more than
one unknowns are involved, then try to formulate as
many equations as there are unknowns.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
~
cili elevdt~; o~~ebi".S. ascrew fall from the
!A speeder, moves:at a ~qns~antJS ,;,;.sin a scho;(zdti-;A
. tceUirlg; Tlle'ceiliflgis•3 mabove:theJlopi: ·
· , : , . . /police car startsfrom rest)ust,as:the:sp_eeder passed it.·Tfr~
m~st~ridi!lt iii.
(a.ii Ift/te,eleva'toi'it,inaving upwai-q-w{th a sp~ed of 2.2 ,nls,
t' _;Jiqw l~,ng clqes,'it.t'*for-_th~ scr~iti to, hit the floor? ! . :
!police. car acce.lerqtes at.2 m/s 2 u_ntil it reaches its.maximum!1
l~e!ori_ty .o! 20_,m/s. Wh~re ·and, when does the ·speeder _ge;I
(!,) J-1:fqw long is.the so:ew m mr_if the:~lev~torstarts from rest
k@g/lt? '
l ·acceleration ofA~4.0mls
Solution: When two particles are involved in the same
problem, we use . simple subscripts to distinguish the
variables, as shown in Fig.1E.75(a). The motion of the
I,~..·:when_the,screwfalls,,_and.moves.
upwards with a constant
2
? .•
·
· ··
·.·solution:,, (a). We consider the elevator. floor to be
origin., •The elevator floor moves with constant velocity.
Equation for floor, y f ·= v ft = (2.2m/s)t. The screw falls
with acceleration due to.gravity.
..
, ..
h.. (22)'
l
2
,.. ,
,· Y, = _+ . t,-;:-2gt
.--~
"-X, =yf
· ·c2.:d~
=,,..:..
c2.2Jt - .!2 gr 2
- .
fg·
9.81
y·:'.~'~P .y· -i- ~, ~~~C:(~x
Xs
Cb) Eqtiationfor floor is·:
.
' ' Equation for ~crew ~
.J
- IPl: '
t=L\t1'
0
: t= ~=~ 2( 3) =0.78s
or ·
~
,;
!Pl· ..
0--
-~s
[;l·--·
·'
t = ~t1+dt,2:
police car has two phases: one at constant acceleration and
one at constant velocity. In such problems, it is convenient to
use M instead oft in the equations. The police may or may
not catch the speeder during the acceleration phase. This
has to be checked. We set the origin at the police lookout,
which means x 0 s = x 0p = 0.
Acceleration Phase: Let us say this takes a time·
interval Llt 1 .
From v=v 0 +at, we have 20=0+(2)M1 , thus
M 1 =10s.
At this time, the positions are given by
1 2;
x = x +v t +-at
, Note tlianhe screw's position at tirrie t = 0 was h and
due .to inertia' of inotion at the instant ofrelease its velocity
will b~ slririe.a~ that.of elevator.'
,.-~ .
[!J ~
~ Vs:= 15 mis
' . _ _ _ _ _ _ _F..,,,lg.1E.75,(a)
.__
~e.,
,·
0
; ' -
.y,
1 .2
= h --gt
2
When screw meets the floor, .
1
y' "' y f
1
2
3
2 aft = - 2 gt·2
i •
.
:.· - ~
t. '.:'
af ';"
'' . :
-,
''
vg+
2(3)
(9:81 + 4.0) ..
·= 0.66 S·
,
Note
iri: this situation the elevator starts from rest;
.._ ·. so initial velocity of screw is zero. Time of fall of screw was
independent of the speed of. the elevator as long as it moved·
with constant, velocity.. When the elevator accelerates, we
can say .. that screw experiences effective acceleration
y' = g .j.'
If acceleration of elevator af = - /J' the time of
fall becomes.1nfjnite, i.e., it appears to be weightless. ·
rll~t
0
= (15) (10) = 150 m;
2
Xp
1
= - (2) (10) 2 ~ 100 m
2
The speeder is still ahead.
Constant Velocity phase: Let us say this_ takes a time
interval M 2.
: Given: Xos = 150 m; x 0p = 100 m; Vs = 15 m/s;
· ·vp = 20in/s; as= ap = 0
·unknown: x~ = ?; Xp ;;;; ?; L1t 2 =?
The cars meet when they have the same position, that is,
X s = X p • However, we cannot find where until we find when
Xs =150+15M 2; Xp =100+20M2
On setting Xs = Xp we find r-·-xc--------,-:,
Llt 2 = 10 s. Substituting into
'(m)
either
equation
gives
..,______,,
300
x = 300 m. The speeder is
,... ,
caught at 300 m after a period
of 20 s. The graphical solution is 1so··
I
i
depicted in Fig. lE.75 (b).
100
1-
'o./.
==:;==:;::->;.,'t(•)I
At1
412
I
~~-F~ig~.1~_{1;!_,)_-=--''-'-I
www.puucho.com
·-·--·-
Anurag Mishra Mechanics 1 with www.puucho.com
DESCRIPTION OF MOTION
;rwocars approach each other.on a straight road. Car A moves]
!at 16 m/s and car B moves at 8 m/s. When they are 45 m
llapart, both drivers apply their brakes. Car A slows down at
2
2
.2 m/s , while car B slows down at 4 m/s • Where and when'
l4q_ th~ collide?
.
Solution: Fig. lE.76 (a) is a simple sketch of the
situation. We choose the origin at A's initial position and
point the positive axis in the direction of its velocity.
-Lx
..
~
~ - --c~I
~
aA<F=
voA voe'
<•l
X (m)
50
-
~ aa
B stays at this position until it is hit by A. The condition
= Xe becomes
16t-t 2 =37
Thus, t = 2.8 s; 13.2 s. We reject 13.2 s since there can
be only one collision. The collision occurs at 2.8 sand 37 m.
,
xA
I
'
(i)
: ,..,
__,,
course, we could have checked this at the outset, but that's
hindsight.
This is a good point to look at the graphical solution.
Fig. lE.76 (b) shows that parabola (i) stops at 8 s, whe~ea,s · ·
parabola (ii) stops at 2 s. From them on the graphs are .
horizontal lines. The solutions 3 and 5 s are the intersections
of the two parabolas. These would have been physically'.
acceptable if the accelerations had remamed constant in
both magnitude and direction.
The graph helps us to find the proper solution. We must
find the intersection of the horizontal line for B and the
parabola for A. Let us find where B stops. At t = 2 s, (ii) gives ,
Xe= 45-8(2)+2(2) 2 = 37 m
I
'
~iE.~qm~
,.
0
3
4
6
(b)
~---~Flg.1E.76
Carefully check the signs. The cars meet when x A = x e,
so we set up general expressions for these quantities using
1 2
x = x 0 +v 0 t +-at
:
2
XA
=l6t-t
2
Xe =45-8t+2t
R
... (i)
2
... (ii)
When we set xA = Xe,
We find 3t 2 ~ 24t + 45 = 3 (t - S)(t - 3) = O
We seem to have two possible times for the collision:
t=3sandt=5s.
Try to find the flaw in the above argument.
Let us look at the velocities to see what has happened?
Att=3s,
VA = VoA + aAt = 16+ (-2)(3) = 10 m/s
Ve =Voe +aet =-8+(4)(3)=+4m/s
Does this give you a hint of the difficulty? v A shows
nothing unusual; A has slowed down. But look at the sign of
Ve· We seem to have found that when the brakes were
applied the car reversed its velocity. You can easily verify
that B stops at 2 s and then stays at rest. This means that (ii)
is not valid after 2 s. Sinillarly, (i) is not valid after 8 s. Of
'-----·-L__
F..
ig_._1E_-·_11_ _ _ _ _ _ _ __
I
, ••• I
Solution: (a) Length of the chord P1P2 = 2R cos 0
Acceleration of bead along v,,ire = g cos 0
•
From equation,
v 12 = vi2 + 2ax
= 0 2 + 2(g cos 0)(2R cos 0)
or
vt
=zJiii. cos0
(b) From equation, , v t = V; + at
=
or
o+ at
t = Vt =
2Jgii. cos0 = 2,/iiJi
.
which is indepe11dent·of0; thus time of travel along any
chord is same.
www.puucho.com
a
g cos0
Anurag Mishra Mechanics 1 with www.puucho.com
[68
MECHANICS-I.
'O_J
I
IA car· is speedi!ig°at 25 m/s. in-~-lli;~pe~d-~one. A poli~e car) IA spaceship
/starts fr~ll}·C~tljust as the sp,eed~fp~ses and ilccele;ates at~
,constant:rate
of 5 m/s 2 • ,,__
.·
,
,
! ..
-. ""
(a) When does the police car ca.tch the speeding, car?
'(bj' Howfast ts·t1tep'o1ice car trav~lling when it catches up
with the speedet?·
-·
· .
· :(c) How far have the cars· trave(l_ed when the p9lice
L __ catches the weeder? · ___ . _ · _ _______ _. __ _
0
. '
.
•
'
'
C 0
!'
cari
x, = v,t
Solution:. (a) For speeder,
: !
.--~~---: j
-
..1,
\..!:~.r--·vO = o·
·
_-.
ap=5m/_s~
r,J first
Wh,ntime?
= 1/re - ~ ""™ """"'j,,,m M,,i fr'
,
1
"'I
(b) How high above the planet's smface will the first meeting
I(cL_takqil_ace?
.
. · ·
. ..
' ,. . , ·
What)s the yeloatyyf eac/J;§pacesh1p when they meet?
,.,l
time (t)
"·rn-,
launched vertically from_ Mars has recicl,~d a
!heig~tof30Oin and a velocity of:80 in/sat time t ·=; 0.f'.tthis
\i!JStant its controls are switched. off. It continues ·to move
upward under the influence of Martian gravity,
approximately equal to 3.72 m/s 2• At the same instant
,anothrr spaceship at height 1500 m is moving downward at
_
,25 m/s and slowing down at a rate of 0.80 m/s 2 •
L~-~---.{ \
:l
.,-.,,-----i !
: ______ : 1
Solution: (a) We write equations for positions of
spaceships 1 and 2.
y 1 = 300 + 80t - l.86t 2 , y 2 = 1500- 25t + 'o.4t 2
---'----"-~...:..:._ _.;•, i
!
--·-·· - (2)
.,
x.=i::•.
_ _;._-"----=----'-----~-.
-t.
l
-T
w,
"_ '(1)
u,=;=-2S'm/s
' ..- : a::: 0 _8-m/sZ•.---=- ~ u,';: 80 mis
Y
'Yo= 300m
:+ve
300m
, a =-gM =-3,72.mls 2
,,,_g
'--L--~-'--....'-__.x
Surface of 111.ars
For police car, x P =
+v~
.! ap1 2
a= O:B m1s2
.2
At time t both the cars are at same position x, = x P
'
1
2
vt=-at
s
2 p .
300m
''
:-:
'·.:·1'
Fig, 1E.79
---------
t= 2v, = 2(25) =lOs
aP
5
(b) . When the spaceships meet,
Yi =Y2
300 + 80t - l.86t 2 = 1500 - 25t + 0.4t 2
(b) Velocity of police car is given by
VP= apt= (5)(10) = 50m/s
(c) Distance moved by speeder= v,t = (25)(10)
=250m
Remark:----------------Distance covered by t~e two cars is same; hence they must have
same average velocity. If the police car waits for I =2 second or
t =4 second, it will catch the speeder after c9~ring a larger
distance as shown by dashed line in graph.
---....._
We get two solutions, t 1 = 19.0s and t 2 = 27.4s
(c) The two spaceships meet twice in their journey. We
require y 1 att 1 •
Y1 = 300 + 80 x 19 - 1.86(19) 2
= 1.14 km.
Spaceship 2 first meets spaceship 1 when the latter is
moving up and a second time when it is moving down as
shown in the graph. Students are suggested to interpret the
curves for a= l.Om/s 2 and a= 0.4m/s 2 •
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
I DESCRIPTION Of MOTION
~exct_!E.~~fsot;>
= (v,; i + Vy.j) + (ax
...v ·;:::: ...vr + ...at·
1
Ball 1 is released from the top of a smooth inclined plane, an~
at the same ins tan~ ball 4 is projected from the foot o};the'
plane with such a velocity that they meet halfway up the
incline. Determine:
- - - - - - - - -1
Simllarly,
2
We may write the above equations for final po_sition
vector.
li
!: ~
-+
-+
2
__,
where
A
J
--·'
(a) Accelerations of both the balls are
2
Yd
A
vi
a t
A
rr =xii+
...
!
1-+
---+
= r; + v, t + -
r1
(a) the velocity with which balls are projected and
·
(b)Jhe velocity of each ball when they mee,,,,t.~------
Solution:
2
1
hi
Fig. 1E.80
1
2
Xf =X·+V
l
XI-t+-aX t
-t+-a
Yi =y-+v
l
yt
2 yt
i
4.LJ"'-------'
i + ay j )t
A
=v.ni+vyij
...
A
A
i + ay j
a= ax
For a particle moving in a plane we may write the
following equations (note that we have assumed initial
position to be origin, i.e., xi = 0, y, = 0).
a1 = g sin0 and a 2 = - g sin8 down the incline.
Ball 1:
.!. =(0)t + .!. g sin8t 2
... (1)
Ball 2:
.!. =v- t + .!. (-g sin8)t 2
... (2)
2
2
2
'
2
V,;f
Adding eqn. (1) and (2), we get
1
l=v,t or t=-
=V,; + a/
Vyt =V_y;+ Of.
vJ = vJ + 2n,(x1 - x,)
v.J.=;; + '2il,0/1-Y1l
v,
Substituting it in eqn. (1), we get
i2 =.!.2 g sine (_l__)
v,
or
or
For ball 2:
[
vJ = vf + 2ax
-+
A'
Since
'1
.
•
I
-+
V
A
A
a= 6i+4j=constant
.........
=U+at
Therefore using
V
,...,...
"'"
= 4i+3j +(6i+ 4j) X 2
...
A
A
=>
v=16i+lljm/sec
The displacement is
-+-+
1-+2
S =Ut+-at
2
A
A
1
A
•A
=(4i+3j) X 2 +-(6i+4j) X 4
2
A
A
,..
...
or
v 2 =0
Two-Dimensional Motion with Constant Acceleration
1. The position vector for a particle moving in the
,y-plane is given by
r =xi+yj
Velocity of the particle is obtained by
...
-·-------.---------------+
,..
,A
Solution.
= g1sin8-g1sin8 = 0
A
-
_constant acceleration is a= 6i+4j m/sec 2 • Find the velocity:
@!d disp)acement_ofihe JLarticle at t = 2 sec. ___ . _ _ j
vf = O+ 2g sine.!.2
v 1 = .Jglsin 8 = .Jii1
v~ = "f + 2(-g sin8) .!.
2
...
-- ---------- -
The velocity of a particle at t = 0 is"u. = 4i+3j m,/sec and aj
v, = .Jg1sin0 = .Jii1
(b) From equation,
For ball 1:
2
A
v=vxi+vyj
Because "i is assumed to be constant, its components ax
and ay are also constants. Hence we may apply equations of
kinematics to the x and y-components, e.g.,
...
A
A
Vt =(V,; +axt)i+(v_y;+ayt)j
'The accel_erqtio.n of a moving body at any_ ·me 't' is giv_en_ by:
-+
"
'-+
'
•
'
!
2 ,.
2
• If ti =·0thenfind the velocity,
a= (4t)i+.(3t
. )j · m,/sec
.
.
I
oftheparticleaJ4:sec. _ ________ .. . ,
u_
t
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
MECHANICS-I
...,
A
1
A
Solution: Since a= (4t) i+(3t 2 )j is time dependent
..., ..., ...,
· therefore we cannot apply v = u + at
We can solve by applying calculus
...,
dv
...,
-+
f
.iuti=
...,
2xl = 1;
-=a
dt
Thus
;::
fo
or
dv=
4
A
A
(4t i+3t 2 j)dt
O
~2 ~t 21= ~2 X (2) X. (1)2
I
'
=1
=.{:J \dt }1; {3Jt dt }J
2
0
0
...,
A
1;1= ~c1) 2 + c1) 2
=..J2 m
A
v = (32)i+(64)j rn/sec
Equations of Motion in Vector Form
The equation of motion in the case of a uniformly
accelerated motion is written in vector form as-
..., ..., ...,
(i)
v.; u+ at
(iii)
-+ -+
V·V
=
(ii)
-+
U·U+
~
-+-+
½°BF ....................... :""'>
1-+2
.s=ut+-at
~~t
tan(j>=
2
.
iut
J·
$
1
lutl
Fig. 1E.83 (b) .
=45°
Displacement is at an angle 45° with the direction of
initial velocity.
~
co}
-+ -+
2a· S
l2...,
2
.:
''
· ,--'(Application only when acceleration vector is constant, motion
' • may be along straight line or along curved path.)
'
·..+iam~ii 03 ~
I
'fA_ pCl!tide 0J'.n1~s 1 kg has a~ velocity· of2171/sec. A cons~ant
'
if9rce, of 2N acts;on· the. particl~fer l sec in a direction
1p'erpe~dicul11",;.:'.t~ .1ts initial ve/qcirg.. Find. the velocity:r/nd
fdisplac~mentoj't/t'e particle at th~ end ofl sec. ' ',, ,' '
F
2
-+
V
..., ..., ...,
v=u+at
...,
-+
'
...
where
ju!= 2rn/sec;
...,
~.
at
. .
V"
r:··i;:····--,;··;··: i '
r
.
=2rn/sec
lvl= 2..J2 rn/sec
or
...
...
8 = tan-1 Iat!
F:~. 1E.83 (a)
'.u
I
iul
1
= tan- (1) = 45°
Hence the velocity of the pa1ticle after one second is 2..J2
rn/sec at an angle of 45° with its initial velocity.
For the displacement equation of motion is
''
a
=10 , Va COS8 =10
10
10
V
5
< , •• "'
')<
Flg.,1E.84
Rate of change of speed = a case = 2 rn/s.
,
latl=2Xl .
...
~u
,··acos8
acos8=-=-=2
-+
!vi=· iul 2 +l.atl 2
'Hence
,""f"B/~;.. ,
AA
= 3i+4j, a= 2i+lj
V·
Since v and at are perpendicular.
-+.
(d):,..g ws~2 - ~ _ , _ _ _ ,
,AA-+
.......
Here acceleration is constant therefore .
...,
(e)_../5_rIJls 2 ·
Solution: vx = 3m/s, vy = 4m/s
ax = 2 rn/s2, ay = 1 rn/s 2
Solution:
Acceleration of the particle a = m = 2 rn/sec
particle is moving in xy,plcine•. :At certain instan~ .the
compon~nts·of its. velocity and'acceleration ,are:as1o(lows
Vx = 3 mis, Vy = 4 m/s, ax = 2 ~ 2 and ay =.1 mls 2'. The
rate.of change of speed at this moment is
,l' r·
·
,
(a) 4 m/s 2 , , •
·
(b), 2 m/s 2 ·
Ee±xgtmtjr->'I~~ 85
------~-~
'·.
~
i
'
•.
"
'I
••. -;,:
The figure shows the velocity and 'the acceleration t of' a
point-like body at the· initial moment of its motion. _The
'direction. and. the absolute value of the acceleration rem1iin
constant. Find,cthe time in secona.fwhen the velocity reach its .
minimum value ?
(Data : a= 6m/s 2 , v 0 = 24.m/s, · q, = 143°)
,,
Flg.1E.85.(a)
•'---..C..·- - - ' - - - - ' - - - - - - - ' - - - - - ' ~ ' - · - '
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
'.DESCRIPTION OF MOTION
1--··----
Solution:
x-components
ux = -v 0 cos37°
ax= a
y-components
uy = v 0 sin37°
Vx =- 4vo +at
ay
=0
V
0
=--
5
3v
5
y
2. Projectile motion: An object is flight after
being launched or thrown is a projectile. We assume that
the distance travelled by a projectile is much smaller than
the radius of the earth so that the acceleration due _to gravity
is constant; secondly air resistance is negligible.
y
7°
/'i:'--"e!----Vo'
A
a=-gj
a
i
when vx
..-·
Fig. 1E.85 (b)
=0
4 24
t = - x - = 3.2 sec
5 6
lliustration 10: Consider a ball initially moving along
the x-axis as shown in Fig. 1.82 (a). At time t O it gets a
constant acceleration ay in the y-direction. At any time
t > t O the x- and y-displacements are
1
2
y=.-ayC
2
The superposition of these displacements is a curved
_,
path. The total velocity vector v at any time is tangent to the
curved path of the ball. Velocity vector is an angle e relative
to the x-axis, given by 8 = tan- 1 (vy/vxl, · which
f-'---.----------,---.-lt-r_.,.,.... __ •
a
~xi
a
1
';'
lfyJ=-Vyli
- -·· ----- - - - .. !~:!·~_(bl _______
_J
Fig. 1.82 (b) shows the path of a projectile with velocity
vectors. Let the launch point be (x,, y ,); y is positive upward
and x is positive to the right. Projectile is launched with
initial velocity vi at an angle e. Since motion of a projecti)e
takes place in a plane we will set up equations for x- artd
y-components separately.
continuously changes with time.
~:~¼
,,.1,:
2 y
/'
Vx
"x = 0
a,=-g
vxf :::vxi+ Cl;.:t
V.>f =Vy;+ G/
=V;COS8;
1
Xt=X-+V
,t+-n•
l
X1
2 "".r"
V2
",, = ., t,
Y2
=½ay t~
Uy1 = ay
1
•1
81
v
Q-2..+,,.....
..•
Y = 0 ta Vx X1 = Vxt1
Strajght-line
vy=O
' '
.t:''
X
:
,1
•
-~
- •.._:-- . .
1:,
v, sme,t--gt
v;, =v~ +
' '
... (4)
=.)'(+v_;.,'.-~'2af
= x, + v, cos8,t · ... (2) =y,+
v
X
2
2 ..
2a1 (y 1 -
I
... (5)
y 1)
:
;
'
X
:
X2·=
Vxl2 X3 = Vxt3
Curvilinear
motion
+av
motion
Yt
. IJ -~
V1
Y1 = 2aY t1 ------·.: '.,
,•
2
ti(,,/ l
2
V
....
'.
---------------·.--
....'
;::; vi sin ei-gt
... (1)
L.:...
"x
atto=O
3. Problem Solving Strategy:
1. Imagine the situation of problem; draw a picture
which shows the object and its possible trajectory.
2. Choose a coordinate system, choice of origin is
arbitrary. Generally point of projection is assigned the
origin. If range is to be calculated along horizontal
level, take y-axis parallel to acceleration due to
gravity. For calculating range along incline take x-axis
parallel to it and y axis normal to it.
0
Flg.1.82 (a) ____ _
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
MECHANICS-Q
3. Identify the initial position, irntial velocity and
acceleration. If irntial velocity and acceleration are
not along assigned x- and y-axis, then resolve them
into x- and y-components.
4: Identify the unlmown quantities and assign them
letter symbols u, v 0, a, etc.
5. Make two set of equations, one for x-components and
one for y-components.
6. Try to identify special condition of the problem that
may define the object's position or velocity at the
point of interest.
7. After writing appropriate kinematic equations in
component form solve them.
8. Always remember that time of flight is same for both
component~ of motion.
·ekam~'. a6 - ~
,·--------------------·----- --------IA bomber ii; moving horizontally at a speed v 1
= 72 m/s at a
height oJ;h = 103 m. An enemy ta_nk is moving horizontally
(x,axis)
constant speed. At the instant .the. bomli is
released a '.tank is at a distance_ X a~ 125 m from origin;
Origin is directly below a bomber at the instant of release ofij
bomb. Assuming the tank to be 3 m high, find the velocity v 2 ,
and. the time of fl.ight of bomb.
~----·-·------ -------7
I
l
V1
!
w;t6
I
I
\
I
i
l
'
I
L-C'---'
"'--~-"
-
Y• =Y; "': Vjif ~
1
j a_,t
=9 + Vo-.,,,,Sill ~J-f gt
;
t
·:
_ X - X; _
Vx
·--- >< (9.8)(1.85)2 ,cC'.
(, -" .·2-~ ,; ,:,~,- .. -.. ·,~,i
t
rv~---"---,--,---'x
32.0m
17.32 _
-,
t
=4.52s
r
-,-:,
="1.sss' ~.
-'-,- ---- - -- - - - - - ' -
.
,
_
.
,
;
.
l
... ..
•
•.. -----··-··-.._.:·... ··
'"S'.,..
· We have assumed y = 0 at 2 m from ground. So height
of ball above ground is 3.73 m.
325 - 125 _
/ i\
- 44 . 25 m S
4.52
1Aboj throws~ ball with v;loqty v0 =10,/2 mis.at an angle
of 45° as ·shown in the figure.'After collision with the ball the
vertical component of ball's velocity is unchanged and the
horizontal.i:dmponent is reversed in d(rection. Where do~ the
ball hit the ground?
··, · ·
·
,
'.
2 _ 100
-4.9
·
2
@exg~~~
i---~~--
------t=?
y=?
',
V2----
,_:. '
=1.73m>
, 32.0
=--
,.__I
. For the tank:
'
,· ;, ''. · 1
'
.•.
. -. -" f26:o}J@30°'d.ssJ;
t=~;.
cy=3m
3=103-(4:9)t
2··
2
•
When the bomb hits the
.. tank,;
~325m
f;eom110nents
, X,=v~t
•
Solution: Initial velocity of the bomb will be same as
that of bomber
vxi = v 1 = 72 m/s;
For the bomb:
Solution: We have to find timet when x = 32.0m. In
the second part we have to findywhenx = 320m or we can
say, find y at the time it crosses the goal -line. We take origin
at the point of release, xi = 0, y i = 0.
• ".
\
L-~---..,.l.~ i g . 1 E : ~ - - - - - - _ J ___
lt:footballer throws d. ball from a height oJ 2.00 m 'aboye_ the
Iground with an initial veloc_ity o/20. 6 m/s at an gngle pf 30°
jabove the hof_fzont~L (a)How lonq,does the ball taki-wcro~
[the goal (ine l2:0mfrom the po~n~of_releas~?{b) Wh~tis the
IJigll's heyghLabove·the grftundASJt cro§ses the.goal lute? -·
.
I
I
-
·~o-/2m/s -.
,
_,,
Fig.1E.88
-~-'----'-----·--'-------~
~
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
.. . -· .• .. ... - ---
P>ESCRIPTIO~ ~-~~OTI_O_N______
---------- -
Solution. We divide the problem into two parts, i.e.,
motion from boy to wall and from wall back to ground.
(i) Motion from boy to wall
x=v;t
.'Y"//'l,,Y"·<··
v0 cos45°_t,= 4 m
y=v0 sin8t-½gt
4
Vo COS
45°
I
16
-3. = - _!
(9.8)t 2
2
Since horizontal distance by policeman is less than 4 m,
the policeman fails to cross the gap.
5
Ball strikes at a height of 16/5;
tnfrom _
I
boy's hand.
·
2
=-sec
5
= v0 cos45° =.10 m/s
vy =v 0 sin 45° -
For Thief:
~;;;;;:;;--;...,c-;-:-;-;-;-;-:--;Y;:'C;:;O:;;m;;:p~o:;;n:;;e;;;n;;;fs;--'_.•·G7
10./2
2
= ---"7,;-10 X :- =6,n/S
i
5
(ii) Motion from wall to ground
We choose point of impact as origin and upward positive
i Hence the ball lands l7.8
i
from the foot of the wall
g
10
t' = 1.78 sec (neglecting -ve va!u~)
··-·---.-
± ]:_
where t 'is time of flight.from waii
to·gr~und
l
6
2
- · · = 6t' _..!_ x lOt':(
5
2
26
or St'26t' - 5 = 0
.
, 6±~36+4(5)(26/5)
. ..
-- . . ,
-----~ - - - - · .
--·-·-1
A policeman is in pursuit of a thief Both are running a_t ;il
·m/s. Suddenly they come across a gap between buildings asj
:shown in figure. The thief leaps at 5 m/s and at 45° while the'
'policeman leaps horizontally.
i
a) Does the. policeman clear the gap?
j
!(bJ By ().QW _mµch _does.the thiefelegr_.th&gap? ___ . .
i
i(
g
=-0.Ss
------'
Jcv,
sin 8)2 - (-6)(g);
.
or
t =1.22s
,- ---
., .. ··-- - '.
!A helicopter is flying at 100 m and flying at 25 m/s at an.
1angle 37° above the horizontal when a package is dropped·
!from it.
~a) Where does the package land?
,
(b) If the helicopter flies at constantvelocity, where is it when,
'i the package lands?
I
-
-- ____,
- -
.-
~
I
'
.
Solution: We choose original launch point with
upward direction positive.
-
!
-·- --
--
'"'
'"'"
}
2
l 2 -vism
• 8t-_-3 = 0:
2.gt
,or
i
ort:
-
'
v1sin 0
t=
~
-265 =6t' _!2 gt' 2
=17,8 m
.
'f '2
-3=v-sm8t--gt
=4.:nm
x-cofup8iiehts
··
<..-,".¼"Fl<'.L·,
_ l0./2 1 78
-..J2X.
1
2
y =V_,.;t + ZU/
x =vnt =Vicos0t
We can find horizontal dis~
tance- travelled covered for
positive value_ -Oft.
X = (5) COS 4,5° X (1.22)
gt
1
-v2
... (2)
_t=0.782s
I
=-m
10..J2 ·(l/./2)
· · - - - - · · - - ,1
1
y=O--gt 2
2
(~)1
sJ1
.Jzsz
4
vx
... (1)
= (5) (0.782)
= 3..91 m
,.IO..J2 -~ _.!_ X10 X
t=--'--
73:
•
For Policeman:
X·GOl!IP0111ll)ls ..
2
. .. .. . ... .I
·- -------
i
I
I
·.·
.
L----'---·-----·
Solution:
Fig.1E.90
·---·-·----·
-··-
Due to inertia of motion the initial velocity
of the package is the initial velocity of the helicopter. We
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
MECHANICS-I
choose origin to be directly below the helicopter, i.e.,
= 0,
X;
y 1 = 100m.
y-components
X
I
t2
Y-Y;
+v_nt +,ay
=:Vxit
=vi
COS
11=100m+ (25)(sin 37')t
37° X t
:Note that in order to calculate
range we must know t. Which
!we can calculate from equations
'of y-coordinate.
- _!c (9.8l)t 2
= 100 + (lS)t -4.9t
~;::::.::'.=-=i31DOrrm;::=~ Traget I
t=-3.24s
Fig.1E.92
We choose positive time
t = 6.30s
;_,. :'
',I -~3"'.2~4-s--+----..6,-,3!-:0-s••
..t:____
2
y = 0 at t = 6.30 s
and
..
u ... ··
2
(: =_(2_0_'.~-~:~fu)
= 126 m
l l
A bullet with muzz/r ··•lncity 100 m/s is to be shot at a target
30 m away in th,
'wrizontal line . .How high above the
•target must the re .• aimed so that the bullet will hit the
'target?
Horizontal range of bullet is 30 m.
2 • 20
u sm = 30
g
. .,,, 30 X 10
sm = = - - -
Solution:
.
For helicopter:
y-components· ·
.
i'
or
(100)2
= 194.Sm
= 126m
Note that the negative time indicates the time when the
package would have been if its motion had started earlier.
~f[:~,ilijpJ~i;J917;>
'A particle is projected from the origin in such a way that it
passes through a given point P (a, b} What is the minimum
rN!J.ir:ed,. sp~ed_ to. do _50;>
Solution:
Equation of trajectory of a particle is
8 = 0.015
Therefore
The rifle must be aimed at an angle 8 = 0.015 above
horizontal.
Height to be aimed = 30 tan 8 = 30(8)
= 30 X 0.015
= 45 cm.
Concepts 1. Relation between maximum possible!
·range and greatest height for any angle 0
!
Rmax
2
y=xtana-
gx
2
sin 20 = 0.03
sin 8 = 8
20 = 0.03
or
For small 8,
i.e.,
=100+(15)(6.30)
,; (20) (6.30)
is attained for
u'
2
2u cos a
Rmax'=g
If projectile passes through (a, b),
b=atanaor
H max is attained for vertical projection i.e., 0 = 90'
ga 2
atanu--(l+tan 2 a)
2
2u
ga 2
2u 2 cos 2 a
I
0 = 45'
ga 2 tan 2 a - 2au 2 tan a+ (ga 2 + 2bu 2 ) =
0
This quadratic equation in tan a must give real roots for
a particle to pass through (a, b).
Thus
Discriminant ~ 0
i.e.,
4a2u 4 - 4ga 2 (ga 2 + 2bu 2 ) ~ 0
u'
H max =2g
-
Rmax ;;::; 2Hrmrx-for same ' u 1•
2. If H 1 maximum height for the angle of projection 0 and:
H 2 maximum height for the angle of projection;
~~-~
I
I
y
:
!
or
u4
or
u4
or
or
-
2gbu 2
-
g 2a2 ~ 0
2gbu 2 + b 2 g 2 ~ b2 g 2 + a 2 g 2
(u 2
-
bg) 2 ~ (b 2 + a 2 )g 2
u
~ ~bg + g~a 2
+ b2
www.puucho.com
Flg.1.83
x\
!'
-- - -- ------ -- -- -·
Anurag Mishra Mechanics 1 with www.puucho.com
IDESCRIPTION OF MOTION
'--------------------- - - - - - - - - - - - - ---- --------- ----. - -- ----·-- -- - u,
In both the conditions the magnitude of velocity of,
I
projection is same so horizontal range will also be'
same and let that be 'R' then
R = 4~H1H2
l ..
V
"2 ..... •....
·-. .·.'.' . . . .~;--..._!:
.
:
. ....
\
...
3. Relation between horizontal range
maximum height
2
u sin2B
u 2 sin 2 0
R - - - and
H=--g
2g
R = 4Hcot8
andl
I
\ \
Fig. 1.86
'
i
... (1) ;
y
-----R---+
751
.. ,v~'--;_-o:;_-···>
hl
Hi = tan 2 8
H2
and
A • :ti;;_··-..
-------
Iwo particles A & B are projected from the same point in,
different directions in such a manner that vertical
components of their initial velocities are same :
(a) Find ratio of time of flight
(b) Find ratio of range.
,Y
X
Fig. 1.84
If
the range of projection is n times the maximum
height of the projectile, then angle of projection is
given by
Using equation (1)
8= tan-
1
(;)
Fig.1E.93
•
Solution:
. 4. Relation between angle of projection e and angle of
elevation <jJ of the highest point of trajectory from the
point of projection
5. (i) If t AD
T1
(b)
= time interval to travel from A to D
g
-
g
,
2
g
(2usin8)
Range = - - - x u cose
g
- = ..:....-'-'--'-'-'-'-"-----'-"--"-
8(h2 - h1l
Rn
g
,h2-h1!/
g
2u sine
= ---
UA sin8 A = Un sin8n
RA
(2uA sinSA)(uA cosSA)I g
t BC = time interval to travel from B to C
2
2
(tAD) -(tBc)
2u
_Y_
So, time of flight same for the two since
_ 2v A sine A T
2vB sin Sn
H
u 2 sin 2 8
2g
tan<j,=-=---x-~R/2
2g
u 2 sin2B
1
tan<j,=-tan0
2
(a) Time of flight=
(2un sin8nun cos8nl/ g
= [UA COSSA]
Un COS8n
'\
B~C
RA = sin Sn cose A = tan Sn
. r ---"
Rn
sine A coseB
tan0 A
Fig. 1.85
(ii) Consider three projectiles first is projected at e angle:
above horizontal, second below horizontal and tfiird!
horizontally.
lu 1 sin0 1 l=lu 2 sin0 2 I
Then .
t 3 _=J_t 1t 2 ___ _
If
:Four cannon balls, S, T, U and V are fired from level ground. '
'Cannon ball S is fired at an angle of 60° above the horizontal
and follows the path shown. Cannon balls T and U are fired at
,an angle of45° and Vis fired at an angle of30° a~ove'the
:horizontaL Wl:!(clt_s:Cinn_on l!all has the largest initial speed?
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
j
MECHANICS,!
176
~I
_.,
-
---~----
Fig. 1E.94
y =(tan0)x-(
The equation obtained is of the form
y =ax-bx 2
.
J
Solution: Expressions for range and maximum height
u 2 sin28
u 2 sin 2 0
are given by R = - - - ; H = - - g
2g
-From figure T, V have same range
sin 2 90°=
sin 2 60°
u¥
Thus
u~
u¥Cl)
~ u}(¾)
which represents a parabola, so path of projectile is
parabolic.
Above equation of trajectory is applicable only when
motion takes place in a particular plane (say xy-plane) and
point of projection is origin. Angle 0 must be measured from
positive side of x-axis and the only acceleration in the flight
must be constant and in the -ve direction of y-axis.
The above equation can also be expressed in terms of
range 'R' of the projectile
y = xtan0 :._
2
or
uv = ,Jgur or uv > ur
Now we compare U and T, (Range)u < (Rangeh
y=x(1-~}an0
2
u S2 sin 2 60° =_r
u2_
sin_
45°
_
I
2g
or
. gxz
tan0
2
2
2u. cos 0 tan 0
y = x(l---gx~_cosO)tan0
2u 2 cos 2 0 sin0
As projection angle is same uu < Ur
Now we compare Sand T, Hs = Hr
2g
2
g
)x
2u 2 cos 2 0
I
u 5 =urx~ or Ur>us
Thus,
Uy > Uy, Uy > Uu
and
ur > Us
Therefore. Uy is m'!l{imum.
Equation of Trajectory
Trajectory refers to path followed by a particle. Equation
of · trajectory is obtained by eliminating time t, from
"!'pression for x-and y-coordinates.
x = (u cos0)t
... (i)
y = (usin0)t _.!_gt 2
ry--_
1., Co~ider a proJe~tile project~d from _a, step/ofi
height h, after following a parabolic trajectoJY it'1ands,
.
A deptMh' below the horizontal plqne ofprojection; tlt~n~y
~J -
--
.........
.. ·
'
? ... .
' :·
_,
y- '
--- ----------···········-····-··
L__
. Flg.1.87
X
I
••••••••••
I_
- ....
J
-
The time, of flight can be obtained from expression't~r y
· ·
•
_1
2·
y =U t+•-a• t
2 ~
' ;. '
2
.
'
½gtf":(usin0)t-h ".' 6,
__..,.:------..'""'P--(x,y)
..
-
• '
1
-h a;_(usin0)t--gt 2
i1
-e
h
'
i.e., time in which y is equal to -_h.
- y
--
A
.
Fig.1.88
After eliminating 't' from equations (i) and (ii), we get
•
X
1
X2
y = (usm0)----g~~u cosa· 2 u 2 cos 2 0
'
1. -. , .-. . .
id!...................
... (ii)
2
"'I
Conc.-,pts: Application ~f--e-q_ua
___ti_o_n__o_'f_tr_a1-·e_ct_f)__
~,
Eroduct of the rootst1t2
- 2h
=--g
Since product of the roots is neg~tive, so one of the roots
is negative other must be positive.
'--''--------'---~--------~
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
iDESCRIPTION
OF MOTION L __ "" _ _ _ --- - - - - - - - -
------ -------- - -
-
'
,
Let negative roots be t 1 then it represents time for part co'
'if motion did exist before t = 0 and positive root t 2 represents
0=tan-1 a
time for part OAR.
-~f--=b
2u 2 cos 2 0
2.
If a body is projected from the ground so that on its
'Wey it just clears two vertical walls of equal height on the
,ground then with the help of equation of trajectory we can
,determine range of projectile.
2
U=~;(l+a )
x,-
- - - - x,,------
(b) In case of ground to ground projection
horizontal range is the x-component of displacement when
the y-component of displacement becomes zero.
Putting y = 0 in the equation of trajectory, we get
0 = ax-bx 2
=>
0 = x(a-bx)
X
- - - - - X2------
Fig.1.89
Path of the projectile grazes the top of the walls, so
coordinates of the top point of the wall must satisfy the
equation of trajectory.
y = xtan0
g
2
x=0
and x = (a/b) = R (horizontal range)
Maximum height is the maximum value of y-component
of displacement, which is obtained at x = R/2
Putting the value of' x' in the equation of trajectory we
get-
a2 a2
Ymax=2b- b
4
x2
a2
2
2u cos 0
h =xtan0-
x1
-
... (iii)
U=~i
y
If x1
1
cos0= r;--;,
vl+a 2
and
g
2u 2 cos 2 0
x1_+ x 2 (s_UTI! of the roots) gives horizontal range.
95
[>
(a) Given equation of trajectory is
y = ax - bx 2
... (i)
Let' u' is the velocity of projection and 0 is the angle of
projection, then the equation of trajectory is given by
y = xtan0
f
x2
2u 2 cos 2 0
Comparing (i) and (ii), we get
tan0 = a
A gun is mounted on a plateau 960 m away from its edge as
shown. Height of plateau is 960 m The gun can fire shells
with a velocity of 100 m/s at any angle. Of the following
choices, what is the minimum distance (OP)xfrom the edge of
plateau where the shell of gun can reach?
I
A particle is projected from origin in xy-plane and its equation
,of trajectory is given by y = ax - bx 2. The only acceleration
;in the motion is' f' which is constant and in -ve direction of,
'y-axis.
( a) Find the velocity ofprojection and the angle ofprojection.
(b) Point of projection is considered as origin and x-axis:
along the horizontal' ground. Find the horizontal range:
and maximum height of projectile. Projectile completes its
flight in ho~onta_l p_~ane of projection.
Solution:
(Maximum height)
I ·----- --- - - - - - - - - r:::::7: ,~~~·~~J<1l~,1 96 I ~
x2
and x 2 are roots of this quadratic equation.
x 2 = d ( difference of two roots gives desired relation)
i-gxg,ijlil~j
H = 4b
or
...
-
'
-----
-
-
960m
960m
0
---.--
Fig. 1E.96 (a)
(a) 480 m
_(c) ~60 m
(b) 720 m
_(d) none
Solution: P is point on
plane for shell when it passes
near edge of plateau for greater
angle of projection. In region AP
shell cannot hit
(ii)
Xmax
= AP
Equation of trajectory for
projectile
www.puucho.com
fig, 1 E96 (bl
Anurag Mishra Mechanics 1 with www.puucho.com
I
MECHANICS-I
2
y=xtan8-gx (l+tan 2 8)
2u 2
For point E,y = 0 and x = 960 m
2
2 S 10 X (960)2
( S)
10 X 960
0
=;. tan
x ---''------'c-- tan 960 + ----=- =
2 X (100) 2
2 X (100) 2
=;.
tan8=3/4,4/3
for
tan8 = 4/3, AP is least
Now, equation of trajectory becomes,
Y
£.o6~Qc..l'.J~]?~J~;>
IA~;oj;c:i~e isfire~-~th·~:;:c_i_ty_v__fr_.-om_a_gu
__n_a_d_if_us_t_ed_fi_or2
0
imaximum range. It passes through two points -P and Q wlio]e1
jheights above . the horizontal _are h each. Show that, the
,separation of the two points is·!l..Q.~v~ -4gh.
I
------·-
=ix- 2~~0( 2:)
4
x2
x2
.y=x-g,
Vo
2
2
-x 2 -v 0
2
x+!l..Q.h=O
g
If x 1 and x 2 are roots of the above equation,
or
!Jwo
particles were projected one ·by ~;~ with the same initiail
1
velocity from the same point on level ground. They follow thel
same parabolic trajectory and ar.e found to be in _the same
horizontal,lev~ts~parated b:y a distan~e.oflm, 2seconcis affy
the second particle was projected. Asst{me that the· horizontal
comporie'nt of their velocities of 0.5 m./s.
Which of the following s·tatements w/ll ·be true about their
·
motion?
(a) The horizontal range of the parabolic path is 3 m.
(b) The maximum height for the parabolic path is 45 m.
I
'(c) Th~ to. al ti7?1e. of flight in the parabolic path for each·
I1 particle = 4s
·
(d). The horizontal nmge offeparabolic I!Qth is 6m ,
g
v2
-X1
+
X2
= _Q_
g
and
I
t.
Solution: Distance travelled by 2 nd
kExam"t,c{!l,eJ
===------~~~
-i,·=4s'11)
)-.
99
t'~
;:-::::,.
.
= ~
~
1--*I
I
6 = 2u sine, u sine= 30
g
u ~in 8 = 900 = 45 _
2g
2x10·
t~-~-"'-t~ e + tan
I
Particle in 2 sec= 0.5 x 2 = 1 m
Horizontal range = 1 + 1 + 1 = 3 m
Flight time = 4 + 2 = 6 sec.
2
·-7
!locate the point of maximum height. Show that
.___~_,_,Flg.1E.97
2
-.
!A football is kicked as shown in Fi1;. ,E.99. The angles 8 and <j,'
i.-1--- 1.
H
h=x--x
2
Vo
--~.....___1~
2
g .
Fory=h,wehave
tLExam~~~e;:f977~~
- :-~
r
.
2
-960= - x - 3
720
solving we get acceptable solution x = 1440 m
AP= x-960 = 480m
-
-·----·
y = xtan 8- gx (1 + tan 2 8)
2v~
Gun is adjusted for maximum range; therefore o. = 45°.
for landing point on plane (P), y = -960 m
=;._
g
Solution: The trajectory of projectile is given by
<P
A
l
,?.1~:::·.·:::::·.J~:::·).C.
______ · .• •·____.,______
Fi_g._1_E_.9_9_
I'
B
_...___..~---·----J
Solution: The equation of trajectory is
gx 2
y=xtano.
Particle will strike the ground after 2 sec.
2u 2 cos 2 a.
= xtan o. [l - 2u 2 co:o. sino.]
www.puucho.com
l
,,.;
... (1)
Anurag Mishra Mechanics 1 with www.puucho.com
-
~
- -
DESCRIPTION OFA\OTIOII
Range of projectile is
R
•
= 2 u 2 smo:
coso:
_.½ g:
gx
Ynmx. =X "
... (2)
From eqns. (1) and (2), we get
or
The coordinates of A are (h cote, h) and range
Ymsx.
=u2g
---
u2
2
_x_ :5 250
2000
or
-500,/2 :5 x :5 500,/2
The fighter jet, can travel 1000,/2 m while it can be hit.
So the plane is in danger for a period of l000,/2
500
r--1-.
k~~Pt!_?.Pl;~J 100·1>
_An enemy fighter jet is flying at a constant height of 250 m
with a velocity of 500 m/s. The fighter jet passes over an
anti-aircraft gun that can fire at any time and in any
direction with a speed oflO0 m/s. Determine the time interval
during which the fighter jet is in danger of being hit by the
gun bullets.
... (1)
=2../2 sec.
A shot is fired with a velocity ~ at a vertical wall whose·
distance from the point of projection is x. Prove the greatest'
height above the level of the point of projection at which the
bullet can hit the wall is
u4-g22
2gu2
Solution: The equation of trajectory of bullets is
1
2
-gx
2
2
Y =xtan 8 - - u2- (1+tan 8)
]
On substituting numerical values, y = 250 m, u = 100
tan 8 + tan q, = tan o:.
-
2
mis, g =10 m/s 2 , we get
tan8cot8 + tanecot<j> = tano:cot<j>
1 + tan8 = tano:
tan cj> tan cp
or
i-·-··
X
-~
2g
tan e = tano: cot<j>
cote+ cotcp
or
g
2
I. gx2.
y:5~--2__
h cote+ h cot<j>
or
4
~
The shell can hit an area defined by
=h cot 8 + h cot<j> . Substituting in eqn. (3), we get
h =h cotetano: (1h cote
)
or
[1 +
1 2
-gx
2
2
y =xtano:[1-~]
2
U
g
R
2
,
x
Solution: Let 8 be the angle of projection.Suppose y
is the height at which bullet hit the wall. We have, from
equation of trajectory.
y
E
wall
0
~
X
Fig.1E.100
For a given value of x, maximum y can be determined
from
1 2
dy
2gx
--"--- = X - - (2 tan8) = 0
u2
d(tan8)
or
u2
tane=gx
On substituting the expression for tan 8 in eqn. (1), we
get
X
Fig.1E.101
y
=xtan8
2
gx
2u 2 cos 2 8
2
2
gx sec 8
=xtan8-~-~2
2u
dy
gx2
- = xsec 2 8- --2sec8(sec8tan8)
ae
2" 2
2
www.puucho.com
=xsec 2 8-gx sec 2 8tan8
u2
... (i)
Anurag Mishra Mechanics 1 with www.puucho.com
. = xsec 2 0[1~ xg~~
8
usin0± lu 2 sin 2 0-4xfxiz
t = .
'J
2
2xf
]
For y to be maximum, dy = O
de
2
xgtan0]=0'
2
xgtan0].=
2 .
.0
u2
or
gx
or
tan0 = -
= xu2
~_!gx2
gx
max
u2
=g
-
2 u2
gx2
2zi 2
[1+~]
-
v
2g
=
2 u4 -g2x2 -u4
2u 2g
case = u
Now from equation (ii), we.have
~
u4-g2x2
or,
Ymax
2gu2
or
pri:~<i!~l;~~
IAn ~ero~lane~ies horizo~taily at:~~;ight ,,-~t.a· ~pe~d-~-"411
'
'
·~·,
:,·,·':,-",,
· Solution:
0
aL
.,· .. . .
'J
'~:
.·'..·· 'I;I
"oWi:: . .r:1 •
'"
Suppose tbe muzzle velocity of tbe shell is
u and it is fired at an angle 0 witb tbe horizontal. To hit tbe.
plane, tbe displacement of shell along tbe motion of plane in
time t is equal to_ tbe displacement of tbe plane. Thus we
haye
... (i)
and
gt
2
umin
2
= 4v 2 + 2gh
V
=-=~=
4v2 +2gh
and
tan0 = ~2gh
V
f~l1~~
f~height
~ci~~l~:~~,:ojectedfro'!'.aR:~~~nth~. level gt~it~J~~d~M
If lirwlien at .ho,;izontgfd~t<ln.ces a.and 2a from ..it<)
P,Oint Of TJJ:01~.'itf®:J'ind the V¢lQClty ofjirojection.Jc;~ "J
Solution: Ifv 0 is tbe velocity ofprojecticin arid a tbe
angle of projection, tbe equation of trajectory is
'
.
1
gx2
y =xtana-- 2
... (1)
2 v 0 cos 2 a.
Witb origin at tbe point of projection,
2
gx - 2v~ sina.cosa. · x + 2v~ cos 2 a· y = O ... (2)
· Since tbe projectile passes ·through two points (a, h) and
(2a, h), tben a and 2a must be roots of equation (2),
2v~ sin a. cosa
.
1 2
h =usm 0t--gt
2
2
or
2
or u -v. 2'2gh
(1-::)2c2gh
cos0=-vumm.
_izbove .the.• gun,,{Shqiy that the iriinimum mU2zle ;ve/qcify
req~ired ~o: Jtfi:. the plc11;,e is::411~ +·2gh at A'! )d~g~
.
2
· ... (iii)
Substituting this value in
equation (iii), we get
a7!ti-ai, craft. Kl!nfires ashell.at t~pldnewhen"it iil!irti.'c:#(ty
. -1.(fiifi'J'.' . ·· .-.
tan .-. -.-·. ;.
... (ii)
= ucose,
v·
x2g2
u2..
u 2 sin 2 0 2' 2gh
u 2 (1-cos 2 0) 2' 2gh
From equation (i),
'Substituting tbis value in equation (i), we get
. y
(u 2 sin 2 0-2gh) 2' 0
t to be real
u.
or
u sine±4u
sin 0'-~2gh
t = ___
.,_____
c_
g
or
u2
2
a+ 2a
•
2v~ cos 2 a. h
g
Dividing eqns. (3) by (4), we get
3a
tana
and
-usin0t+h=O
Solving above quadratic· equation fort, we have
--''--·g
www.puucho.com
... (3)
ax 2a
2a2 =-h-
or
... (4)
3h
tana=2a
Anurag Mishra Mechanics 1 with www.puucho.com
.
v~
From eqn. (4),
ga2
=-,;- sec 2a
ga2
=-
h
= ga2
h
(1
v0
(4ah .+.9h)
2
= 1 (4a
·-,;-+
9h ) g
2
IA man is riding_on q.flat car travelling With aconstan~ speed!
of 10 m/s. He. wishes to ,th,:ow a follthro,igh a stationary
hoop 15 m above the height of his hands ·in such.a manner
that the ball ;,,ill move horizontqlli as)tpasses. through· th~
hoop. He throws the ball with ,a speed· of 12.5 IIVs w:r.t:1
. If. _ . •. .
' ------~,
'
h. 1mse
: .
. ., , I
1------ ---------------- -- r.·,
1
f
Sm
C
·
'"
"'!'
~~! . .
:1
and
5
cos0=~
5
v 0 sine= (12.5) x (~) = 10 m/s
.
.
herg
. h't = 2vo
sin 8.
tak en to reach maximum
~~T1me
g
2x 10
= - - = 2 second
10
(c) Horizontal distance ofloop from point of projection
= (12.5 COS 0 + 10) X 1
=17.5 m
ii~¥cimmi,~~>
I
sin0=~
and
2
Sm
sin2 S = 5 X (2 X 10)
12.5x 12.5
or
4
or
2g
or
+ tan 2 a;)
(1 + 9h2)
4a2
=~
(12.5 sin 0) 2
i:e.,
[,;1=xa-1mf,:;c~3
~""k~ 105
i
----
"----
-·
J.~-
~
ball ~ projected· with velocity vO and at an angle ofi
~fojection a. After what time is the ball moving at right a'!gles!
ltSLthe initial direction? _________ _____ ,, .. _________ · .• i
IA
·solution: Method 1:
If initial velocity v 0 and
velocity at time t are perpendicular, then the final velocity
will be at an angle a; with the vertical.
,.- - - -- --- - - ---
&it1:
;
~:-
i
goo ··--••.
t ~ ·_-~---_.' :f"i:£'--..J
'(a) What must qe th~ verticd'l compQnent of the initia() '
1
[_ .·., _, Fig; 1E.1-05 (a) ______
velocity oftf!e.ball?, _ - _.,
·';::, . _. ·.; J'
i(b) How many,ieconds after IJe teleases t_h~ ball Will it pass
Horizontal component . of velocity is unchanged
1
through the.hoop?' .• ·
· · · - ,· I
throughout the motion.
1
cc) At what horizoutal dista.nce infrqn~. o(t. he fo'op musthel
v O cos a = v sin a
Therefore
I release thl!.11!#1?
_ . · , · ___
, _. .: _·_ • · . · ,
or
v = v 0 cote,;_
Solution: Two important aspects to be noticed in this
problem are:
(1) Velocity of projection of ball is relative to man in
motion.
(2) Ball clears the hoop when it is at the topmost point.
..,
..,
V ball, man
..,
V ball
= V ball
..,
..,
- V man
..,
Vertical component of velocity after time t = - v cosa
From the equation v y = v O sin a - gt
-vcos a= v 0 sine,; - gt
v sin a+ v cosa
t = -0" - - - - or
g
v 0 sina + v 0 cotacosa
·=-"---~---g
= V ball, man· + V man
~ v0
(a) Now we apply the above relation to x- as well as
y-component of velocity. If ball is projected with velocityv 0
and angle e, then
g
..,
x-component of v ball = (v O cose + 10) m/s
..,
y-component ofvba!l = (v 0 sin0) m/s
(b) Since vertical component of ball's velocity is
unaffected by horizontal motion of car, we can use the
formula for time of flight,
= Vo
[sin a_+ cos a]
2
2
sma
coseca
g
Method 2: We choose x-axis along the initial velocity.
If after time t the velocity is perpendicular to initial
direction, v x must be zero after time t,
www.puucho.com
.
.·~..
Anurag Mishra Mechanics 1 with www.puucho.com
'
..
. ..,.,_
1a2 ..~.~···~·=-~_. ._:._J_-~'-":f__ ~-· ..~-~ -·.t·· ~ir~i:_.~-':-~---~-----~~
r--..-, _ .,;~:~~).·
The equation of trajectory of a projectile is
gx2
l~,i
L_
Flg.1Ei105 (b)
i.e.;..._.----
0-v 0
or
t
j
-_
g smat--
=~
gsina
Method 3: Slope of trajectory at the point of
projection,
m1 = tan0
--~--_--7~-.;--,
.:7
'
I "
- - -.
. . ''
. ._, --~
Point (R cos p, R sin I}) must satisfy equation (1).
gR 2 cos 2 1}
Hence R sin P = R cos!} tan a 2u 2 cos 2 a
gR 2 cos R
or
R(tan a - tan P) = ~c--"2u 2 cos 2 a
gli cos p
or
cos a cos P 2u 2 cos 2 a
2
'
L,_~Fig.~5(c)·· _, ,
g cos 2 p
,
i-_---------- --·
Slope of trajectory after time t,
'
dy dy/dt
m2 =tana=-=-.dx dx/dt
Vy
R = 2u sin (a - I}) cosa
or
Method 2: We take axes along incline and
perpendicular to incline as shown in Fig. 1.9L '1n this
coordinate system, components of velocity and acceleration
along the incline and normal to incline are
ux c:a u cos (a - !}), ax = - g sin,P
uy = u sin (a - !}), aY = - g cos p
,'
. t~
___ (l)
y = x tan a - - ~ -22u 2 cos a
\;
\,
,
'
u
y
'
I
'
' u, = u sin (a-~)
v sina-gt
0
=-=~---
X
Slopes are perpendicular,
(
vosina-gt)ctana) = - l
v 0 cosa
_
or
When projectile lands at A, its y-coordinate is zero.
1
2
O=ut+-at
y
2 y
t=~
gsina
PROJECTION ON AN INCLINED PLANE
A particle is projected ry, ·: . ---;--:-;--- , ,. r
from point O on the foot of
an· inclined plane. The [ '. ~- · ,,u ~ath :fR~i~c\i;e : :
::-.< ,;
/A'' . :_,'.. ,·
velocity of projection is u,
"'angle of projection a with
- - : ,. ':· ( - .5'
,, R• ·,.. :i
,.
,,.Cl)
x-axis, angle of incline P
,,
.0:::
~: ,[see Fig. 1.90]. We wish to
,·
B, ,,
_x
determine range along .,:;o.,,
'_:,;· ~, Rcos
13----.
I
incline, . time of flight, 1 -·~'
Fl 190
vertical height at which Lr_·_·_---~·--·-~---~
projectile strikes.
(a) Range Along Inclined Plan·e
Method 1: Point A where the projectile lands has
coordinates (R cos p, R sin p}.
i
l
,
or
0 = u sin(a - J})t - .!_~cos J}t 2
2
2u sin (a - P)
t=----~
g cos p
This is the expression for time of flight from O to A.
For motion along inclined plane (x-axis),
1
2
or
X
= Uxt + -
2
axt
, =UCOS(a-J})t,-.!_gsinl}t 2
2
Substituting expression for time of flight, we get
2
R = 2u sin (a-P) cos a
g cos 2 p
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
DESCRIPTION OF MOTION
Method 3: We revert
back to a new coordinate
system with x-axis in
horizontal and y-axis in
vertical direction [see Fig.
1.92].
x=u coso:t
2u sin(o: - Pl
= u coso: - - - ~ ~
Projection Down the Inclined Plane
y
y
-xFig. 1.92
gcosp
R = ~ = 2u coso: sin(o: - P)
cos P
g cos 2 p
2
and
x"'
Fig. 1.93
From figure we have,
ux =ucos(0+o:), ax =gsino:
"Y =usin(0+o:l, ay =-gcoso:
(b) Vertical Height at Which Projectile Strikes
Method 1:
From the equation y = u sino:t - ~ gt 2,
2
on substituting time of flight t, we get
Time of Flight
.
2u sin(o: - Pl - -g
1 (2u sin (o: - PlJ
y =usmo:----~
g cos p
2
g cosp
2
As displacement become zero along y-direction in time
'T'.
1
2
O=uyT+-ayT
2
2u 2 coso: sinp sin(o: - Pl
Method 2:
g cos 2 p
y = :nanp_
u cos o: x 2u sin(o: - P)
A
=--------'---'--'-X tanp
gcosp
2
2u coso:sinP sin(o: - P)
=----~=----'-g cos 2 P
or
0 =usin(0 +o:)T- ~(g cos·o:)T 2
or
T
2
g coso:
Range Along Inclined Plane (R):
1
R=uxT+-axT
(c) Angle of Projection for Maximum Horizontal
Range
Range R, is given by
2
R = 2u sin (o: - P) coso:
2
2
1 . [2usin(0+o:l]
=ucos (0 +ex l[ 2usin(0+o:l] +-gsma
----g coso:
R=
2
u [sin(2o: - Pl - sinpi
g cos 2 P
g coso:
2
2
u
g cos 2 a.
[sin (20 + al + sin a]
or
(20 + o:l
= 90°
or
0 = 45°-_<:
2
1l
2o: - p = -
m~
p
0:=-+4 2
The maximum range,
u 2 (1 - sin Pl
Rmax.
2
R
2
Jt
or
= u (l+sino:l
2
g cos o:
= u 2 (1+sino:l
2
g (1 - sin o:l
u2
Rmax
=- - - g (1- sino:l
- -- - '
= ----:;--_;___
2
k.:~-?5::~tD.l?}e
g cos P
_ u 2 (1 - sin Pl
- g(l - sin 2 Pl
... (ii)
For maximum range sin(20 + al =+ 1
For R to be maximum sin (2o: - P) must be maximum.
or
2
After simplifying, we get
g cos 2 p
Hence
= 2u sin(0 + o:)
uz
g(l + sin Pl
~06 _.;-
'A heavy particle is projected from a point at the foot of a fixed
plane, inclined at an angle 45° to the horizontal, in the
vertical plane containing the line of greatest slope through the
point. If$ (> 45° l is the inclination to the horizontal of the
initial direction of projection, for what value of tan $ will the
particle strike the plane:
(i) horizontal
(ii) at right angle?
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
Solution: Let the particle be projected from O with
velocity u and strike the plane at a point P after time t.
Let ON = PN = h; then OP = h-Jz.
(i) lf the particle strikes the plane horizontally,
then its vertical component of velocity at P is zero.
Along horizontal direction,
· h = (u cos <!>)Ct)
... (1)
Along vertical direction,
0=usin<!>-gt
... (2)
or
usin<!>=gt
·-----
or
or
2tan<j,-2=lttan<!>
tan<j,=3 ·
~~~~~
[0e a~la(,,dkation ofan enem,y's_P?Sition on a_hil('h\~
\;!Jecdle~in1:!r:~:~%~; ~~e t;;e:i1mum ;efo~1trjo{i'tfe,j
Solution: 'O' is the point of projection of the shell
and 'A' is the position of enemy at a height 'h' above the
level of'O'.
I
. _.
·'. F . .
.·
•,Flg.1E,106 (a) .
-,~+-·•-
=---''--'-'
.
1 '
h=usin<!>t--gt
2
Using eqns. (1) and (2) in (3),
and
2
... (3)
(ucos<!>)(t) = (usU-:<1>)(t)-:! (usin<!>)t
2
tan<!>= 2
(ii) If the particle strikes 'th~ plane at right angles
at P, then the ·component of velocity parallel to the plane is
0
zero.
Along perpendi9ular to the plane,
r5~~~~p~;~:-~-;,J .
~~~,r~ ·:: · ·,~/:;'4'·,
(• , . . · v, .g cqs, ..
!t-$fn:45; · ·-~
i{~~}\~~-~.Q/ .: :.• '
1• ,· "',
· ·
-._..,,-,:,,,.-'/," '
w:~~-,_ ,.,,- .U·
. I~,-:,· .;> •
~\J,
•
'
;'(4J-_45_o,}
,_
i
:"
___
De:-::--~:=·~~-·-,
,;:;E•""rU!l:Oi!Wf1 ei - 108 t,:,~
~--~---~~~-
j
0 = usin (<j,- 45°)t - .!g cos 45°t 2
2
zJzu , .
t = - - sin C<i> - 45°)
g
Along the plane,
u cos_(<!>- 45°) = (g sin 45°)t
_JzJzu sin (<I> -
= -~_
,Jz
or
or
l
~_}~,
0
'
C
'
'
'
'
'.,
,_.-,7',--· -:--r ,;,
id'
N· -
f?/:f:;-. 1;0
;_·_·.r.:~-~> :··.:, ,.
b:fu,.;.L...::_ Fig: 1E:tOI! (b) , ' .
... (ii)
C
,'
,,
-:~ prOjeft(le /,t t/-irQWn, at ar( G71$/e ;~With an inf/i~ed plci.rtfOJ
rlric(irtqtion ~- as .shoWrt.
f:ig.: ).~:108.. Find ,the r~ldtio~
.bet:)l(een~q11i:!Bj(:. . _ ,,,, • ,,:',.'
,',, ';'/ C·
. f(a) p"oJectf,le,stri~es the inclinedpCane,perpendicular1y1, {'
ll;,),p:r:._ojectiie,~@<es· the inclinefiuilqn~h'g_rj.;Qntab,,.::i;;;:/~.Solution: (a) If projectile strikes perpendicularly.
.h
~'
From 6.0AB, OA = h coseca
From eqn. (i) and (ii),
u,;, ~gh (coseca + 1)
~ - : ; _ , , •.,,,,-, -~:-~'""".'"
.'
<ti·-· - , .450 ; ~-i:
If 'u' is the minimum initial velocity of the projectile to
shell the enemy, then 'QA' must be the maximum range up
the inclined plane of angle a .
·u
... (i)
So
OA=---g(l+ Sina)
45° )]
g
ucos (<!>- 45°) = 2u sin(<!>- 45°)
.! = tan (<I> - 450} = _tan_<l>_-_l
.2
l+tan<!>
www.puucho.com
i -i:
'
,
.
7
1\·. :.:'.
">X·8X1Sl
'
Anurag Mishra Mechanics 1 with www.puucho.com
I _D~~~RIPJION O_F MO_TION
ssl
-------- - - --------------------~
2u sin9
T = --g cosp
u cos9
2u sine
gcosp
g sinP
as
or
we also that
=>
=> 2tan9 = cotp
(bl If projectile strikes horizontally, then at the
time of striking the projection will be at the maximum
height from the ground. Therefore,
~:.
____________
0
u
g sin 60°
10-./3 =2s
x -/3
10
2
(bl Initial velocity along y-axis is zero. The velocity
along y-axis after 2 s;
Vy= Uy+ Uyt
= O-gcos60°x2
1
2
tf1(_______________________ . . . . _____ _
=-10x-x2=-10m/s
(cl We have, v; = u; + 2axs
Since
and
2u sin9
tap=---
gcosp
=>
t=
=>
'
Fig.1E.108 (b)
=>
Vx=ux+axt
0 = u -g sin60°t
2u sin(9 + Pl
tap=---~2xg
2usin0 2usin(0+Pl
g cosp
2g
or
vx = 0
ax = g sin 60°, u = 10-./3 m/s
O = (10-./3l 2 -2xg sin60°x (OQl
OQ =
10
2
X
3
= l0-./3 m
-./3
2xl0x2
Distance
2sin0 = sin(0 + Pl cosp.
l -~-~~':Tl-r. "'7 I~cjJ.->
PO = O+ .!:. g sin 30° x (2) 2
2
1
1
= - x lOx- x 4 = 10 m
2
'lwo inclined plane.s OA and OB having inclination 30° and
60° with the horizontal re.spectively intersect each other at 0,
as shown in Fig. lE.109. A particle is projected from point P
with a velocity u = 10-./3 m/s along a direction perpendicular
to plane OA. If the particle strike.s plane OB perpendicular at
Q.
2
Therefore height h of point P,
h = PQ sin 30° = 10 x .!:. = 5 m
2
PQ = ~P0 2 +OQ 2
(dl Distance
= ~(10l 2 + (10-./3l 2 = 20 m
l :!.~~G9!!11?J,f7: .G10 1;>
u
30°
0
Fig.1E.109
Calculate
(a) time of flight
,(b) velocity with which the purticle strikes the plane OB
(c) height h of the point P from point 0
( d) distanc~ PQ.
nvo guns situated on top of a hill of height 10 mfire one shot,
each with the same speed 5-./3 m/s at some interval of time.,
One gun fire.s horizontally and the other fires upwards at an'
angle of 60° with the horizontal. The shots collide in air at a,
point P. Find
(al the time interval between the firings and
(bl the coordinates of point P. Take the origin of coordinate
system at the foot of the hill right below the muzzle and
trajectorie.s in the xy-plane,
Solution: Consider the .motion of particle along the
axes shown in figure. We have
ux = u,
ax= -gsin60°
Uy= 0,
ay = -g cos60°
(a) As the particle strikes the plane OB·
perpendicularly,
www.puucho.com
u,
---~
_,;..---ll,un 1
P(x,, Yr)
10 m
(0,0)
P(x,y)
xFig.1E.110.(a)
Anurag Mishra Mechanics 1 with www.puucho.com
Solution: Let gun 1 and gun 2 be fired at an interval
Llt, such that
t1 = t 2 + M
... (1)
where t 1 and t 2 are the respective times taken by the
two shots to reach point P.
For gun 1: ·
X' -~
X, -·.'Xt,
Method 2: We take point of firing as origin and xand y-axis as shown in Fig. lE.110 (b). Equation :of
trajectory of '\ projectile is
2
y =
tan0-
X
gx
2v 2l cos 2 0
For gun 1 1 . 0 = 60°.
y
anent
=11,t q>S 60° ~l
. .
./3 . . l i'
· Y=Y·+-v-t 1 -~•t1
. . .
X
1
= )(.+,.;....·V·t1
\' 2
l
: •
2
I
I
z9•.
2gx2
(a) Now we can equate x- and y-coordinates of
shots,
i.e.-,
... (2)
=x-./3-v?l
For gun 2, 0 = 0°.
-gx2
y = 2v 2 ·
l
and
... (3)
or
Two shots collide at point P; therefore their coordinates
must be same;
i.e.,
On substituting t 1 from eqn. (2) into eqn. (3), we get
-./3
1
2.
. v; (2t2) + Z g(-3t2) = 0
2
or
or
t 2 (-./3v,
t2
=0
-1
=0
gt 2 )
.
and
t2
2
·Or
collide are
x=xi+vit 2
= 0 + (5-./3)(1)
= 5-./3 m
and
3gx2
2v 2
= --
2v?!'
l
2v2
2(5/3)2
- -,J3g -
./3(10)
x--' --=~-
= 5-./3 m
and
-gx2 _ (10) (SV3)2
y=--=--~~
. 2vr
2(5-./3) 2
=-Sm
· If originis assigned at ground the coordinates of point P
will be (5-./3 m, 5 m).
Now We consider x-component of displacement for both
the shots.
Gun 1:
x = 5-./3 m = v,t = (5-./3 m/s)t
or
t, = 1 s
2
y=y,-2gt2
= 10 - .!_ (10)(1)
=Sm
x=O
V·
t 1 = 2t 2 = 2(1) = 2 s
M = t 1 - t·2 = 2 - 1 = 1 s
(b) Tbe coordinates of P at which the . two shots
2
v2l
-
-./3 g
Therefore,
1
x-./3 = - - -
=- -'-
=lx(7o3)=ls
and
or
2
2gx2
~gx · = x-./3 - - 2v2l
v?l
2gx2 gx2
Gun 2:o
2
X
= 5-./3 m= V;
COS
or
t 2 =2s
Time interval between two shots is Llt
www.puucho.com
5,/3
60° t2 = - - t2
2
=t 2
-
t1
=1 s
Anurag Mishra Mechanics 1 with www.puucho.com
[ DESCRIPTION OF MOTION
. a~J
0~( .->
lJ~.j~}J~J~·~
->
V panicle, box
1
A large heavy box is sliding without friction down a smooth
plane of inclination 8. From a point P on the bottom of thebox, a particle is projected inside the box. The initial speed of
the particle w.r.t. box is u 1 and the direction of projection
makes an angle a with the bottom as shown in the figure.
->
V particle, ground
->
= :V particle, ground
->
->
- V box, ground
->
= V particle, box + V box, ground
Applying above equation to x-components,
o =u cos (a + 8) - v cos8
ucos (ex+ 8)
v=
or
cos8
Method 2: The above
y
condition can be meet if the box
covers exactly the same distance
as the range of particles,
i.e.,
Fig. 1E.111 (a)
I
(a) Find the distance along the bottom of the box between the
point of projection P and the point Q where the particle
lands. (Assume that the particle does not hit any other:
swface of the box. Neglect air resistance.)
(b) If the horizontal displacement of the particle as seen by
an observer on the ground is zero. Find the speed of the'
box w.r.t. the ground at the instant when the particle was,
projected.
Solution: (a) Motion of the particle will be
reference frame of box.
0
2
Fig.1E.111 (c)
or
or
Relative Motion
Fig. 1. 94 shows an observer on ground, a balloon and an
airplane, we denote them by G, B and A respectively. At any
instant position. vector. of airplane for -an observer on
ground, on balloon have been represented.
uy =u sin ex
ay=gcose
a,=gsin0-gsin0=
y
1
=U/ --g/
2
2
Put y =0 for calculating time'
ar flight.
'
2 1
x=u,! =ucosat
=U COSCX
J=v( 2gu:~~0cx)
1 . e(Zu sina)
+-gsm
2
g cose
u sin0sina
ucosa= v + - - - cose
cosa
cos0 - sin a sine)
v = u( - - - - - - - cose
u cos(a+0)
cose
in
x-component
or
( u; : : :
I
2u sin ex)
( g cos a
Q::;::usincxt _!gcos0t
2
u2 sin 2a
or
gcosa
t
2u sin a
g case
(b) According to problem the horizontal displacement
of the particle as seen by an observer on the ground is zero.
If we analyse the situation in the reference frame of ground,
resultant velocity of particle in x-direction must be zero.
X
Fig. 1.94
....
rA/B
.--J
~
object observer
Q
....
rB/G
.--J
~
object observer
Fig.1E.111 (b)
....
rA/G
.--J
~
object observer
www.puucho.com
....
r;y8
position vector of airplane for an
observer on balloon
....
rB/G = position vector of balloon for an
observer on ground
....
rt\'G= position vector of airplane for an
observer on ground
Anurag Mishra Mechanics 1 with www.puucho.com
_,
From figure
_,
(al
or
Thus,
or,
VA/B
=
_,
_,
VA/G
V,rG
.j,
.j,
.j,
VP/E
_,
Rate of change of position vector is velocity.
_,
_,
_,
_,
rA/G = rA/B+ rB/G
Velocity of Airplane Velocity of
Velocity of
as observed by
A for observer balloon for
observer on balloon on ground
observer on
ground
When we say velocity of airplane w.r.t. balloon or
velocity of airplane in inference frame of balloon. it means
VP/G
_,
= VP/G-VE/G
_,
_,
= VP/E+VE/G
which implies that absolute velocity of the passenger is
the vector sum of his velocity relative to escalator and
_,
_,
velocity of .escalator relative to ·ground. v P/E and v E/G both
pciint towards right as ~hown in Fig. 1.95 (b)
~T~~,-Ir·~-~,
r- :
\, ..
!
i
}
i
.I
I
,JA/ 8 referred as relative velocity.
Application of Advanced
Concepts of Relative Motion
l
·I
River Condition
Consider a swimmer in still water. The swimmer can
generate a velocity due to its own.!'ffort. We call this velocity,
velocity of swimmer in still water.
_,
Velocity of swimmer relative to water = v s/w
Next consider a person with a life jacket in a river
flowing with a velocity. Person makes no effort to swim, he
just drifts due to river flow. Velocity imparted due to river
flow is called velocity of water relative to ground, i.e., it
denotes the rate at which water flows.
Velocity of water flow relative to ground
_,
= v w/o
----,_~,,.-----~------· ~,·- ~-~-·------,--~
'.(a) "Find but the motion of t;Je, ql~d (I~d ~Id IJlan(lS ~e~n by
I . boy: ;'
(b)
re)
..
·
· ,. ·. ·· •
·
··
= Vsjw+vw/G
Esc,ilator Condition
Here is an analogous
treatment. Just
..
_, imagine an
·•; ...
.
Firi4 out,m~tioft oltree, bird, boy as seen.by,oid man,
Find'qut11uitio[l.Pftre.§, DOYcartd old man as seenbjzoira.
Solution: (a)With respectto boy:
Vtree =4m/s (~) .
vbinl =. 3 m/s (I) and O m/s c~)
Next consider a swimmer applying Iris effort in flowing
water. In this case swimmer's net velocity resultant velocity
will be decided by two factors
Ci) his own effort
(liY water flow.
Thus resultant motion is obtained by vector sum of two
velocities imparted to swimmer..
Resultant velocity of swimmer relative to ground =
velocity of swimmer relative to water + velocity of water
·
flow relative to ground.
_,
_,
_,
VS/G
i'
;(b)
Fl~.1,.95
(b) With respect to old man:
=6m/s(~)
= 2m/s (~)
vbird =.6 m/s
l and 3 m/s (I)
(c) With respect to bird:
vtree = 3m/s (.J,)
and 4 m/s (~)
and
3m/s(.J,)
Void man= 6m/s (~)
Vboy = 3 m/S (.J,)
escalator moving horizontally with velocity v E/G. A person
_,
.
.
begins the run with velocity v P/E fa the same dii;ection as .
escalator. What do you think about resultant velocity of
passenger?
We assign a latter to each body P, passenger; E,
escalator; G, ground.
www.puucho.com
vboy
Viree
c~
Anurag Mishra Mechanics 1 with www.puucho.com
89,
, _DE_SCRl~TION OF MOTION
~-113 ",;>
f_-~?f9:'.T\P Ie
.
.
A helicopter is trying to land on a submarine deck which is
moving south at 17 m/s. A balloon is moving at 12 m/s with
wind into the west. If to the submarine crew the helicopter is
descending vertically at-'5 m/s, what is its speed? (a) relative
to the water and (b) relative to the balloon. See fig.
y-.
~E =iA/P
+4J__,,
Vp/E
Vp/E
8
1-----v='-------'=illF--,1+._-+•
EX
Fig.1E.113
Solution:
Velocity of ant relative to paper
= Vsub/water + Vhel/sub
= l7j + (-S)k = (17j-Sk) mis
(b) V hel/ballodn = V hel - V balloon
(a)
Vhel/water
Fig. 1.96 (b)
=(17j- 5k)-12i
=(-12i+17j-5k)m/s
Ant Moving on an Ruler
Fig. 1.96 (a) shows an ant scampering along a ruler. The
Girl Moving in a Train
Illustration 11. Fig. 1.97 shows top view of a girl
(G) walking in a moving train (T). Two observers one in the
train and the other on the ground (E) determine the position
vector of the girl._,
_,
_,
_,
ruler has been displaced w.r.t. Earth by
_,
SR/E,
the ant
undergoes a displacement SAfR w.r.t. the end of the ruler.
The net displacement of the ant w.r.t. Earth (i.e., w.r.t. a fixed
point P0 on the ground) is given by the vector sum
_,
SA/E
_,
SAjR
=
->
+ SR/E
rG/A
V A/E
_,
••. (1)
The position of the girl walking in the train relative to
frame of reference of A is different from her position relative
to frame of reference B (Fig. 1.97). Time derivative of eqn.
(1) gives the ralation between various velocities.
... (1)
Taking time derivative of eqn. (1), we get the
corresponding velocity expression
_,
=rG/B + rB/A
->
VG/A
_,
->
=V G/B+ VB/A
... (2)
Ya
->
=V A/R+ VR/E
... (2)
G
Train
~t:;z=::==:f---Xa
Reference B frame
fixed to train
~------+XA
QA
Reference frame A
fixed to Earth
Fig. 1.96 (a)
Eqns. (1) and (2) are valid irrespective of the direction
of two vectors. Fig. 1.96 (b) shows the motion of an ant
walking across a sheet of paper, that is itself being moved at
a speed .f P/E. The ant is carried along with the paper so that
it actually moves north-east w.r.t. Earth.
(a)
- - - - - - - - v'G/r---tVelocity of girl
relative to train
1Grr+~
-JTIE__,, Velocity of train
relative to Earth
VT/E
VG,E---t Velocity of girl
'--------'=--'
relative to Earth
·1 ~
(b)
www.puucho.com
Fig. 1.97
Anurag Mishra Mechanics 1 with www.puucho.com
,·
Velocity of particle G relative to reference frame A
= velocity of particle G relative to reference frame B
+ velocity of reference frame B relative to reference
frame A
If the girJ, walks across the compartment, her resultant
velocity will be as shown in Fig. 1.97 (b).
...,
V G/T
...,
or
V G/E
i
.
I
,_
. -··t:e
~
•••••
••••
\
H
•
->
VAJG
:
/
_;:f, ------(..
•••
Fig.1:98(b) ____ , ,
•
->
->
.=vs+vw
Case (ii) Swimmer moves opposite to river flow
(upstream)
when swimmer moves upstream.
->
->
...,
lvs;G l=lvs;wl-lvw/G I
A floating object like a wooden log move with the
velocity of river flow.
' •
Step 1: Problem Solving strategy:
Assign the initial point as origin of a coordinate system.
r-:~---·y . ---,--,--·1
I,
-+--.........,.~--,-
:
.
'
.f
!
River flow.I
Iv s/q l=lv s;w l+lvw/G I
-•••
•••• ••••• !
.
_-
~
~
Vw/G ._vstG
Note:]------=-'-V£.s_-_v_,w"----------
Plane in which
airplane moves
''
'''
~--1I
...,
Iv s;w I=vs =velocity of swimmer relative to water
...,
_I v W/G I= v w =velocity of river water flow
V AfW•
•
------·--:
Vsr,,
___,___
...,
i.e., velocity of girl w.r.t. earth (reference frame of
ground) is vector sum of its velocity relaitve to train and
velocity of train relative to earth.
Airplane-wind Condition
..., Consider an airplane moving in still air, with ve!O(;ity
------- ____ /
•• -,
~
1
...,
...,
= V G/E-VT/E
...,
...,
= V G/r+Vr/E
~AJW
MECHANICS,! ]
!_...... -· ...-_· .. '· -··----------.--. .x::,··· ........,
i
' · Ground plan~
->
.
W~E'
Vsr,,
swimmer
begins here
~--,
#/
·''"O
s '
Flg.1.98 (a)
~--- -·--- -- ··---· ...,------·- --- -
.
--·-
0
-
Fig.1.99
Wind· flows with velo~ity v W/G due east direction,
Resultant velocity of air plane will obtained by equation
->
V
->
A/G =
V
A/W +vw/G
~---- -------- ---V object/ground
.,
_____ ·-----
j
velocity of medium
·
• :Position
P6sition
where man
where man
heads
actually reaches
I
A
B
..··
II
Initial direction
of motion of
man
----,
j
= V object/medium + V medium/ground
Resultant velocity • velocity of object _
relative to medium
~---- ---
__
----_---....·-=-===;~-:::-7
->
In all the previous real situations there is an object that
moves on a moving medium
Object
Medium
River condition
Boat
Water
Swimmer
Escalator condition
Passenger
escalator
Ant-ruler condition
ant
ruler
Girl-train condition
girl
train
Concept
Step 2: Draw vector diagram
0
----
.
-··---~ ----- - - - - -
River Condition Revisited
Case (i) Swimmer moves in direction of flow (down
stream)
X
Flg.1.100
When swimmer reaches Bits x-component displacement
is x whereas d represents of y-component of displacement.
Step 3 : Apply component method of vector addition.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
DESCRIPTION
OF
-- -- MOTION
·- __ ,~,_.,
---· ·-------··-··----
-·
,
Concept Motion in x-direction is due to x-component of
resultant velocity, similarly motion in y-direction due to
y-component of resultant velocity.
...,
A
A
- - ·-
(vs;a\ = Vs sin0.
Time taken to cross river
d
t=--VsSin0
Vs;w = Vs cos0i+vs sin0j
...,
------
Drift= (vs -vw cos0) x
A
Vw;a =vwi
...,
A
Vs;a
d_
Vs sm0
Concept: Note that drift can be zero if
Vs= vw case
In this case swimmer moves along shortest path.
But above condition can be satisfied only ifvw > VsIfvs > vw, drift can be minimized but it cannot be zero.
For minimum drift
d
-[vwcosec e+vs cot0] = 0
dw
vwcosec0cot0-vscosec 20_= 0
Vs
cos0=or
Vw
A
=(vscos0+vw)i+vssin0j
x-component of resultant velocity
(vs;alx =vscos0+vw
x-component of displacement
x = (vs cos0+vw)t
Similarly y-component of displacement
y = Vs sin0t
Thus time taken to cross d width of river
It= Vs ~n 8
Drift during crossing of river
x= (vs;alx t
(VsCOS0+vw)d
x=
VsSin0
A boat moves right across a river with velocity 10 km h-1
Concept: How to obtain time taken to cross river?
y-component of displacement
t=--"---'-----'------''---y-component of resultant velocity '
relative to water. The water has a uniform speed of 5.00
km h- 1 relative to the earth. Find the velocity of the boat'
relative to an observer standing on either bank. If the width of
river is 3.0 km, find the time it takes the boat to cross it.
...,
Solution:
What is drift ?
Distance the swimmer is carried away along flow while
crossing river.
Xdrift
=[vSJalx
v B/R -, velocity of boat w.r.t. river
...,
v R/E -, velocity of river w.r.t. earth
...,
v B/E -, velocity of boat w.r.t. earth
--t
---),
--t
VB/R =VB/E -VR/E
x time
--t
Position
Position
where man
where man
heads
actually reaches
A
B
•
...
·· ...
,,
·- ..
.
Actual direction
of motion of man
..
Ay
······ ...
->
->
VRIE
VR/E
River flow
lniUal direction
of motion of man
Man begins
atO
0
X
D
Fig.1.101
What happens if swimmer moves opposite to flow ?
---),
,,._
A
,.,_
Vs;a = -vs cos0i+vw i+vs sin0j
(vS/G lx = (vs -vw cos0)
www.puucho.com
---),
--t
or VB/E = VB;R+vR/E
Fig.1E.114
Anurag Mishra Mechanics 1 with www.puucho.com
Angle at which boat starts is given by
Hence,
VB/E = ~V~jR
• ·
+ V¼E
·= ·/i0 2 + 52 = 11.Zkmh-1
The direction of v B/E is
8 = tan-:(VR/E)
....
....
.
Note that v B/E is resajtant of v B/R and v R/E • Effective
....
VB/R
cross
If x-component of resultant velocity vanishes the boat
will move straight, along y-axis.
Hence,
VB/R sin8 = VR/E
.
or
velocity of boat in y-direction is v B/R •
to
river
is
Crossing River Along Shortest Possible Path ;_e.,
Moving Perpendicular to Flow
In this case, x-component of resultant velocity is zero.
·---- -·,,. B. • -----1
cft=~-,J£1X0,mi.:i~5
··' ·
~r}\'.!,~~
[vs;alx
= O'
.
-v 5 cose·+vw = 0
.
V
cos8=_.!!',
i.e.,
Vs
From Pythagoras' theorem,
VB/E
~-~lg.1;1~~----·-_j
-)
= VB/E
I
- x,
:_.·,.,A
Solution: Method 1: The boat must h~ad at certain
angle upstream so that vector sum of ·velocity of boat
relative to river and velocity of river relative to earth must
be directed right across,
-)
I
~7&G,.:
(relative to the riv.er and is to be iowards fight across, il(w/laq
iiJir{ctiol!!.s/;lo.!!lsl.Jtb,ead?
-':;,.:::.:,_
' .
. '/:_j
-)
v.··
:
, _.
Wthe b;;t-i>j'i,receding ,;:m,pl~ ,"ttd\>et_; with s~me 'jp~eij !
VB/R+VR/E
VR/E
sm8=--·
VB/R
the
lQ
i.e,
2
vx =vB/Rsine-vR/E
-1(105) =tan-1(1)2
Hence time taken
_!!__ = ~ = 18minute.
vB/R
= VB/E
Vy
VB/R
....
1
or
8 = 30°
Method 2: , Consider point O to be. origin of a
coordinate system x,y.
....
=tan
VR/E
sm8=--=-
Direction of swimmer's velocity relative to flow
direction is (180° -8).
= ~rv-:~/-R___V_¼c--E= ~10 2 - 5 2 '= 8~66 km/h
->
·
~--
/2
2.
lv's;al=vssm8=Vsv1-l-;;';) =,iVs-Vw
I
Time taken to cross the river
d
d
.d
t = - - = - - - = ,====
Vs/G
t
I.
v 5 sin0
~---~-0 .. -'""
E
y"
~v~-v~
'
,
.
~·
-~-·--1
nod
C!>nce~t: What happens if vs < v w .swimmer ca~
cross along sfiortest.path because v5 ,cos8 < vw•diift,will'.bej
always po1iff,ve .swimmer can . mqv~ ,right across qnly
if/
'-"-..'-'!-dre~----·-----~. ·- -·---·~----.'.....---·-··-~I
v 5 >vw,
i'.
·
.•
,
Crossing River in Minimum Time
If swimmer begins at angle 8 with µver bank, time taken
to cross river will be given by
d
t=--Vs sin8
For tmin,
Sine must be maximum
d
tmin = -
www.puucho.com
Vs
Anurag Mishra Mechanics 1 with www.puucho.com
DESCRIPTION OF MOTION
93
Drift in this case
N
Concept: Swimming in a desired direction:
Many times the person is not interested in minimizing the,
time or drift. But he has to reach a particular place. This is'
common in the cases of an airplane or motor boat.
B
Flg.1E.116
______ ....,. . - ----------------·
.:.-:::Vmf ." _-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_ '
Solution: If bird is to move along AB, component of
velocity of bird and wind perpendicular to AB cancel out .
3
(a) 4sina = 2sin37°:; a= sin- 1
)
0
3
=}
37°+sin-1
with east.
(i
........... ·0················.
...,
v,
Fig.1.103
The man desires to have this final velocity along AB in other'
'words he has to move from A to B. We wish to find the
direction in which he should make an effort so that his actual'
velocity is along line AB. In th~ method we assume AB to be'
the reference line the resultant of v mr and v, is along line AB.
Thus the components of vm, and v, in a direction:
perpendicular to line AB should cancel each other.
:
(io)
(c)
.
...,
vm =vmr+vr
or
I,
-:J m = [vm, cosai + vmr sina]J +[v, cos8i- v, sin8j]'
and,
~
Vmr
5
t = lOOx 5
8 + 2ffi
.,
..
[·.- cos a= ffi]
10
5
=
250 sec
4 + ffi
··--,
.
A woman is running through rain at a speed of 5.00 m/s.
:Rain is falling vertically at a speed of 20.0 m/s. (a) What is
the velocity of the rain relative to the woman? (b) How far in
front of her would an umbrella have to extend to keep the rain
'off if sh~ hold_s the umbrella 1.50 m above her feet?
Fig.1.104
...,
2foi = 8 + 2v'91
[,g-?::$<:l_-!'.!'.PJ_'::_j 117 !;>
·:::::::::: Ymr::· .::::·.::::::·
...,
...,
=~+
10
X
B
...,
...,
(b) vb = vbw+ v w = vw cos37°+4cosa
'
...
...
-
Vw/E
VR/
sina-v, sin8 = 0
Vmr sinCX = V 1 sin8
20.0 mis
14.0~
1.50 m
5.00 mis
...
l'
(b)
(a)
Wind is blowing in the el!St direction with a speed of 2m/s. A,.
bird wishes to travel from tree A to tree B. Tree B is 100 m ·
away from A in a direction 37° north of east the velocity of ·
bird in still air is 4 m/s.
(a) Find the direction in which bird should fly so that it can:
reach from A to B directly.
(b) Find the actual velocity of the bird during the flight.
(c) Find the time taken by the bird to reach B.
d
VW/E
Fig.1E.117
Solution: We assign the following letters: W, woman;
...,
R, rain; E, earth. We have to find v R/W.
--+
(a)
V
--+
--+
--+
--+
R/W = V R/E - Vw;E = V R/E
From vector diagram,
-------
--+
I --+
2
--+
I VR/w I= \f(VR/E) + (-vw/E)
= ~(20.0) 2 + (5.00) 2
= 20.6 m/s
www.puucho.com
2
+ (-vw!E)
Anurag Mishra Mechanics 1 with www.puucho.com
...,
...,
and
1-vw/E
tan0
I= (5.00) = _!
_, I
20.0
I VR/E
0 = tan-1 .! = 14°
·4
or
Solution:
...,
-4j= VR/a-(2i + 3J)
...,
VR/G = 2i- j
when man starts running downv'M;a·= -(2i + 3J)
...,
...,
'
V R/M = VR/G-V 'M/G
d = (1.50) (tan 0)
.
1
= (1.50)
X-
.
= 2i-j+2i +3j= 4i +2j
4
Speed =!.;:~Ml= .J16+4 = ..J20 mis
= 0.375 m = 37.5 cm
. ~77.:7~
g~~.im~~~~~
'dE-x~~""le·,j
120 ~-'lk•·.
i§_ .. -=--~.:
~~=:c.::t.'i!.
~-~----·-------
~.
•"-~
[A boat is movi~ towards eastwith v~lodty 4in/s w_ith re,1p~c9
Ito still water _dnd river is flowing towards north w'.th veloc\tyi
2 fn/s and the wind is blowing: towards north wzth vel9c1tyi
6m./s. ..The. -.di~.ec···t.iqn.. oftheflagblqwn_·_by
. "er,by
the win.dh·o.~t
_ Jd·.·
on
the . boat IS •
.
>'
,
•
.· ,1 ••
(a) northswest
.
(b) sbuth,east
. · . ·. · ·
c,JJaTL'Jl/2.Jwith east~CdJ _,nr>rJL_·__
. _ _ .\;:c;,
Solution:
-
IAn aeroplan~ A' is flying horizontally due east at a'.speed oJi
km/hr. Passengers in A, .observe another aeroplane Bi
\moving [)er:p.eridicular to direction of motion at A. Aeroplane!
,B is actu:a(lY_:moving in a directicm 30_ north of east,il) thej
same horizontal plane as. slzown m. the Fig. 1R120.,,
Determine
.tlodty of B.
_____
·· ·
1400
0
tlje
·,,I
,,
'
Solution:
-+
-+
A
A
-
,,
-+-+
""""'"
. Vw/B = Vw-VB = 6i-4i-2j = 2i-2j
Direction will be north·west.
be=a~{ifie--.11119
~• ,;;,-~~-::::.::..-.~~£::!l
j~
[r~-a marz ru_ntu_:ng Up.wa~ds on ~he_ ill; the ra~ appea_ rs tof_a !pl! .
Ii.
vertically dol1'.71wards w,th 4 fn/s; T!te ve/oc,ty vector pfi;tlze
man w.r.t. ear:th- is (2!+ 3J) fn/,s.·Jfthe man starts rr:lrzning
down the hill with the same speed, then determirze tile reldt/ye
lspeed of the ralrz_ll!,r.t. 111arz.
·
·· ·
I
:...1__
Fig.1E.120
Vw/G = 6i
'Y
0
EI
1,
,
,
~~=--=-.,,_~-- - ' -
,VB/G = VB/R+VR/G = 4i+ 2j
-+
.. --3~
450
~I__F_ig_:J_l::J18."
...,
_ _ _. . -· ~;+,.; II'
' .-::.J.......... ~7' .. >
1B
i.,
-+
~- .
~ - - - - , - - - ----- - ............... --- -·------- ----.··-- I
:f~-4:··-·-
I
...,
VR/M = VR/G--.VM/G
4
(b) From Fig. lE.117,
-
...,
.......
....
www.puucho.com
... (i)
Anurag Mishra Mechanics 1 with www.puucho.com
' DESCRIPTION OF MOTION
A river is flowing with a speed of 1 km/hr. A swimmer wants
to go to point 'C starting from 'A'. He swims with a speed of 5
km/hr, at an angle ew.r.t. the river flow.' If AB =BC= 400 m.
At what angle with river bank should swimmer swim ? Then
the value of e is:
30°
Fig, 1E.121 (a)
400 m
-),
Solution:
"
=
VR/M
-),
Xj
=
--t
VR-VM
t
;M = 2v'3[cos30i+sin30j] = 3i+v'3J
...,
=>
,0
A
vR
rx
A
!
i-/3
Concept: Resultant path of swimmer is at 45° with
bank therefore x-and y-components of swimmer's resultant'
velocity must be equal.
(c)
Fig.1E.121
5=~3 2 +(x-v'3) 2
16= (x-v'3) 2
VR
A
Solution:
(b)
...,
C
Fig. 1 E.123 (a)
~
=>
400m
=-3i+(x-v3)j
'f
- Jo
B
A
=>
Condition for reaching the point C
4+v'3 =
-->
X
VM
A
=-3i+4j
=>
tane = 3/4
0= 37°
·A pipe which can be swivelled in a vertical plane is mounted
on a cart (see Fig. lE.122). The cart moves uniformly along a
horizontal path with speed v 1 = 2 m/ s . At what angle a to the
horizon should the pipe be placed so that drops of rain falling
plumb with a velocity v 2 = 6 m/ s move parallel to the walls of
the pipe without touching them? Consider the velocity of the
drops as constant due to the resistance of air.
Fig, 1E.123 (b)
Vy
tan45°= - , Vy=
Vx
Vx
(VR
+ VM COS0) = VM sin0
l+ Sease= Ssin0
On squaring,
1 + 25cos 2 e + lOcose = 25- 25cos 2 e
socos 2 e + lOcose - 24 = o
e = 53°
On solving, We get
1__.§!',~~P'~ ·_124
Fig. 1E,122
Solution: Rain drops will move parallel to the walls
of the pipe if their velocity relative to pipe is along the pipe.
...,
First we find v Rain. Pipe·
--t
--t
V Rain, Pipe
= V Rain
--t
- V Pipe
-),
=V
-->
According to condition of problem velocity vector v
must coincide with axis of pipe. This will occur if
v,
tana=-=3
V1
L>
The minimum speed with respect to air that a particular jet
aircraft must have in order to keep aloft is 300 km/hr.
Suppose that as its pilot prepares to take off, the wind blows
eastward at a .ground speed that can vary between O and 30
km/hr. Ignoring any other fact, a safe procedure to follow,
consistent with using up as little fuel as possible, is to:
(a) take off eastward at a ground speed of 320 km/hr
(b) take off west.ward at a ground speed of 320 km/hr
(c) take off westward at a ground speed of 300 km/hr
((1.) take off westward at a ground speed of 280 km/hr
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
----------·=·:__: MEC~~i~
veloci~:01
for
~Soluti~m"~!>nc;pt: Fin~i,,;;;:ae of
l~irc~aft relanve to w~d. Iv'A,lwl> 30~~: .____...,:.. ____ ~J
Umin•
f(elmax = 5
·
Umin=
'
16
'
_,
(3 sine - 4cose) = 16
5
VA =-VAi
.
3 sine - 4cose = 5
sin(e-a) = 1
e-a= 90"
e = a+ 90° = 53°+90°
e = 143° with river flow
_, A/W = (x+vA)t'
V
_,
I(v A/W) I= x + v A
~ 300
=>
=>
VA~ 300- X
x varies between 0 to 30 km/hr means v A
in westward direction.
'
'
--
~
300 km/hr
----··- • -
' ._,,·J
IA man wants to !'.each poind3 on the opposite bank ofa :rivet/
in
l,zowing "at a.1peed. 4 mis as shown the Fig. lE.125 (a~.\
What minimrim:speed relative to water should the man ·have
!.$othcit he can reafhl?oint B'.directly6yswiniming? ln which
~30
•...m..:_~---!
·.·. ::·
ldirection_shouldhe,sWim?
.~
• i.
-.,
!
.I ,r---~==:"')--ll. I
i
:
I
I
i''. .
I
\
,
I
.
I
-
!
• .1
.J
.,
l
4m/s
40m
I
I
,
JC--_ __,__ _ __
lL-,,====""""'-----""'""'""-~,.,.L--_-:_--::::-_-_-.:_::::
I
_.-~~
Fig. 1E.125 (a)
:
~
-.. :so.both
luti. ~---!l.
.•,.-~--n.~e.pt:we
.N.o.have
te th·_·.a_ to
·.t.. s•_.·peed
~-,j swimme:
_ ati._d.
angle
are: I(nknown
determine
function
of
~.peed and maximize· ·;t .time ta~en 'far x-component qf,j
(.i.~~in1W~.fml>
oc~C:.Qllll!onent of::di:splace1JJ.l!.~-~l!.'ll!2,. ·
I
___ :. ,.,..
30';;;-·"·s'
.
fcrossing time: 'During the second. ,;tossi1!-&, his goal ~-Ni
lminimize the distance that the boat is carryed downstream;}rt1
[the first case, the crossing time is-1'0 • -In the second cdse,<thej
!crossing time is 3T0 • What is the speed of the ri~erflo'A;?; 'I;jn~l
I,,L_I/OSSL
11 · : ·b1-~'_l!L
a wer.
· · J ,.,..JC .."J'
.... - .- ·--·- ... - --« ~..~-------'-'-
-~ol!J!io!l: ... --·--- __ ..... ·, , -·,- ______ , .....,:oc,
Concept: Case (iJ : If v 1,.. <.v B, boat can cross ~~r'I·
along a path perpendicular to flow.,, . .
, . ;,;: ...
Case (ii) lfv~ < v R drift can ~otb~ zero appl_y caicul~t~!
lt./Ji.Lc.ase, ___ -··-- --···. _ --··--··· . . ·:
· · · '1·, J
Case-I: If
vR < vB
f
Shortest Path:
VB
Quickest path:
d
sine
d
VB
:v•,J
-
-
·case-II:
=ucosei+usinej
x
30
40
u cose + v 4sine
3usin8=4ucose+16
3u sine - 4u cose = 16
16
U=----3sine-4cose
VB
dx
-=0
de
vB(-cosec 2e)+vR cosececote = 0
For min. x,
cose = v 8 fvR .
Time taken in this case is given by
www.puucho.com
,_,..,.__.,.s,' 4'" ' ~
= ( . d_
.
or
__
If vB < vR
VB
~s;E = (ucos8+v)i+usinej
----=--
Fig. 1 Ec126 . )
I
· )cvR -vB-cos~)
szne
.
d
.
x = -.(vR cosece-vB cote)
-
VRiE=Vi
=>
.... (ii)
-=To
= j__ '11- (1/9) = 2./2d
To
:ITo
_,
.. :. (i)
:ITo
also
v R - v B cose = O for shortest path
... (iii)
Thus, sine= 1/3fromeqn.(i) and (ii)
· ·-- --~-'"'~
or
VR = VB cose
--~1c--.-.
·~-·-"·...
-
40m
....Vs!R
.
1
¼
ri~~r has ~..·width. d. A fish_ ~nnan _in a b;~t eras;_ th;-_·.rlJ~;,_j•
!twice. During the first crossmg, his goal 'ts to mmzmize~~he
·_1_
0 --- - -
16
-m/s
5
_, =Xl
Vw
-~
J(e) = 3sine-4cose should be maximum
,,·
,.,,
Anurag Mishra Mechanics 1 with www.puucho.com
' DESCRIPTION OF MOTION
VP=
[60 2 + 120 2 ]1/ 2 = 134.16km/hr
VJ\
120
tan8=-=-=2
V1
60
8=tan-1 2
Hence
An airplane i.s observed by two persons travelling ilt 60:
km/hour in two vehicles moving in opposite directions on a:
straight road. To an observer in one vehicle the plane appears·
to cross the road track at right angles while to the observer in:
. the other vehicle the angle appears to be 45°. At what angle:
does the plane actually cross the road track and what i.s itsi
speed relative to ground?
--;
Vp,
A
hailstones relative to first car is v - v 1 as shown in Fig.
lE.128.
--;
--;
Vp
8
Solution: According to observers in cars hailstones
bounce in vertical direction which implies that the angle of
reflection is 81 as shown in Fig. lE.128, which is same as
angle of incidence in the cars' reference frame. Velocity of
.... ....
C
v,
'lwo motor cars have their wind screens at 81 = 30° and
6 2 = 15° respectively. While moving in a hailstonn their,
drivers see the hailstones bounced by the windscreen of their,
cars in the vertical direction. What i.s the ratio vifv 2 of the
velocities of.the cars? Assume that hailstones fall vertically.
0
Fig, 1E.127 (a)
Solution:
-:!_-
....
Let v p be the velocity of plane relative to
....
- -;
v,
the ground, at angle 8 to velocity v 1 of observer in car 1.
VJ\
=Vp
-V1
....
or
Vp=VJ\
Fig, 1E.128
....
....
In case (i),
From figure,
....
a+ 28 1 = n/2 and
+ V1
--;
Vp,
C
Hence
or
c•
v,
V
tan a=-
v,
tana = tan(1t/2-28 1 )F cot 28 1
V
- = cot 28,
v,
.
Similarly for second car,
cc>
Vp,
--;
-:
V
-
45'
= cot 28 2
V2
B
B
(b)
(c)
Fig.1E.127
Vector diagram is shown in Fig, 1E.127(a). Note that
according to observer in car 1 the plane crosses the road at
right angles.
Similarly, in case (ii)
....
....
Vp =Vp +V2
2
We can combine Figs. lE.127 (a) and (b).
From the velocity diagram,
tan 45° = AC
AB
v!\ = (v 1 +v 2 )tan45°
= 120 X 1 = 120 km/hr
, -
Therefore ratio of velocities of the two cars,
v 1 cot 28 2 =
3
v2
cot 28 1
r ---
!,, E:.x_qmpJ~
, 129
, ..... - ~,.,- - ... ----
-----
An annoured car 2 m long and 3 m wide is moving at 13 m/s
when a bullet hits it in a direction making an angle
tan-1 (3/4) with the car as seen from the street. The bullet
enters one edge of the car at the comer ·nd passes out at the
diagonally opposite comer. Neglecting any interaction
between bullet and the car, find the time for the bullet to cross
the car.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
lss
MECHANI@
Solution: Letthespeed -.=·2m-.
of the bullet be v. Velocity of '
~
bullet relative to car along
Jf: / _,
x-axis = (~ cos 8 :- 13) _and _m j,/ ,lvea,1 "-13 mis
3
along y-axis = v sm 8. Smee · ·
bullet appears from diagonally ·: . ,: :'
opposite
· corner,
its
"----'
displacements relative to car @
• .'
along x- and y-axis are 2 m .__ Flg.1E.129
and 3 m respectivel}j
i.e.,
2 = (v cos 8 - 13)t
... (1)
f
_.
. · ./,' /
1
and
3 = vsin 8t
On eliminating v from eqns.
·
, t =
3
)~
0
_!_ ( -- - 2
13 tan 8
... (2)
and (2), we get
=
13
....
Vrel
--+
1-+
--+
2
Vrel
2
= Urel
+ 20 ret 5 reI
Consider yourself standing in an elevator which is moving
with an upward acceleration a: A coin is dropped from res
om the roof of the elevator, relat:ive to you. After what time
the coinyil[ strike the base of the,elevator?
'
Solution: Here, we need
,, 11,,,
. to apply the rr.=e;==a1
1 -> 2
->
->
•
•
· · +a
formula s = ut + - at for the com
relanve
. a, r•1,ff
9
t srawh
2
Sre1
1 --+
--+_
= Ure1 t +- arel t
2
-h] = _ .!_2 (g + a)t 2. 3'
This yields
t-
· Concept: . If
~ g+a·
2h
the
elevator
= (-g]J-{-ai) = -(g - a)j.
--+
arel
·-~-+
'
-
= ;.._(g,;- a)J,, urel = 0
,· ·
_[!.nd
REh
=_
.
--+
.srel
_,
a,.1 = -(g + a)j_
•
"""?'
--+
= Urel t +
m Srel
._ ..
,For freely falling ·elevato~
g-a .
. ·
!Find the velocity of the coin relative to ground when_· it ~trikes
/the /!_ase..]Jf.the elevator.
•
--+
--+
0
= -hj. Now, substituting
I
A
Solution: Substituting ure1 = 0, are1 = -(g + a)j, we
have
v~1 = -2(g + a)j. (-hJ)
This yields = Vre1 = .J2(g + a)h. Since the coin moves
down, (relative to the elevator), we have
...
V rel
-
= -_i2(g + a)h j
As the coin strikes the elevator after a time t
~.
=~
2
h ,
,(g + a)
the velocity of the elevator at that time is
_, _,
V
= at =
av{gtt)J
Then
-; = a~
substituting
; rel
2h j, in the equation -; ,
(g+a)
; , =
(a~
2
h -.J2(g + a)h
(g +a) .
'
= -.J2(g + a)hj
=;
)j
and
rel+;, we have
=-
~ 2gh
1+!!.
j
g
b~~~~RJ.~.J 132 ~
As the coin moves down with a displacement of
-
'
= 0,
_-·- _ - . '
velocity at the time of release of the coin. Hence u rel = 0.
sre1
downJ
'
Substituting 1
accelerates
ci,e1 = "ii,- "ii,
1
....
_,
formula
h
2
Flg.1E.130
2
_,
_,
where t = time of fall of the coin, u re1 and are1 are the
initial velocity and the acceleration of the coin· relative to the
elevator, respectively. Since the coin was attached with the
elevator, both coin and the elevator would move with equal
magnitude h,
the
2
keoxa!m}:B).s.\,~c
----~·~J 130 ~
~
-+
in
a= g. Th~n, i = =. Hence the coin will never touch" the
surface. _It just ha~., below t/ze roof of the freely falling
elevator.
Sometimes a body moves in a moving reference frame
or we have to analyses motion of a particle from point of
view of a moving observer. In such situations above
equations prove useful as illustrated by following examples.
to the elevator
1
...
-hj
sre1 =
=Ure1t+2arelt, we ave
Sret
2
....
= Ure1 t +-2 a,e1 t
--+
and
1....
= Urel + are1 t
s,e1
,.
are1 = -(g + a)j
-a,.1t 2 , __ we- have t
= 0.15 s
Equation of Moti!)n for Relative Motion:
....
-+
_,
ure1
= 0,
IA lift is movi~ with uniform downward acceleration of 2 /
Im/s2 • A ball is dropped from a height 2 metre from the fl_ oor ofi1'
rlift. I:ind the time after which baU will strike the floor.
www.puucho.com
.
Anurag Mishra Mechanics 1 with www.puucho.com
DESCRIPTION OF MOTION
ggj
Solution: Initial velocity of ball with respect to lift
k:~gm~'fJrnl>
....
=0
....
ab= -g = -10
....
Ure!
u, =-2·
acceleration of ball with respect to lift
--+
--+
--+
2
.are! = ab- a 1 = -8 m/s
Displacement of ball with respect to lift till it strikes the
floor
Cann~; A is loca;d cin a pl-am-._a_d_is-ta-nce L from a Wall o.fi
height H. On top of this wall is an irlentical cannon (cannon
~BJ. Ignore air resistance throughout this problem. Also ignore
the size of the cannons relative to L and H. The two groups of,
gunners aim the cannons directly at each other. They fire atl
,each other simultaneously, with equal muzzle speed v 0. What/
;is the value ofv 0 forwhich the two cannon balls collide just as
/they hit the ground?_ _ __
'
B
..····
·,
t··
·-·----
IA toy train moves clue north at a constant speed 2 nVs along a
.....··
H
L
L____ _
...._Fig.1E.1~ (•..
)_..c.._________.
Solution:
straight track which is parallel to the wall of a room. The wall
is to the east of the track at a distance 4 m There is a toy dart
gun on the· train with its barrel fixed in a plane perpendicular
to the motion of the train. The gun points at an angle 60° to
the horizontal. There is a vertical line drawn on the wal~
stretching from fl.oar to ceiling, and the dart gun is fired at the
instant when the line is due east of the gun. If the dart leaves,
/the gun at speed BnVs relative to the gun,find the distance by[
1which the dart misses the vertical line. That is, find how far 1 ,
· (north or south of the vertical line is the point at. which the
0 sine)
0 sine)
2v~
v 0 cos e(- +v 0 cos e(2v
-~
- =L
g
.
g
H
sine
L
H
cose=,====
v2
i
0
!~~h~ts-the wal~) 3m ---· (c) lm .. __(d) Sm ---- _j
Vo=
~H2 +L2
gL
L
Fig.1E.134 (b)
4sinecose
g(L2+H2)
4H
Solution: Consider east as x, north as y and
vertically upward as z velocity of dart w.r.t. to train at firing
¼platform is moving upwards with a constant acceleration ofl
, ·r ~y(N'. )~all
,..· .··
I• d •• -··
!2tn1sec 2 • At time t =Q a boy standing on the platform throws
/a ball upwards with a relative speed of Sm/sec. At this instant
;platform was at the height of 4 m from the ground and was
. :moving with a speed of 2tnlsec. Take.g =l0m/sec 2 • Find
( a) When and where does the ball strikes the platform?
'(b) Maximum height attained by the ball from the ground;
'(c) Maximull!- distance_ of the ball _fr2m the platfg_rm.
__
l ~ I
x(E)
.•
I
I
'I
--·· __ Fig. 1E.133 •.••.• 1
;d, =8cos60°i+Bsin60°"ic= 4i+4J3k
velocity of dart w.r.t. ground at firing
---t
•-)
--+
A
Ar;:;A
Solution: (a) We solve the problem in reference frame
of platform
....
A
A
ud = ud,+vr = 4i+<tv3 k+ 2j
V Ball/platform
-)
Time taken to strike wall t = df 4 = l sec
Displacement along y = 2 x t = 2 m (North)
aP/E
....
by
www.puucho.com
=8 j
A
=2j
-)
and
....
aB/E
A
are!
=aB/P =-12j
Srel
= Ure! t + 2 arel t
1
2
A
=-g j
Anurag Mishra Mechanics 1 with www.puucho.com
I 1·00·
MECHA~~
LI..=-.;c.__~.::;t:...._.._.._:_... _ _ _ _ · ~ - - ~ - - - ~ - - - - - - - -
For vertical motion
.!gt 2 + (200sin0)t _.!gt 2 = 1000
2
2
sin0t = 5
From (1) and (2)
1
sine
1+ cos0 - ../3
On solving,
0 = 60°
0 = Sxt-_! X 12t 2
2
4
t = - sec
3
.
4 10
Tota1 tune = 2 + - = - sec.
3
3
.
f 1 ~ .. lO
.
1
dtsp acement o p auorm m - sec.
3
2
(ii)
=4+2x.i+.!x2x(i)
3 2
3
76
=-m
-,
(b)
-,
= lOj
V B/E
2
by
30-8
= -lQj
2
v = u + 2as w.r.t. earth
(0) 2 = (10)2 - 2(10)s1
s1
:
= Sm
· I
:
----~.... ..
1 km
:
i
:
'
;
------------·· - ..1
·
··n. _ ~
.
'
••
t
i
5 .... !
0
,
I
Flg.1E.136 (a)
-=-=-_.,...,,.~,.,--=--' -------~-
• • ·~,_.,,._,...._...,,.,,.... __
~
Solution: (i) Suppose shell destroy the bomb at time
't ' then for horizontal motion
t(200 + 200 cos0) = ../3 x 1000
t(l + cos0) = s../3
... (1)
=I-~~~I=½
AB= 2km
BP = minimum distance = AB sin(30° -0)
BP = 2[sin 30° cos0 - cos30° sin0]
=
2[½ ls-~
X
X
Js]
= 2-../3 km
..Js
r·--.
1A balloon
1
·...
•
:-.\'\'\'\\.'\)$'\'-"\~\\,,\\\\,,\\\\'\.'\'\~\.,-W.........
.v3km
I.
-,
= (200 + 120) i - 1'60 j
,IAn aircraft i§ .flying. horizontally with a constant vefocity
i= 200m/s, at a·height =lkm. ciboye the ground. At the.
tmoment shown, a. bomb is released from the aircraft and the
jcan~on-gun ~elow fires a shell with. initi~l speed =·~O~ m/s, at
,some arigle.0. ·For what value_of'0' will the proJectile shell
1destroy
the bomb in mid,air?• If the
value of0 is 53°, find
the.
I .
.. • . .
, •
..
:minimum distance between the bomb and.the shell as they.fly
:past each other. Take sin 53° = 4/5. ' _ .... __
!_ , __ - - -
-,
tan0
-- --·
-~
-·--
VA/B =VA-VB
8
3
.
·
= -120i+160j
-,
s=-m
'
,.
v3
Fig.1E.136 (b)
= -2oox~i+ 2oox.iJ
5
5
~g:xam.Rle6~
'
;
[_____
Hmax=5+4=9m
(c) Also platform frame
. v2 =u2+2as
or
(0) 2 = (8) 2 + 2(-12)s
or
A
r-------A -- ·.. --:·1
rL' ' ~ !j
A
& aB/E
A
v 8 = -200cos53° i+200sin53° j
9
-,
... (2)
"
-~--,
is moving vertically upward with constant
!acceleration (g /2) in upward direction Particle .'il was
!dropped from the balloon and 2 sec later another particle 'B'
'was dropped from the same ballooni Assume that motion ofi
the balloon.remains unaffected. Find the separation di.st.ailce
between 'A'.and.' B ', 6 sec after dropping the particle 'B ', !'JO[!e
of the particles reaches the ground during the time interval
urrd~Lf.Oll§igqggo1i (g-=.10 m,ls_gc_~) __ . __ .
Solution:
! ·- c~~;~pt: s;-:c~- ~articles are being dropped fro~-~
:moving body i..e., a moving reference frame, we used reference/
. ,frame of balloon itself for both the pa_rtz_·c_les_._ _ _ _ _ _)
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
I,DESCRIPTIQN Of l\'IOTION
'1.01
Acceleration of platform relative to the ground
Motions of particles '.A' and 'B' are
w.r.t. balloon, so balloon is reference
point therefore it is assumed to be at rest.
We denote balloon by b
For A
u,01 = 0
_,
->
and
For B
urel
Seel
-48g
= 0;
arel
1
= Ure1t +-are1t
2
->
=- 2
A hori2ontal platform is moving vertically upward with
constant acceleration' a'. When velocity of the platform is
'v ', a particle is projected from the platform with velocity' u'
relative to the ground at an angle 0 with the horizontal. Find
the hori2ontal range and the time.of flight of particle on the
platform.
Concept:, In case of projection from a moving platform
entire motion takes plane on platform, always use reference
!frame of platform.
----.----
Velocity of projectile relative to the ground
g+a
t =0
T= 2Cusine-v) (time of flight)
g-1:a
2ucos0(usin0-v)
Range CR) = (u cos e)T =
g+a
Projection of a Ball in a Horizontally Moving
Trolley
A trolley is moving horizontally with a constant
acceleration' a'. When velocity of trolley is' v', a particle is
projected with velocity' u' at an angle 0 above the horizontal
from the position which is
at distance 11 from the •y
I
front wall and 12 from the ;
rear wall. This velocity and .
angle of projection are
hf
v .
relative to the ground.
a
1·
_,
->
+'---,.,r-""'--.,,.,.-'----+!
X'
->
A
I
acceleration vector a are in
the same vertical plane.
Fig.1.106 _ --·--Find the time of flight and
the horizontal range of particle on the trolley. Also discuss ·
the condition for whicli particle will fall
(i) in front of point of projection
(ii) at the point of projection
(iii) behind point of projection
I__ _
la upward
v j
moving
~~,,~:~J
_,
Acceleration of projectile relative to the ground
aP/gr
2
2(u sin0-v)
Velocity vector u, v and
= (u cos0) i + (u sin 0) j
->
2
or
and
Projection of a Particle in an Accelerated
Elevator
l
1
2
+ 2 ayre/
=(usin0-v)t-.!.(g+a)t 2
t
= ISA/balloon 1-1 SB/balloon I
= 480-270 = 210m
y
t
'
.
A
UYrel
0 = (u sin0-v)t - .!.(g + a)t 2
2
1 3g
2
· '
2 2
SB/balloon= -27 g = -270m
Separation distance between ' A' and 'B'
u P/gr
A
Yrel
3g
s,el =SB'balloon = 0---(6)
->
A
= -(g + a)j
a,.1
Y rel =
-480m
=
A
At the end of flight y-component of displacement of
projectile relative to platform becomes zero.
2
(A falls off 8 sec)
Srel =SA/balloon=
.
u,.1 = (ucos0)i+(usin0-v)j
1
2
= Urelt +zarelt
(8)
A
Velocity and acceleration of the particle with respect to
the platform
3g
s,., =0-21(3g)
2
_,
= (a) j
ap~gr
a,.,= -2
Sret
I
uP/gr
A
= (-g)j
.'
->
and·
aP/gr
A
->
www.puucho.com
A
= (ucos0)i+(usin0)j;
A
= -(g)j
Velocity of platform relative to the ground at the time of
projection
and
I
A
a,1gr = (a) i
Anurag Mishra Mechanics 1 with www.puucho.com
.. ~·
102 .. ,..,-:o<
->
A
l----Concept:
A
u,., = (u cose-v) i+u sinej
->
and
A
·'A1
We assume that the flight completes on the floor of
trolley. It does not strike the roof or the front wall or the
back wall.
1
= UYrel t + 2 aYrei t
Y rel
2
If the di;ection of relative velocity is
,
.
!through
.(ii) theWhenposition
.direction of relative velocity does_ not pass
of' A' then perpendicular' AN' f,wn the
1
':
.
· ·i
_4
Iposition of'.,!:' on the line of action of relative velocit_y(vB/A)
!gives the m
_ ·. _i_'nimum possible .,d_ista. nee between 'A_.·, a. nd 'JB'
iduring their'_motion
· , .
·. ··
,
-···--•.. - _ _
AN= dsino: ·
·
.. ,
0= (usine)t-~gt 2
2
2u sine
T = ---
Thus time of flight is
(i)
:directed towards , th~ position of;A' then the body \Bl meets
A
= -(g)j-(a)i
a,el
MECl-l~NICSi!j
->
a is the angle which v B/A forms with y-axis.
g
For range
->
First we will determine velocity and acceleration of
particle in reference frame of trolley.
Horizontal range as observed from the trolley
1 2
Rre, =(ucose-v)T--aT
A
vB/A =-(vBsin0 2 +vAsine 1 )i
+(vB cose 2 -vA cose,)j
(vB cose 2 -v A case,)
tan ex = --'-.;...---"'---"--"-- (v B sine 2 -vA sine,)
2
As observed from the ground
u 2 sin28
Rrel
g
From tan a determine
sin a and· cos a
Time required to come closest is given by
BN
dcoso:
t=--=--
Closest Distance of Approach Between Two
Moving Bodies
->
lvB/~I
->
lvB/Ai
->
, Two bodies are moving with constant velocities v A and
->
v:i,
as shown in Fig. 1.107.
y ~
Ll:E:~p~l~~
r1wo roads •_z_~.;erse~~ at righ/a_ngle-.s-._C_a_r-.A-.-is-s-itu_a_t_e_-a.a:·-;i
which is 500 m from the intersection O on one of the roads.I
Car B is situated at Q which is 400 m from the intersect/.6n on
the other road. They start out ,at· the same time and, ti:aVel
towarcls the intersection at 20 mis and 15 m/s respectively.
What is the}ninimuni distance between them? How'long' do
they tdke to reach it?
_ _ _ _ _ · ··
·
:'
A_,;
::
.
E.
vi;~~-,~·
' '
:
..... i
:.
:
:-' .
...
.'
.
..
.
:
.
~
••••• ')
~
:fd
.
. . . .S t .
J,
Vs :
..._.... _... --· i
\
-'vl\.
a ,
'a
i
2
· vA
'i)J),JIV
500m
~ 20 mis ~ L _ - - - ! 0
P Car A
.
..
. -·········----····>x
400m
CarB
B
Flg.1·.f07
->
A
A
v A = (v A sinei) i+ (v A cose,)j
and
->
A
Fig.1E.138 (a)
A
sine2H+ (VB cose2)j
Motion of' A' relative to' B', is along a straight line in the
VB= -(VB
->
direction of relative velocity (v B/A).
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
DESCRIPTION OF MOTION
Solution: First we find out
A
20
the velocity of car B relative to A.
e
As can be seen from Fig.
lE.138 (a), the magnitude of
15
velocity of B with respective :
vA = 20
m/s, vn = 15 m/s,
OP = 500 m; OQ = 400 m
Fig.1E.138 (b)
15 3
4
tan0=-=-· cos0=-·
20 4'
5'
. 0 = 3sm
0
B
5
3
= ADtan0 = 500x- = 375 m
4
.p
500 m
BC =OB-OC
e
=400-375=25m
BD = BC (cos0)
625m
OC
4
o
c 375m
=25x-=20m
5
a e
Shortest distance = 20 m
..,
Fig.1E.138 (c)
_j
PD= PC +CD
= 625+15= 640
Therefore,
relative
acceleration between them is
zero i.e., the relative motion
between them will be straight A
line. Now assuming A to be at
rest, the condition of collision
..,
.., ..,
will b e that V CA = V c-V A =
relative velocity of C w.r. t. A
should be along 0\.
..,
VA=
C
,n'
-
Vn=-5i-5v3j
VBA
..,
:. VBA
.. .
r.;:;;:;l •
l}=~g,tp:12;1~ 11391.>
-
C
Am
Di
--- 30;60°·
A
~
1
r
10m
'T
r;;'
= -15i-5v3j
--'ss~=d=~o
10
d=l0-./3 m
Two towers AB and CD are situated a distance d apart as:
shown in Fig. 1E.139 (a). AB is 20 m high and CD is 30 m
high from the ground. An object of mass mis thrown from the
top ofAB horizontally with a velocity of 10 m/s towards CD. ·
VaA
=-5i-5..J3j-10i
:. tan60°=
Bi
Fig.1E.139 (c)
-
5.Jam/s
10mis
lOi
..,
l~ABl=25m/S
640
t == 25.6 sec
25
'
D
Fig. 1E.139 (l>)
Fig.1E.139 (d)
l~c:;;;~m21~8> ,On a ftictionless horizontal surface, assumed to be the X·Y
.plane, a small trolley A is moving along a straight line.
:parallel to the y-axis [see Fig. IE.140 (a)] with a constant:
,velocity of ( ,Jj-1) m/s. Ata particular instant when the line'.
:oA makes an angle of 45° with the x-axis, a ball is thrown
ialong the surface from the oriiµn 0. Its velocily makes an'
·angle <I> with the x-axis and it hits the trolley.
·y
Fig: _11:.139_ (a)
_..nA
Simultaneously another object of mass 2m is thrown from the
·top of CD at an angle of60° to the horizontal towards AB with
the same magnitude of initial velocity as that of the_ first'
.?bjec~. Tl'.e two ob!ects move in the same vertical plane, collidej
m mtd-mr and suck to each other.
:
Calculate the distance d between the towers.
Solution: Acceleration of A and C both is 9.8 m/s
downwards.
2
.· •'45°
X
0
Fig.1E.140 (a)
:Ca) The motion of the ball is observed from the frame of the
trolley. Calculate the angle 0 made by the velocily vector
· of the ball with the x-axis in this frame.
(b) Find the speed of the ball with respect to the surface, if
• _ <I> =_40/3_.
--- __ _
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
F ·'
,
. 1)1~(1!A~ld-il,S,I
- - - ~ · - - - - :::'::::::====::==· ·~';:;::;::==:'.:::::::::,~-_:'.:::.::..:J:::;
Sol""•".' [,) ""A ~ml, fu, ®lley rnd B fud•lL
~· 1
I
.
Relative velocity ~f B with respect to A(~_;!A) should be
alongOAfortheballtohitthetrolley.HencevBA will make
450 th
an angle of
wi positive x-axis.
(b)
tan0 = vBAy = tan45°
or
Further
or
v BAy
v BAx
= v BAx
... (1)
vBAy
=vBy -
v Ay
v BAx
= v Bx
O
VBAy
= VBy -(-J3-l)
tan<jl =
-
2
... C )
3
... ( )
or
vBy =
tan<jl ·
From eqs. (l), (2), (3) and (4), we get
("3 -l) and
tan<J>-1'
vBx
j
Fig.lE.
141
(•)
j
,
. "'
.
(a) find ihe?listance,_a_lo_n""g-th-e"bo-tt-o;;;~f the bo)( be·i:ween the
.· point of projection P and the point Q.where the particle
_' la!1ds'.(Assunie that tlie pafticle does not hib:a!lS{·other
· surfate,of the box. Neglec(i:Lir resistance)
'" ~
~
J. h• o.rlz···on···t.al
disp.la
..c.em.· .nt_o.if
pa. r.tide··.as·····
. •·.s.". of
en the
by
. (b) If
ant·h.observer
on the
ground•is
zero,th.e
find
the speed
bo~- with respect to the ground 'at the instant When the
,__,,.P.f!!:tigle \'!IJ§..PI9jected. ,
,,
. .
.
. • --- •
' ·..·
'·--··/",,.1
· ~·.. ·i ·1J !
:
g'.'.s.in.e..
..
8
"
i
'
•~'-----·-F_i~g;~:1E:1M (b) · - - ,
"3-1
=---sec<J>
tan<jl-1
Acceleration of particle with respect to box
= Acceleration of particle - Acceleration of box
= (g sin0i + g cos0j)- (g sin0)i = g cos0j
Now motion of particle wlth respect to box will be
projectile as shown in Fig. lE.141 (c).
,· ·.·. 7
Substituting <jl = 60°, we get
VB= 2m/S
Alternative: Relative to frame of A
ex
~;r--
v~,)\·
e
·
15•
R.:.
gc~s·e'
Fig, 1E.1ll1 (S)__
The only difference in g will be replaced by g cos 0.
·
·u 2 sin2a
PQ = Range(R) = - - .
g case
'
08
45
0
PQ
• ~['_··._.·--~_i;"'~-'1=E:140j~:__c__ _·~
Resultant velocity is along OA, so perpendicular
components = ·O
VB X5m/s= ("3-l)COS45°
VB=
.
8
::
("3 - l) · tan<j,
tan0-l
·
vBy
Speed of ball w.r.t. surface vB = ~v~ +v~
lv sin 150.
,
... (4)
<I>= 40 = ic4so)
3
3
r-- ._.... -- -
,',"'
·
P . a'
Solution: (a) Acceleration of particle and box both are
shown in Fig. lE.141 (b).
vBy
vBx
vBx
or
·
u 2 sin2a
gcos0
(b) Horizontal displacement of particle with respect to
ground is zero. This implies that initial velocity with respect
. to ground is only vertical, or there is no horizontal
component of the velocity of t;he particle.
y/;in (cxu+8)
~1
t ,.·
(-J3-l) ,_!_ = 2m/s
Sm/s ,/2
1
Gi.:1.·
~~A~FJI~ 141 ~ .
'!-_<--
L4 large h~~ box ~ sliding witho~;fricti~n °dow.n J:~;,;;,:J
/Jilane <if inclination 0. From a pq{nt.P on• the bott6,,r,ofth;1
lbox; a particl~ is.projected inside:tJie box. The initigl sp~ed p}j
'!th·e· par.,tz.'¢1e :W. .·. .ith respect to th~ bo>1.. is u.'.and the, d.ir,ectt.;o·'·il····.:'o··1
.pro;ectton makes an ·angle a Wlth,.the' bottom as s/town·m 'the
jFig. 1E.14lfa). _ _" _
, ' , _
·. · , ., :·,,',:
--
ucos(a+0)
·
..
/VCOS8
,j
. . 8 ,.J· ,.
""--"'-'-=---,1(,
f'----·"-''~'_·---~Fig. 1E;141: (~L ______,
Let up, b(Hl is. component of velocity of particle w,r.t. box
in horizontal direction.
up,b(Hl =ucos(a+0)
If vb is .speed of block along the incline w.r. t. ground.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
I DESCRIPTION OF MOTION
1051
vb H = component of box's velocity in
'
horizontal direction
VbH
Now,
We find arbitrary constant c by employing initial
conditions v = 0 at t = 0 which yields
mg
c=ln--
= -Vb COS0
k
-ub,g(Hl
up,g(H) = up,b(HJ +ub,g(HJ
=ucos(o:+0)-vb cos0
But, as up,g(Hl = 0
~ u cos(o: +0)-vb cos0 = 0
u cos(o: +0)
vb=
cos0
up,b(H) =up,g(H)
On substituting the value of constt. c in eqn. (3), we get
1n(mg -v)-lnmg =-!t
k
k
m
or
1n{mgfk-v}
mg/k
---,,."
---··----·-------- - ·-
···~
l
!A small sphere of mass mis released from rest in a large vessel
i_filled with oil where it experiences a resistive force
;proportional to its speed, i. e, Fd = - kv.
'(a) Find the law according to which the'ball's speed varies.
'(b) After a certain time the sphere reaches a terminal'speed;
: find it.
1
( c) Time constant, is the time it takes the sphere to reach
\
632% of its terminal speed; find it if m = 2.00 g and
I terminal speed is 5. 00 cm/s.
'(d) Determine the time it takes the.sphere to reach 90% of its
j terminal speed. 1 , , , . - - - - ~ ·
i
.I
m
v = mg (1-e-Cl;'m)t)
k
= mg (1- e-<f,)
k
or.
,------ -
= _!t
where T = m is called time constant.
k
(b) When the particle reaches terminal speed, the
acceleration of the particle becomes zero. When the
magnitude of the resistive force equals the sphere's weight,
acceleration is zero and from then on the particle continues
to move at constant speed called terminal speed.
mg= kv, or v, = mg/k
(c)
k = mg = (2.00)(980) = 392 g/s
v,
5.00
m 2.00
Time constant,
T=-=-k
392
= 5.10 X 10-3 S
. (d) Speed of particle as
function of time is given by eqn.
(4).
Fig.1E.142 (a)
Solution: (a) Force acting on sphere = mg - kv
where k is a constant. We have assigned downward direction
positive and upward negative.
Acceleration of ball
dv
k
... (1)
dt=g-mv
Separating variables, we obtain
dv
=-!dt
mg -v
m
"')>
v
;,_0/
~0/
Vt •••••••••••••••• - •••
i
0,63v1 ••• ••_:;-'
or 0.900v, =v,(1-e-C</<l)
1- e-,1, = O. 900
or
or
or
k
ln(m:-v)=-:t+c
I
.
... (2)
On integrating the above expression, we obtain
... (4)
e-</< = 0.100
I
l
.
:
O
~
Fig. 1 E.142 (b)
t
l
I
-t/T=ln(0.100)=-2.30
t = 2.30T
= 2.30 X (5.10 X 10-3 s)
= 11.7 X 10-3 S
In the graph of v versus t for the ball, the slope of v versus
t graph gives a. At t = 0, v = 0 and a= g. Ast becomes large,
... (3) · v approach es v, and a approach es zero.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
110s
. ' ,MECHANIE~J
_,
_,
_,
9. The resultant of a and b makes angle a with a and~
_,
i.
It is possible to add five. unit vectors to get an unit
vector, The statement is :
(aJ True
(bJ False
2. If a vector is rotated by angle 0 then it is necessarily
changed. The statement is :
(aJ 1hie
(bJ False
3. It is possible to add n vectors of equal magnitude and
get zero:
(aJ True
(bJ False
4. It is possible to add n vectors of different magnitude
and get zero.
(aJ True
(bJ False
--+
--+
--+
--+
= a . b for some suitable
--+
-+
--+
-+
selection of a and b. For example a = 0. The statement
5. It is possible to have a x b
is:
(aJ True
--+
6.
-+
_,
-+
IflAI= IBI and A,;, ±B then angle between the vectors
--+
-+
--+
--+
(A+ BJ and (A-BJ is:
(aJ 0
(bJ it/6
(cJ it/3
(dJ it/2
7. A vector of magnitude a is turned through angle 0. The
magnitude of change in the vector is given by:
(aJ l2asin01
(bJ l2asin0/2J
_, _,
10. Let C = A+ B.:
(aJ
(bJ
(cJ
(dJ
ICJ is always greater than IA 1.
-+--+
--+--+
It is possible to have IC l<IAI and ICl<IBI
--+
-+
--+'
ICJ is always equal to IA[+[BI
_,
_, _,
[C[ is never equal to IAl+IBI
_,
.
_,
11. Let the angle between two non-zero vectors A and B
_,
be 120° and its resultant be C. Then:
--+
(aJ
(cJ
(dJ
--+
-+'
ICJ must be equa!I IAI-IBI I
_,
_, _,
ICJ must be less than I IAI-IBI I
_,
_, _,
ICJ must be greater than I IAJ-IBI I
_,
_, _,
[Cl may be equal to I IAI-IBI I
12. Which of the following two statements,,- is more
'·
appropriate?
(aJ Two velocities are added using triangle rule
because velocity is vector quantity.
(bJ Velocity is a vector quantity because two velocities
are added using triangle rule.
_,
8. Which of the sets given below may represent the
magnitudes of three vectors adding to zero?
(aJ 2, 4, 8
(bJ 4, 8, 16
(cJ 4, 8, 4
(dJ 0.5, 1, 2
,_,
_,
(bJ
(b) False
--+
.with b, then (a, b represent magnitudes of respective
vectors):
(aJ a < ~
(bJ a < ~ if a < b
(c) a <~if a> b
(dJ a<~ if a= b
_,
13. Vector ais increased by/!,. a If increment in magnitude
_,
.
of a is greater than_,magnitude
_, of increment vector
then angle between a and /!,. a is:
(aJ greater than it/ 6 (bJ exactly it/ 6
(cJ exactly it/ 2
(dJ <I>
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
c...D_E_SC_RI_PT_IO_N_O__
F_M_OT_IO_N_ _ _~ - - - - - - - ' - - - - - - - ' - - - - - - - - - - - - - - - _10i]
14. A motor car is going due north at a speed of 50 km/h.It
makes a 90' left turn without changing the speed. The
change in the velocity of the car is about :
(a) 50 km/h towards west
(b) 70 km/h towards south-west
(c) 70 km/h towards north-west
(d) Zero.
15. A person moving on earth's surface starts from north
pole & moves 500 km towards south and then moves
1000 km towards east and then again moves 500 km
towards north and stops. The displacement of the
person is:
16. A person moves 20 m towards north-east then moves
20 m towards west and then again moves 20 m
towards north-east and stops. The magnitude of
displacement of the person is:
(a) 2W5-2./zm
(b) 20 m
(c) 2W5 + 2-J2m
(d) None of these
-, -, -,
17.' If A, B, C, are mutually perpendicular vectors then
which of the following statements is wrong?
-,
(a) C X (AX B)
J
=0
is:
(a) 2
(c) 1/2
(b) 3/2
(d) 1
--+
--t
--+
(¾)
1
(b) cos- 1 ( ~ )
(d) sin-1 (~)
minimum force, then the force is :
(b) _SN and SN
(a) 6N and lON
(c) 4N and 12N
(d) 2N and 14N
23. What is the component of 3 i + 4 j along i + j
00
!d+J)
(b)
2
w ~d+b
2
--t --t
--t
--t
-,
-t
-,
_j&
(b)
_I aJ2
--t --t
--t --t
a.b
a.b
(AA2+B2B2)
~d+J)
2
--+
--+
a.b
.
- B2 )
2(A2 +B2)
19. A plane is inclined at an angle 30° with horizontal. The
-,
(AA2-B2
+ B2)
2
2(B2 -A2)
-,
-,
25. The resultant of A and Bis perpendicular to A. What is
-,
-,
angle bet)veen A and B ?
1
(a) cos- (;)
1
(b) cos- (-;)
-1(- A)
B
26. A particle moves through angular displacement 0 on a
circular path of radius' r'. The linear displacement will
be:
-,
component of a vector A =- lOfc perpendicular to this
plane is: (here z-direction is vertically upwards)
Ca)
s..!z
--t
(d) cos-1 ( A 2 + B2 )
.
(d) sm
(d) None of these
-,
JaJ 2
(b) cos~!
2 -
-, -,
(c) 5
:
~) !d+j)
2
(c) cos-1 ( A 2
law V = a+ b t where a and b are two constant
vectors. The time at which velocity of the particle is
perpendicular to velocity of the particle at t =0 is:
(c)
-,
22. The sum of two forces acting at a point is 16N. If the
resultant force is SN & its direction is perpendicular- to
18. The velocity of a particle varies with time as per the
(a)
--+
-,
so that the resultant is ~ A 2 + B 2
(c) A.B = B.C = C.A =0
-, -,
-,
(d) (B +C) is perpendicular to A
--t
--+
and 3 units respectively the angle between A and Bis :
(a) cos-1
--t
--+
= B +C and the magnitude of A, Band Care 5, 4,
21. If A
--t
C
--t --t
--t--+
24. At what angle the vector (A+ B) and (A - B) must act,
-,
--t --t
--t--t
the value of (a1 - a 2 ). (2 a 1 + a 2 ,)
(c) (~)
(d) Zero
-,
--t--t
Ia 1 + a 2 = ../3, then
(a) ~os-
(a) 1000 km eastward
(b) -Jiooo 2 + 500 2 km towards south-east
(c) ~1000 2 + 500 2 km towards
-,
-,
-,
20. If a 1 and a 2 are two non-collinear unit vectors and if
(b) 5F3
(d) 2.5
(a) 2rsin(~)
(b) 2rcos(~)
(c) 2rtan(~)
(d) 2rcot(~)
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
I
108
ME(H~l'U(S-1
~
- - - - · - - - - - - - - ~ - - - - ~ ' _ , ; , __ _ _ _ __,:_ _ _ _ _ _ _-~-------'-'--'-_CJ·
_,
2 7. If a vector A makes an angle a, ~ and y with X, Y and Z
axis respectively then sin 2 a+ sin 2 ~ + sin 2 y =.....
(a) 0
(c) 2
(b) 1
(d) 3
_,
'
'
28. The X and ¥-component of Pare 7i and 6 j. Also, the X
--+
-+
A
A
and ¥-components of P +Q are lli and 9j respectively.
_,
Then magnitude of Q is :
(a) 7
(b) 6
(c) 5
(d) 13
29. Two vector
such that the component of B along A is zero. Then the
value of x will be:
(a) 8
(b) -4
(c) +4
(d) -8
-+,_
A
30. Two vector A = 3i + 8j - 2k and B = 6i + 16j + xk are
_,
_,
such that the component of B perpendicular to A is
zero. Then the value of x will be :
(a) 8
(b) -4
(c) +4
(d) -8
31. A blind person after Walking 10 steps in one direction,
each of length 80cm, turns randomly to left or right,
After walking 'n' steps, the maximum displacement of
person is 16-/2 .Then value of'n' is :
(a) 20
(b) 30
(c) 40
(d) 60
_,
_,
32. Two vectors A and B have magnitudes 2 and 2-./2
-+ -+
-+
-+
. respectively. It is found that A. B =IA x BI , then the
_, _,
(a) 5
(b)
.rs
-./2 + 1
-./2 - 1
(d)
-./2-1
-./2 + 1
33. If the resultant of two vectors having magnitudes of 7
and 4 is 3, then the magnitude of the cross product of
the two vectors will be:
(a) 28
(b) ../65
(d) zero
(c) 53
34. The adjacent sides of a parallelogram is represented by
vectors 2i + 3j and i + 4] . The area of the
parallelogram is :
(b) 3 units
(a) 5 units
(c) 8 units
(d) 11 units
35. The maximum magnitude of cross product of two
vectors is 12 units and the maximum magnitude of
their resultant is 7 units, then their minimum resultant
·
vector will be a:
(c)
_,
(d) F2
=-~~N~A~C!:~
,, _,
3 7. The quantity
J
t1
V dt represents:
(a) Distance travelled during t 1 to t 2
. (b) Displacement during t 1 to t 2
(c) Average acceleration during t 1 tot 2
(d) None of these
J''
_, = Vx 'i + Vyj' + vz..,
i'. then ,, Vydt represents: (for
38. Let V
the
(a)
(b)
(c)
(d)
duration t 1 to t 2 )
Distance travelled along y-axis
Displacement along y-axis
Total displacement - displacement along y-axis
Total distance travelled - distance travels along
y-axis
39. A particle has a velocity u towards east at t = 0. Its
acceleration is towards west and is constant. Let x A
and x 8 be the magnitudes of displacement in the first
10 seconds and the' next 10 seconds then:
(a) XA < Xn
(b)
(c)
value of ~
_, _, will be:
A-B
_,
(c) vector of magnitude between IA I and IBI
(d) nothing can be said
36. Six forces are acting on a particle. Angle between two
adjacent force is 60°. Five of the forces have
magnitude F1 and the sixth has magnitude F2 • The
resultant of all the forces will have magnitude of:
(a) zero
(b) F1 + F2
(c) F, -F2
A= 2i+ 3]-4k_, and B_,= 4i+ 8] + xk are
--t,_,_,_
(a) unit vector
(b) null vector
XA = Xn
XA > Xn
(d) The information is insufficient to decide the
relation of xA. with x 8 .
40. A stone is released from an elevator going up with an
acceleration a. The acceleration of the stone after the
release is:
(a) a upward
(b) (g - a) upward
(c) (g - a) downward (d) g downward
41. A person standing near the edge of the top of a
building throws two balls A and B. The ball A is thrown
vertically upward and Bis thrown vertically downward
with the same speed. The ball A hits the ground with a
speed v A and the ball B hits the ground with a speed
v 8 • We have:
(a)
(b)
(c)
VA >Vn
VA<Vn
VA=Vn,
(d) The relation between v A and v 8 depends on
height of the building above the ground.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
109!
DESCRIPTION OF MOTION
42. A body traveling along a straight line traversed one
third of the total distance with a velocity 4m/s. The
remaining part of the distance was covered with a
velocity 2m/s for half the time & with velocity 6m/s
for the other half of time. The mean velocity averaged
over the whole time of motion is :
(a) 5m/s
(b) 4m/s
(c) 4.5m/s
(d) 3.5m/s
43. 'l\vo bullets are fired simultaneously, horizontally and
with different speeds from the same place. Which
bullet will hit the ground first?
(a) The faster one
(b) The slower one
(c) Both will reach sirnultanetmsly
(d) Depends on the masses
'l\vo
projectiles ,A and B are projected with angle of
44.
projection 15° for the projectile A and 45° for the
projectile B. If RA and RB be the horizontal range
for the two projectiles, then:
(a) RA < RB
(b) RA= RB
(c) RA> RB
\
(d) The information is insufficient to decide the
relation of RA with RB.
45. In the arrangement shown in
figure, the ends P and Q of an
inextensible string move
downwards with uniform
speed u Pulleys A and B are
fixed. The mass M moves
upwards with a speed :
(a) 2u cos0
(b) u/cos0
(d) u cos0
(c) 2u/cos0
46. The accelerations of a particle as seen from two frames
S1 and S 2 have equal magnitude:
(a) The frames must be at rest with respect to each
other.
(b) The frames may be moving with respect to each
other but neither should be accelerated with
respect to the other
(c) The acceleration of S2 with respect to S1 may
either be zero or 8mh 2 •
(d) The acceleration of S2 with respect to S1 may be
anything between zero and 8 m/ s2 • .
47. A train passes an observer standing on a platform. The
first carriage of the train passes the observer in time
t 1 = 1 sand the second carriage in t 2 = 1.5s. Find its
acceleration assuming it to be constant. The length of
each carriage is: l = 12 m.
2
(a) 3.3m/s 2
(b) -3.2m/s
2
(c) 24m/ s2
· (d) -24m/ s
48. The position vector of a particle varies with time as
-+
-+
~
= r 0 (1-atJt where r 0 is a constant vector & a
is a
positive constant then the distance covered during the
time interval in which particle returns to its initial
position is:
r
(a) r 0
Id.
(b) ro / 2a
(c) ~r;
+
~
(d)~
49. A point travelling along a straight line, traversed 1/3
of the distance with velocity v 0. The remaining part of
the distance was covered with veloc,ity v 1 for half time
and with velocity v 2 for the other half of the time.
Then the mean velocity of the ·point averaged over the
whole time of motion:
(a)
Vo+v1 +v2
v 1 +v 2 + 2v 0
(c)
3(v 1 + v 2)
v 1 +v 2 +v 0
(d)
3v 0(v 1 + v 2)
· v 1 +v 2 +4v 0
50. A point moves in zy-plane according to equation x =
at, y = at (l - bt) where a and b are positive constants
and tis time. The instant at which velocity vector is at
I 4 with acceleration vector is given by:
1t
(a) 1/a
(c) l/a + lib
(b) 1/b
(d) (a+ b)l(a 2 + b 2)
51. A particle starts from rest at A and moves with uniform
acceleration a m/ s2 in a straight line. After 1/a
seconds a second particle starts from A and moves
with uniform velocity u in the same line and same
direction. If u > 2m/s then during the entire motion
the second particle remains ahead of first particle for a
duration:
=--(a) 2 -Ju(u - 2)
(b) !!.-Ju(u- 2)
2
a
(d) None of these
(c) ~ -Ju(u - 2)
a
52. A particle is moving in x-y plane. At certain instant of
time, the components of its velocity and acceleration
2
are ·as follows. "x = 3m/s,v, = 4m/s,ax = 2 m/s
and ay = lmls 2 • The rate of change of speed at this
moment is:
(a) ..Jf.o m/ s 2
(c) 10~/s 2
www.puucho.com
(b) 4m/s
2
(d) 2m/s 2
Anurag Mishra Mechanics 1 with www.puucho.com
I 110
; MECHANICS-I
53. Two cars start off to race with velocity 4 rn/s and 2
rn/s &'travel in straight line with uniform acceleration
1 m/ s2 and 2 m/ s2 respectively. If they reach the final
point at the same instant, then the length of the path
is:
(a) 30 m
(b) 32 m
(c) 20 m
(d) 24 m
· 54. The instantaneous· velocity of a particle moving in
-->
A
A
:IJ(-plane is : V =(ay)i+(VoJj, where y is the
instantaneous y co-ordinate of th'c particle and V0 is a·
positive constant and a is a negative constant. If the
~ro-~:1cr·
, , 6B. ,~ 00
•
'
-->
respectively such that angle between V1 and line ABC
.
-->
and V 2 and ABC is 8. If point A and both the particles
· · are always in a straight line then :
(a) aV1 bV2
(b) avi2 = bV}
2
2
(c) a ½ = b V2
(d) aV2 = bV1
=
57. A point source of light is
(Source)ro
rotating in a horizontal
plane . at a speed of OJ
Jb.- .'.· '
.. ••r\J
radians/second. There is
,,.· .. :' d
- P.-···
:
a wall at a distance d
M11mnuin11uim11111n11/llii11111i1111111 N
from the source. At some
instant the focus of the light is at P and LSPN = 8 (see
figure). Speed of the focus at this instant in terms ofe
is :
(a) rod/ cos8
(c) rodtan8
(b) ( u - gt) downwards
. t .
(c) ( 2 u - gt) upwards··
t
(2u gt)
·
(d)
downwards
t
59. A block is kept on the floor of an elevator. The elevator
starts descending with an acceleration of 12 m / s 2 •
The displacement .of the block during 1st one second
with respect to elevator is: ·
(a) lm downwards
(b) lm upwards
(c) Sm downwards
(d) Zero meter.
60. A point moves rectilinearly. Its displacement x at time t
is given by x 2 = t 2 + 1. Its acceleration at time t is :
(a) 1
.
(b) .!_ _ _.!_
..
x
x3
t
(c)
(d)
- x2
55. An open lift is coming down from the top of a building
at a constant speed v = 10 rn/s. A boy standing on the
lift throws a stone vertically upwards at a speed of 30
rn/s w.r.t. himself. The time after which he will catch
the stone is : ·
· (a) 4 sec
(b) 6 sec
(c) 8 sec
(d) 10 set
56. Three points A,B, C are located in a straight line AB = a
and· AC' = b. Two particles start from points B and C ·
and move with· uniform velocities ½. and V2
x2
t2
x3
61. Two particles start moving from the same point along
the same straight line. The first moves with constant
velocity' v' and the second with constant acceleration
'a'. During the time that elapses before the second
catch the first, the greatest distance between the
particles is :
v2
(a) -
a
v2
(b)
2a
v2
2
(c)' 2v
(d)
a
4a
62. A ball is thrown up with a certain velocity at angle 8 to
the horizontal. The kinetic energy varies with height h
:,~
~,~
(c)~·
(d)~
·
63. A ball is thrown up with a certain velocity 'at an angle 8
to the horizontal. The graph between kinetic energy
: h=E:;~·-=.t2: .
horizontal
displatement
(b) rodjsin8 .
(d) ro a/sin 2 e
58. A body is thrown up from a lift with velocity u relative
to lift. If its time of flight with respect to lift is t then
acceleration of the lift is :
(a) (u - gt) upwards
t
I
~~
KE
(c)
www.puucho.com
.horizOntal displacement
, horizontal
-d_isplacement
(d)l~_I
displacement:
Anurag Mishra Mechanics 1 with www.puucho.com
-
j DESCRIPTION OF MOTION _____________ _
l~El / •
v2
v2
,_
~
. ' I'
65. The velocity of a
particle varies with
time as shown below.
The
distance
travelled
by
the
particle during t = 2s
andt=6sis:
(a)
~
~~hl•
!Cl
12m/s·
>
.
k
0
·
1ime !
. .
...,
,
Time ,
projectile in vector form is v = (6i + 2j) (the x-axis is
horizontal and y-axis is vertically upwards). The angle
of projection is: (g = 10m/s 2 )
(a) 45°
(b) 60°
(c) 30°
(d) tan-1 3/4
70. A point moves in x-y plane according to the law x = 4
sin 6t and y = 4(1- cos 6t). The distance traversed by
the particle in 4 seconds is: (x and y are in meters)
(a) 96 m
(b) 48 m
(c) 24 m
(d) 108 m
71. A swimmer crosses a flowing stream of width 'ro' to
and fro in time t 1 . The time taken to cover the same
distance up & down the stream is t 2 • If t 3 is the time
swimmer would take to swim a distance 2ro in still
r~·- -
water, then :
i_~
(c)
1 ffi
(d)
Displacement
69. At a height of 0.4 m from the ground, the velocity of
6s 1ime in second'
ili:;h
~- _
(d)
Displacement
:
2s
0= tan-1 2a
Circular
O
(c) . ~
Displacement
v2
(c)
:a51.Qml~
Time
(b)
:~
66. From a high tower at time t = 0, one stone is dropped
from rest and simultaneously another stone is
projected vertically up with an initial velocity. The
graph between distance between the particles and
:.:·
¾
Displacement
,
l
8= tan- 1
;KEl /
(b) (2n + 40) m
(d) 40 m
(a) 2n m
(c) 4n m
68. A particle moves with constant acceleration a in the
positive x-axis. At t = 0, the particle is at origin is at
rest, then correct graph between (velocity) 2 and
displacement is :
(b)
(d)
Time '
67. A particle moves with constant acceleration in the
positive x-axis. At t = 0, the particle is at origin and is
at rest, then correct graph between velocity and
displacement is :
= t 2t 3
(b)
tj = ti( 3
(d)
(a) tf
72. The trajectory of a particle is as shown here and its
trajectory follows the equation y = (x-1) 3 + 1. Find
co-ordinates of the point A on the curve such that
direction of instantaneous velocity at A is same as
direction of average velocity for the motion O to A:
y
, -----
(a) 'v~~me'nt
(c)
J. / .
'.~·
--- 7
111 '
64. A particle is thrown up with a certain velocity and at
an angle 0 with the horizontal. The variation of kinetic
energy with time is given by :
·
(a)
--
.... , . . . _ .. _ j
(d)
0
1
_Disp!acem~nt
(a) (3/2, 9/8)
(c) (3, 9)
www.puucho.com
X
(b) (2, 2)
(d) (5/2, 35/8)
Anurag Mishra Mechanics 1 with www.puucho.com
112
73. A bird flies for 4sec with a speed of It - 21 m/s in a
straight line, where t = time in seconds. It covers a
distance of :
(b) 4 m
(a) 2 m
(d) 8 m
(c) 6 m
2
74. A particle has an initial velocity of 9 m/s due east and
a constant acceleration of 2· m/s2 due west. The
distance covered by the particle in the fifth second of
its motion is :
(a) Zero
(bl 0.5 ni
(c) 2 m
(dl None
75. From the top of a tower, a stone is thrown up and
reaches the 'ground in time t 1. A second stone is
thrown down with the same speed and reaches the
ground in .time t 2. A third stone is releas~d from rest
and reaches 'the ground in time t 3 then :
(bl t3 = ~t1t2
(al t3 =.!ct1 +tz)
2
·w l=l-l ·
t3_
t2
t1
oo r:=~-r:
76. A hollow vertical cylinder of radius R and height h has
. smooth internal surface. A small particle is placed in
contact with the inner side of the upper rim at a point
P. It is given a horizontal speed v O tangential to rim. It
leaves 'the low~r rim at point Q, vertically below P. The
number of revolutions made by the_ particle will :
(al h
(bl ~
21tR
~2gh
(cl
?-:
(dl
along _the line y = x with such a speed that all the
three always stay in a straight line, then velocity of the
·third particles is:
·
,.;--;;:(bl V1 +V2
(al ;rV 1V2 ·
;:ii (t)
77. Two particles move in a uniform gravitational field
- with an acceleration g. At the initial moment the
particles were located at one point and move "1Yith
velocities v 1 =3.0 m/s and v 2 =4.0 m/s horizontally in
opposite directions. Find the distance between the
particles at the moment when their velocity vectors
. become mutually perpendicular: .
(al 5 m
.
(bl 7..J3 m
7
(cl ../3 m
(dl 7/2 m
5
78. A particle is projected vertically upwards from O with
velocity 'il and a second particle is projected at the
same instant from P (at a height h above Ol with
velocity 'v' at an angle of projection 8. The time when
the distance between them is minimum is :
(a)
h
(bl
h
2vsin9
2vcos9
(cl h/v
(d) h/2v
79. Three particles start from origin at the same time: one
with velocityv 1 along positive x-axis, the second along
the positive y-axis with a velocity v 2 and the- third
(cl
(d)
V1V2../z
~vf +v~
80. A particle is projected from the ground at an angle of
60° with horizontal at speed u = 20 m/s. The radius of
curvature of the path of the particle, when its velocity ·
makes an angle of 30° with horizontal is :
(g = 10 m/s2)
(a) 10.6 m
. (bl 12.8 m
(cl 15.4 m
(dl 24.2 m
81. Two particles are projected from the ground
simultaneously with speed 20m/s and 20/../3 m/s at
angle 30° and 60° with horizontal in · the. same
direction. The maximum distance between them till
both of them strike the ground is approximately:
(g = 10 m/s2l
(b) - 16.4 m
(al 23.1 m
(c) 30.2 m
(dl '10.4m
82. A rod of length I leans by its upper
end against a smooth vertical wall,
while its other end leans against the
I
floor. The end that leans against the
_,XI
wall moves uniformly downward.
Then:
(al The other end also moves uniformly
(bl The speed of other end goes on decreasing
(cl The speed of other end goes on increasing
(dl The speed of other end first decreases and then
increases
83. A body throws a ball upwards
with velocity v 0 = 20 m/s. The
wind imparts a horizontal
acceleration of 4 m/s2 to the
left. The angle 8 at which the
liall must be thrown so that the ball returns to the
boy's hand is (g = 10 m/ s 2 ) :
(al tan- 1 (1.2l
(bl tan "1 .(0.2)
1
(cl tan- (2l
(d) tan-1 (0.4l
84. Positio_n vector of a particle moving in zy-plane at time
f~~G,_·.
-_
i"'tv."·
',, ...
--->
A
A
tis r =a (1-cosootli+asinootj. The path of the
particle is :
(a) a circle of radius a and centre at (a, 0)
(bl a circle of radius a and centre at (0, 0l
(c) an ellipse
(dl neither a circle nor an ellipse.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
DESCRIPTION.OF·MOTION
- ,, .,_.,_
ss;
A particle moves in :,y-plane. The position vector of
--+
,,_
"
particle at anytimet is r ={(2t)i + (2t 2 )j}m. The rate
of change of 8 at time t = 2 second. (where 8 is the
angle which its velocity vector makes with positive
x-axis) is:
2
1
(a) - rad/s
(b) -rad/s
17
14
6
i(c) j rad/s
(d) - rad/s
7
5
86. Velocity versus displacement graph of a particle
moving in a straight line is shown in figure.
Corresponding acceleration versus velocity graph will
be:
,: 12') .-
,~2)·.
!10
·····,
l
(a) :
:
I
'
.
'10 ·-·--
(b) I · ·
.
<
· ___ _10 v(m/s)
2
i
!
I
I
I
i
110E (. m
· )_/.s.
CcJ
,
:
(
'
'
! _ .. !0 p(mls)j
(d)
:r;·2) .
'10
'
·_ - · - v(m/s)
10 v(rnls)'
..
'
I
. k.
1. Which of the following graphs cannot represent one
dimensional motion of a particle?
~:-
(a)
"
•
1;l·o · ---
7
.
(b)
~-
/l
I ·
1
L _______ . Time __ ~
(c)
J, (
L.~-~
1 (d)
lL
-
..lime
_____11
2. A lift of very broad - floor is moving vertically upward
with a constant retardation equal to 'g'. At an instant a
stone is projected from a point on the floor of the lift at
angle of elevation 9. Then the trajectory of the stone is:
(a) A parabola in the lift-frame
(b) A straight line in the lift-frame
(c) A parabola in the ground frame
(d) A straight line in the ground frame
3. An aeroplane flies along straight line from A to B and
backs again to the same point. There is a steady wind
speed v. The distance between A and B is l still air
speed of the aeroplane is V, then:
(a) Total time for the round trip, if the wind blows
along the line AB, is
2
Vl
2
V -v
(b) Total time for the round trip, if the wind blows
perpendicular to the line AB, is
.Jv
-v 2
(c) Total time for the round trip depends on the
direction of wind
(d) Total time for the round trip is independent of
direction of wind,
4. For a constant initial speed and for constant angle of
projection of a projectile the change dR in its
horizontal range R due to a change dg in value of
gravitational acceleration g is governed by the
relation:
dR dg
(b) dR = -dg
(a)-=R
g
R
g
(c) dR = dg
(d) dR = -dg
g
R
R
g
5. Two particles, 1 and 2, move with constant velocities
-;
-;
v 1 and v 2 , At the initial moment their radius vectors
-;
-;
are equal to r1 and r2 . How must these four vectors be
interrelated for the particles to collide?
--t
--t
r 1 -r2
(a)
-;
-;
lr1- r2I
-;
-;
r1-r2
(c)
2
21
2
-;
-;
lv2-v1I
www.puucho.com
--t
=
--+
v 1 -v 2
-;
-;
lv1-v2I
-;
--+
r 1 -r2
--t
--t
lr1-r2I
--t
-;
lr1-r2J
--t
= v 2 -v 1
--+
--+
lv2-V1I
-;
= V2-V1
-;
--t
(b)
(d) None of these
Anurag Mishra Mechanics 1 with www.puucho.com
,. ~.:"cl: pc!:! !
l~. .·: ·'~
, ·: ::<
hemispherical bowl. It
·1
passes the point A at t =
,:~:; , .,.· · -: ~ _.· '.: ':
O. At this instant of time, ·"·---~·-"·-- - ·---~-.:..:l
. the horizontal components of its velocity ate v. A bead
Q of the same mass as Pis ejected from A at t =O along
the horizontal string AB, with the speed v. Friction
between the bead and the ·string may be neglected. Let
t P and t Q be the respective time t_aken by P and Q to
reach the point B. Then :
(a)tp<tQ
(b)tp=tQ
(c) tp > tQ
(d)2- = length of atcACB
t Q length of cord AB
7. Two partides ·ate thrown from the same point in the
same vertical plane, as shown in figure simultaneously.
Then indicate the correct statements :
·r~-~ :-~---~:/·~---~: 'f-----~-'
~:<.:J
-·. \j
10. An aeroplane moving horizontally from west· to east
with some velocity and with an acceleration 5 m/s2
drops a food packet at some instant. Then:
(a) The path of the packet is parabolic with respect to
ground
(b) A person sitting on the aeroplane shall see the
packet is always vertically below the plane.
(c) With respect to plane the packet travels in a
straight line making an angle tan-1 (1/2) west of
vertical.
·
(d) With respect to plane the packet travels in a
straight line making an angle tan - I (1/2) east of
vertical.
(e) The packet moves in a parabolic path with respect
to aeroplane.
11. Two balls are thrown from an inclined plane at angle
of projection a with the plane, one up the incline and
other down the incline as shown in figure (I' stands for
total time of flight):
!-;~.~-~:.--· -, .. - -.-~ ·----~::·;-;·-·1
l·. ' 1/ A'. ,'t ~,c·,.'·JJ
· .. f' _
82 . 1 . ·
'· - •_ ••. · . ·
I-- ---,------·
/ --------~ i
t~--~-:--2-~:{:.~~L~J
(a) Tiine of flight for B is less than that of A
(b) Projectic:m ~peed of B·is greater than that of A
(c) Horizontal component of velocity for Bis greater
than that of A
(d) The vertical component of velocities of both ;\and
B are always equal throughout the duration for
"1hkh both the particles in air.
8. A particle of mass m moves on the x-axis as follows : it
starts 'from rest at t = O from the point x = 0, and
comes to re~t at t = 1 at the point x = 1. No other
information .is available about its motion at
intermediate time (O < t < 1). If a denotes the
acceleration of the particle, then:
(a) a cannot rem~in positive for all t in the interval
0 ;,, t 2' 1.
(b)
IcxJ cannot exceed 2 at any point in its path
Ial must be <C4atsome point or points in its path
(c)
(d) a must change sign during the motion, but no
other assertion can be made with the information
given.
··
9. The magn.itude of acceleration of a particle as seen by
observer A is am/s2and that observed by Bis b m/s2. If
m:agnitude of acceleration of A with respect to B is x
mls2 then indicate the correct statements is :
(a) la 2 -b 2 I :S x :S la-2 +b 21
(b) l<i-bl:Sx:Sla+bl
(c) la-bl<x<la+bl
(d) O:S x :S la-bl or·x;,, la+bl
(b) Ti
= T2 = 2v 0 sina
gcos8
(c) R 2 -R1 = g( sin8)T/
(d) v,2 = v,1
12. A particle moves in the zy-plane according to the law
x = asin(cot) and y = a(l-coscot) where 'a' and 'co'
are constants. Then the particle. follows :
(a) a parabolic path
(b} a straight line path, equally inclined to x- and
y-axis
(c) circular path
(d) a path such that distance moved by it is
proportional to time
13. Mark correct statements.
(a) Two particles thrown with same speed from the
same point at the same instant but at different
angles cannot collide in mid air.
(b) A body projected in uniform gravitational field
follows a parabolic path
(c) In projectile motion, velocity i~ never
perpendicular to the acceleration.
(d) A particle dropped from rest and blown over by a
horizontal wind with constant velocity traces a
parabolic path.
www.puucho.com
V
Anurag Mishra Mechanics 1 with www.puucho.com
e-
.t ,_,J
\·1.:.·-, :."\·
\·.:-ti.\::, 1 1.hr;,.r"t~l,;
cLuivc :.c-;:1.: c. t:~, . .
· erLfr.:il d:rer.:r::;r'~
,,'"
1
:
.( ... } 'i"
--
1··
14. An aeroplane at a constant speed releases a bomb, As
the bomb drops away from the aeroplane,
(a) It will always be vertically below the aeroplane
(b) It will always be vertically below the aeroplane
only if the aeroplane was flying horizontally
(c) It will always be·vertically- below the aeroplane
only if the aeroplane was flying at 11n angle of 45°
to the. horizontal- •. •...
·
. .
(d) It will gradually fall behind the aeroplane if the
aeroplane was flying horizontally
15. Two straight lines 11 and 12 cross each other at point P.
The line 11 is moving at a speed v 1 perpendicular to
itself & line 12 is moving at a speed v 2 in the similar
fashion. The speed of point P is :
• -- . ·~-
·,·· . ' -····· --···1
I
I
aE
!I '1B
(c)
,.a
~
!,..
.,
l2_ -
a
'
'
tl;ebr??.' . . 'Ir.~:
(b) ;,
I
'
-·
i_
- - --
:
17.. A ball is dropp;d_ fro~. C~rtain height on a horizontal
floor. The coefficient of restitution between the ball
and the floo~ is 1/2. T~e displ~cement time graph of
(a)
'
,_
I
I
•
L. _.:. ___ ½
_t
_
r
• i.
;
-·
--·-J
(a)
(b)
(c)
18. The speed-time graph of the ball in the above situation
is :
cosa
~v~ +v~ + 2v1v 2 cosa
sina
(v 1 +v 2 )+~v 1 v 2 cosa
(d)
cosa
16. The velocity-time graph of a particle moving along a
straight line is given as below. The displacement time
curve for the particle is given by :
.
.
-I
!
--------
________ ___
- ----
--- - - - - .
IE
'E
•
(a) ,
i
l Cl
I
.____ .. -1...
:
I
2
I
·I
i
_,
• s
.
J
__
I
:
J
:
10 1fm_e, l
l ... - - - ·..· - - - - ~ - ~ " - - - - - - ~
I
I
i~
I«
(b)
--·--'
..
:,~a'k-----,--.;:.•-7"'-,'-=o. 11me I
,_
I
-1
.· · . --,.
(--~··v·c:·
~' .
(a)
I
:
-·~
:'
•
' .-
'k·v·
·-· --;
'
. (b) !'
i
'
i
I
L..~: _ 1!
: : ''' .t'
---- - -
:L2''
:.- --_-,I
!· __ _; - .-· ti
(d) .
19. In a car race, car A takes time 't' less than the car B and
passes the finishing point with a velocity 'v' more than
the velocity with whicn the car B passes the point.
Assuming that the cars start from rest travels with
constant accelerations a1 and a 2 , then :
· (b) a1 < a
(a) a 1 > a 2 ·
2
(c) v = ~a1a2 t 0 •
(d) v =(a,+ a2 )t 0
20. Two particles are projected with
;
4m/s
1
speed 4 mis and 3 · m/s
simultaneously' ·from same point
' \ ._k.·3mlsl
.·
as shown in the figure: Then :
'
5 •
i
31•
(a) Their relative velocity is ·- ----------along vertical directiol}
(b) Their relative·acceler~ticni. is non-zero and it is
along vertical direction· ·,
(c) They wjll _lJ_ittn,;\ii~~ce ~imultaneously
(d) Their relat;i".~. velocicy is /constant and has
magnitude 1,4 m/Jk · , , ,
j
1
,·
I
I
1\
www.puucho.com
~--.
•J\-) \·:";,
\ . .,
\
"·:,
·-:
./'
:·r,;
·-, --...._~ ...
I
Anurag Mishra Mechanics 1 with www.puucho.com
21: The motion of a body falling from rest in a resisting
medium· is: describ~d by .the equation dv = A - Bv,
.
.
.
dt
where A and B are constants. Then :
(a) maxiriuun possible veiocity is NB mis
(b) initial acceleration is A m/s 2
24. A particle moves along x-axis with constant
acceleration and its x-positio11 depend on time 't' as
shown in the following graph (parabola); then in
interval O to 4 sec
!~-~1
.(c) ~elocicy a~:iiny time t is v = ; (1- e-B,)
_(d) velocity_ ~t_:ty time t is V = ; (1- e-At)
•
.
I
4 -t(sec)
22. vVhich ·of -the following statement is/are correct 7
(a) .Average speed of a particle in a given time period
· · is never ·1ess than magnitude of average velocity
....
(b) _it. is. possible
to have situations in which ~ * 0,
. ..
dt
't -
(a) relation between x x=t-t 2 /4
coordinate & . time is.
- -----.~-
(b) maximum x- coordinate is l m
(c) total distance traveled is 2 m
(d) average speed is 0.5 m/s
25. The velocity versus time of two particles moving along
x-axis varies as shown in the following two plots.·
Then:
·
--+
•
(~) it is pcissible"to"have situations in which d Iv I * o,
.\ _'·,;; : :
:-
,,., 4
dt
dv
but--= 0
. dt .·
(d} 'fhe'.averag~ velocity of a pa!'(icle is zero in a time
interval. !!' is possible that the instantaneous
. ve\ocity is' never zero in the interval.
'-.
. - .
23. A particle is moving with uniform acceleration along a
sfi'.aight liJ1~
Its speed at A and Bare 2 m/s and 14
ni,)'s. respectively. Then :
.
(a)'·, its spied ·at the mid-point of AB is 10 m/s
(b) its spe~d at a point P such that AP : PB = 1: 5 is 6
.
m/s. :: ,
.
.
.
Ali.
m / ~·-v
r
·:it,~:iJ,.
.t ~.mis --~-~
.
-----
2
;
4.
(a) maximum separation between the two particles is
2m
(b) maximum separation between the two particles is
2.5 m
(c) maximum separation between occurs after time ·
t =2 sec
(d) maximum separation between occurs after time
t =3 sec
,•:
(c) the :ti!Jle to go from A to the mid-point of AB is
·double of that to go from mid-point to B
. (d) . hone of µiese
,,·
,.
'
. ',;
__ , . 1:
.- ,:n!l-:r:
www.puucho.com
,
..,
..;
~
_
Anurag Mishra Mechanics 1 with www.puucho.com
... ,
_DE_SC_R_IPT_,IO_N_O_F_M_OT_IO_N_ _ - - - - - - - - - - - - ---- ·---·
1
r
IL.
_
PA s's'A G'E ,;
'
'
.• ----
3 __
Comprehension B~~~-~~~b_i~_-~_s_·_____
:. ' ' -
A swimmer wishes to .cross a river 500 m wide
flowing at a rate 'u'. His speed with respect to still
water is "v'. For this, he makes an angle 0 with the
perpendicular as shown ip. the. figur~. ..
': •.. _B ___ ----- - -I
Based on the above information, answer thel
I
following questions.
' - - - - - - ---•·•· -·--·-··•--..-···---~~=
·-
--
.. •
: v~ 0] .
, ....::s!
2.
3.
3
(d) x = - m
3
(a) 3~4 m
56
(c) -
3
(b)
4.
936 m
(d) 36 m
m
5.
5. Magnitude of the relative velocity of the two particles
when they meet for the first time is :
(a~ 16 rn/s
(b) 12 rn/s
(c) 20 rn/s
(d) 18 rn/s
6. Magnitude of the relative velocity of the two particles
when they meet for the second time is :
(b) 32 rn/s
(a) 16 rn/s
(c) 36 rn/s
(d) 28 rn/s
7. Variation of velocity uf the particle B with time is best
represented by :
ilLv--·---1
(a) :
I
~ ___ t:
i
i k : "tl;
(c) '
I.
•
L.'___ ,_____ . ..:J
(d)
i~" '
:
[_,
t
·-----~-----;
!d
=
sooml
I
1. To cross the river in minimum time, the value of 0
56
4, Total distance traveled by the particle B when it meets
the particle A for the second time is :
u
.. -'~---- -:_ ,__ ,..____: ---....c.'_______,
... I
1. Particle B will stop again at the position x equal to :
(a) 72 m
(bl 36 m
(c) 3 m
(d) 6 m
2. The two particles will meet twice in the due course of
their motion. The time interval between these two
successive meets will be :
(b} 4 sec
(a) 6 sec
(c) 2 sec
(d). 8 sec
3. Position where the two particles will meet for the
second time is given by :
128
(a) x = 72 m
(b) X = - m
= 36 m
J
PASSA'lfE
,,.,j '-'
A particle \4.' starts moving frorri point A with constant
velocity 4 rn/s along x-axis. Another particle 'B'
initially at rest starts moving along x-axis after (8/3),
sec after the start of A, with acceleration varying ~s,
1
a= 4 (3-t) rn/S 2 •
(c) x
'•
:, '" ~:,
._I
6.
7.
should be:
(a) 0°
(b) 90°
(c) 30°
(d) 60°
For u = 3 km/hr and v = ·5 km/hr, 'the time taken to
cross the river in minimum time will be :
(1,) 6 ·hr · ..
(a) 3 min
(c) 6 min
(d) 3 \1,r
For u = 3 km/hr and v = 5 km/hr, the swimmer :
(a) can reach to Bin 7.5 min
(b) can reach to B in 6 min .
(c) can reach to Bin less than 6 min
(d) can never reach to B
For u = 5 km/hr and v = 3 km/hr, the swimmer :
(a) can reach to Bin 7.5 min
(Ji) can reach to B in 6 min
(c) can reach to Bin less than 6 min
(d) can never reach to B
For u = 3 km/hr and v = 5 km/hr, the swimmer can
reach to B if e is :
(a) 37°
(b) 53°
(c) 60°
(d) can never reach to B
Foru = 4 km/hr and v = 2 km/hr, and to minimize the
drift, the swimmer must follow a path in which 0
should be:
(a) 30°
(b) 60°
(c) 0°
(d) 45°
For u = 2 km,,'hr and v = 4 km/hr, and to minimize the
drift, the swimmer must follow a path in which 0
should be:
(a) 30°
(b) 60°
(c) 0°
(d) 45°
www.puucho.com
ptu: s. -1..'1:..1 with www.puucho.com
Anurag Mishra Mechanics
• -, 1
~..,-:-:i ,~ 'l ~! '·{,.
·,·,-•v.iifl
,.-,·,~,::-:~_,.,_.,
!:•·).I
,,
.{,,, ;:-•:-;,,;·•:"'·,
.~ .
'' -
·, ~; {-.f.1if?J.';u_.l1 d.:~'lJ 1ic1r\1
,,;.
I
.1,
..... ,. -~-
.
----
'
'
- ••. ,1
l"" \
'
L...;..::,
....-.. .: ..;i
, •
L
,
~
0
•
1· :
',
• ' ~
·-·... .
,, '··
,
- ·--
"1
.,,
'
' ..
www.puucho.com
. .,..'
;,
j , l,
''
Anurag Mishra Mechanics 1 with www.puucho.com
-- ---7
~-DESCRIPTION OF MOTION
C-
-- ------ -- - - - -----
-
,- - - ---
-·---
119
--- . - .. - - -------:
· - - - - - - ----r,-..---..,.
c::· .: '··..: .: ',. ~- ' _.'.J \;·~~:;_
MATCHING TYPE PROBLEMS
-----~--~-~-~--~£-··; -~-·.
-,>.-(,
1
_:Column-2-.,:~:\',,bzt1§',
1. A dart gun is fired towards a Squirrel hanging from a
tree. Dart gun was initially directed towards Squirrel.
P is maximum height attained by dart in its flight.
Three different events can occur. (Assume Squirrel to
be a particle and there is no air resistance) .
.·~
-~·
Two projectiles are projected
from a height such that they
strike ground at the same time.
..· :
.... :
'
(B) :u 1 > u2;81 > 82 (Q)
v/:_f--::t.~"".'Tra)ectory of dart
0
J::-_..
rs"··· ...
d
Colurnn-1
'-'-''------- --------·
'
•
Two projectiles under standard'
ground to ground projection
such that horizontal range is
----------------~-
(A) Event-1 : Squirrel drops
itselfbefore the gun is fired.
(P)
same.
When dart is at P
Squirrel may be at
A
(B) Event-2: Squirrel drops '(Q) When dart is at P
itself at same time when the
-gun is fired.
Squirrel may be ·at
,B
Two swimmer starting from
'same point on a river bank such
that time of crossing is same. u1
and u 2 are velocities relative to
(CJ :Event-3: A strong wind imp- (R) In gravity free
arts same constant horizonspace dart will hit
tal acceleration to Squirrel
Squirrel.
and dart in addition to
gravitational acceleration.
Squirrel drops itself at the
same instant as the gun is
fired.
river.
(S)
(SJ :Dart cannot hit Squirrel in presence
of gravity.
2. Column-1
shows certain situations with certain
conditions and column-2 shows the parameters in
which situations of column-1 match. Which can be
possible combination.
www.puucho.com
Person moving downward along
slope in rain such that he
·observes rain vertically.
Anurag Mishra Mechanics 1 with www.puucho.com
--120
- - -- - 3. Figure shows a graph of position versus time graph for
-
y
a particle moving along x-axis.
Parabola
X
Straight
Line
t,
A
---!<'--•:x
Straight
. Line
'
a
(C) ,
Parabola
arabola
"
'\(Q) ,....+a= .:.acos0i+asin0j
a
'
Y
i (R) 'i:. = -asin0i- acos0j
_ _!U-"l---•x:
'
(A)
(B)
Slowing down
(P) t1 -->t2
(
y
(D)
;(S)
Returning towards origin (Q) t2-->t3
'
-
I
,ts--, t5
4. Trajectories are shown in figure for three kicked
footballs. Initial vertical and horizontal velocity
components are uy and ux respectively. Ignoring air
resistance, choose the correct statement from
column-2 for the value of variable in column-1.
6. Consider an object at point P along each trajectory
shown in column-1 in the direction of arrow shown.
Column-2 gives algebraic sign of v x, v Y, ax and ay-
(A) .y
~
> 0, Vy > 0, Clx > 0,_ ay < 0
,(Q)
:Vx
> 0, Vy= 0, a, > 0, ay < 0
'
'
'
I
l
...,1--,,~~-~-x·
;, _ Speed constant ,.
y
' (P) greatest for Aonly
p
i
(B) ,uy/ux
: (Q) greatest for C only
(C) ux
: (R) ,equal for A and B
'
, (S)
'equal for Band C
'
...,S_p_e-ed-is~in-c-,e-a~si-ng•x,
-
(C)
-
-.
y
A . _ w a vector 'i:. at angle 0 as shown in the figure
column-2. Show its unit vector representation.
p
--'---'-----•x
(A)
1
\Vx
I
(B)
(D) ux !l.y
',·(P)
I
p
.o.
(A) Time of flight
.
e
a
(S) t4 ->ts
'(t)
.
- - - - J . - -..x,
(C) Moving away from origin (R) t3-->t4
(D) Speeding up
_,
,a·= acos0i-asin0j
,
Speed is decreasing'
- ------ - - I
ia
'., (P) _, = asin0i. + acos0j•
' > O,vy >0,Cix > 0,ay>0
(S) .lvx
7. A particle is moving along a straight line. Its v-t graph
is as shown in figure. Point l, 2 and 3 marked on graph
are three different instants. Column-1 has fill in the
blanks, which are to be filled by the entries in
column-2.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
-
-
--
! DESCRIPTION OF MOTION
~---·----~--- -------~
i (S) :Magnitude of velocity
->
V
(D)
di rl
I
I
'
. dt
i (T) :None
--l-------r
10. For the velocity-time graph shown in figure, in a time
interval from t = 0 to t = 6s, match the following :
Column-1
.. -v(mls) .
(A) a1 is .......... a 2
(P) 'Parallel to
(B) v1 is .......... V2
(Q) Anti-parallel to
i0
0
(C)
I
: (R) ,Greater
•
V3lS .......... V1
than
(in
magnitude)
'.
(S) ,Less than (in magnitude)
(D) a1 is .......... v 1
(A) Change in velocity
8. Figure shows a cube of edge length a.
·y
Ht,1------,,G
!' (Q) .:_ 20 SI unit
(C) Total displacement
: (R) :- 10 SI unit
'
X
_,,;/ ..-.,
'
. ,' .:·9;
. ...,,,r; ·.:r.olumn-1
. -..i · .. . · Columr1,2)':
!-----·~---~-------~-------·~;....;.,;..,:.;:JL
(A) The angle between AF and x:CP)
-axis
·_ 5 SI unit
i1:~,z,:.,_ .£'."- ..'*t~}J.l
.· -~~9.lu"!n-2
60°
(A) ,M
'
'
(B) Angle between AF and DG : (Q) 'cos-1 _!.
3
(C) ,Angle between AE and AG , (R). 'cos-1
'
= 3 s I' (S)
s·
N
velocity is positive increasing. '
A -I when velocity is negative
and increasing. R when velocity
is positive and decreasing and
R- 1 when velocity is negative
··-·
and decreasing. Now match the following two tables
for the given s - t graph :
C
D
'
11. Let us call a motion, A when
"-'A'-_ _,__ _...J..CB'-
z
i- 5/3 SI unit
'
(B) !Average acceleration
(D) Acceleration at t
E ,----'f---1',F
; (P)
'
J_
f
i
(Q) IR-1
(B) ~N
I
(C) 'p
'(R) 1A
'.Q
I (S) iR
(D)
-.J3
! (P) iA-1
'
'
(S)
9. Match the following :
-:::-:-
_,
(A)
i (P)
Column-2
.
. '¼~;~
~-~~
Acceleration
I~
dt
->
(B)
:di vi
: (Q) 'Magnitude of acceleration
' dt
->
(C) dr
, (R) 'velocity
(C) Maximum height
, (R) '45°
(D) 1Horizontal range
I
,
I
I
'dt
www.puucho.com
;
I
' (S) tan-I
( 1)
2
Anurag Mishra Mechanics 1 with www.puucho.com
122
----~-~--..;....________.;.;.___
M_ECHANl(S'\J
= lp + 20 t
13. In the s-t equation (s
following:
·_~tf:c~_-r:,·
~:x_·'.'.?-ii:_<-~_~~~ ~i~_·ti_.,_,_-,f.
·:k-,~Jt}!n?n,;,•(1:±; -7:~- ·
- 5t 2 ) match the
- 'i
(B) !Dispiaceinent in ls
I
•
:' (Q) ilS
unit
i
-
I
1
i
i (R) 125 unit
! (S) ;--10 unit
(C) 1Initial acceleration
(D) !velocity at 4s
14. A particle is· rotating in a circle of radius lm with
constant speed 4 m/s. In time· ls, match the following
(in SI units) :
'it.
Column,1f,_
I
(A) !Displacement
'. (P)
. I
!s sin 2
!
I
I
J _-,
i
;
(B) Distance
! (Q)f4
(C) Average'velocity
, (R) •2sin2
'
I (S) 14 sin 2
(D) [Average acceleration_
I
(A) :·co~stant' positive accele-/ (P) speed may increase
1
ration ,r
'
'
(B) Constant. negative accele-1 (Q) !speed may ~ecrea~e ·
l
•
.
. . 1rat1on .
,
_
I
I
'
I
'
(C) !constant displacement ' , (R) ,speed is zero
(D) \constant slope of
a-ti
/graph
I
(S) Jspeed must increase
I
=
(T) 1speed must decrease
10
16. A balloon rises up with constant net acceleration of
m/ s2 • After 2s a particle drops from the balloon, After
further 2s match the following : (Take g = m/ s2 )
10
I
I
(Q)
'
(D) /A~celeration of particle
;::::.;;;;:;ut!.\.,.-.
ASSERTION ~r,,!!)"REAS.£>J!
-""-':.ifA,
Directions : Read the following questions and choose
(A) If both assertion and reason are true and the reason is
correct explanation of the assertion.
(B) If both assertion and reason are true, but reason is not
correct explanation of assertion.
(C) If assertion is true, but the reason is false,
(D) If assertion is false, but the reason is true.
(E) If both assertion and reason are false.
1. Assertion : A body can have acceleration even if its
velocity is zero at a given instant of time.
Reason : A body is momentarily at rest when it
reverses its direction of motion.
2. Assertion : A body having uniform speed is circular
path has a constant acceleration .
Reason : Direction of acceleration is always away
from the centre.
3. Assertion : The two bodies of masses M and
m(M > m) are allowed to fall from the same height if
the air resistance for each be the same then both the
bodies will reach the earth simultaneously.
Reason : For same air resistance, acceleration of both
the bodies will be same
4. Assertion : A body is momentarily at rest when it
reverses the direction.
Reason : A body cannot have acceleration if its
velocity is zero at a given instant of time.
5. Assertion : A particle in motion may not have
variable speed but constant velocity.
Reason : A particle in motion may not have non-zero
acceleration but constant velocity.
6. Assertion : A particle in .zy-plane is governed by
x = a sin rot and y =a-a cos rot, where a as well as ro
are constants then the particle will have parabolic
motion.
Reason : A particle under the influence of mutually
perpendicular velocities has parabolic motion.
(P) /Zero
(C) Displacement of particle
**h.
: 'l":r1,s. ~~'%
~_,: o umn""'
(A) ;Distance traveled in 3si (P) - 20 unit
.
6-;:::;-·
rn
~e. .
~~-·~~~·i~-i-;-:~·
.I
,._,.«:.;:;:,µ;;q,..w,::;>iim:s;.a:c.
/10 SI units
(R) 140 SI units·
I (S)
''
J20
SI units
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
\
[ DESCRIPTION OF MOTION
--
,-·-·--·
i
·123
···--.-:----·
--- ---- ·-··-----~~--------
----~- -
AN8WER9
- ----- ---- - -- - -- -----------------
-
-------- -··
=
7
··_J'
------·
i;;vel-1: §niy One Alternativ£is Correct.~
1.
(a)
I
9. : (c)
17,
(b)
25. ' (b)
10.
18.
26,
42.
49,
50.
57. ', (d)
65.
1
(b)
73.' (b)
81.' Cal
=-
3.
(b)
11.
(b)
I
(a)
I'
27. i
19.
(a)
'
''
58.
I
(b)
35,
43.
(b)
51.
''
(c)
4,
(c)
12.
20.
(b)
I (a)
5.
(!;,)
6.
(d)
7,
(b)
8.
(c)
I
(b)
13.
(d)
14.
(b)
15.
(d)
16.
(a)
(c)
21.
(b)
22. : (a)
23.
(d)
24
Cal
30. i (b)
31.
(c)
'
I 32.
(d) I
39.
(d)
46. : (d)
54. I' (a)
47.
(b)
''
(a)
63.
(a)
71.
I (a)
(d)
79,
(d)
I
I' 28. !' (c)
:
'
(c)
I·' (a)
(a,b,c,d)
7,
I
13.
19.
25.
:
I
2.
(d)
45.
(b)
(d)
53.
(d)
II (a)
(c)
61.
(b)
62.
69,
(c)
70.
i
76. I (d)
84. I Cal
, 77,
(c)
78.
Ca)
86. I Ca)
44,
I
(c)
52.
59. ; (b}
60.
,
' (b, c)
.I
I
!
I
I
68.
3.
i
I
(a, d)
14. I (a)
I
(a)
20. j (a, d)
: 21.' (a, b, c}
:
t' - -
I
I
Cb, al
I
I
' 85. :
I'
'
I'
''
i
55.
I
40,
I
I
I
I
(b)
I
(d)
I
(b)
56.
(b) '
(d)
(c)
64.
(c)
' 48.
72.
I
'
(~)
'
80. I, (c)
! '
9. I {b)
15. I' (c)
I
'
4,
(b)
10.
(a, c)
16.
Cc)
22.
I'
(a, b, d)
I
i
'I
I
5.
11.
'
'I·
1
I
l
(b)
(a, b,'c)
I
17,
'
23. \' (a; b, c).
'i
I
I
(c)
'
3. (b)
4. (a)
5, (b)
6. (c)
7, (c)
Passage-2:
1. (a)
2. (c)
3. (a)
4, (d)
5. (a)
6. (a)
7, (a)
Passage-3:
1. (a)
2. (c)
3. (b)
4, (d)
5. (b)
=sMatchhl!!~!Ype P_rob~-~~~
2. A-P, Q, R, S; B-Q, R, S; C-Q, R, S
4.A-R;B-P;C-Q;D-S
6. A-P; B-Q; C-R
8. A-R; B-Q; C-P
www.puucho.com
: ,6.
I
(a)
'
i
;
'' 12. i,I (c, d),
i
.I
I
I
I
I
2. (a)
A-P, Q, R, S; B-R; C-R
A-R; B-P, S, T; C-Q, R; D-P, S
A-S; B-P; C-Q; D-R
A-P, R; B-P, S; C-P, R; D-P
I
I
'
Passage-1:
1. (a)
1.
3.
5.
7.
I
I
-~
(a, b, c)
' 8. !' (a, c)
(b, c, d)
'
38. '' (b)
(b)
(c)
83 . . Cd)
'
37.
i
67. ' (a)
'
75. j. (b)
I
(c)
36.
L~vel-~: Mcir~!han o~~-Aii<:rn-at~'!e~ a_re Co~re
1.
29,
·I
I
74, J!' (b)
82. : (b)
I
(a)
I
(c)
66.
'
I
i'
34.
33. ! (d)
41. : (c)
(d)
'' (b)
2.
18.
(b) •
24.
(~,b,c,d).
'
I'
, II
I
I
I
Anurag Mishra Mechanics 1 with www.puucho.com
,,
· MECHANicf.f'\
8. (c)
For resultant to be zero the given magnitudes must
form a triangle. When lengths are 4, 8, 4 a triangle is
formed with height zero.
9. (c)
Resultant is inclined more towards vector of larger
magnitude .
nue. Take a hexagon whose all the
sides _are of unit length.
--,,
--,,
--+
--,,
--,,
--+
BA= BC+CD+DE+EF+FA
.-2;
(b)
'·
r. _ .False., When 0 = 2mt the vector remains the same
10.
,, 3, (a)·
. ,\, nue. Consider a regular n-gone
4. ·.(a)
True. Consider a non-regular n-gone.
(b)~i
and
11. (c)
5. ~)
--t
-+
"--t--t
-),
False. a x b is a vect_or quantity while a. b is a scalar
·,
-+
-+
--t
--t
quantity. Therefore ~ x b can never be same as a. b
=
6. (d)
-+
--t
--+
-+
=>
(A+B).(A-B)
cos 0 =--),- ---j,----t--t
IA+BI IA-Bl
-+
=
2
-+ 2
--j,
-+
--j,
7. (b)
-+
Bis
-+
--j,
,_, _,,
-+--t-+
-+"-+
--t'-+
I
-),
-+
-+
-+
-)
--j,
~ la/ +11ial +21a/llia/cos0 >la/+llial Squaring
both sides.
.,
-+
The magnitude of change in vector -+
_a is
2
2
-+-+
--t·~
=> 2 la/llialcos0 > 2la/llial.
=>COS0 > 1 =>0Eq,
lb:_~~ ~lbl 2 +1 ;l 2 +2lhl/_~ cos(it -0)
= /2asi~9/2/
I·
-+I
180°. Since 9 = 120°, /C_, /> I_,
/A/-IBI
I
-+
= a.J2c1 - cos0)
= a~4sin 2 0/2)
-+ --t
13. (d) Given a+li a/-1 a/ >Ill a/=> a+lia/>I al+llial
:. I a/=lbl= a and angle between a and bis 0
14.
~,1 . r.~-1
.,•
,._,·.
ICI> /1AI-I~
-+
-+
-),
-+
Let a is rotated through angle 0 to get b
;
-+
IIAI-IBll 2 +[A[IB[
this minimum is achieved when angle between A and
-+
0 = 7t/2
---t
--t-+
2
-+
LA+BIIA-BI
=>
--t
Aliter : The miniJllum value. of IC I is IAl-I Bl and
=0
IAl -IB1
--t
ICI= IAf+IBl +2IAIIBlcosl20°
www.puucho.com
:21_~"
vt -:~
south ~west _ .
l
s - ----- -- ---- -___ I
Anurag Mishra Mechanics 1 with www.puucho.com
l DESCRIPTION OF MOTION
Change in velocity = final velocity- initial velocity.
-+
I!,. V
A
,..
,..
= -vi-vj = -v(i+ j)
...,
20. (c)
[ I!. V[ = v..fi. towards south-west
=
~
··1
I
;
equator!
~-
..., ...,
-+--+
-+-+
2
-+-+
15. (d)''
'
2
I
[ __ -- . s _____ -
---+-+
. ·- _,. ,
0
2
21. (b)
As obvious from the figure
4
cos e = -
-
22.
8
'L_____. ....
B•
(a)
As shown in the figure
xsin9=8
16.
..., ..., ...,
Net displacement = St+ S 2+ S 3
= (20cos45° i + 20sin45° j) + (-20i)
+
(20cos45° i + 20sin45° j)
= (20,J2 - 20) i + 20,J2j
-+
...,
-+
, x cos·a
...,
...,
-+
-+
-
-+
A x Bis parallel or antiparallel to C. Hence C x (Ax B)
-+ -+
-+
...,
...,
the plane of B and C , it is perpendicular to A . But
-+
-+
-+
AxB. not necessan·1c.
---1s
y-, 1t may b e equal-C
to-.
-+
-+
-+
-+
JA x BJ
[Cj
[CJ
' '
Ji+ jJ2
")
1+ J
=Zc1+Ji
2
-+
= O. Also A .B etc = 0. Again B +c being a vector in
...,
.,
16-*· ·
.
_ (3i + 4j). (i + j) (:
-+
~
1.... ...8. ,... . .
JBJ2
17. (b)
-+
.xsin~:-s
..., ...,
= A.B 8
.
= 2oJs- 2..fi. m
-+
i
Component of A along .B is given by
[St+ S2+ S3J =. ~(20(..fi.-1)) 2 + (20,J2) 2
-+
·-
rjJ:
- ·-·..
xcose = 16-x
Solving, we get
X= 10
So the required combination is
lON, and 6N.
·
23. (d)
C
A
5
-+
2
+ (at. a2)- 2(at. ;i2)--: a2
-+
2
= 2at2-- +
(at. a2)- a2
= 2-1.1..!.-1 = l
i
~
(at- a2).(2a1+ a2) = 2at
'
'
:
·-, i>
Since Jat+ a2J = .,/3, and if _angle between them is 9,
then (../3) 2 = 1 2 + 1 2 + 2.1. leas 9i.e., e = 60°
=SM
70 km/hr towards south-west
_____ N_ _ _ · - ·
I~-'io
Component perpendicular to
the plane has magnitude 10
COS 30° = 5.J3
A
24. (d)
( ~A 2 +B 2
r
= (A+B) 2 +(A-B) 2
+2(A +B)(A '--;,ll)cose.
18. (b)
-+
-+
-+
-+
-+
art= 0, Vt= aAtanytimet, V2 = a+bt
-+
-+-+
-+
when Vt and V 2 are perpendicular Vt. V 2 = 0
~ ;,(;+ht)= 0
~t
19. (b)
As shown in the figure
.. ,
'·
25. (b)
As .shown in the figure, A
= -(~~:]
= B case
B
Required angle
www.puucho.com
.
J
= lt - e
= cos
i!iJR
....B
A
cos e = -
a.b
-1(- A)
B
....... ~··· ....
·.
!Brose .
''
'1."I
Anurag Mishra Mechanics 1 with www.puucho.com
j12s.
MECHANICS,(
26. (a)
As shown in the figure for.
angillar displacement 8 the
linear displacement AB is
equal co 2r sin (8/2) .
I
32. (b)
-->
-->
IAl=2, 1B1=2'12
--t --+
--+
--+
A.B=AXB
=>
8=45°
~
- - - ---+--+
---+--+
2+1B1 2+2A
IAl
.B
- = ,~~~~----+--+
--+--+
--+--+
--+--+
27. (c)
We know if a vector makes (al+ bj + ck) an angle a, J3
and 'Y with x; Y and Z respectively then
a
b
cosa. =
cosJ3 = --;=====
~a2 +b2 +c2
~a2 +b2 +c2
--;=====~=,
C
cosy= ,===ea~==
~a2 +62 +c2
2
So, sin a.+ sin 2 J3 + sin 2y =
1- cos 2a. + 1- cos 2J3 + 1- cos 2y
= 3 - (cos 2a. +cos 2J3 +cos 2y)
A+B
=..J5
33. (d)
Let angle between the two vectors be 8
3 2 =7 2 +4 2 +2x7x4cos8
cos8 = -1
=>
8 = 180°
Cross product will be zero.
34. (a)
--t
-->
•
--+
A
A
A
--t
A
P+Q =lli+9j
~
A
A
-->
--+
IQl= 5 .
--+
--+
--+
--+
AxB=12
A+B=7
A=4, B=·3
A=3,B=4
--+
Given A .l B (Le., component _of B along A is 0)
--> -->
A.B=8+24-4x=O
X=B
-->
... (i)
... (ii)
-->
:. Minimum resultant· is A- B (when· they are
antiparallel)
30. (b)
-->
BIIA
-->
--+
IA X Blmax= IAIIBI= 12
29. (a)
-->
--t ' .
Let vector be A and B
givenA+B=7
(when they are parallel resultant is maximum)
Q=4i+3j
--+
-->
Area of parallelogram = IA ,x BI = 5 units
35. (a)
•
P =7i+6j
--+.--+
-->
-->
IA-Bl min= 4 - 3 = 1 i.e., unit vector
-->
=>
B= kA
=> 61+ 16j:+-xfc = kC3i+ sj-2kJ
=>
=>
A
-->
28. (c)
Given
A
A= 2i+3j, B= i+4j
=·3-1=2
=>
IAl 2-t1B1 2-2A.B
A-B
36. (c)
Consider a hexagon with all
sides equal
k=2
X=-4
E
F1
A
F,: B
0
31. (c)
The displacement will be
maximum if he walks in
the way as shown after
walking
20
steps
displacement is sJz
:. He will walk 40 steps for
displacement 16../2 m
,I'
/4~_;.-,/
'..
,.,,/e_....-·/
('"
:.-···
IA..
··-,··..
(
Bm
8m
8 ni . .
/
Sm/
•
, ,10s!eps /
,
•
//
--+
--+
--+
--+
--+
-:7
(By polygon law)
Resultant of the five vectors F1 will in opposite sense of
F2.
.
Therefore resultant of all the given vectors
i.e. (SF1 and F2) will be F2 -F1 or F1 - F2
AB +BC+ CD+ DE+EF=AF
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
- - - - - - - - - - - - - _________ fu]
D.ESCRIPTION OF MOTION
=KIN§Atj~
=>
·21 dl = 2x dx + 2b db
dt
dt
dt
39. (d)
Let magnitude of acceleration be a. Also let west to
east be positive direction. Hence
=>
2l(u)
xA = Ju(l0)-½(a)(10)
2
= 2x(:) + 0
=(:)
I= JlOu - SOaJ
l
u
cos0
=-U=--
Xn = Jcu -10a)(l0)-½ (a)(10)
2
X
I= JlOu- lSOaJ
(lOu - 50a) = 11
Let
xA = l11I, Xn = 111 -lOOaJ
Hence x A may be less than equal to or greater than x B
depending on 11 and a.
[For example: If O < 100a < 211 then xA > Xn, if
100a = 211 then xA = Xn, if 100a > 211 then xA < xB ;
under the condition 11 is a positive quantity. However if 11
is a negative quantity then xA.< xB]
41. (c)
Let both the balls be thrown with speed v O and let
2gh and
height of the building be h. Hence vi =v
Alternative :
Let the veloc/ty of the block be v
upwards. Hence velocity of the
block along the string is v cos 0
and perpendicular to string v sin 0.
Hence
VCOS0 =U
46. (d)
_, _,
d
\
T
of
aP-
a, 2
= 4ft 2
... (ii)
_,
as,- a,1
= 4ft 1 -
I;,,-;,, I=
4ft 2
~4 2 + 4 2 + 2(4)(4) case
= 8 cos~
.
d
2
=>
Here 0 is the angle between ft 2 and-ft 1.
47. (b)
-+-
6
43. (c)
Since both have same initial vertical velocity (zero in
this case) and displacement along vertical axis is also
same for both when they strike the ground therefore
time of flight is same for both.
44. (d)
2 . 300
2 . 900
R - UA Sill
d R - UB Sill
A 2g
an B 2g
Hence RA and RB depends upon initial velocity of
projection which is not given Le. , information is
insufficient.
z2 = x2 + b2
... (i)
= (4-Jz).Jl + case
So average velocity=~ = ,4 m/s
45. (b)
= 4ft 1
_,
-x2+-x6=-=>T=2
2
3
6
12
a,1
where ft 1 and ft 2 are unit vectors.
Subtracting (ii) form (i)
d
2d
aP-
_, _,
m=--=-sec
3x4 12
Let body travels for next T sec then
T
_,
_, _,
v~ =v5-2gh =>vA =Vn,
law
u
= -COS0
Let a p, a,1 , a,2 be accelerations of the particle, fr3.!11e
S1 and frame S 2 with respect to ground. Hence
5-
[Note that v A = v B also follows from
conservation of mechanical energy]
42. (b)
Suppose the total distance be d.
Time taken for first d/3
=>V
;_u_ --------:_i
12 = u(l) +~(a)(1) 2 = u +~
2
2
... (i)
12 = (u+a)(¾) +½(a)(¾r
3u 21
=-+-a
2
8
a= - 3.2 m/s2
... (ii)
Solving
48. (b)
__, __,
_,
r = r 0 (t -at 2 ). At t = 0, r = 0. Hence the particle
returns back to initial position if
velocity of the particle = dr = r0 (1 - 2at)
dt
So, particle will come to rest when v = 0, i.e., after time
www.puucho.com
,,
Anurag Mishra Mechanics 1 with www.puucho.com
1
t=2a
· Positio11 of particle at this m9ment
= r0(
\,so distance travel
=>
¾[(u +~u 2 - 2u J-(u-~u
= 3.,Ju(u -
!-
=:
-2u )]
2)
a
a x (~)2 )
2
52. (d)
-+
= double of the above distance
= _:9_
·-:
'-+
'::
/2
2
V =Vxl+VyJ =>lvl=\/Vx +Vy
dl;I = ( 2vx ~+2vy~)
2a
dt
.,fv2+v2
~
y
X
"7
div I= Vxax +vyay
dt
fv2+v2
\I X
y
=3x2+4x1= 2 m/s 2
.JJ2 +42
53. (d)
50., (b)
,_,' (dxJ·'i (dyJ· '
v = dt
"7
s=4t+.!.(l)t 2 =2t+.!.(2)t 2
2
· 2
4t + 0.St 2 = 2t +t 2
''
+ dt j = ai + a(l- 2bt)j
'
Solving we get, t = 0 and t
'
A,= 0 i + (-2ab)j
...
,_
'
-
s=4x4+.!.(1)4 2 =24m
So,
-+
·· · Hence acceleration A is along
. ·, ·, .- ·,
··:'
.
,-;,
'negiitive y-axis. Hence when A
= 4s.
.
2
-54. (a)
"7
'
'
= (ay)i + (V0 )j
Vx = ay and VY =.V0
v
"7
,,and· v enclose it 14 between
them the velocity vector makes
·.angle ·It/ 4 with negative y-axis. Hence
dx
-=ayand
dt
dy V0
-=-=>·
dx ay ,
1 . .2
-ay =V0 x+c
· tan 2: =
a
=> [1- 2bt[ = 1
. · 4 1ac1-2bt)I
.
<.~-:' .
1
=> '
1- 2bt = ±1 => t = - or 0
b
But when t = 0 the y-component of velocity is along
positive y-axis, hence t = 0 rejected.
. 51. (c)
. Let at .~y time t the displacement of first particle b~
S; and that of second particle be S 2 •
2
·. S1 =½at and S 2 =u(t-~)
For required condition S2 > S1
1 2 =>t 2 --t+-<0
2u
2u ,
.=> u ( t--1 >-at
a
2
a
a2
J
dy.
dt
-=V0
f aydy= s·V dx
.
·
0
.
2
1 ay 2, = "v x,
0
(·: (0, 0) satisfies)
2
' . 21'
y=± __o_x
---!!,_,,
negative
-~2V
0x ·
y.-,
a
· Also for y to be real x must be negative.
55. (b)
0= 30t+.!.(-10)t 2 =>t = 6
2
=> ¾(u-~u -2u) < t < ¾(u+~u 2 ~2u)
2
Hence the duration for which particle 2 remains ahead
of particle 1
www.puucho.com
'
Anurag Mishra Mechanics 1 with www.puucho.com
_.-F:'o'ES(RiP.Tfo":q~LM9YI~'.,.,,...___.....,.
. · ~~-- .. , · <:~~~~-----··
.... _· ..
"
62. (a)
· 56. (d)
ABP and
triangles.
ACQ are similar
V2t
.
v;
r
= :!m((u cos0) 2 + (u sin0) 2 - 2gh)
V t .a
-1= -
Hence
~
KE= ½m( ~V} +
2
b
= :!m(u 2 -·2gh) = (-mg)h +:!mu 2
2
2
The graph will be straight line, which will retrace the
same graph after it reaches its maximum height. Also
kinetic energy is not zero at the highest point.
bV,,1 = aV2-
.. · 57. (d)
a(~-e)
OJ
·dt
Speed of focus
.
= ldxl = d[(d) cot0]
dt
dt
2
= l-dcosec 0
d0
dt
=--
63. (e)
KE= ½m( ~V} + VJ
r
= :!m[u 2 cos 2 e + (u sin0-gt) 2]
:~1
2
= :!m(u 2 + g 2t 2 -2ug sin0t)
OJd '
= idrocosec el = - ,
sin 2 0
2
2
2
2 2
= (½mg } - (mug sin0)t +½mu
58. (e)
But horizontal displacement x = (u cos0)t. Hence
-> 1 (->g-a->) t 2
0=Ut+
2
-> 1 (->g-a->) t
0=u+
1
KE=-mg
2
-> ->
->a= (gt+2u)
2
a = ( u ; gt ) upwards.
59. (b)
·_1/vith.,espect to: ~ievator the initial velocity of the block
i~ ·zero ·and the block stans accelerating upwards with
acceleration of 2 m/s 2. Hence
1 .
S = 0(1)+- x 2x 12 = lm upwards.
an
2
Let _x be:the distance between the particles after t sec.
Then
,.
1 2
x = vt--at
. dx
V
-=0 ~v-at=0ort=dt
·
a.
Substituting the value of x, we get
.
v2
X=-
2a
2
= :! m[(u cos0) 2 + (u sin0) 2 + g 2t 2 + 2ugt sin0]
2
1
.
KE= -m[u 2 + g 2t 2 + 2ugt sin0]
2
'
. 61. (b)
x to be maximum
2
KE= :!m( lv.2 + v.2 ) = :!m(V.2 + V.2)
2 ~ X
y
2
X
. y
2
2
= :! m[(u cos0) + (u sin0-gt) J
65. (b)
Distance traveled by a particle is equal to area under
speed-time curve. Hence
d = lOx 4+:!1t(2) 2 = (40+ 21t)m.
=t2+1~ dx =
t
.
dt ~t2+1
d2x
1
1
dt2 - (t2 + 1)3/2 = xs
For
2
Parabolic graph.
60. (a)
2
X
64. (e)
t
x2
2
) -(mgusm0)-.
•
X
1
--+-mu
u cos 0
ucos0 2
2
2
2
=( ~ 2 )x -(mgtan0)x+:!mu
2u cos 0
2
2
.
2(-2 - 2
2
.
66. (e)
At any time t the distance d between the
particle is :
2
2
. d = l.(h-½gt )-( h
-½gt
\~t
J[
=i(-u)tl =Ut
\:
Alternative : ·
,''-;._ ·
Let us take particle 1 us observer.. Hence till both the
particles are' in air the relative a~celeration is zero.
I
Also the relative velocity of particle 2 with respect to
particle 1 is u. Hence d "c ut
www.puucho.com
\\
Anurag Mishra Mechanics 1 with www.puucho.com
~;:;.
·\:;::~0:;:
,it ::S.~·
=~~--~z•/'~;,j:,
=> (2X1 + l)(X1 -1) 2 = 1
67. (a)
v 2 = 2as => v = ±,/2as => v = +,/2as
u; = 2
2
:. A is
(~,i)
d
d
..f0,
1
tan0=-=--=ux
6
../3
(dy)2
+ (dx)2 dt
dt
dt
= ~(24sin 6t) 2 + (24cos6t) 2 dt
= 24I: dt = 96m
71. (a)
'
the river velocity and u the velocity of the
Let v be
swimmer in still water. Then
t,
= 2(
ro
.Ju2-v2
ro
)
ro
1
4
Distance = )(dx) 2 + (dy) 2
=
--I',,
=I
(2-t)dt +
I:
(t - 2)dt = 4 metre.
2uro
Is I9-2t Idt
(9 - 2t )dt + Is (2t - 9)dt = I_ m
2
1..,1
V dt=
45
= 30°
0
I:
74. (b)
ux = Vx = 6ml s
so,
70. (a)
=I I-;; Idt =I: It - 21 dt
=
+ 2(10)(0.4) = 12
Uy
3) = 0
73. (b)
= ..f0.m/s
uy
=>
2xf - 3xf + 1 = 1 => xf (2x1 3
X1 = 2
=>
68. (c)
v~ = 0 2 +2.(a)s => v 2 = 2as.
69. (c)
Let u be the initial speed of the particle
v2=u2-2gh
U2 =v 2 +2gh
u; +u; =v; +v; +2gh Cvx =ux)
·
u y2 =v y2 +2gh
.
4
45
75. (b)
Let h be height of building. Hence
1 2
-h =ut 1 --gt 1
2
1 2
-h = Ut2 - -gt2
2
1 2
-h = --gt3
. 2
From (1) and (3) :
1 t2
g
-g2-=-u+-t 1
2 t,
2
Time taken for one complete rotation =
And It is obvious from the above that
2
t1 "'t2t3
Total time taken to reach the bottom =
'[d~]
cJx
= 3(x1
-1)
So, number of rotations =
,' ---·--" ·-·--·"
'
2
( · - - - y·---·
l
at
(xi,y 1)
But this tangent passes through origin. Hence
-Yi =-3x1Cx1 -1) 2 =>y 1 =3x1(x1 -1) 2
=> (x1 - 1) 3 + 1 = 3x1 (x1 -1) 2
2
rrR
t
Vo
~~
fg
·2rrR
77, (c)
Let the particles move perpendicular to each other at
time t.
X=~l
Hence equation of tangent
(y-y 1) = 3(x1 -1) 2 (x- x,)
... (3)
From (1) and (3) :
1 t2
g
-g2.=u+-t 2
2 t2
2
u
=>
... (2)
Adding above two questions :
76. (d)
t2=--+--=~-~
v+u u-v u 2 -v 2
2ro
t3=-
72. (a)
Clearly A is the point such that OA
is tangentto y = (x-1) 3 + 1 at the
point A. Let point A be (xi,Y1).
y = (x-1) 3 +1
... (1)
'
,'3ml
.! .-···
is
! '!
;
1
·.,•. . mis
.' I
. ·
h
·:\
i
i '-
xJ
'- •., - . Q_'~--•·--~ "
,.I
Hence
(4i - gtj).(-3] - gtj) = 0
www.puucho.com
-···7
Anurag Mishra Mechanics 1 with www.puucho.com
,- " {
DESCRIPTION OF MOTi_O~N_ _-~··_·_-'_'~·-·-~--~--:i,~·:_;', .. ·
-12+g2t 2 = 0
=>
=>
t = r12
·) => t =
vtjociJ
f 7_·-:.:::2,~;~·-··-'-,_,.,_"i'_'___
+.,J··---,_.-~------•--_-.>-'t"-3-'-"1~
1
82. (b)
.
,.
. !. ••.\ -~) :;
x2+y2=z2,
../3
s
dx
dy.
2x-+2y-. =O
dt
dt
dx = -yvy = -yvy
dt
X
~12 -y2
=>
=>
1:1 = J1
2·;~2 _
1
2
2
Smee y is decreasing ~ I / y - 1 is mcreasing
78. (d)
Relative acceleration between the
particles is zero. The· distance between
them at time t is
s=~r{h=-~(=v-_-v_s_m_0_)t-}~2-+_(_v_c_os_0_t~)
2
continuously.
83. (d) ·
The ball returns· back to,boy's. ·
,fZOm/s
hand only if the path of the I _
a_/
?~11- is a strai~ht liJlt, 1-le_nce , _IAm/s2
minal velocity and net 1- - - - - - •
acceleration must be albng' .- i:
.·
the same line. Hence
·
· : !· / 9
: , •• ·
10m/s2
tan0 = ..±. =>0 = tan-1 0.4. t~. - - - - - - - -
ds2
-=0
dt
2{h - (v - v sin0)t}(v si~0 - v) + 2v 2 cos 2 0t = O
h
t=2v
79. (d)
At time t the positions of the
particles are shown ' in the
figure.
Slope of AB = Slope of BC
v,t- v 3 t
0 _ v 3t
../2
--~_ v 3t
0
. 10
84. · (a)
_, - .l"
'
"
...
r = aO.-_msrotJ i-i;P sincotj
x = a(1- coscot) andy = a sin cot
(x- a):=, -a cos cot andy = a smcot
'
C
•
(x- a) 2 + y 2 =·a 2 '.
=-t - -~
=>
=>
=>
../2
t · ~.
85. (a)
../2
2:f
,
J.
y 2·2t 2
80. (c)
Let v be the velocity of the particle when it makes 30°
with the horizontal. Then
v cos 30° = u cos 60° => v = 20/ ..J3m/s
So,
j..
'
=>V
"'dy = 4t
· .•
-~dt
...._~· ::._.Y
,-,
\
- -·
'V
·,
4t
+-='--2 = 2t
·v;
'
Differentiating with respect to time we get,
d0 •· .(sec2 0)-/= 2
,__ dt ,, .. .
,.
2 ' d0
.
d0
=>.
(l+tan 0)-=2=>(1+4t 2)-=2
dt
-·
dt
, ._ ,,
. dB-;;· ... z' .'
=>
· ,r,·" ;. ·dt · • 1;+:4t 2
R
15.4m
81. (a)
Components of the velocities of both the particles m
vertical directions are equal. Therefore, their time of
flights are equal and their relative motion is in
horizontal direction only. Thus the maximum distance
between them is the difference between their
horizontal ranges.
,,
· - , · tan0 =
v2
gcos30°=v2
R=--g cos30°
:;
,,,_,.,
1·:;r dx
. ·x = 2t;=> V = - = 2
X '
dt
=>
Now
decreasing
continuously.
or s ={h-(v-vsm0)t} 2,+(vcos0t)2
s is minimum when
2
=>
is
Hence
2
-rad/s
p ,[::1:2 ~r+ ~2) 2 17
=>
f!6.
11
,
C~? ,:.. ~-
.,~ ·;.~ ·-:::
11
·,1·L
From V·S gti;'PP. '
"'
1
...:·nn
'.
1 ;__.
'
1 ... ,
.. .
'·duds
·• V =·S
l
www.puucho.com
.... l'. l11;_'
•
=> ....,.....-= - =>a= V
"dt dt
···~t''
l
Anurag Mishra Mechanics 1 with www.puucho.com
; MECHANIC5:~ .
--+
r1-r2
--+
'.!.· ca, b,
c, dJ
Graph (a) ~dicates two c\lsplacements at a given time,
'. · '
which is imposs)b!e. .
Gr~ph ·(b) 'iridj~tes. two velocities at a given time,
which is impossible.
<;rraph; ,(~) hi.di~i!'t~s speed can ·be negative, which is
_impossible.·_ . '
.Graph. (.d) _inµic~tes distance-travelled increases then
'ciecre's,.ses, whi~ 'is impossible,
2.
'o;, ~] :'' ' .
,.;. ;~ . <...:,. . • '
•
'
'
t~¥,~rd
·:.'
.
l
l
21V
T=tAB +tBA = - - + - - = ~
-. .
V + V V - V v2 - v 2
(ii),):.eft,· >}\7\nd,,:. is
blowing
perpen#i!;U[ilr to AB. Hence
ii=Vsin0=>sin0=~
V
.
l
tAi, =-.-·'
Vcose
'
r-·,,··-:~
Alternative :
Let us assume that the· reference
frame is rigidly fixed with particle
1. Hence
; windv cos
;,A
8
Bl
I
LY§in~Q-~·-'
l
.tBA =-;====
.Jv2 -v2
'
'
21
T=tAB +tBA =-;====
.
.Jv2 -v2
Hence·
-,
Vz
w.r.t.l
-, -,
= Vz - V1
4
--+
--+
--+
Cr2 -r1 )
_
--t
--+
--+
6. [a]
Since Q moves along a smooth
;
'""ijJ.';--. ----·:--· '
horizontal rod its velocity ,
i -•• - ~ - ~
.
• -- ------>vsin B
.
.
remains constant. But as P '.•
.·'
moves downwards its speed l . ••
increases.
Therefore
its __v.cos.B_! _____ _J
horizontal component of
velocity v sine increases and becomes maximum at
lowest point. Afterwards it decreases gradually &
becomes minimum at B; but at B, the. ho.rizontal
component of velocity is equal to that at A. f!ence
horizontal component of velocity of P, is never less
than velocity of Q. Since horizontal ·displacements of
both are same, therefore, P takes less time o'r t p < t Q.
Hence (a) is correct. ·
.,.
7. [b, c, d]
Both A ·and B have same hmax· Hence
(uA siir0A)
2g
2
= (uB sin0B)
2
' 2g
=>
UA sine A = UB sjn0B => (Uy)A = (uyh
Hence option 'd' is correct.
Again time of flight
,
= 2 ( vertical velocity of projection)
2
dR=-(u s:20)(~)=>~=-~
g
[b]
If the partic;les collide at time t then
i)+v; t =i-'2+1½ t => (ri-r;)= (v-;-v;}
--t
(v 2 -v 1 )
/v 2 -vd
/r2 -'r1 /
R = l!2 sin20 => dR =·(u2 sin20)(-l)dg
'
g
g2
s.
-,
Fpf the particles to collide with each other, the
particle 2 must be moving towards particle 1. Hence
4;. [b]
=>
-,
-,
r2 w.r.t.1 = r2 - rl
!
.Jv2 -v2
~. V·I-~2
....
v2
.
1
I 8"· ..... .
l
Similarly . · .
ii'2- ii', 1
11, - 121
Tii~:lift is /ICC~lerating downwards with acceleration g.
. · !ieic~. a~~~ler'!,tion of sto~e in lift frame is g - g = 0.
3. [a,'~, ·er .
.
.
· (i)
is blowing alongfIB. Hence total time T for
the f<;!\l!ld h:iP is
--+
--+
= Vz-V1
... (i)
Hence time of flight of A is equal to that of B. Hence 'a'
is wrong .
Since range of A is less than that of B and time of flight
of A and B are equal, therefore
(ux)A < (uxh·
... (ii)
Hence 'c' is correct.
Speed of projection= ~Cux ) 2 + (Iiy) 2
Since uy is same for both and (ux)A < (ux)B,
therefore, speed of projection of A is less than that of
B. Hence 1b' is correct.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
~ DESCRIPTION OF MOTION
8. [a, c]
Since the particle starts from rest and finally comes to
rest therefore the particle first accelerates and then
retards. If it is assumed that the particle accelerates
uniformly only once and retards only once in the
entire journey then the velocity-time curve is two
straight lines forming a triangle with time axis. Also
area of each of these triangles is one unit since
displacement of the particle is 1 m. Hence in each of
the
above
,- - ____,_ -- - -- --~- - -·- '
mentioned
!
motions
the
I
maximum velocity
of the particle is 2
m/s. Again, if it is
assumed that the
1s A
particle
accelerates for.! s and retards for .!s then the v-t curve
2
2
is represented by the triangle OQA and magnitude of
slope of the lines OQ and QA are both 4. That is the
acceleration and retardation are both of magnitude
4 m/ s2 • Also from the figure giyen below it is obvious
that if acceleration has magnitude less th~n 4 m/ s2
then retardation has magnitude greater than 4 m/ s2
and vice-versa.
Again if acceleration is not uniform · still then the
magnitude of acceleration or retardation of the
particle has to be
Q
greater than or
equal to 4 m/ s2
tane =4
at some points in
the path. One of
such
possible
motion is shown
in the given below.
Note that it is not necessary that v max should be 2 m/ s .
The only essential condition is area under v-t curve
should be 1 unit.
9. [b]
Let accelerations of the particle, observer A and
observer B be
p, A, Bm/s
2
with respect to ground.
Hence as per question
=>
X = -Jb 2 + a.2 + 2ab COS0
=>
la-bl:,x:,Ja+bJ
.
Hence 8 is the angle betwe~n ft 2 &·-ft,.
10. [a, c]
If the velocity of 1 ·
·
pacl<et
!
I
aeroplane is u m/ s when
g
.
the packet is dropped
then path of packet is IWest•-····:.
ground- ···>Eastj
_
parabolic with respect to
ground as shown in figure. .
2
With
. respect
to
4m~s
. ···-·>4m/sZ-7
l
2 .
aeroplane the initial 1',· ..
••• O~s
\
•
I
velocity of the packet is I'
1West
c-'-----·-----······>East
,- .
gro~risJ._ _ ___,
zero and acceleration is
as shown in figure.
-1 5
-1 1
. a]'.
west..of vernc
8 = tan - = tan 10
2
11. [a, b, c]
2
h
_ (v 0 sina)
h2max => (a) is correct
lmax 2g cos8
rs
1
... (i)
p-B= bn 2
... (ii)
2v 0 sin a
(b . · ·
T1 = - ~
- - = T2 => ) 1s c.orrect
g case
.
R, =(VoCOSa)T, _.!gsiilBT,~
2 '
R2
=. (v 0 cosa)T2 + .!2 g sinBT,,2
(R 2 -R1 }= g sinBT,2
=> (c) is correct
v ,, & v ,2 are the velocities of the particles at their
maximum heights. Let the particles reach. their
maximum heights at time t 1 and t 2 respectively. Hence
0 = (v 0 sin a)- (g cos8)t 1
v 0 sincx
t, = ~
-gcos8
v sincx
·
·
t2 = 0
. Hencet 2 =t1
Similarly
gcos8
Hence
v,, =v 0 cosa+(gsin8)t1
v,2 = v 0 cosa+ (g sin8)t
2
.
·
vt1 -:t=vt2·
12. [c, d]
•
·Hence ft 1 and ft 2 ate unit vectors depending upon
direction of acceleration of the particle with respect to
respective observers.
Subtracting (ii) and (i)
I
rs, . ,
=>
...,p-A
...,
= an
A-B=bn 2 -aft 1
JA-BJ= Ibft~ +ac-11 1 )[°
=>
smrot
=>
x2+(y-a)2
=-Xa and cosrot
=1..-· ·a-y
·=a2,
which is a circle. Hence (c) is-correct.
www.puucho.com
Ul-'·
Anurag Mishra Mechanics
1 with www.puucho.com
, '7._,>;;:,~ (n .•
1
,•,·. c1.1
1
"
-,ci
L..,.,,.
~fJ' \'
·c1x ., '
dy
.
vx = - .. = arocosrot and vy = -.-·=
arosmrot
1
dt.
~<--;l?-:-,1
dt
.
·
I 2
2
··
v=vvx.+-py =aro
Hen~~ particl~ 'is in~~ 'b'i(/i\cirtular path with
constant speed am. -Hence distance. travelled' by it on
circular path _in time 'i is"a,~t; _Heft~_e.,(d) is· con:ect.
13. [a, d]
_
: f'='i--..:_Ji/1~t•'"',;:-i-1 ·-,\·:, .
(a} is correct because
.
. , particles
. . ', .. - have a non-zero
relative velocity (always) and acceleration .
iela_tive lo one
is 1Zerci (both are falling
withg).
!--) T·;;
(b) is wrong because if a body is psqjected vertically,
it wil\ not follow•pataboliclpat4,
(cl is wrong because at higl!e~t point ,of projectile
velocity is perp~~<liC1!1.\lr t_q,a~celeration.
(d) is correct since particle has, uniform horizontal
velocity._and constant'<i\cce\eration (down~ard)
Le., it.will trace.a par~bolic<paJ]i. .
·
14. [a]
·'-'(: ,:: :1;,P.q ::~Jr"l,, '
Only (a) is correct because aeroplane wilf'provide
same velocity with which it is flyingand·in·the same
direction of its flight. c
15. [c]
:nu ,,., -:i " ·
· At time t, position of.lines are shown in the diagram.
\
anoth~r
.
·--·
r"-;-..--~il;,''",_,i'i'~',;>-=.-~-:-.,
';·~:·:~_~N:;:A
···v2t
!.1 - , :
, , __
•.
.~~~·:1_ :--~~~- ~
',F~t, '
.
t
;··1
:_::·__ - - · ~ - " ; . } : ·
1
: .....
. a_ ·z::::-t:::<.t.-i;.: ...: . a ·,>"ir,'',.,' ',· >·
· i- ,__·-.{ .....tv 1_t
,, •
·-•• -:."9: . ~: ·......
<
I
•
'
L~
· ·:~.2:~vt-~:. ~~--:t,'--vt··
f_~-J~:~1~ _j·
··
PA= -
2
-
· sin a'
PP'= ~~A 2
PB'= -
1
-·!'Hence
+:P~;:+ 2PA.PBcosa-;
= ~v; +v~ t2111v, 2 coscx(~~J
·
Velocity of point
s1na)
I
•
· dv
t:1~2~v>0and-·· <0
. dt
2
ds
,
d s
.=>
- > 0 and - 2 < 0
dt
dt
=> s-t curve is increasing _and lies below its tangent.
·· ,
'
P' = p_p
t .
~rv~f_+_V~~-+-2-'-V-1_V_2_C_O_S_CY.
sinu
·
dv
ds < 0 and d2s > 0
dt
dt 2
=> s-t curve is decreasing and lies above its tangent.
' · ·
·,
.
dv
t:9~. 10~ V > 0 and->
0
. dt
ds
d 2s :
->0and->0
dt
dt 2
=> s-t curve is increasing and lies above its tangent.
17. [c]
The ball will stop after a long time. The '.final
displacement of the ball will be equal to, .the height.
The motion is first accelerated, then retarded, then
accelerated and so o~.
=>
The velocity of the particle first increases linearly and
then at the point of collision it suddenly changes its
direction and then starts decreasing·in magnitude and
the pro~ess is repeated again and again. Also every
collision decreases the speed to half its value before
collision. Hence graph given in option 'a' is v-t curve
and th.at given in 'b' is speed,time curve.
19. [a]
v 1 =a2 (t:t-t 1 ) (forcarB)
V +v
a t
Vi
a2 t + t 1
a1 .. ·(V+v 1)(t+t1 )
-= > 1 ==>a1 >a2
1 =1- -1---
vfinal
=s •
U2
For case B
=V1,T=t1 +t, distance=s
(V+v 1 ) = a1t 1 ·
(forcaseA)
. ,vfinal
dv
t:5~ 9~v < 0and- < 0
.
dt
16. [c]
For case A we C'l,Il write . , ,
=v.+v1
T = t 1 , distance
.
. t:2 ~ 5 ~ V < 0 and - < 0
'
dt
ds · · d 2 s
-<0and-<0
dt
dt 2
=> s-t curve is decreasing and lies below its tangent
18. [b]
.. sin ex
r·. - ,• ._-:-~
dv
t:0~ 1~ v > 0and-= 0
'
'
dt
ds
· d 2s
·
->0and-=0
dt
dt 2
=> s-t curve increasing and a straight line.
.
v1
20. [a, d]
So, velocity of first particle
' ..:.::' '
www.puucho.com
t1
Anurag Mishra Mechanics 1 with www.puucho.com
I DESCRIPTION OF MOTION
,,. ., ,~ . '
23. [a, b, c]
zero.
.
1 .
el oCity=---=-=.
.
16 9 7 14 m/s
Therrreanvev
5
5
21. [a, b, c]
dv
a=-=A-Bv
dt
"""7 max. possible velocity is terminal velocity (i.e.,
when a= 0)
=>
A-Bv=O
"""7 initial acc. is when t = 0, u = 0
a=A-O=Am/s 2
f
f,
dv
dv- = dt
-=A-Bv
=> v- dt
OA-Bv
o
..!1nA-Bv =-t =>1-.!l.v=e-"'
B
A
A
A(l
-Bt)-e
-v
B
22, [a, b, d] ·
v 2 =2 2 +2xax~
2
=> v2
= 4+ 142 -
v 2 =4+
v = 10 m/sec"""7 if AP=.! AP=~
·
PB S
6
Let velocity at P is v 1
2
2
'
d .
142 -2 2
v 1 =2 +2xax-=4+---
6
Let time taken to reach mid-point from A is t 1 , and t 2
be time taken to reach B from mid-point.
6= 2+at1
••• (i)
14 = 6 + at 2
.:. (ii)
t
4 1
·
...!. = - = - => t 2 = 2t,
t, _8 2
24. [a, b, c, d]
Since the graph is like a· parabola
:. let x(t) =At+ Bt 2 + C
(dx)
dt
Put in (i), we get
_,
dj vi = tangential acceleration
dt
_,
dv
' dt
*0
6
=> v 1 = 6m/sec
nme
=0
192
=100
2
x(4) = 0=> 16B+4A = 0
distance > displacement
:. Average speed > Average velocity
dt
"""7 In uniform circular motion
22
2
From graph x(O) = O => C = O
x(t) = Bt 2 + At
Total distance
Total time
.
displacement
Average velocity = - ~ . - - -
Average speed =
d!vl
dt
I
j
14 in/secl
142 =2 2 +2xaxd
at mid0 point let velocity is v
So, relative horizontal velocity is zero. So their relative
velocity is vertical only. Since both particles are
moving under gravity, so their relative acceleration is
_,
.1
~ - " - - ~ -B
IJ. 'A- - -'p
=>
_,
13sl
~sec
= s 1 +sJ
dj vi = net acceleration
r
- •.·' d '
= 3cos30°1+3sin30°j
12: 9:
=-1+-J
5
5
velocity of second particle
= 4cos 53° i + 4sin 53° j
12: 16:
5
t-,.
[Q]
"""7 In circular motion from pt. A to Pt. A again
Average velocity = 0
(at any time)
lnstaneously velocity ;e 0
o
... (i)
=1 =>(A+ 2Bt),=0 =1
'
1
4
B =-- -
t2
X=t-4
max. x coordinate·= 1 (from max. and min._)
"""7 Since motion is a straight line motion
"""7 total distance traveled = 2 x 1 = 2m(
.
2
.
Average speed= - = O.Sm/sec
4
25. [b, d]
Separation between. them will be maximum when
both particles have same velocity. This situation come
at t = 2 sec, but just after it, first particle comes to rest
and second 1 m/s. So first particle will again gainthis
velocity in next one second. So, maximum separation
will ocCIIr after 3 seconds.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
j'---1 c.:3~6-·-------~---~_;_________ ~._ __,___ __o..',.,_.-...;·.__..;___._··...:.M_EC'--H-'-A.,__Nl~Cs_:1..:J! .
Maximum separation
= Displacement of seconds particle - Displacement
of first particle in first 3 seconds
= (2x 2+ lx 1)-(.!. x 2x 2+.!.x1 x 1)
·
=5 -
2.5
2
2
= 2.5 m
• . ·&;e1.'i;:c;;;;"'eh=e=n~""i;-n""'"ii=--se"'"d""''i,-;-;;-b""j;=;"'~,;------.
lll,::e::--z:r-,rc::::::e:r::!::'t·-n·
•
4
c;-=ur
~ j@&s,
Passage-1
·
dv
= 96-128
=-32
Relative velocity = 4 - (-32)
= 36 mjsec
7. [cl
2
V = 12t-2t
a= 4(3-t) =>- = 4(3-t)
dt
=>
f:dv= J;4(3-t)dt
=>
v=l2t-2t 2
2t 3
2
x= 6t - -
=>
5. [bl
JA ~ 4 ,m/sec .\
At t=2sec
(from Q. 2)
[!~16m/~cj
They meet for 1st time
v(2)=24-8=16
Relative velocity = 12 m/sec
6. [cl
At t = 8 they meet for 2nd f
A.,_-.,:4m(sec;
time
!32.m/sec
I
B
1
1
v(8) = 12 X 8 - 2 X 8 2
3
= 18 -
1. [al
For particle B to stop v = 0
=> 12t - 2t 2 = 0 => t = 0,6
2(t 2 - 6t + 9)
Passage-2
1. [al
2x 63
x(6) = 6x 6 2 - - -
500
t=-vcos8
3
5
=63(1-¾)=6: =2~6=72m
t mm
· = 00 for 8 = 0°
V
2. [al -
For particleBx(t) = 6t 2
2t 3
3
For particle A x 1 (t) = 4(t +
2
For u = 3 km/hr, v = 5 km/hr
500
t . = - - ' - - 360 sec = 6 min.
mm
5x1000
J)
For particles to meet x(t) = x 1 (t)
6t
2. [cl
--
3
3. [alu = 3 km/hr, v = 5km/h
to reach exactly Pt. B
vsin8 = u => 5sin8 = 3
. 8 3
Slll = -
8)
2t = 4 ( t +3
-3
t=8,t=2
time interval = 8 - 2 = 6 sec
5
I .
~¥-·- --- - -··· --
5
= 0.5 hr = 60 X 5
4
4
.
t =30
- =75
. mm
4
V
= 12t- 2t 2
4. [dl
x(6)=72m
x( 8) = 128
U
Total distance;;,'d + d1 = 72 + ( 72 304
=-m
3
= 5 km/hr, V = 3 km/hr
again to reach Pt. B
vsin8 = u => 3sine = 5
3
1 8
~ )
sin 8 =
~ not possible
3
:. The swimmer can never reach to Pt. B
www.puucho.com
··.··11
~ ·, .· :,
500
0.5
t=--=-vcos8 5 xj
3. [bl
Required position
x(t) at t = 8 or x 1 (t) at t = 8
1 8
=4(8+¾)= ~ m
4. [al
For particle B
GZJ,
r-~
-
....,.
,·
. .
Anurag Mishra Mechanics 1 with www.puucho.com
[_•._DE_SC_R~IP.J~l~~-O~F_'~~O~T!~O_N_._ _
~--·-·-----------------,____-·_··~~--,3IJ
5. [a]
From question 3 required angle '0' is
. 0 =3
sm
4. [d]
At the top of trajectory speed = u cos0
= 5../s m/sec
5
i.e.
5. [b]
u 2 sin20
Range=--g
0 = 37°
6. [a]
U
= 4 km/hr,
V
= 2 km/hr
=
d
vsin0
t=--
drift
vsin0
l
=> Speed time curve will be
,- . -----7
~
!·~
-
----·
v = -J250 = 5-Jio m/sec
usin0
--
r··--·1
t
t /
!
i~~~. '---~~-·
-----~_:'.'.~_~j
!
_,s-[fo "
u 2 sin 2 0- 2x !Ox 12.5 = (5-J5) 2
u sin0 = .J125 + 250 = .J375
and
u cos0 = 5../s = -/125
tan0 = ~375 = F3
.
125
-t 4 )
'
(B) (x) ( : ) < 0 to return (P,S, T).
(C)
x(:)
> 0, i.e., in (Q, R)
(C)
Slope of v 1 is+ve }
Slopeofv 3 is-ve
anti-parallel
(D) Slope at sis +ve
a1 > 0 :. parallel
V1 > 0
a1 and v 1 cannot be compared.
-+AA-+
16. AF=ai+aj+ak
AA
DG=ai+aj-ak
b= i
(A)AF · b = AFcosa
1
cos a= F3
=> a=..ffacosa
0 = 60°
->
. ->
(B) DG ·AF= (AF) (DG) cosp
3. [b]
· and
3
(D)Speed in increasing in (t 1 -t 2 )(t 4 -t 5 )
15. (A) Slope of a1 and a 2 is +ve
parallel
Slope of a1 > slope of a 2
R
(B) Both v 1 and v 2 are +ve
parallel
v 1 < v 2 (obviously)
PQ = .J15 2 + 20 2 = 25
15'[5-·---··-- ·'"
4
· - - _:,______:__i
'-··
v 2 sin 2 900 v 2
25=----=g
10
2. [c]
l~Ol
·
I 1,i,(1 1,; I,, i speed decreasing in·(t
------ u=2I
Let speed at P = v
Range
25.)3 m
n/C',u,.l
From this we get sin 0 = .!. => 0 = 30°
.
lOOx sxF3
2xl0
11. (A) Corresponding w graph will
x=(4-2cos0)x~
vsin0
.
.
.
For x to b e maximum or m1n1mum -dx = 0
d0
Passage-3
1. [a]
10
Matching Type ·Pr~em?i;;--:--_,
= (u - v cos0) x _d_
7. [a]
Here
v>u
Minimum drift = 0
=>
4sin0-2= 0
0 = 30°
=
(10v'5)2 sin 120°
ucos0 = 5../s
0= 600
u=
5../s = lOv'S m/sec
1
2
a 2 = F3a F3a cosp
->
•
,.
(C)AE =aj+aK
->
www.puucho.com
->
.
->
•
•
AG=ai+aj
AE ·AG= (AE)(AG)cosy
cosp
= -1
3
Anurag Mishra Mechanics 1 with www.puucho.com
...,_
Tl
\,~
\\;@
r
,. -·
"
~
,
\:
\
\
}
\\.,(''V.:'_y ::~--,>':'~=r-·
: .-.:":;:
~:b"'"."'t:' · ..;.;..._.
:·
'iI{ff,!j{{fliil:;;,
>:
' "G:l~~-...
""" ~,
<~~ ;( , -.:
~
d
u ..
:OI
I
I
FORCE ANALYSIS
i
M
1
/
•
....... ,
Important Concepts
THE CONCEPT OF FORCE
Force may be defined as action of one body on another.
In order to completely specify a force its magnitude,
direction and point of application should be specified. Effect
of force depends on magnitude of P, the angle 0 and the
point of application as shown in the Fig. 2.1
·::.~~--n
,,,,,
[ : ~ ...
1
'
~
:I
p
+
-----~I'.,.,_-~
Fig. 2,1
Forces can be generated through direct physical contact.
They may also be applied through distant action of fields,
e.g., gravitational force keeps objects bound to the eartb, a
bar magnet exerts force on a piece of iron, etc.
Force is a vector physical quantity that is a measure of
the mechanical action exerted on a point particle or a body
by other bodies or fields. A force is defined completely if its
magnitude, direction, and point of application are given.
The straight line along which a force is directed is called the
line of action of the force.
The action of a force results in a given body changing
the velocity of its motion (it acquires acceleration) or
deforming.
1. The various interactions known in modem physics
can be classified under four headings.
(a) gravitational interaction appearing between all
bodies in accordance with the law of universal
gravitation.
(b) Electromagnetic interaction-between bodies or
particles having electric charges.
(c) Strong interaction existing, for example, between
the particles which atomic nuclei consist of, and also
between mesons and hyperons and
(d) Weak interaction characterizing, for example, the
processes of transformation of some elementary
particles.
2. In problems of mechanics, gravitational forces (forces
of gravity) and two varieties of electromagnetic
forces - elastic forces and friction forces are taken
into consideration.
3. The forces of interaction between portions of a
system of bodies being considered are called internal
forces.
The forces exerted on bodies of a given system by
bodies not included in this system are called external
forces.
A system of bodies on each of which no external
forces act is called a closed (isolated) system.
4. If several forces act simultaneously on a point particle
(F1 ,F2 , ... ,Fn), they-can be replaced by one force F,:
called the resultant force and equal to their sum :
www.puucho.com
n
-F,:
= IF,1
i=l
Anurag Mishra Mechanics 1 with www.puucho.com
·- - · " ___ , ____
The components of the resultant force onto the axes
of a Cartesian coordinate system equal the algebraic
sums of the corresponding components of all the
forces:
n
LF,
=IF·1r·
w
I
1391
FORCE ANALYSIS
n
=IF..,,,
w
LF
y
n
LF,
=IF;,
w
5. Mass is a measure of the inertia of a body; i.e., the
mass of a body is a measure of the body's resistance
to acceleration. Mass is a fundamental property of
matter just as length is a fundamental property of
space and time is a fundamental property of
existence.
6. Every object on or near earth's surface experiences at
least one force acting on it, its weight mg.
7. In order to study motion, we must specify system
first. A system is a collection of bodies or a single
body under consideration, whose motion is to be
studied.
8. A system in mechanical equilibrium has zero
acceleration. Acceleration is rate of change of
velocity, hence zero acceleration implies the system
has constant velocity, whose magnitude and
direction do not change with time.
9. Total force, net force, resultant force mean the same
thing. A system in equilibrium has zero force on it.
System in equilibrium
a= om/s
2
<=:> Zero total force on the system
->
F1otal
=0
System not in equilibrium
3;e om/s 2
<=:> Non-zero total force on the system
->
F1ota1
¢
0
Reference Frame
The laws of dynamic can be stated the same way only
for the inertial frames (system) of reference which are in a
uniform rectilinear motion relative to each other. Suppose
that there are two frames of reference (see Fig. 2.2) one of
which, denoted 1, is regarded as being at rest (Le., as being
fixed) while the other, denoted 2, moves relative to the
former with a constant velocity v O. Then all the bodies
which are in a state of rest with respect to the latter frame of
reference will move with velocity v O relative to the former
and the bodies moving with velocity v 1 relative to system 2
will obviously have the velocity v = v 1 .+ v 0 with respect to
system 1 (assumed to be fixed). The velocity v O being
constant, the acceleration of a body relative to the moving
frame of reference coincides with that relative to the fixed
frame of reference and vice-versa.
Concept: 1; In all systems of reference which are in
uniform rectilinear motions relative to each other the
acceleration of a moving body is the same.
2. Experiments show that the forces acting on the bodies
and the mass of the bodies are independent of the choice of
any of these systems of reference relative to which the motions
of the bodies are considered.
3. The forces depend on the distances between the bodies,
on their relative velocities, and on time, all these quantities
not varying when we pass from one system of reference to
another system of reference which is in a uniform rectilinear
motion with re~pect to the former.
If we choose an arbitrary set of frames of reference
which are all in uniform rectilinear motion relative to each
other and if, in addition, it is known that the laws of
dynamics hold for one of these frames then the first and the
second laws of dynamics are stated in the same manner for
all the frames of reference we have chosen. All such frames
are referred to as inertial (or Galilean) frames of reference
and the Galilean inertia law is valid only for such frames.
This is the proposition we call Galileo's relativity principle;
the transformation from one inertial system of reference to
another is called a Galilean transformation.
Concept: A frame of reference which is in an
accelerated motion with respect to an inertial frame of
reference is spoken of as a non-inertial frame of reference.
Which of the systems of reference we deal with can be
regarded as inertial one ? However, the investigation of
motions whose velocities are small in comparison with the
velocity of light indicates that the coordinate system whose
origin is connected with the centre of mass of the bodies
forming the Solar system and whose axes have invariable
directions relative to the "fixed stars* can be taken as an
inertial frame of reference. The experimental data obtained
both in the study of the motion on the Earth and from the
astronomical observations confirm the validity of this
assumption.
www.puucho.com
z'
v9_,•..........-......... v,
~----·,"
·,
y'
x'
y
X
Fig. 2.2
Anurag Mishra Mechanics 1 with www.puucho.com
.
(
·-~;,,
'.'
As to the frames of reference connected with the Earth,
· they can be considered inertial only approximately. This
·. approximation involve§ some errors which will be analysed
later .on. ·
Newton's first law· of motion 'is related to the state
·of equilibrium. If a system is in a state of eqtiilibrium it will
remain in equilibrium unless compelled to change that state
by a non-zero force acting on the sy~tem.
•
1
'\,·, . •
. -- . •
, .
•
··7:-::··'"1
Concept; 1. N~to!l's first law; establishes thefqct oft,
the existen,e J!f inertial reference :fr;ames an:d describes the\
11ati.tre of,th¢tnotiort of a free poi~t'p(lrtic/~ in'.ari iner)ti~f!
. riiferenr;!?.fra,rne. .
.
: '. '. .
. .
. .
2. Referehctframes in which dfr/er pointparticlds;i~ its
state pf rest or ofµniform moti.on/n;'d'straightline are defined
inertial'refeferice frame.
. ·• :. ' .
· · '
.
frdmes in ~hich lifree point particle or.fre~
body.does.not
retain a constant.veloeir:j
(i11.insinertial
motiori). ·,
f..,
' ' ' \• ,, ' . .:
' .-.
'
•
..
~are ~fifre~ ~ rt/!ll·iner:tiiI{ referertfe:-frqmes. · . • . ..\ · . ·
as
- a: ke.te~ert¢j
-~~ral
·e
i.• ·_,_1: 4. :" r!'f1_"_~n. ·.ce fr.wn. trµyell.~g_ _;w_J.th.·_a._c_ ·c.el.er~tio. n_ r_ei_ci.··tt·
to. an merµal reference frame is .a• tt.on-merttal one; ;In.
I1norl'iilertialfrq,!lf4,
·even ajree bogY,,cq.n.perform nori!irierti.al,
f!!:.£.t!en, Le,. tratel with. acceleratfoni:,
·
,
CONCEPTUAL EXAMPLE-1: A stationary cart carries
a vessel with water in which a wooden bar floats (Fig. 2.3).
Describe the behaviour of the bar in accelerated rectilinear
motion of the cart to the right using two reference frames:
(1) a stationary inertial frame associated with the surface
over which the cart travels [the coordinate axes OX and OY
of this frame are shown in
Fig. 2.3 · (a) and (b)l a
non-inertial reference frame associated with the
~ccelerating cart. [axes O'X' and O'Y' in Fig; 2.3(c)]
J
, fill-. 1
-~~Iro-----:·i .. '.:'7
Y.
•,
,
:.
'
·
..
·~"
,.·,,.!',•.
'.;'."' -'f,.
___ --
~
-
0
, __ '
. •" <•1 ..
",
,
I
•••
••••
_8 :
•.·, ... ~
. n 1nmii,71111fiJ11,piFAm1m ~ ·: : .\
•
. · er·
!~,
-~
v
.
o,--
·I
I
·i
!===-="\"···
L • • •'
',
,V.
,
T~•,•••
!
1·
_ .- -
,x·
#·
•.
The bar can be considered as a free body because the
force of gravity of the bar is balanced by the buoyant force
while all the ,other actions on the bar may be ignored.
It is known from experiments that when the cart moves
in this way the bar will approach the left wall of the vessel.
In the first case, the behaviour of the bar is interpreted
on the basis of Newton's first law: the free bar continues in
its state of rest (its unchanged position in the coordinate
system XOY), whereas the cart together with the vessel
travels to· the right (the lefr side of the vessel approaches the
bar with acceleration).
•.
·- ·-"':""-·"·-------~-----,,..- .---..----·:'
. ·,,, _1,7
. Ct>nc!lph In the second,. the bClrmoves iyith acceleration
(non-/nertiallyJ to the left without 'any actions whats¢eyer on
it in this.dire,tiort, while the carf\,,ith' the vessel is ai·i~t'.iri ·
the coord,inate system X'OY'. Here.Newton's fi~st latv 'is "not
observed for 'the bar (the bµr performs non-inertialn1ptiori
although_ii'mqy be 'considered. as ajree body)..
. j
Newton's second
__, law : Acceleration of a system
depends on total force F1o1at acting on the system. According · ·
to Newton's second law of motion a system of mass m, .
__,
subjected to force
is given by
F1o1at
experiences an acceleration 1.which .
__,
~
__,
. i.e.,
Ftotal
a=
-m
__,
L F,x1 =ma
Vector sum of forces on the system (action taken by
external agent) =Response of system
Newton's third law : According to Newton's third
law; if a system A exerts a force on another system B, then B
exerts a force of the same magnitude on A but in opposite
direction, which implies that forces always occur in pairs .
(a) Forces that constitute a pair act on different bodies: ·
The two members of a given. force pair point in opposite·
·
directions.
(b) Each member of a given pair of forces has the same
magnitude .
(c) While applying Newtonls second law; consider force
exerted on any system by other bodies. Thus only one force
of the pair is involved in applying the second law of motion,
e.g., if we are studying system A, then the force on A by B is ·
relevant. Force of A on B will try to accelerate B.
·'
www.puucho.com
.~., ·. ',!:
:
,.:_
Anurag Mishra Mechanics 1 with www.puucho.com
.-
L~CE ANALYSIS
--- ---:
----- --------7
141
!.
: One system:
I
;you
1
GravJta.tipOal f9fce: •
Force of the •
surface on yoti.;',--,f-11~
of t~e
.------ .. -
E•rt_h on you
1
/
Gravitational.force
of you on the Earth /
, .
Another system
the surface
Force of you - - = . i
on the surface
N
(b)
(a)
I
Fig. 2.4
L.
-
.
- ---- -
---- ..
- ----··
---·--· ·-------
. ..
..
""
------- ~---~-- -·
(a) Collision (b) Boxer
(c) Tennis
(d) Attraction
(e)Gravitational (Q Block
struck by
ball struck
of billiard
between
attraction
attached
opponent
by racket
two magnets
to stretched
balls
between
skydiver
spring
and earth
,
C
0
:;:,
A
A•B
" G}G
~
Q)
.!:
~ A1B
<(
Q)
!:!
~
0
LL
m
C
0
Q)
!:!
0
LL
B
A
C
0
A
ll'CSDE:S
-
lll,"
~
B
~
A
A
---e
NI::$-
r
~
•
i "iiiiii~"-il ii :
~
~r ~
•
B
B
-N:=l!
Fig. 2.5
CONCEPTUAL EXAMPLE-2 : Let us consider a
weight lying on a man's palm (Fig. 2.6). The force exerted by
the palm on the weight is FWP; it is applied to the weight and
is directed upwards. The weight, in its tum, acts on the palm
with the force Fpw which is applied to the palm and is
directed downwards. Now imagine that the man lifts his
palm or lowers it. By the third law, in all the cases, we have
Fwp+Fpw=O
•
This equality always holds irrespective of whether the
palm supporting the weight rests or moves.
The third law does not characterize the magnitudes of
the forces and only asserts that they are equal. It is also
important to stress that the forces of which the third law
speaks are always applied do different bodies.
Let the palm move in a certain way. It is required to
determine the forces acting on the palm and on the weight
_.and find the acceleration of the weight.
B
+{]
I
1
____________ ._. - , I
Besides the force Fwp with 1- - - - - - F--,,-- ·7
which the palm acts on the weight, i
· wp
the weight is acted upon by the· ;
l
force of gravity, that is by the force i
!
generated by the interaction
between the weight and the Earth;
we denote this force as Fwe· Now
; -:, .
.. . " .. ,IJ.•
we can determine the resultant
·' .. ,
force acting on the weight and find
the acceleration of the weight
which is the sum of the two forces
Fwp and_ Fwe. According to the
resultant is equa\ to the product of
the mass of the weight by its - - - - - -Flg.2.6
- - - - - --- _J
acceleration:
~+~=~¾
.
Hence, if the magnitude of the force Qf gravity Fw, is .,,
,t\:~
..
greater than that of the force of the palm Fwp, the ,1·
I
acceleration of the weight is directed towards the Earth; if
www.puucho.com
i
'
I
.,.l~
~·
I
•.
Anurag Mishra Mechanics 1 with www.puucho.com
MECHANICS-fl
________ ,
17,.
otherwise, that is if the magnitude of the force of the palm
exceeds that of the force of gravity, the acceleration is
directed upwards.
'
Concept: The magnitude and the direction of the acting,
force determine only the acceleration but not the velocity and;
,therefore we cannot find from the direction of motion of the:
weight. For instance, when Fwe > Fwp the weight can be eitherl
in a downward accelerated motion or in an upward:
decelerated motion. More precisely, when the acceleration is·
directed downwards the velocity can have an arbitrary,
'direction; it can go upwards or downwards or even form an
angle with the vertical. The direction of the velocity at a given;
moment has no direct connection with that of the acceleration:
,while the acceleration itself is completely and uniquely
:determined by the acting forces.
F,p
.
'
~
If the acceleration of the weight is equal to zero, the sum
of the forces acting on it must be zero; in other words, in this
special case the force Fwp of the action of the palm on the
weight is equal in its magnitude and opposite in its direction
to the force of gravity Fwe. In these circumstances the weight
can be in a state of rest or in a uniform rectilinear motion
with any constant velocity.
CONCEPTUAL EXAMPLE-3 : A weight is suspended
from a spring attached to the post placed on the table, we
consider the interaction of three bodies: the weight, the
spring, and the Earth (as has been said, the Earth together
with the table and the post form one body). The forces
taking part in this interaction are shown. The earth acts on
the weight with the force Fwe (the force of gravity of the
weight) and on the spring with the force F,e (the force of
gravity acting on the spring). The weight acts on the spring
with the force F,w and the post (considered as one body
together with the Earth and the table) acts on the spring
with the force F,p. According to the third law, we always
have the equalities
Fwe +Few= 0, F,w +Fw, = 0 and Fsp +Fp, = 0
Assuming that the magnitude of the mass of the spring is
negligibly small (and only under this assumption) we can
write, on the basis of the second law,
F,p +F,w = 0
Condition shows that the force of tension of the
("massless", i.e., 11inertia-free11 ) spring is in all the
circumstances the same at both ends of the spring. In this
approximation the magnitudes of the forces acting on the
ends of the magnitudes are equal to those of the forces
acting on the ends of the spring are regarded as being
precisely equal. Further, by the third law, these magnitudes
are equal to those of the fores Fp, and Fws with which the
spring acts upon the bodies stretching it.
"
' '"
nmnmm mmmm
Fig, 2.7
Concept: Thus, an "inertiafree 11 spring 11 transmits 11 a
force without changing the later irrespective of whether that
'spring rests or moves. Any body whose mass is negligibly
small possesses this property; for instance, in our discussion
we tacitly imply that the threads connecting the bodies in
1
,question are 11 massless 11, 11 inertia-free" and possess the
indicated property. That is why when speaking or a tension of
a spring or of a thread we mean the magnitude of the
,stretching force which is considered the same for both ends of
the spring or of the thread.
1
The force Fwe with which the Earth acts on the weight
(the force of gravity of the weight) is no longer equal to the
force Fw, with which the spring acts on the weight. The
difference between these forces determines the acceleration
of the weight. It should be noted that if Fws > Fwe at a certain
time this does not necessarily mean that the weight moves
upwards; this only implies that the acceleration of the
weight is directed upwards. The force of the spring Fws and
the force of gravity Fwe are not equal to each other
(according to the second law). It is the difference between
these forces that produces the acceleration of the weight.
When the weight and the spring are at rest their
accelerations are equal to zero; aw = a, = 0. Then the force
Fw, with which the spring acts on the weight is equal in its
magnitude to the force of gravity Fwe and, by the third law,
to the force Fsw with which the weight stretches the spring;
in the state of rest the force Fsw coincides with the force of
gravity of the weight. Thus, in the state of rest the absolute
values of the three different forces Fwe (the force of gravity
of the weight), Fw, (the force of tension of the spring) and
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
Fl
FORCE ANALYSIS
------
--
-- ---
F,w (the force with which the weight stretches the spring)
are the same. The force of gravity of the weight and the force
Fsw are simply equal to each other:
Fwe
Nonna!
component
N ---·-
= Fsw
Force of the
string on girl
Force of the string
on the wall
: Horizontal
: component
. - . ----Ffric.
Fig. 2.10
The system
is the wall
push is the normal force of the table upon your hand. The
component of force parallel to surface is friction, discussed
later in this chapter.
+ Normal force in perpendicular to the contact surface
as shown in Fig. 2.11
The system~
is girl
''
(a)
Force of the wall
The system
Force of your friend
:n thi strtn~- -----.--_J_ _------~~-t_h_·-~rg
8
.'!\: _- - - - - - - - - - - . - - - - . - - - .• '
(b)
Fig. 2.8
A
Ideal String
An ideal string is considered to be massless (negligible
mass), inextensible (does not stretch when pulled), pulls at
any point in a direction along the line of the string, can pull
but not push. The force with which one element of the string
pulls on its neighbouring element is called tension in the
string.
A girl pulls a string tied to a wall. The string will exert a
force on the girl in a direction opposite to the force the girl
exerts on the string. The string exerts a force on the wall in a
direction opposite to the force exerted by the wall on the
string (Fig. 2.8).
8
B l!---+---+-+Ns ----"t---t--+-+Ns
A
+
Whenever two surfaces are in contact they exert forces
on each other. Such forces are called contact forces. We
resolve these contact forces into components, one parallel to
the contact surface, the other perpendicular to that surface
Fig. 2.10 shows contact force on finger by a tabletop as it
slides on it. The component of force perpendicular to the
surface is called normal reaction. The force resisting your
A
~~
--~
Fig. 2.12
+
+
+
Contact Force
Fig. 2.11
If direction of contact force cannot be determined, it
should be shown as two components (Fig. 2.12).
Ideal Pulley
An ideal pulley is assumed to
be massless, frictionless. Action of
the pulley is to change the 7B;•,Tlcieal pulley
direction of force. The ideal pulley
.
does not change the magnitude of
tension in the rope. Tension is
same in the string on both sides of
Fig. 2.9
the pulley. If there is no stretch in
the string, the speed at which rope comes onto the pulley is
equal to the speed at which it leaves the pulley (Fig. 2.9).
A
When contact between two bodies breaks, the
normal reaction vanishes.
The weighing equipments measure the normal
reaction.
Normal force is a variable force; it can very in
magnitude as well as direction. In Fig. 2.13, normal
reaction passes through centre of gravity of body in
the absence of any external force. Line of action of
normal reaction shifts to the right when an external
force is applied, as shown in the Fig. 2.14. At the
instant the body is about to overturn, it passes
through the edge of the body about which
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
'
L..f1~44_,~'--'_·,_·'_~'------------~"-~'--'~~---~----ME_C_H~
overturning talces place. For a block kept on an
incline, N = mg cos0. If angle of incline is gradually
. · increased, the normal !eaction decreases.
;:c
'~.~---,~
. :·1 .N
earth to be separate systems, the weight -is, external
force on both the .bodies.
+ Internal forces always act in pairs.
+ Vector sum of all the internal forces on a system is
,,'
zero .
7
L, Fiilternal
System
=0
Problem Solving Tactics
By Applying Newton's Second Law
mg:
mg
(c)
, (a)
Fig, 2.13,,
r-.., '
. ·_ _
L
0
_·
m!!
Fig. 2.14
- ,I
L Identify the object you are considering; make a simple
sketch ofthe object.
2. Draw arrows on your sketch ·to show the direction of
each force acting on the object. Arrows are drawn to
represent direction of forces acting on' the body. This
diagram is called direction of forces acting on the body.
This diagram is called a free body diagram. Only
external forces (forces exerted by the other bodies)
acting on a body are shown in the free·body diagram..
.·,··:,-·
Concept of External and Internal Force
Consider a boy pulling two toy cars A and B connected
through a string. In Fig. 2.15 (a) our system includes A and
B; the· pull of the boy comes from outside the system,
:<,_YL
;
<.,
X
,..
-
p
.
f
"
'
'
-c~i
'",
A
-
w'>
\
'
,mAg
'
~'
' .
.
(b)
,. '. (a)
C
B
,
'
~.
,
.
.•0
,
A
A"
. Wsyre~·-.-.
-'., . >.~~\::!.~¼~.' ,. Pull_
"'~-_-:•.'!;.:.--:,
'c;E'NA
·, '
,,
. msg
''.Ns} .. _-
-1,
,·
..
[
/
.. ,
"•'
Groun!l,.-"..,.,- . - - •,,,
_F_i9~·_'2_.1_s~--------~
-,..
•
,,
. r -· :_:- '
'
'
.
(b)
(c)
.fig.
2.11 ·
:,
internal force. Note that this tension is paired and acts on
both the toy car as well as B. In Fig. 2.15 (b), the pull of the
string on the toy car B is external force, because string is not
part of the system.
Now .. consider. a ball projected upon the .surface of
earth. If we include the ball and
in our system,
then weight is internal force. If we consider ball and·
earth
www.puucho.com
In the Fig. 2.17 (a), force Pacts on block A; it.must
be shown only on A. Block A presses the body B
with certain force, which is represented by a normal
reaction NA , which acts on both the bodies. Neither
weight of A nor force P should be shoWn on B.
Whatever force A exerts on B is communicated
through normal reaction. Similarly body B presses
the ground with normal reaction NB downwards. ·
Anurag Mishra Mechanics 1 with www.puucho.com
C
FORCE ANALYSIS
---·- -
3. Assign a coordinate system to your free body
diagram. Coordinate axis is assigned according to
convenience in resolving forces and accelerations
into components. For exampk in Fig. 2.17 (a), x
and y axes point in horizontal and vertical direction
respectively. In Fig. 2.17 (b) it is along incline and
normal to incline. For a particle moving along
curved path tangential and normal axes are
assigned as shown in Fig. 2.17 (c):
4. Resolve all the forces acting on a body into its x aod
y- components.
5. Apply Newton's law in component form as
LFx == max,
.EFz == maz
LF'y == may,
:_dL-= o·-·;:r;-x
Pulley System
A pulley system allows you to lifr an object while
exerting a much smaller force in a more convenient
direction and with greatly improved control over the object's
motion.
In a single pulley system (Fig. 2.19), the rope exerts
equal tension force at its two __,ends. __,At one end, tension
1
T(x)
L
rJ 1
j_
X
j
w
= Mg x
Fig. 2.20
When a pulley is used to change the direction of a rope
under tension, there is a reaction force on the pulley. The
force on the pulley depends on the tension and the angle
through which the rope is deflected.
A string with constant tension T .is deflected through
angle 28 0 by a smooth fixed pulley. What is the force on the
pulley?
At the other end,
......
Block
Element of
rape at hand .
(b) Fre"i9:-body diagrams,'
(a)
r
_Reaction Force on a Pulley
Fig. 2.18
Object being lifted with
vL
__/~
At the bottom of the rope the tension is zero, while at
the top the tension equals the total weight of the rope Mg.
v...
= - F.
,,
:~.,,
L
'y
balances the force you exert: T
The force diagram
for the lower section of
the rope is shown in the
figure. The section is
pulled up by a force of
magnitude T(x), where
' T(x) is the tension of x.
The downward force on
the rope is its weight
W = Mg(x/L). The total
force on the section is
zero since it is at rest.
Hence
T(x)
6. Solve the set of equations for any unknowns.
l:F
145·
--··. -- ··---
T
/18/2
the aid of a single pulley.
Fig. 2.19
-+
-+
---+
-+
tension balaoces the objects, weight, T = - W. Thus, F = W.
The single pulley is useful because it allows you to pull
downward rather than upward, but it doesn't reduce the
necessary force.
Tension in a Hanging Rope
Fig. 2.21
Consider the section of string between 0 and 0 + t.0. The
force diagram is drawn below, center. t.F is the outward
force due to the pulley.
The tension in the string is constant, but the force T at
either end of the element are not parallel. Since we shall
A uniform rope of mass M and length L hangs from the
limb of a tree. Find the tension at a distance x from the
bottom.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
. MECHANICS-I
2F - (M + m)g = (M + m)a
(i) Minimum force is required when box moves
with constant velocity, i.e.,
a= 6, thus
shortly take the limit t.e -, 0, we can treat the element like a
particle .. Fo.r equilibrium, the total force is zero. We have
Af/ -
2T sin t.e = 0
For small t.e, sin(t.e/2)
= t.e/2
2
Af/ =
and
2T t.e = Tt.e
Fmin
2
Thus the element exerts an inward radial force of
magnitude T t.e on the pulley.
The element at angle e
exerts a force in the x
direction of (T t.e) case. The
Tt.0
total force in the x direction is
LT case t.e, where the sum is .... ~- ....
over all elements of the string
which are touching the pulley.
In the limit t.e -, 0, the sum
---~ig.2.2_2_ __
=
(M+m)g
2
(ii) If F > FmJn, then acceleration of the system is
2F
a=---g
M+m
(iii) For calculation of normal reaction will have to
consider FBD of man. Considering the free body diagram of
the man, we have from Newton's Second Law,
F+N-Mg =Ma
F +N - mg= m[_l!__-g]
or
M+m
N=[M-m]F
or
becomes an integral. The ' -- -
· total force in the x direction is therefore
Tcosede = 2Tsine 0 •
J-•o•o
M+m
b~§~~P{!?:
[~~2ci'iR!~~.J>
;A
ma~ of mass.M stands on a.box of mass mas ;
1shown in the Fig. 2E. l (a). A rope attached to
'the box and passing over an overhead pulley :
,allows the man to raise himself and the box by ,
!pulling the rope downward.
'
°(i) With what minimum for¢e should the '
man pull the rope so as to prevent himself ,
fromfalling down.
·
(ii) If the man pulls the rope with a force F : '---'"'--""'
greater than the minimum force, then : Fig. 2E.1 (a)
determine the acceleration of the ·- -· --- -· ·
.
(man + box) system.
..
\
'.(iii) Determine. the normal reaction between the man and the
.. trqlley.
. .
. .. ·- ....
• ..
.
,12-le>
1· ..... ... ..
.
. ...
.
.. ..
·- ..
,A heavy block of mass M hangs in equilibrium at the end of a
:rope of mass m and length l connected to a ceiling. Determine
1tlie.. temiq_n in_ the rope qt aAisJance xfrom the _ceiljng. ....
Solution : Procedure: When a rope has mass, due to
force of gravitation it tension in it will vary, separate the part
of string and block on which tension is required :
. --
.
-
-
-
-
--
)
T
l~
(e-x)g
lMg
'j, ..
Free body diagram of the block anii rope of lengtt, (f- x)!
,j \ ·
Fig. 2E.2 (a)
'
Using the condition of equilibrium,
:EFy
=0
m
/
T--(l-x)g -Mg= 0
l
or
(M + rn)g
(b)
~,
m
e
Solution: Procedure: Draw free body diagram of
box and man apply Newton's second law separately to them.
Let the whole system moves upward with an acceleration a.
Applying Newton's Second law,
·
f'
~--
X
I
.
'
(c)
www.puucho.com
(/- X)
T=Mg+mg - -.
1
Anurag Mishra Mechanics 1 with www.puucho.com
r FoRce ANALvs,s
I-·----- - - -
- - -- --
•-----
--- - - - - - · - - • - - - - -
T
T
Mg 1 - - - - - - ~
x•l
0
X 0
X
I
Tension is constant along
the length of a massless string.
(c)
Variation of tension
T as a function of x.
(b)
Therefore ITx = - Ta cos 8 + Tb cos <j> = 0
... (1)
ITY = Ta sin 8 + Tb sin <j> - mg = 0 ... (2)
_Tacos8
From eqn. ()
1 , Tb - ~ - cos <I>
On substituting Tb in eqn. (2), we get
Ta cos 8 sin <j>
Ta sin 8 + - - - - - - - mg = 0
cos <I>
mg
or
Tb = - - - - - - sin 8 + cos8 tan8
,----
4
Fig. 2E.2
-- -
Tension in the rope is minimum at the bottom, at I = x
i.e.,
T
= Mg,
and the tension is maximum at the ceiling, at x = 0
i.e.,
T = (M + m)g
Let us consider an idealized case of massless suing
When the weight of the string is already small compared
with the other force involved, we consider the suing to be
light.
For a light suing, tension is constant throughout its
length.
T = constant O :S: x :S: 1
If the block would have been suspended from a light
string, then the tension would be
T = Mg, constant everywhere.
---
Fig. 2E.4 (a) _shows a block of mass m1 sliding on a block of
mass m 2 , with m 1 > m 2 • Find
(a) the acceleration of eadz block;
(b) tension in the string;
(c) force exerted by m1 011 m2 ;
(d) force exerted by m 2 on the incline.
\;:\
Fig. 2E.4 (a)
Solution : Fig. 2E.4 (b) shows free body diagram of
each block. We will apply Newton's second law along x- and
y-axis shown in free body diagram. Block m1 is heavy, hence
it slides down whereas m2 slides up.
A bucket is suspended by two light ropes a and b as shown in
Fig. 2E.3 (a} Determine the tensions in the ropes a and b.
-
-
y
Fig. 2E.4 (b)
mg
(a)
(b)
Fig. 2E.3
Solution: Light rope implies that weight of rope is
negligible as compared to the force it exerts.
Since the bucket is at rest, its acceleration is zero. Thus
Newton's second law gives
ITx = 0 and ITY = 0
Block 1:
ITx = m1g sin 8 - T = m1 a
ITY = N 1 - m1g cos 8 = 0
Block 2 :
ITx = T - m 2 g sin 8 = m 2 a
ITY =N 2 -N 1 -m 2 g cos8= 0
From eqns. (1) and (3),
.
m1g sin 8 - m 2 g sin 8
a= --'-"-------"'CC--m1 + mz
And
T = m 2 a + m 2 g sin 8
www.puucho.com
... (1)
... (2)
... (3)
. .. (4)
Anurag Mishra Mechanics 1 with www.puucho.com
,"~
,+,.
\.
i
I
.
8
= m2(m1g sin. e - m2g sin 8) + m 2g sm
I
'
I,'
=
From eqns. (1) and (3),
m 3 a= m 2 g
m1 + m2
2m1 m 2g sin 8
a= (m 2 /m 3 )g
or
From eqn. {4),
F=T+M 1 a+N2
m1 + m2
From eqns. (2) and (4),
'
= mzg + m1 x (m2) g + m2 (. m2) g
.
m3
m3
,
m
= (m1 + m2 + m3) -2 g
m,
Constrained Motion
In unconstrained motion the moving body follows a
path determined by its initial motion and by the forces
which are applied to it from external sources .
.'
'' . '
Equations of block m3 :
T=m 3 a
iv~= m 3 g
Equations of block m2 :
T=m 2g
N 2 = m2 a
Equations of block m1 : .
F-T=m 1 a-N 2
N 1 =N 3 +m 2g+T
... (1)
... (2)
... (3)
... (4)
lllustration-1
... (5)
... (6)
Remark:------------------If m 3 has to be at rest relative to m1, they must have
same acceleration.
->
->
->
am:,m, = am, - am, = O
_,
an73
->
= elm,
In constrained motion, the moving body is restricted to a,
specific path i.e. the path of the •body is governed by the·
restraining guides e.g. a train moving ,tlong its track; a ball
tied to end of string and whirled in a circle a lead gliding on
a fixed wire frame.
Kinematic Constraints: Kinematic constraints an
equations that relate the motion of two or more· bodies. B)
differentiating the kinematic constraints for the position 01
the particle in a system, the corresponding kinematic
constraints among the velocities and accelerations of th,
particles may be obtained.
·
In the figure shown the masses are
attached to the inextensible string. At
any instant, let the positions of m 1 and
m 2 be x 1 and x 2 respectively as showri
in the Fig. 2.24.
then, x 1 + x 2 + 1tR = l
(length of the string) = constant
Differentiating with respect to
tirile, we 'get-
.
m,!i ~
"<::_,,;_"<',< ·.---.~'.-:·j·.. )
!
~ -"..c/.!!11::'.:·24·_, :,,
dx1 + dx, = 0
dt
dt
v 1 +v 2 =0
or v 1.=-v 2
Again differentiating w.r.t. time, we get
.I
I.
'
>\;
... (i)
... (ii)
I
.-,~I
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
[!:ORCE ANALYSIS -
---- - ----- --
The equation (i) and (ii) are constraint relations for
velocity and acceleration. Negative sign denotes that their
directions are opposite to each other.
llfustration-2
In the Fig. 2.25 the blocks 'A' and 'B' are connected with
an inextensible string. The block 'A' can slide on a smooth
horizontal surface.
VB
X9
{
J_7
.
-
r-1-.
1491
___,.1___
_,_
L ~~~!'.DP}":::!~_>A block of mass m1 on a smooth, horizontal· swface is:
connected to a second mass m 2 by a light cord ovbr a light,:
frictionless pulley as shown. (Neglect the mass of the cord and'
of the pulley). A force of magnitude F0 is applied to mass m1 ·
as shown. Neglect any friction.
'
'
IB
{
h
Fig. 2E.6 (a)
a
(a) Find the value of force F0 for which the sy!tem will be in
equilibrium.
'
(b) Find the acceleration of masses andtensio~ in string if F0 :
Fig. 2.25
has a value which is double of thatfoun,d in p_art (a).
Since the thread is inextensible, its length remains
constant i.e.
)xi+ h 2 + xB = constant
Differentiation w.r.t. time, we get,
XA
dxA + dxB
)xi +h2 dt
dt
/
=0
As the ball moves, xA increases and xB decrease with
time.
dxA
dxB
--=VA--=-Vs
Therefore
dt
dt
XA
and
--;===== = cosa
)xi+h2
hence
v 8 = v A cosa.
Concept: If blocks are connected by an extensible string,
,component of velocity along the length of the thread of the
any two point of the thread must be same, otherwise either
length of the thread will increase or thread will get slack.
Solution : (a)
F0 =
zr = 2m 2g
i.e.,
(b) Concept: Movable pulley is massless therefore
forces on either side of it must be equal.
r--r-:,.,___
T'~T
T-Zf'= Oxa
although pulley is accelerated
_
For
2T
F'o
T=m2g
m,g
Fig. 2E.6 (b)
For m 2 : T - m2g = m2 (2a)
Component
of
For m 1 :
F0 - Zf = m1a
v,
velocity perpendicular
4m
e,
2 g - 2T = m1 a
to the length of the
Solving eqns. (1) and (2), gives
thread changes the
T = m 2 g[m 1 + 8m 2 ]
angle of the thread.
m1 +4m 2
If the thread is
attached to a sliding
2T
constrained body then
at
the
point
of
T
attachment
of
the
Fig._2.26
thread, component of
velocity of the body along the length of the thread is equal to
the component of velocity of every point of the thread along
its length.
Fig. 2E.6 (c)
v 1 cos8 1 = vb sin8 2
1· 1 sin 01 changes the angle of the part 'AC '?f the thread
www.puucho.com
and vb -:.,0 2 changes angle of the part 'BC' of the thread.
... (1)
... (2)
Anurag Mishra Mechanics 1 with www.puucho.com
\
\
ITTo- --\ ---- -- .. .
~
am, = 2a =
4m24g
+
m1
m2
2
=
m 2g
m1 +4m 2
4
mzg
acceleration of m 2 =
\
m1 +4m 2
ace ~ration of m1
i
Fig. 2E.B (a)
'.
.
.
_m___,"'g.:.[m-'1--_+_B_m~2""]
Tens10n m strmg = 2
·
m 1 +4m 2
1
.
;.app ly equatwn
-....., . . -~·-1
I. Exci.m::JP. l.e
I 7 · __..
I
L:T- .- .•..
Solution:
·-,------------~--'-----!
Concept: If a.body slides on another accelerated .mrjacej
S rel = u rel .+ -I arel t i
I
·
2
. ~--
-~reJ_::= Oi are1_=:__~_- °:.
7 S9S 3_7_0 =-: N__ __
ma
A sm~;l ~:b~ca\~l~ck is p~ac:n a triangular block M.so thati
,they touch each other along a smooth inclined contact plane:
'as shown. ThJ, inclined surface makes an angle 0 With the I
horizontal. A hqrizontalforce Fis to be applied o.n the block mi
so that the two\ bodies move without slipping against each
other. Assuming the floor to be. smooth al.so, determine the
I
·--'
N
,,.l·
m(g- a) cos 37° mg
37°
ill~~//
.... Fig. 2E.. 8 (b) __
Fig. 2E.7 (a)
or
.(a) normal force with which m and M press against each'
other and
(b) the magnitude of external force F. Express your answers;
in terms of m, M, a and g.
_i
or
4
N=7x- =5.6N;
5
lxarel =7sin37°
3
arel = 7 X - = 4.2
5
1
2
2.l=-X2.1Xt
2
Solution:
Concept: When' there is no sliding at any contact
lsu,jace we may take c~mplete system as a single body.
Considering motion of the system
A particle of mass 10 kg is acted upon by a force F along the I
1line of motion which varies as shown in the figure. The initial'
;velocity of the particle is 1oms·1. Find the maximµm velocity'
;attained.by the particle before it comes to instantaneous rest.
~F---,a
£mg
FBDofm
Fig. 2E.7 (b)
F
t=lsec.
or
.,. =-·-------- ---,
: F(N) •
= (M +m)a
... (1)
,, 20---
From FBDofm
N cos0 = mg
F -N sin0 ~ ma
:' (0, 0)1---l--- - - -...
, t (sec) I
10
... (2)
... (3)
and
N = mg/ cose
From eqn. (2)
Solving eqns. (1), (2) and (3), we get
:
'15N--1---'----Fig. 2E.9
F = mg (m+M)tane
. F = 20 (0 ~ t ~ 10)
a=F/m=2m/s
Max. velocity will be attained at t = 10sec. because
after that force stan acting in opposite direction
Solution :
M
:---1
~~~gmplg ! a : ._.-__,.
=======->\._____ ~
f
.A block of mass l kg is kept on the tilted floor ofa lift moving;
jdown wfrh 3 m/s 2• If the block is released from rest as shown,
.what will be the time taken by block to reach the bottom.
'¼'hat is the normal reaction on the block during the motion? ;
·'
- -.
- - .
- . .. .
'
www.puucho.com
dv
dt
10
V
or
=2
fdv=f2dt
10
0
v
= 30m/s
Anurag Mishra Mechanics 1 with www.puucho.com
r-Foiice ANALvs1s
L
- . -
-
-
- - 151'
___lj
!J~~.Gtnf:>!·~ r: 10~'->
A homogeneo!'S and flexible chain rests on a wedge whose side
·edges make-angle a and p with the horizontal [refer Fig. 2E.,
,lO(a}]. The cehtrarparcuf the chain lies on the upper tip
the wedge. With what acceleration should the wedge be pulled
,to the left along the horizontal plane in order to prevent the,
displacement of the chain with respect to the wedge?
[Consider all surfaces to be smooth]
o/
T
Nsina
Tsin a
·~
P,
'="·A
P1 cosa
.....B..+
..o
J',s,0
'I-
Fig. 2E.11
Solution:
,
Concept: Draw neat and clean FBD of fixed wedge and
•blocks. Let reaction at comer on wedge is R.
Fig. 2E.10 (a)
. I
Equation of wedge:
Solution:
Concept: Consider the parts of chain on either size of.
incline as two different element, draw FBD. Apply Newton's
law or these parts separately.
R+Tcosa=Nsina
R =Nsina-Tcosa
Equations of blocks :
N
=P1 cosa
. .. (3)
P2
T-P2 =-a
... (4)
g
.
P1
- T + P1 s1na=----:-a
g
(J,
mg/2
Fig. 2E.10 (b)
... (5)
-1)+(~ -sina)=o
T
Taking comp. along incline
... (1)
... (2)
=
p1p2 (l + sin a)
P1 +P2
R =Psinacosa- PiP2 (l+sina)cosa
P1 +P2
mg sina-T = m acosa
2
2
T- mg sinp = m acosP
2
2
g[sin a - sin Pl
on, solving we get
a=~---~
cosp + cosa
= Pi cosa[(P1 +P2)sina-P2 -P2 sina]
P1 +Pz
R = P1 cosa(P1 sin a -P2 )
P1 +P2
L~~~a~~~!~".f12 [>
A body A weighing P1 descends down inclined plane D fixed of
·a wedge which makes an angle a with the horizontal, and,
'pulls a load B that weights P2 by means of a weightless and'
inextensible thread passing over a fixed smooth pulley C, as:
,shown in Fig. 2E.ll. Determine the horizontal _component of,
:the force (in Newton) which the wedge acts on thef/.oor comer
E.
,
The pull P is just sufficient to keep the 14 N block in,
equilibrium as shown. Pulleys are ideal. Find the tension (in
.N) in the cable connected with ceiling.
Upper
cable
p
'
Fig. 2E.12 (a)
Solution:
www.puucho.com
T1 =P
T2 = 2T1 = 2P
T3 = 2T2 = 4P
Anurag Mishra Mechanics 1 with www.puucho.com
MECHANICS-I
j
Upper
cable
a~ei
Tw~bl~1k.A ~;d·B h~ving ·;;:.;;;s-;~-~1-kg, ~: =4kg
arr.anged as shown in thefigµte. The pulleys P imdQare light,
~and frictionless. All the blocks are resting on. a horizontal\
'floor and the pulleys ate held such that strings remains just!
taut.
.
At.momentt = OaforceF = 30t (N)starts acting on thei
ipulley p along .vertically upward direction as sho,.;m ii'). thei
[figure. Calculate.
· · ·
j
'(i). the time when the blocks A and B loose contact with!
I ground.
,
'
(ii) the. velocity of A when B looses contact with ground.
1
(iii) the height raises]. by A upto.this instant.
·
(iv) the work done by the force F upto this instan(.
T,
'
p
Fig. 2E.12 (b)
..
For equilibrium of block
T1 +T2 +T3 =14
7P = 14
P~2K
=
=
--:
.. - . . ~7-.
v·
~"-~~~~R!c~ ·J 13
'
},
.
t,
r"
1
--------------
'
,For the equilibrium situation shown, the cords. are strong,
enough to withstand a maximum tension 100 N. What•i$ the 1
largest value ofW (in NJ that J/)!lY can support as slwwn. ?.
.
•
=---53°
·,
'
·---·-
F -30t(N)
i
t
!
I
j
I
..
Fig. 2E.13 (a) __.,...:. • _,
Solution:
-
- " ' .. \<.-·
,,-,;~
...
1~0N : .
rft, :.-·~- ·-
?_....-.i~J~~
··:YL:·:
·~
.·
53° :''
x,;
. . .~.~:-·;
..
w
i
I
.'
(b)
.... Fig. 2E.13
---- ------·
_,.;:;;-
or
i
,
~.:.~-"'-,<'--"""''",· .,...~----·'- - _, ·--' --- - !
- ·· Solution:
<
~~-½'·I
Concept: Consider the Mint at the function ofstrlngs;
'as string element at function is massless and in e'lµilibrium,I
sum of forces in x · and :l .direction . must .be ,-e9uaL ,
1
:EF,.. = 0 ' - , ..., ......... , .- -.
---. - ___ ,,.,_" .,,--"'.
I
" ....
-"
Flg.2E.14(a)
- ,,w
Tsin53°-100cos53°= o
T = lO0cot 53°= 300/4 = 75 N
:EFy =0
100sin53°-W-Tcos53°= 0
W = 100sin53°-Tcos53°
= 400 _75x~
5
5
= 80-45
=35N
1,--~~~ll~;;;:- Mien
w:;t;;-;~;;~
bldcks loose--contact
_normal rea':~?~_'?..n _t_1!_em bec;ames ze!:.o:._., ___ .···- . .
(i) When A looses contact T = 10 N
F= 3T
30t = 30
or
t = 1 sec
When B looses contact
.
,.
2T = 40
T=20, F=3T
or
or
30t = 60
or
t=2sec
(ii)
T-10 = a
and
3T = F = 30 t
For getting velocity we have to use
calculus because acceleration is variable
O
(iii)
=
(iv)
www.puucho.com
2
V
V = Jdv = 10J(t - 1) dt
= 5 m/s
I
J· dx = J10[.c-r
+~]dr
I
2
2
X=
W
5/3m
= fFdx
_J
• • - · · "301 = F ,
p
T
T
T
Anurag Mishra Mechanics 1 with www.puucho.com
--
' FORCE ANALYSIS
2
= J30txl0
J
[t
l] 175 ·
2
--t+- dt=-J.
2
2
6
i-'=X~9Dl';Pl_!2- [ul;>,
In the figure shown, friction force between the bead and the
15 ,_.' ·In the given figure find the velocity and acceleration of B, if
instantaneous velocity and acceleration of A are as shown in
the Fig. 2E.15 ( a)
light string is mg. Find the time in which the bead loose
4
contact with the string after the system is released from rest.
Im
•
I
I
Fig. 2E.16 (a)
Solution:
Fig. 2E-15 (a)
Concept: Only interaction force between string and
bead is friction. Tension in string is due to friction.
Solution:
Concept: We use the fact that string is inextensible and
length of string is constant.
11 + 12 + 13 + 14
..
Tension in the string,
T=f=mg
4
mg
mg--
Acceleration of the block,
a2
= constant_
d1 l d2l d l3 d l 4
+- +- +_ =o
dt
dt
dt
dt
vA +vA +(-vc)+(-vc)= 0
differentiate to get -
a2
= II_ .1.
2
vA=-1m/s
aA= -2m/s 2
f"
mg
mg-3
ab = - - ~4- = g .1.
m
4
Similarly
ac = a A
ac =
2m/s
Now,
14 + Is
dl
dis
dt
dt
-Ve+ (va)
2
.1.
va
. Similarly,
4
= 2a 2
mg
Fig. 2E.16 (b)
[downward]
Now,
=-+g
4
7
a,,1 = ab, =
Now, apply eqn.
Sret
or
= 16
!
=
t =
1
2
-a,,
1t ;
2
~ 21
ab,
=
1
l = -ab,t
2
2
/]f _
1J7g
-~-
d!6
-4+ - = -
Va
mg
Relative acceleration of bead with respect to string
.
3g
Fig, 2E.15 (b)
or
V C = VA => V C = l m/s 1'
where Ve is velocity of pulley C _
2
m
[downward]
Now acceleration of string, a,
a, = g 1'
[upward]
[where bead is placed]
Thus equation of bead
--
=
L,S-~fil-t,TIP J,!=!-_ I 17
dt
= -va
Ve
=2
In Fig. 2E.17 (a) shown, both blocks are released from rest.
Length of 4 kg block is 2 m and of 1 kg is 4 m. Find the time
they take to cross each other? Assume pulley to be light and'
string to be light and inelastic.
= O.Sm/sl
:. aa
1.;:>
= lm/s 2 .1.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
154
~---------- ---- ··· 1 -=c..=-=
=-=--1-- ·-- -- -- --- -·1
,11
! mjQ0!4m
!2
L
Solution :
_,
_,
v p/g
or,
1kg
ig. 2E.17 (a)
•-,.,----_.,..- --=-~----....,__ ____ --···- --
--- --··- ---·-
C
¼bead c~n ;;,ove fre~ly ~n a h_o_nz_·-o-n-tal ro~Th;bead ~
I.connected by blocks B and D by a string as shown in l:he/
ifigure. If the velocity of B is v. Find the velocity of block D. /
r
T
I
!•l
I
T
2m!
-·
i. :9
I
t
A
1g
L_ - - - - - · - - - - - - - - - - - - - - - - · - · -
'
!
Solution:
. [---~o.:~e_p_t_:_a_lo_n_g_thelength of s;ing ~~ocity component/
I
Fig. 2E.17 (b)
-----
.)
-
From FBD of blocks A and B solve acceleration of each
block
... (1)
4g-T=4a
T-lg =lxa
... (2)
3g
After solving eqns. (1) and (2), a=acceleration of A w.r.t. B
_is sa'!!:.f!'!..!11!..!1!:.J!Oi!_1-ts_onstrl11g. ~--
.I ~. l
.I
5
6g
aA/B = - = 12m s2
,
I
l
!
·i
'---
~
Vp
(b)
~
.,.
Ve COS
(c)
--- -------- -- --·----VB
PULLEY CONSTRAINT
· lllustration-3
In the Fig. 2.27 shown pulley moves 1 -- with acceleration P. Let acceleration of
lI
i!0'······.....'•,,
tD
Fig. 2E.18
A
__
37°
I
I
____ _J
= Ve COS 53°
Ve cos37°= v 0
from eqns. (1) and (2) we get
·
vB cos37°
VB(4/5)
Vv =
cos53°
(3/5)
4
2
, t = 1sec
6= o+I.xl2xt
2
.
blocks m 1 and m 2 w.r.t. ground are v 1
and
I.
t""
Ve
I
6m
_,
...-o
,. ,.
I
5
If A will cross B then distance travelled by A w.r. t. B is
v
I
4g
4m
I
.a,
a
4~g
I'
_,
v1/,+v~,
=
2
Lltxamr.:.l·e
. - -r,;_=~1--~_':---..
its I ~
r--- -- - ----- --I
j
MECHANICS-I
... (1)
... (2)
VD =-VB
3
i _,
! v,f
i;:g~me1~ai::>
i m,
¼-ii{; goes ~~-with lOmj~.A-pulley P ~-fa<;d~~ th;·;;m,;;J
I
m2
Il. _ Fig. 2.27 __ _
--------- ------ .. ------ - ---- --------- ----- ..I
Concept: According to string constraint for.an observer
bel
on pulley the length of string that approaches pulley must
released way form the other end of pulley, Le., relative to!
pulley velocity of both the blocks should be equal in\
magnitude but opposite_ in_ direction. _____ _ _ _____________!
Jthe lift. To this pulley other two pulley P1 'and P2 are attach~i:l.
·p1 moves up wi-th velocity 30m/s. A moves up with velocity
10 m/s. D is moving downwards with velocity 10 m/s. at same
!instant of time. Find the velocity of B and that of C at that
~t~_n!:_~_s~:1!_~ that all :_'~l'!..C!_t!es are relati~e to the gr<_J'!_nd.
l
www.puucho.com
'
Anurag Mishra Mechanics 1 with www.puucho.com
I FORCE /\NALVSIS
-1
1551
r----
.
i
[
a,
I
II
.
3
2
I
Solution: Apply constraint on pulley P
->
->
V P1 /P
--+
--+
= -V P2/P
-t
-t
Vi,-Vp =-(Vp -VP)
_,
2
_, _,
v Pi , v
P,
v
P
2
are respective velocity w.r. t. ground,
_,
· V
_,
P2
'imrrln1TT1rmr!rn'11111TT1TTT
__,
=tJll- ~~Q:,=I
--+
M
--+
VA-Vi, =-(v.-vi,)
_,
__,
= 2[30jJ - [10J1 =
i
Fig. 2.30
~--------
soj
---- _,
1
lllustration-6
Normal Constraint
Consider two blocks moving on a
surface and always remaining in
contact. In order to maintain contact
component
of
velocity
vector
perpendicular to contact surface must
be same.
__,
__,
sinlilarly
I
7[9ne
~cline
Va =2Vi,-VA
i.e.,
J
In the Fig. 2.30 shown plank 1 and wedge 2 are free to
more obtain relation between their acceleration procedure is
similar to that of previous illustration.
a1 = a 2 sin El
'
__,
:
a,
= -10j
--+
a, sine
If wedge (1) and (2) are to remain in contact
component of acceleration perpendicular to contact surface
must be same.
a1 sin"B = a 2 cos9
= tan9
Apply constraint eqn. on pulley P1 to get
--+
\
'----------------------
= 2[-10JJ-[-10J1
Vc
I
I
Fig. 2.29
c = 2v,,,-vD
__,
\
·mustration-5
--+
--+
--+
--+
Vc-v,,, =-(VP-VJ>,)
_,
I
'
a1
= 2[10Jl - [30.fl
= -lOj
Now apply constraint eqn. on pulley P2
v
I
a1 case\ ·
a1
!
\
a,
_,
= 2Vp-V "1
_,
a,
\
I
_J
hl
&a 2 cosa
'2:.-···
.,:re'
I
~ - - - - - - - j _____
F1_g.2,_e_.1_9_ _...,__ _ __
s\fi-~.-
.i
,~e;J~C
I,;:
Fig 2.31 shows three identical cylinders, cylinders are
released, find relation between accelerations of cylinders.
~;~--,7
I
L
Conta._ct ~u,rface
_!:!g. 2.28
V1 =V2
_,
a1
_,
Frontview
= a2
82COS
lllustration-4
In the Fig. 2.29 shown find
acceleration of wedge 1 and 2
,....
60°
, ___
relation between
-
Fig. 2.31
------
•.
60°
------ ---
_J
Constraint equation relates component of acceleration
perpendicular to contact surface shown in figure.
a1 =cos 30° = a 2 cos 60°
a1
1
a, = ..J3
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
r1ss -
- ·- MECHAN!cs:1 -I
---------
L-- ----- -
' - - - - - - - - - - · - - _,;__j
k~$~~~J~,T-wl>
--·-
---·- ,-, --- ---- --- - ··---,-,
1n the situation given, all surfaces arefrictionless, pulley is/
;ideal and string is light. If F = Mg/2,find the acceleration ofl
,both the blocks in vector fonn.
,
'
I
·
.
F
I,
r':
~
:~
l
y
Fig. 2E,21 (b)
Now write constraint equation for pulley to get
I
Fig. 2E.20 (a)
-+
-+
~
-->
-->
-+
vn-Vp = -(vc-Vp)
Solution: First consider both
the blocks as system force that we
apply at one end of string is tension
in the string.
For system block (A + B)
Mg =2Ma
also
·
•
I
llA
=g/4i+g/2]
-->
•
-->
I
I
Thus we get
vA
J
-- -
-·. ·-
' --· -
(-12.Sg)m/s
r·-· - ---- - --- -·
.
,
iSystem is shown in figure. All the surfaces are smooth. Rod is
lmove by e.x:temql agent with acceleration 9m/s 2 vertically
:downwards. Find the force exerted on the rod by the_ wedge1 ;
mg
Fig~ ~E.20 _(c) __
i
__j
1
"'cn
l
I
1
E
a,
I
10kg
[,"sng!'.l?Rl,~J21l>
. -
.
VB=
a, = g/4i
!
•
= (37. Sj) m/s
-->
_ Fig:..2E._2_0_lb)_
:.1$-,
a,= g/2j
Vp =-VA
I
:
- -.----
'
-->
A
,
1
2
Thus,
B
l
2
a=g/4
Thus,
a= g/4i
For system block A:
Mg-Mg =Ma
or,
'.~~Lsysterii
---~=Mg~,]
-
37'
- -- - -
-----
--- .
-·--
•Three blocks shown in figure more_ vertically with constant!
!velocities. The relative velocity ofA w.r.t. C is 100 rri/s upward!
'.and the relative velocity of B w.r. t. A is 50 m/s downward. 1
:Find
the velocity. of C w.r.t.
ground. - All.l the. string are ideal.
I
.• ·- •
~
I
I
•
I,
I'I
I
!-
Fig. 2E.22 (a)
Solution : Constraint equation
a2 sin37°= a 1 cos37°
or,
a 2 = a1 cot37°
= (9 x 4/3)m/s 2 = 12m/s 2
~
· ~. A~a
2
I
I~
!a 1 sin37° a
i
Fig. 2E.21 (a)
Solution:
Let velocity of blocks, A, B and C are
-->
-->
-->
-->
•
-->
-->
•
vA,vBandvc
VA-Ve= 10Qj
Vn-VA=-50j
cos 370
~
·:ct a sil137"
2
_ _ !ig, 2E.22_ (b)
--- - - .=-,e_-,.-1_
-->
a1
cos3r-
... (1)
From FBD of wedge we can see that
N sin37°= Ma 2
Thus force enerted by rod on the wedge is
N= Ma 2
10x12
sin 37°
(3/5)
...(2)
www.puucho.com
=200N
-
Anurag Mishra Mechanics 1 with www.puucho.com
15_!i
; FORCE ANALYSIS
L.-.
On solving equations we get
T
4mg
1
Find the tension T needed to hold the cart equaibrium, if there
.is no friction.
T
= 3,,J3
= 2mg
3Jj
2
a=__!_
3Jj
T
Concept: What is cause of a acceleration of bob?
,Resultant force on ball in x direction is (T1 - T2 ) cos 60° it
cause acceleration in bob.
30°
•
•
Fig. 2E.23 (a)
Solution:
Nsine
Rl*N T
-t'-i'(:-
w case
_•• J;,:.·
··)B.....
T
w
Fig. 2E.23 (b)
·A block of mass 10 kg is kept on ground. A vertically upward
force F = (20 t )N, where tis the time in seconds starts on it at
t = o.
(a) Find the time at which the normal reaction acting on the
block is zero.
(b) The height of the block fr~m ground at t = 10 sec.
Solution:
N =Wease
Nsin0=T
(a) When
or, T=W[.}372x~]
Wcos0sin0=T
mdv
20t-mg = - dt
r- - - ...
.
2~J>
v(t)
B
Fig. 2E.25
lO)dt
0
v(t) =lt 2 -10tl~
v(t)=t 2 -10t+25
h
10
0
5
f dh = f (t
60'
A
mg
C
Jdv = J(2t -
A steel ball is suspended from the ceiling of an accelerating,
carriage by means of two cords A and B. Determine the
'acceleration a of the carriage which will cause the tension in-A
,to be twice that in B.
60°
N
t = 5sec to 10sec
(b) from
9
[J~~fl~J?..I e
F =wt
t == 5 sec
or,
2
T=Jjw
-
N =0
20t = lOxlO
2
-
lOt + 2S)dt
.il.,.
10
t3
h= - - 10t2
- + 2 5 t1
3
2
l
5
125
=--m
3
Fig. 2E.24_ (a)
Solution:
Concept: When force is variable always apply calculus .
T1 cos 60° - T2 cos 60° = ma
T1 sin 60° + T2 sin 60° = mg
T1 = 21'2
T1
y
Lx·
-
.. . (1)
... (2)
.
... (3)
'lwo mass A and B, lie on a frictionless table. They are
:attached to either end of a light rope which passes around a
,horizontal movable pulley of negligible mass. Find the
,acceleration of each mass MA= lkg,M 8 = 2kg,Mc = 4kg.
:The pull_ey P2 _is vertical._
mg
Fig. 2E.24 (b)
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
ffsa
,---
MECHANICS:::i-J
its equilibrium position. If we hold the block in position x,
form Newton's second law,
B
JL
•
__Flg'._2E~6
'-...···
··,....
: •
-+
4
-+
->
->
->
'
(a) ap = aA + a 8
->
(c) ap = aA- a 8
(BJ Acceleration of A is :
(aJ 3g
(bJ 4g
->
': Fexternaf
Equilibrium
position
(a)
(cJ 2g
5
5
--¼
\..
/ Frictionless
•·· -· surface
->
(d) a, =2(aA+a 8 )
· (CJ Acceleration of B is :
(a) 3g
(h) 4g
5
'
5
(DJ Acceleration of C is :
(a) 3g
(bJ 4g
I
1
i
System
!•] __
(AJ Constraint equation for pulley A is :
5
. -· . -- ....... - - ·r
1·-- _ .............. --- -··· ···-···
A
...
(c) 2g
System
j x>o,
Fspring
5
:
-· -· .. 5- . . ---~ ---
2
(cJ g
(dJ ![
__ 5 ________ _5
, ,
. ,
,. .1
........... EqJ_ilibrium
position
Equilibrium
position
Solution :
... (1)
... (2)
... (3)
mcg - T = mcac
,
I
, ,'t.xtemal
J a .£me· I
,--f·1
"t.pnng /.. ··-\
(b)
(c)
Fig. 2.32
->
F external
->
+ Fspring
=0
->
Robert Hooke experimentally found that F external is
proportional to x.
...
Fe?(ternal
x>O
x<O
Constraint equation is
-+-+
,, Spring compres$ed
Spring stetched. _ _ _~:---+x
-)-+
aA-a, =--(a 8 -a,)
2ap
= aA + aa
... (4)
on solving eqn. (1) to (4) we get
4g
aA=-
.5
2g
aB=-
5
T= BN
~ig.3,33
3g
ac=-
->
5
Fexternal
Where k is called spring constant and has unit N/m
Elastic Force of Spring
->
Spring shown in Fig. 2.32 (a) is stretched or compressed·
by applying a horizontal external force on spring. We choose
origin of coordinate system at equilibrium position where
the spring has its normal length. In horizontal direction
there are two forces acting on the system:
->
(1) Fextemal
= kxi,
->
(2) F,pring •
When we pull the block to stretch the spring, force of the
spring is opposite to out pull [Fig. 2.32 (b)]. Ifwe push the
block to compress the spring, force of the spri_ng is again
directed opposite to our push [Fig. 2.32(c)]. Force of the
spring is restoring force since it acts to restore the block to
Fstring
= -kx i
Therefore force of spring on block is proportional to the
amount of stretch or compression of the spring. It is always
directed towards mean position. It is independent of mass m
attached to spring. An ideal spring has negligible mass as
compared with mass m attached to it.
Series Combination
Elongation or compression in different spring may be
same or different but tension in each and every spring is
'
.
same.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
---159
\FORCE ANALYSIS
~- ---
Fig. 2._34
x represents is the total extension produced in all
springs.
x 1 , x 2 , x 3 , ... Xn are extensions produced in individual
springs.
If this spring is replaced by a single spring and same
elongation is produced and tension developed is then this
single spring is equivalent to combination and its force
constant is equivalent force constant of combination. Even
total energy stored in combination will be equal to energy
stored in this single spring for same deformation.
X1 + Xz+..... ... +Xn = X
k1 X1 = k2X2 = .... : .. . = knxn
.
.
.
_ 1 . 1
. 1
X1,X2, ......... Xn - - . - .....•... .kl k2
kn
1
X
k1
X
l
Fig. 2E.27 (a)
Solution:
Concept: Force of spring does not change
instantaneously so find spring force at initial instant,
Initially
m1g =kx
When support is removed, spring force does not change.
k1
1
1
1·
k1
k2
kn
kx
-+-+........+-
T = k1X1 = k,qX
1
k1
l
l X = k,qX
kx
M2g
FBD litially
FBD when support
is removed
- + - ........+k1
k2
kn
1
1
1
1
- = - + - ........+k,q
k1 k2
kn
(b)
NewFBD
For m 1
or
For m 2
k
Equivalent force constant is smaller than smallest
individual force constant.
or
Parallel Combination
Tension in different springs may
be same or different but direction of ·
tension in each spring is same. Even·
elongation or compression produced
in each spring is same.
Total tension in this combination
k,.x
and that produced in single
equivalent spring must be same.
k,q x = k1x + k 2 x+........ knx
Fig. 2.35
k,q = k 1 + k2 +..... , .. kn
k,q =:Ek
Equivalent force constant is greater than greatest
individual force constant.
(c)
Fig. 2E.27
1 -:E(l)
k,q
1
_,
The system of two weights with masses m1 and m 2 are
connected with weightless spring as shown. The system is
resting on the support S. The support S is quickly removed.
Find the accelerations of each of the weights right after the
support S is removed.
----c--~---,---X
I -
- -
:
m1g -kx = m1a1
=0
m2 g + kx =m 2 a 2
a1
(m1 + m2lg
az = - - - - m2
An object of mass mis suspended in equilibrium using a string
of length l and a spring of constant K(< 2mg/!) and
unstretched length !/2. Find the tension in the string. What
happens if K > 2mg /! ?.
Fig. 2E.28 (a) ·
Solution : The string is under tension and the system
is in equilibrium, if Kx < mg
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
l
[herex = -]
K(½) < mg
i.e., for,
2
K< 2mg
l
T=mg-Kx
l
=mg-K-
i.e., if,
2
2
acceleration of 3 m will be zero.
If K > ~g, the spring force is more than
mg, for x
=.!. Thus the system will ai:c~lerate
2
! -
#- _';-~ -;_::-,-~~ --- . --;.",-f0:~1
·· •
[The mass in the Fig. 2E.30. can slide on a fr.ictio11/~sJ
lsurface.TIIe mass is pulled out by,a,distance x. The ..sp,;ifzgj
)constants are k1 and k2 · respectively. l:i".d the force. pulluj;
. . · .l
•back on the mass and force on the wall.
I
-~\·
;
. -·· ___ ..•. ...
Fig. 2E.3~ ·"'
.J ._
·)
''
Solution : Springs are in series
Hence
k = k,k 2
eq
k1
+ k2
and
Solution:
/. •.:<;:oJ?-Cept: Sp~ngforcedoes 710t,~haMe insta~;~ne~~;~J
~:c.afirststepfi'i!!:;,:f:'.'!frm in all the springs'._ : ___ :__J
·
r+·i<,;,0:~-: .· Kx;
•
Ki,
,rb:\::. ·: J; ' . rt
,r··
13~~ . .,·,--r
'·
L: •
Kx3+2rng
.Fig, 2E.29 (b)
··T
.
!
t ,
: Kx2 +rng:
,c.~___J
Form FBD of blocks we get
... (1)
3mg = Kx 3
Block C
BlockB
2mg+Kx3 =Kx2
2mg + 3mg = Kx 2 ~.Smg = Kx 2
... (2)
Block A
·Kx1 = Kx 2 + mg.
...(3)
when spring 2 is cut spring force in other two strings
remain unchanged, at that instant.
Kx1 -mg= ma 3
~
aa = Sgt
Kx 3 + 2mg
..,
..,
..,
ail/•· = aA/8 + a 81• _
= 2ma 2
[aA/g]x
www.puucho.com
= [aA/B]x
+[aB/g]x
... (1)
... (2)
Anurag Mishra Mechanics 1 with www.puucho.com
FORCE ANALYSIS
----------
From FBD of A it is clear that Block A cannot accelerate
horizontally.
i.e., in x-direction because there is no force in
x-direction.
Block A can accelerate in y-direction only.
Solution: Constraint relations ;
LetX A•XB and Xe ate the positions of pulleys A, Band C
respectively at any instant with respect to a dotted line
shown in Fig. 2E.32 (b). The total length of string
[aA/,lx = 0
Therefore
[a A/B ] x -- - [aB/g ] X
That means for an observation on wedge block moves
only x > 0.
For block A;
mg-N = m(asin8)
... (3)
ForblockB; (N + mg)sin8 = ma
... (4)
On solving eqns. (3) and (4), we get
.:·;·w···.·····~··1
'
2gsin8]
a= [ l+sin 2 8
:
:
I 2g sin8]sin 8 = [ 2g sin
''
''
1g+ T
:
0
1
Displacement of block A in 1 s is
1
2
S = O+-aAt
2
_19't_T ···-···········
T:
:T
'
2
C
= .!:_ x [ 2g sin 2 8] x (1) 2
l+sin 8
= [ /+s:;:288]
[.. §.X-Ql}JPI~
:
TGJ'.
....~-··;·····:
2
8]
l+sin 2 8
l_l+sin8
2
A
''
''
The acceleration of block A,
aA = asin8
=
1s1:
--"··-_j
1g
---- - --. ------. -- -.,. --·
i 32 :_>
Fig. 2E.32 (b)
In the pulley system shown in Fig. 2E.32 (a) the movable,
pulley A, B and C are of 1 kg each. D and E are fixed pulleys.
The strings are light and inextensible. Find the acceleration of
the pulleys and tension in the string.
2XA + 2Xu + (XB -XA)+Xc +(Xe -Xu)+ lo= l
or
XA +Xu +2Xc+l 0 =l
... (1)
Where 10 is the length of pan of string over the pulleys,
which is constant
Differentiating equation (1) w.r.t. time, we get
dXA + dXB +zdXc
dt
dt
dt
=o
or
also
Let
and
vA+vu+2vc=O
aA + aB + 2ac = 0
aA = a upward
aB = a upward
then
ac = ( a A ; au ) = a downward
... (2)
l-).
Since string is same throughout and uniform, the
tension in it will be same every where. Thus
For pulley A :
Zf-(T+lg) = la
... (!)
Fig. 2E.32 (a)
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
For pulley B::
... (2)
2T-(T+lg)=la
From pulley C :
... (3)
lg-2T=la
Solving above equations, we get
a~
-g
aB =-3'
=-t
'
g
ac .=3
i ------·- ."'-,,-··----·----- --
bL~mi:;,J~,]
----;;:,;-~__,:,
33
r~
3
~'
.---··-w<7
----~ 1<:"M~~--
'lwo identii:ill, blo_cks each hav.ilig a mass of 20 kg are
connected to each other by a light znextensible string as shown
and are pla~er! _over a rough ·surface. Pulleys are connected "to
the blocks:· ·,:! · · .
: ,
.
. .' .
Find :a"'eleration of the blocks ;ilfter one second,' after the
applicqtibn of·the time va,yingfotteof 40t N, where t'is- tn
second. · •. ~-~
·
··
,.· J:
~~kg
u=O.
e~
---~
·
Solution:
a1
I
6 kg
!
·
•
a1
;L·=_-.~"-4-,·--i·~--+~-7
Fig. 2E.33 (b)
-~---'-·-·.
1lll/771i7lli'7i'C.'fiL;,'fi-'7i-'T,-'T,-rd~i~
2a2
F - 3T - f
Considering block Ii
u=0.4 '· ·
l
i
==-~~-
Solution:
11 + 12 + 1, + 14
a
. '
F
~=:~~(~)- ~- ~~
-=-' - -- -" - ' ---~,
2T
Put T
= 15N in eqn.
30
6
... (2)
Put aA
+ 1501 )- f = 20a1
... (3)
2
... (3)
... (4)
... (5)
= OOA
(5)
= aA = 5m/s
2
= 5 m/ s2 in eqn. (1)
2
aB = -9m/s •
T-MBg =MBaB
15-M8
When motion ;;tar-r's t = µmg = 8 N and a 1 > 0
•, 40
1
t = -s
2
Solving we get,
~
... (2)
from eqn, (4)
= 65a1 +-f
#
... (])
T-MBg =MBaB
2T=MAaA
·--· -· -··---
5
;
=0
o
18-3 = T
T = 15N
= 3a1
Motion of blocks will begin at t
_cc.....:,.
-aB-2aA+l=O
aB + 2aA = l
3g sin32°-T = 3
3
3gx--T=3
5
·
... (1)
!
Z1.,+ Z2"+Z3"+Z4"=
: a~:I
= 20a1
• .....
3 kg
B
;:·7
2T- f = 20a 2 = 30a1
Solving Eqns. (1) and (2)
i
·-· '
~/s2:
A
Considering. block I
F- 3(
·-- --- ...... • -
~:1, ,. _l=<-~=-™ -
~~--··-~-~O
kg ---F = 4
l 4 \ \ \ \ \ \ \ \ ~ \ \ \ \ ~ \ t .:
'----'"'--.-:."~-- ~ig.:;;..:.~ .
__j_:.c·
·
. --
is:
T~2g
and
----- ·---·- --
:Three blocks" 4, B & C are arranged as shown. Pull'!Yf andi
!strings are· idea( All surfaces .are frictionless. If block C
observed.mbving down alongthe:incline at 1 m/s 2 .}',ind, masst
of block. B, tension in string and accelerations of A. B as the
system is ~ele(!5ed from rest.
1
=- s
2
www.puucho.com
X
10 = M,;
X
(-9)
15 = MB(l0-9)
MB =15kg
MB = 15 kg, T =15,
2
2
aA = 5 m/s -c>, aB = 9m/s .J,
Anurag Mishra Mechanics 1 with www.puucho.com
:f~iJ
L~~~CE ANAL-'-)'S-'-1S_ _ _ _ _~ - - - - - - - - - - -
}i::q~.@!\')f?J~j35l:':>
·· - · · · - · · ..___,,,_.The -;,;;e-m-sh~-:Vn.
Fig. 2E.35 ( a) is given an
;acceleration 'a' towards left. Assuming all the surfaces to be
jrictii°':l_ess_,_ji~d_t~e_J!)_rc~ _on ~~1!'_5_P.~"!_e._____ __ ·--.
i~ --;h;
i
g
I
':_N\
I
i i
....,,,_.,..(a_)_._.i.,
I:
i
I
·
aBA
a
Jj
i
;BG = ;BA + ;AG
Takethe~~~~:t;~in_Fig._~:~~-~)~.
N,
(b)
N,
=
mg
cos 30°
N 2 =ma+ N 1 sin 30°
=ma+ (1.15 mg) x (1/2)
= m(a + 0.58g)
l_ .. _. -----~
TI1e block B st-;;,;.fr~~.~~; and slides ;n-th~ ·,:;~e A~hich1
Block B:
!
move on a horizontal surface. Neglecting friction,
1determine (a) t/ie acceleration of wedge, (b) the acceleration
;af! t!Je _bJoctrelati)'Ll'Q tli<!....IY~ggg._,__ ··;;·\. __
~
'
8 eA
,
mg
2_E~~J~L--------'
:r.Fx = N 1 sin 9 = MA
:r.Fy =N 2 -N 1 cos9-Mg = O
:r.Fx = mg sin 9 = m(a - A cos 9)
};FY = mg cos 0 - N 1 = mA sin 0
MA
N,=--
...
...
...
...
(1)
(2)
(3)
(4)
1
I
or
->
A
9
__
Fl~g.
mgcos9- MA =mAsin0
sin 0
-+--+aA
! ...,,.,...,...,.__....,
A·mgc~·se
sin 9
Substitute N I into eqn. (4) to get A.
7
'
-+--+•A=A.
1 8
mgsn
Wedge A:
lcoJ:E~p:~,w,~·J36~
ican
AN,
-+
x-component of acceleration
aB, aBX = a - A cos 9
•
-+
y-component of acceleration aB, aBy = A sin 9
= '1.15 mg
From eqn. (1),
N,cos
!
: -- .
I
:--···----~~--~:61~!--~-----·- . .J
~
Solution : This problem involves two branches of
mechanics: kinematics (which deals with motion) and
dynamics (which deals with cause of motion).
>,\;--. First we shall analyze the. accelerations of wedge and
"·'.',·block.::·.·.. ·
·
Wedge A : It moves on horizontal surface; we assume
its acceleration towards right.
mgcos0=mAsin0+ MA
sin 0
A= mg cos0sin0
or
msin 2 0+M
. a
I __ Fig. 2E.36Jc)
From eqn. (3),
ma= mgsin0+mAcos0
~:
vn
l
.
a=g sin0+A cos0
.
mg cos 2 0sin0
= g sm 0 + -''-----m, sin2 9 + M
(M + m)g sin 0
=
M+msin 2 0
Note that axes of x and y can be assigned in another
manner, as shown in Fig. 2E.36 (c).
www.puucho.com
1
AG·
Fig. 2E.35
Thus
..
-
For sake of simplicity we drop the subscript G. Therefore
resultant acceleration of wedge is vector sum of its
acceleration relative to A and acceleration of A on ground.
-- ~olu~:;~~-F~rces actin;:n -;~er~ are ~ho_w_n i~-~e
Fig 2E.35 (b).
:r.Fx =N 2 -N 1 sin 30° = ma
... (1)
:r.Fy = N 1 cos 30° - mg= 0
... (2)
,
= aBG
where B stands for block B, A for wedge and G for
gro~~s
~
N,·..
I +J!.
Bloc:ic B : ~o acceledrationsd~re sudpealrpose~ clino':1 it: itsd
acce1eranon re1anve to we ge aBA rrecte oni:jm e an
acceleration of :edge.-+
-+
.
Anurag Mishra Mechanics 1 with www.puucho.com
:_164___ --- . --- --- ·x-componenc of acceleration of B = a cos 0 - A
and y-component of acceleration of B = a sin 0.
Now force equations for block B are
D'x = N 1 sin 0 = m(a cos0-A) ... (5)
D'y =mg-N 1 cos0=masin0 ... (6)
We can arrive at tbe same result by considering eqns. (5)
and (6) instead of (3) and (4).
-
On substituting expression for Nin eqn. (4), we obtain
mg cos0- MA = mA sine
sin 0
. 0+MA
mg cos 0 = mA sm
sin 0
A= mg sin0cos0
m sin 2 0 + M
or
-- - --·--- - -- -- r - r
LJ=,~t;pHJ~J
37
l__>
'' In the F~. 2E.37 (a) shown, mass ;m, is being pulled on the
incline of a wedge of mass M. All the surfaces are smooth.
Find the acceleratiqn of the wedge._ .
A rod 'A' constrained to move in vertical direction rests on a
wedge B, as shown in the Fig. 2E.38 (a) Find the accelerations,
of rod A and wedge B instantaneously after system is released,
from rest, neglecting ftiction at all th_e contact surfaces.
'
. _JmL
m
M
F
B
A
Fig. 2E.37 (a)
B
I Solution : Fig. 2E.37(a) shows force diagram of tbe
wedge and the block. Let acceleration of block relative to
wedge be
a'.mM
=
or
-+
Solution: In the Fig. 2E.38 (b) dotted line shows initial
position of rod and wedge. If the rod is displaced vertically
through y, then tbe wedge moves a distance x.
y=xtan0
Therefore tbe relation between accelerations of rod and
wedge is
... (1)
a=A tan0
-+
= am - aM
-+
-+
= amM + aM
(amlx = a - A cos 0
, Cam\ = A sin 0
·_r
X
B
Fig. 2E.38 (a)
a'. and acceleration of wedge on ground is
-+
amM
-+
am
or
and
M
N,
+-·f
<ill-. - - - - - -
N sin 0
o:• 000
Y~~x~•
:8
'OZ
X
c.Pe:, ,·
<i'°' ,·' 0
/JI.
N sin Oco
N
z
F
'9,s,,.
~0
mg
AcosO
"·.
?,..··
8
A
••
,i·..
N'
N cos B
'\.a
y
r xf
N
:
A sin B
Fig. 2E.37 (b)
Equations of wedge:
D'x = N sin 0 = MA
D'y = N ' - N cos 0 - Mg = 0
img
... (1)
Equations of block :
D'x = F + mg sin 0 = m(a - A cos 0)
D'y = mg cos 0 - N = mA sin 0
MA
From eqn. (1),
N = -.sm 0
a
Fig. 2E.38 (b)
... (2)
Equations for wedge:
... (3)
... (4)
www.puucho.com
D'x =Nsin0=MA
D'y = N' -N cos0-Mg = 0
... (2)
... (3)
Anurag Mishra Mechanics 1 with www.puucho.com
-1651.
[ FORCE ANALYSIS
Equations for rod :
:r.Fy = mg - N cos e = ma
From eqn. (2),
N
... (4)
=-MAsine
On substituting expression for Nanda in eqn. (4), we
obtain
MA case
mg----=mAtane
sine
A=
mg sine case
or
m sin 2 e + M cos 2 e
mg tan e
=--''---~-
M +m tan 2
e
and from eqn. (1),
a=Atane=
mgtanze
M + m tan 2 e
FRICTION
A friction force arises when one body moves on another
and is always opposite to the motion. Friction plays an
important role in many transmission mechanisms, such as
belt, friction, rope drives, the motion is transmitted with the
aid of friction. In other cases friction opposes the motion and
leads to a useless expenditure of work.
Two types of friction are distinguished, depending upon
the form of motion: sliding friction, kind, and rolling friction.
As experiments show, friction is a complex phenomenon.
Here is a simplified explanation of sliding friction.
The surfaces of any contacting bodies have irregularities
[Fig. 2.36 (a)]
When one body moves on another the asperities of one
surface will interlock with those of the other. Causing their
deformation. As a consequence, tangential as well as normal
forces will develop at the surfaces in contact, as shown at
one of the points of contact in [Fig. 2.36(a)]. The friction
force is the resultant of these tangential forces. If the
asperities of the surfaces are in direct contact. We have dry
friction. When the surfaces are lubricated, it is fluid friction
[Fig. 2.36(b)]. Fluid friction is always much lower than dry
friction.
The Laws of Sliding Friction
I
1
Friction depends on a series of complex mechanical,
chemical and other phenomena. The laws of sliding friction
are the result of generalization of a great body of
experimental data. The basic laws of sliding friction are
presently formulated as follows:
1. The friction force is pmportional to the normal
pressure.
2. The coefficient of friction depends on the nature of
the bodies in contact and the physical condition of
the surfaces in contact. ·
3. Friction between similar bodies is generally larger
than between dissimilar bodies.
4. The friction force does not depend upon the a\ea of
contact, except at high unit pressures.
5. The static friction force is greater than the kinetic
friction force for most bodies.
6. The friction force depends on the relative velocity
the bodies in contact. In practice the friction force is.,
often assumed to be independent of the velocity is
the range of velocities encountred usually.
7. Coefficient of static friction depend on the material
of the bodies in contact, on the quality of machine
of contacting surfaces.
Analysis of Friction Forces
I
I-
(a)
.
-
I_ _
:
I
I
I
Fig. 2.37 shows a block of mass mg resting on a rough
surface. A horizontal force 'P' is applied to the block force P
is gradually increased from zero.
* When applied force P is very small, the block does not
move. From condition of equilibrium,
l:F'x = p - F friction = O; Ffriction = p
:r.Fy = N - mg = O;
N=mg
!
~-...;:::
~ -
iW!
I
(b)
L__________ _:~9: ~-36 __
m_
F'--'---<'---'--rough
J
I
(Fraction force)
mg
p (applied force) [
I
]
I
1
Fig. 2.37
____ _J
www.puucho.com
I
Anurag Mishra Mechanics 1 with www.puucho.com
166,,
+
+
- ..
MECHANl~S-1
Friction force counter balances external force, till
the block is static. This friction force is referred as
static friction (F, ). As external force is increased,
static friction also increases to its maximum value
f,max.
As applied force P is gradually increased, a limiting
point is. reached where friction force F, (maximum
value fsmax.) is not sufficient to prevent the motion
of block. When the block is about to move, the state
of motion is called impending state of motion.
At this point friction force has maximum value.
F,max. = µ,N
where µ, is defined as coefficient of static friction,
N . is, normal contact
· ¢,, is maximum angle of friction if If>,.; ¢,, then block is
static.
+ If applied force P is greater than F,max (µ ,N), the
block will have a resultant force F - fx on it.
Where A is kinetic friction force.
µk
Fig. 2.3s
· F=µk N
A =µkN
~
~
coefficient of kinetic friction
N
Normal contact force.
The block will accelerate in the direction of
resultant force.
Fig. 2.40 shows a
N
block of mass m,
kept on an incline
plane whose angle
of inclination can be
varied: At certain
value of ¢,, just
sliding of block
starts. At this instant _ _ _F)_g. 2.40
friction force at its
maximum value F,max. the equilibrium equations
I>
if~~:~e Pis greater than
IF
Fmax., the block will have I
a
resultant
force "·
P - Fmax. on it. The block
-----··;1.·,~,----Dynamic
will accelerate in the iL
direction of resultant ~· ta·
force
when · sliding
45°
"-'-'--~--..,.P
mdtion ensues.
a
are:
+
where µ 1c is d¢fined to be coefficient of kinetic
friction.
·
·
Fig. 2.38 shows variation of friction force versus
external force graph.
When condition of impending motion or sliding is
not known. To determine friction force we assume
static equilibrium and solve for the friction force F,.
The possible results are:
(a) F, < µ 8 N (maximum value of friction): Body is
in static equilibrium. The value of F, can be
determined from the equations of equilibrium.
Jb) F, = µ, N: Body is in impending state or about.to
move assumption of static equilibrium is still valid.
(c) F,,> µ, N : This condition is impossible. Friction
force cannot be greater than F,max. (µ, N). ·
+ Normal contact force N and friction force F are two
components of the resultant contact force R of the
surface on the block. Angle between resultant
contact force R fill.d contact force N is called angle
of friction(¢,).
+
r.Fx = mg sin¢,, -µ,N = 0
Friction force opposes relative motion between two
surfaces. In order to decide the direction of static
friction, try to imagine the likely direction in which
the· body will tend to move; friction force is opposite
to it. In the figure, force P pulls block B towards left
and A is pulled towards right. Friction force on B is
towards right and on A is towards left. Important
point to notice is that for two contact surfaces
friction force is in opposite direction. It is intern.al
force for two contact surfaces, so it must be an
.. ,
...._+-_,_- fe
i
.
ma9'
_Fig. 2,41_;_____ · - - - - - - "
~
f
,___ _F_;;ig. 2.39
'
'
.-,.~-"--IA
T
tan¢,='~; R=~f 2 +N 2
When block is in impending
state maximum value of
static friction force is acting
on block.··
f=F,,;, .._ =µ,N
l
+
Consider a conveyor belt moving with velocity v A. A
small block is gently lowered on it.
-+
--+
--+
-+
VBA =VB -VA=O-VA
Velocity of block B relative to surface A is towards left;
friction force is opposite to VBA, i.e., towards right.
Due to this friction force, the block accelerates
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
- - . ----.;:::;-,
FORCE ANALYSIS
~
1671
Solution: Force of friction = 0
=>
N =0
towards right and the belt retards. Finally the block
acquires the velocity of belt and moves with it.
.. ·1-:-7
J~='~g~_J;~-l~,cl
39
1..>
=>
f=tan0 =>a=gcot0
a
I
A block weighing 20 N rests on a horizontal surface. The:
coefficient of static friction between block and surface is 0.40
and the coefficient of kinetic friction is 0.20.
;
How large is the friction force exerted. 011 the block?
(b) How great will the friction force be if a horizontal force of
( a)
5 N is exerted on the block?
(c) What is the minimum force that will start the block in:
motion?
(d) What is the minimum force that will keep the block in.
motion once it has been started?
(e) If the horizontal force is 10 N, what is thefrictionfprce?
F = ma = Mg case
k.J~.x:g_t;.ti.l!i!l:;;;.--::Q_':-,..._
=
·- .
~~~
·····--··7
A black of weight W rests on a rough horizontal plank. Thel
slake angle of the plank 0 is gradually increased upto 90°. ;
Draw two graphs both withe along x-axis. In graph show the:
ratio of the normal force to the weight as a function of 0. ,
In second grapl~ -show ·the ratio of the friction force ta the!
weight. Indicate the region of 110 motion and where motion,
exists.
Solution
Solution: (a)
I
N
When
- I
Fmction
i
=0
From condition of equilibrium,
P=F=O
(b) First we calculate
mg cos a
_F_lg. 2_1=:~1
= (0.40 X 20)
Till block is static mg sin 0 =
mg
P=F=SN
I
f,; ~ = sine
mg
As incline angle is increased, if block does not move
friction force has balanced component of weight down the
incline
In impending state of motion
mg sin0 0 = µ,mg cos0 0
tan0 0 = µ,
r.Fx = pmin. - Fkinetic = 0
or
Pmin. = F!cin,ti, = µk N = (0.20)(20 N) = 4 N
(e) Since P > µ, Nin this case, the block accelerates.
From Newton's second law,
:r.Fx =P-µkN=ma
Therefore
F = µk N = 4 N.
40
increased
:
I
(a)_
Fig._2_E.39
(c) When the block just starts to move, it is in
impending state. From condition of equilibrium,
:r.Fx=P-F=x_=O
or
.P=F=x_=µ,N=BN
(d) When block is in motion, F = µ kN.
Minimum force will be required to move the block with
constant velocity.
From condition of equilibrium,
lc~S~-S¼W,B!iJ
angle of incline is
being gradually
----mg
F=,.=µ,N
=SN
Since P < F max., block is in static
equilibrium, i.e.,
. I
..!'!.
mg
- l f - - - - ~ - - - . . . 1 . . . ._
goo
Fig. 2E.41 (b)
[>
-
Th~
A wedge of mass M m~es an angle 0 with the horizon~!.
wedge is placed on horizontal frictionless surface. A small'
block of mass m is placed on the inclined surface of wedge. ,
What horizontal force F must be applied to the wedge so that!
the force of friction between the block_ and wedge_(s ,._ero ]__ ;
www.puucho.com
-
-.
----
_.e
1,'
Anurag Mishra Mechanics 1 with www.puucho.com
-·-·· -- . -
j168
--- ·-····
--------------- - -------------- ---...!...
Impending state
mg
MECHANICS-I '
Tmin=40N
:50N-[I
1----,,L-'--4~
I
- -
,
'/ µsCOS8o ,i---·-----:a,,/
'
I µ,cose 0
-
r
(f,)max•40N
(c)
. mkcos00 j
Static
T•40N
i
sine
motion
occurs
50~
f, •10N
e
90°
(d)
Fig. 2E.41 (c)
I.
Fig. 2E.42
When block begins to slip
fk =µkmg cose
fk
- =µk case
w
Thus block A remains static
Force F can not pull block A
1-····-- -------r-i-
E._f;~~'~'~,P.-'~- i 43 1>
[~>f~'L'gfg~J42!>
r
--- -
- -
---,
:Find the acceleration of the block and magnitude and:
'direction offrictional force between block A and table, if block I
:A is pulled towarq~ !eft iyith a forq, pf !jO /'{._
1'
.
And friction
force is (10~ N
I
}--X
I
!
;11ie 10 kg block is resting on the horizontal surface when the
force 'F is applied to it for 7 second. The variation of 'F with
;time is shown. Calculate the maximum velocity reached by the
lblock and the total time 't' during which the block is in1
!motion. The coefficient of static and kinetic friction are both,
;a.so.
'
µ•0.8
g•10m/s2
F(N)
100 ...... .
B
4 kg
I
Fig. 2E.42 (a)
L -- --- -
-·-'
Solution: Case (i) If block moves down, maximum
possible tension T = 40 N is attained when it moves with
constant velocity. In this case
N
---··
[
50 N
Fig. 2E.43 (a)
I_
A
0
'---'---'------+ t(s)
4
7
l
/
40N=Tmax'
Solution : Block begins to move when
F=µN
40N
= O.Sx lOx 10 = SON
50 N
i . - --- --
From
t
=O to t =4sec
F = 2St
Fig. 2E.42 (b)
Tmax can not over come apposing forces of 90 N,
therefore it is not possible.
Case (ii)
If F=SON force can pull block A to left, mm1mum
tension in string Tmin = 40 N if B moves with constant
velocity.
From
t=4tot=7sec=40N
Block begins to move at t =2 sec.
after that
F-µN= mdv
dt
2St - SO = 10 dv
dt
:41:NF:
'
µ
'
'
mg
Fig. 2E.43 (b)
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
' - FORCE ANALYSIS
169
V
4
0
2
f dv = f (2.St - 5)dt
or,
V
Concept: Kinetic friction is opposite to relative velocity
it opposes relative motion. When horizontal component force
is reversed, relative velocity is not_changed therefore, direction
of kinetic friction does not change.
-0 =12-~t2 - 5{
Fcoh-'
Fsin37°
= (2.5 X 8 - 20) - (5 -10)
= 5m/s
4 sec.; block retards due to greater friction
V
After t
force
=
Stage 2:
v = .Jl 12 m/s
µ,N
a
----+Uinitia! = 5m/s
,._
F cos 3 7° is reversed, block continues along original
direction, but due to retardetion created by µ kN and
F cos 37° block travels till it stops.
-(Fcos37°+µkN) = ma
-(20x .:': + (0. 25 x 8) = 2x a or, a= -9m/s 2
- - 0 - F a 40N
µ,Na 50N
Fig. 2E.43 (c)
a= 50- 40
5
= lms-2
Displacement of block in this phase
10
0 =v
Velocity of block at t = 7 sec
at
t = 4sec, v; = 5m/s
v 1 =v;+at
= 5 - 1 x 3 = 2 m/s
2
v2
-
2as;
s =-
2a
(112) 56
=--=-m
2x9
9
Stage 3: Which block returns its a acceleration is:
Fcos37°-µkN = mg
2
a= 7m/s
Fsin37°
~
LE~ff~J!tl:?l~ .~
44j;._>
'
A force of 20 N is applied to a block at rest as shown in figure.
After the block has moved a distance of Bm to the right the
direction of horizontal component of the force F is reversed in·
direction. Find the velocity with which block arrives at its.
starting point.
Velocity of block when it
returns to original position
v2 = 2as
56
)
=2x7x(s+
9
Fco~
~
mg
Fig. 2E.44 (c)
16-.fi
v=--m/s
3
-~7'
-
µ•O.~
-
-
-
.-~
lE?ffl~BL~ ! ~__;>
Fig. 2E.44 (a)
,Find the contact force on the 1 kg block.
() ':,
Solution: Stage 1: Motion till force reverses its
direction
N = mg -Fsin37°
Fsin37°
= 20-20X~= SN
5
Fcos37°-µkN
= ma
2ox .:':- 0.2Sx 8 = 2x a
5
2
a= 7m/s
• • Ijj
. µ,N
'
Vs;;.;
'<
.
Fig. 2E.45 (a)
Solution : fk = µN
Fig. 2E.44 (b)
Velocity of block after
displacement of 8 m
v = .J2 x 7 x 8 = .J112 m/s
N
4
=0.Sx10X-=4N
5
mg•20N
N=lxlOx.:':=sN
5
Contact force = ~ fk2 + N 2
= .J16+ 64
= 4-JsN
www.puucho.com
37'
10
Fig. 2E.45 (b)
Anurag Mishra Mechanics 1 with www.puucho.com
MECHANICS-i .
- - - - - - ,M~-----<•<
Concept: Contact force is res.ultant of normal reaction 1
'"s
m,,C:J::F
and friction force.
l~~~q~Fl~,·~
Fig. 2E.47 (a)
Solution : In the impending state
,Blocks A and Bin the Fig. 2E.46 .(a) are connected With a
'string of negligible mass. The masses are placed on an inclined
plane of inclination 30° as shown inFig. 2E.46 (a) . If A and
B each have mass m and µ A
= 0 and µ n =
J1, where µ
N
f,
A
and
50 N
}"espectively, calculate the acceleration of the system and
tension in the string.
Fig. 2E.46. (a)
F = kt =µ,mg
2k=0.4x5xl0
or,
k = 10 N/sec
When force F is further increased, block accelerates
kt -µkN = ma
Sa= lOt -15
or,
a= 2t - 3m/s 2
Solution : If system is moving down with acceleration
a for block A
I~~
~
mg
2
'
Fig. 2E.47 (b)
µ n are the coefficient offriction between plane and the bodies
.
.F
a
f=-yf'mg
~ o s 3 0 ° = l.mg
300
;/2
1 m/s2
t----,
t= 2·sec
Fig. 2E.47 (c)
Fig. 2E.46 (b)
mg -T=ma
2
mg +T_mg =ma
2
-.fz
Concept:
... (i)
... (ii)
Static friction:
l
The direction and;
magnitude both are self adjusting such that relative motion is 1
opposed.
!
( a) Direction: It acts always tangentially to the contacti
surface.
solving eqns. (i) and (ii)
a=½(1- Jz)
Fig. 2 42 (a)
1
. l,
This example is to show that friction acts against _the:
tendency of relative motion._ .
'
T= mg
2-.fz
,~,lg®:
~J~~f!-~~J~1~
iln the Fig.' 2E;47 (a) shown a time dependent force F
·expressed as F = kt is applied on a block of mass 5 kg.
.Coefficient of static and kinetic friction is µ, = 0.4 and
·µ k = 0.3. Motion begins when t = 2 sec draw a acceleration·
'vs _time graph for block. (m = 5 kg,µ, = DA µk = 0.3)
www.puucho.com
f =~F2+p2
'
'
y
Fig. 2.42 (b)
Anurag Mishra Mechanics 1 with www.puucho.com
-
-
171
FORCE ANALYSIS
(b) Magnitude: Maximum strength of the joints formed is
directly proportional to the normal contact force because
higher the normal contact force higher is the joint
strength i.e., f, max ~ N
It al.so depends on the roughness of contact surface.
f, max (al.so called fumufr,g) =µ,NJ
·
Magnitude of static friction is self adjusting such that
relative motion do not start.
Find unit vector in direction of friction force
acting on block
-)
AA-+
Vp=7i-2j, VB=3i+j
-+
A
Fig. 2E.49A
A
A
Solution: v 81 p = 3i + j- (7i - 2j)
•
rk
=
•
-V B/P
4:
=- - l
5
3:
+- J
5
It is not self adjusting as in static friction.
fixed.
m = 20 kg,µ, = 0.5,findftiction on block.
/4100
~--
(1)
Fig. 2E.48 (a)
J-so
Solution: N+60-mg=0
N = 140
f,max = µ,N
f, max = 70, hence answer is 70.
A
itN
mg
Fig. 2E.48 (b)
~~Vp
AA
(2)
(3)
(4)
(5)
A = µ kN. It is
Concepts:
Value of µk is always less than µ,(µk < µ,) from
experimental observation.
If on(y coefficient of ftiction (µ) is given by a problem,
thenµ, -µ k = m (assumption for)
Value of µ, and µ k is independent of surface area it
depends only on surface properties of contact swface.
µ k is independent of relative speed.
µ, and µ k are properties of a given pair of surfaces i.e.,
for wood to wood combination µ 1 , then for wood to iron
µ 2 and so o_n.
, --E-xample i 50' -~
~---' --:_: ,:_: -" -~-' '··-·---·~
Find 0 at which slipping will start. µ, is coefficient of static
ftiction. (Angle of repose)
_
L·
Blocks are given velocities as shown at t = O,find velocity and
position of 10 kg block at t = 1 and t = 4.
4-12m/s
10
Fig. 2E.49 (a)
Solution: N - mg cose = o
f, max =µ,mg cos0
when slipping starts f, = f, max
Thus mg sin 0 = µ mg cos0
tan 0 = µ,
tan-1 µ, is called angle of repose.
g=10m/s2
l =0
µs = µk = 0.4
Fig. 2E.50 (a)
Solution: How a student will approach making FBD.
-·
Fig. 2E.49 (b)
Direction of Kinetic Friction
It acts when there is relative motion between two
surfaces in contact.
Direction: It acts always oppositely to the relative
velocity.
-
mis
5m/s
!.-..0._fk
w.r.t.B
~
~
Fig. 2.43
Jii.
~
__Fig. 2E.50 (bl__
40+T=l0a; 50-T=Sa; a=6m/s 2
u
= 12;
a= -6
v = 12-6xl = 6m/s;
S = 12xl-3xl = 9 m
But it is wrong. Since velocity has changed the direction
during motion friction would also have changed thus
direction and acceleration will change.
u = 12; a = -6
(till velocity becomes zero)
v=0 => t=2sec; S=2x2-3x4=12m
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
-·
NowFBD
50-T= Sa
T-40=l0a
10
f
JB:oN
~T
= 40 •
a=~m/sz
(4) Now check if this acceleration is possible by
verifying f ~ J1 i.e., make FBD of
2
4
u=0,a=-,t=2,v=3
3
S=
3
3
3
I rn_l~2:
f ~ 30
3
30- f = 5 x 2
or
f = 20 < 25
Concept: Friction oppose rela_tive velocity not relative
f = 10x2= 20
L~~~~~:~-~- r-;;-t>
acceleration.
.
a
Find acceleration of blocks F
f •
-2
.!. x ~ x 4= .'.!; Total displacement= 12-.'.! = 10~
2
60 N
= 40 N
0
µ 1 =o.s-
a =4 m/s2
Fig. 2E.53 (a)
. --~·
Solution: Assuming same acceleration
, µ, = 0.2 .........
Fig. 2E.51 (a)
.~(
Solution: (1) First of all find values of limiting friction
at all contact surfaces. CJ, max)
0
60-f=Sx4 or J=l0x4=40
f = 40> 25
hence our assumption is wrong.
fsmax=25
"'l"~~,J~,"
'
= 30
,, 60
•2E • \
(2) Maximum force upper surface of 10 kg can
experience is 25 N so it will not more relative to ground.
(3) Hence only 5 kg will move.
a1 = 7, Oz = 2.5
Fi!!: 21t,53 (b)
t -
-· - -· ·- - - -
~
k:~0:9"''22,~l~ .l~~-•>
aA=3,a 8 =0
Solution:
·
Fig. 2E.51 (c)
25-m
l' !e;?~f~-riiel~. r;-27>
_30~~!;:t
-ill-
:: :
Fig. 2E.54_ ....
A
if they are moving together a 1 = az = a
30 N
smooth~
F-f-30=l0a; f=5a; F-f-30=10x1. ·
5'
F = 30 + 3/ maximum f is 25.
F=30+75=105
,
Fig. 2E.52 (a)
Solution :
\
-
Find m(l,!Cimumforcefor which they can moyf_tgget!Jer..
25____[D--- 40
1.:7 .
I 10
25
25••-'-----'1
5
~ 25
Fig. 2E.51 (b)
, µ, = 0.S
a= 30 = 2ms-z
15
Fig. 2E.52 (b)
Fig. 2E.50 (c)
3
MECHANICS_;i-j.
IT] /1 max = 25
Two Block Problem
..
- ------.
~ fzmax =0
LJ=:;~~g.fD,~J~ i 55 L->
,
(1) 10 kg block must move because some force on upper
surface will act on it.
(2) B can either move with same velocity and
acceleration as A or it can move relative to A.
(3) Always assume it moves with A and solve.
,.
as:
_<;:onsider two b.locks with friction coefficient and mass
shown in Fig. 2E.55 (a).
.
www.puucho.com
I
Anurag Mishra Mechanics 1 with www.puucho.com
I~--------------------FORCE ANALYSIS
- ~_71]
r-- -_ .:·""""·-""'"' ""'·: -_ - ----- --·- -
SE
I
I
I
I
'
I
· mmm
13
I______ __!,._,=,..
I
a
0.2, mA a 2kg
1
F-6=2x-
'
Solution: Force is applied on blockA
!qfl--A
I
A
F
'
= 6-5 = "I.m/s 2
meg
L_ - - --
. - ··-
5
NA
and N 8
6-/=2x"I.
5
f1 max =µ,mAg=4N
f2 max = µ2Na = µ2CNA + m 8g) = 15N
(i) If motion of B is to take place
.-"--~ ·- .. 1
8
f 1 > Ji which is not possible in this : 12 ,
j
case. Therefore block B cannot be :
Fig. 2 E.SS (c)
moved by applying any force on A. •
- - - Thus only A can slide, it just begins to slip when F = 4N
(ii)
Let µ 1 = 0. 3 and µ 2 = 0.1
now
f 1roll = 6N
f2 max = SN
Now, f, mM > f 2,,,,,; blockB can be moved find force F for
I
f-+t,
----
which slipping occurs at any surface.
- -
'
I
1
A f---+F :
B
f= 6-.~
or,
5
= 28N
5
c:tt->5!im,lilg ~i58l>
Coefficient of friction between 5 kg and 10 kg block is 0.5.
!applied on 5 kg. The fio.9r is frictio11le_s~, _
----~
20•41~--~
10kg
Fig. 2E.55 (d)
As force required to cause slipping of A is more than that
at B, slipping starts at B. Blocks A and B. move in
combination. For slipping to start
F=f2max =SN
- -
-
-·
ta) In example 55 what is maximum possible acceleration of,
j(b) fn°example 55 what is mcu.imumforce F for whichblocks1
_ ______ _J
Solution: (a) Maximum force that A can exert on B is
/1
max
= 6N
Thus,
a8
max
= 5 - 5 = "I.m/s 2
3
3
110kg , • :
2 m/s2:
Fig. 2E.58
I
F-20=5X2
F = 10 + 20 = 30 N
LE,~o.r,q~~~J
~~~111riJ~~v
L_rr,pye)n_cqm/1iJ'!..a_t(on.___
·
Solution : First compare friction with force with µN;
f < µN implies.
20 N is static friction so there will be no relative motion
between blocks and acceleration of both will be same
J '
1-
If:
'friction between them is 20 N. What is the value offorce being,
I~~,
I
5
From Newton's second law on blocks A, we get
-
Find f max that exists at each surface and
-----
J57 [>
I
Fig. 2E.55 ( b)
--
Solution : For this force both the blocks move in
combination acceleration of system
I
I
I.
.I.11_exqmp/e !i!i_fil_lg.frictiol_l_fosce b~tw~~n_ !,locks if F = 6N._
NqfJ!
Na
B
mAg
I
Fig. 2E.56 (b)
3
I~E:~F~t11-~l.~
_,__,..,,_,.,_._,,""=--. __,.,._---=-~"--'-
! 6
F= 20N
or,
,
Fig. 2E.55 (a)
::u=:,__--,
i ~ F-;
3
.,----µ•0.3,m 6 •3kg,
mh,fir
II
l
.,----µ,
59
b>
!An object is given a quick push up an inclined plane. It slides:
:up and then comes back down. It is known that the ratio of
·the ascent time (t up ) to the descent time (t dawn) is equal to the I
·1coefficient of kinetic friction (µ). Find the angle e that the,
inclined plane makes with the horizontal Find also the range!
1
,ofµfor which the situation described is possible. Assume. that;
-the_ coefficients_ of static_and_ kjneti~ fric_tion a~e_equal.
_i
Solution :
aup = g sine+ µg case;
= g sine - µg cose
L = "I.[g sine+ µg cose]t~;
ad,wn
Till this moment blocks A and B more in combination.
(b) Fmax can be obtained by applying Newton's second
law on upper block
www.puucho.com
2
Anurag Mishra Mechanics 1 with www.puucho.com
-MECHANiC:S:fj
'"~---,-'
So, Resultant force = ~ fk + N 2
2
L = .!_[g sin0 - µg cos0] tJ,wn
2
(sin0+µcos0)µ 2 = sin0-µcos0
µ(1+µ2)
tan e = '--'--'--c--'(1 - µ 2)
;E
=)(µN)2+N2 =N)1+µ2
= mg cos0~1 + µ 2= 12-F, N
.
.
,---,,
iJ;;lff½~:EiJ~ I. 62 J.>
.
µ <1
. . ,- . r.:7 ...._
hi,~~~m?Ii?i~
.1
60 f ~
A time varying for F = ·10../2 t starts acting on the
3 kg block kept on a rough hori,;ontal surface (µ = 0.2) at
t=0.Find
1
(. a) the moment of time when the blocks leaves the surface
I'
.
'(b) the moment of time whm.the<horizontal motion begins. :
'
·~;._ '
!
,µ=0.2~
'
. .
.....
I
Solution:
Concept: Car must stop within the maximum ·vi.sible!
,safe distan_ce.
a= -µg, v f = 0, s = I
2
2
2
VJ-V;= as
Fig. 2E.60 (a)
Solution: From FBD of
block calculate N
'. ~ N
Fsin45°=10t
N = 30-lOt
I
•
(a) The
block
leaves \
Fcos45~=10t
the contact with surface,
'. f
l
.
I
when N=0
W=30
I
t = 3 seconds
Fig. 2E.60 (b)
(b) The
block
begins
horizontal motion, when
F cos45° = f max
l0t =µ(30-lOt)
lOt = 0.2(30-l0t)
1
t = -sec
.
A car has headlight which can .illuminate a horizontali
straight road in front upto a distance L If coefficient offriction
between. tyres & road is µ. Find the maximum safe speed of the
car during a night drive neglect the reaction time of the;
driv.q.
.
._ .
.
.
.
0
2
-u!ax = -2µgl
i
2
1
·A block of mass 3 kg slides on a rough fixed inclined plane of
1
37° angle having coefficient offriction 0.5. Find the resultant,
force exerted by plane p_n the blocf5..
/
'
:
f
¥)
, mg sine
37° mg case
If angle
of incline is greater than angle of
:repose friction force is kinetic in nature.
I
'
•
'
-
-
-
--
· -
&
tanq, S: µ,
A =12N
-
" " " " ' _ ,_ _ _ _
--
--
J
·o-!
·--
Solution: External force = 7 N
External force is smaller
maximum friction force.
f, = 7 N
Hence,
I . .. . .. . . .
·r,-:i-
.
•
1>
7N
; f5
than
mm~.
•mm,
Fig, 2E.63
'--•'
- . ..
···- - I
Block 1 sits on top of block 2. Both of them have a ma,ss bf 1'
kg. The coefficient of friction between blocks 1 and 2 are
µ, = 0.75andµk = 0.60. Thetableisfrictionless.AforceP/2[
is applied on block l to the left, and force Pon block 2 to the.
right. Find the minimum value. of P such that sliding occurs:
between the two__blof.¾.?.,. ..
Solution:
Solution
Conc!!pt:
- -
Max. friction force= 0.4 x 2 x 10
=BN
Fig. 2E.61
'
-
a/
.
•
,,-'.">
.
= .J2µgl
A block of mass 2 kg is placed on the floor (µ = 0.4). Ai
;horizontal force of 7 N is applied. on the. block. The force
:.frictiqn};,etweeri t~e bJgck & floqr 4 J,. Find .the.J, ~- __
t:. ,_!;=..~,f¼.~~i~'j 64
~;i;;~!i~~L~ ..1617::__>
N
Umax
P- f = la
f-P/2= la
P/2 = 2a
as f is static
f = 3P/4 5 µ,mg
µ,mg= 0.75 X 10 = 7.5
4
Thus
P 5 -µ,mg
3
P=lON
www.puucho.com
Fig.2E.64
Anurag Mishra Mechanics 1 with www.puucho.com
FORCE
____ ANALYSIS
,...,,_,·"·~------
- .. ··--- -·-·- ---- -----·----
1
".,,.__.
A block of mass m rests on a rough fl.oar. Coefficient offriction
,between the block and the fl.oar isµ.,
(a) Two boys apply force Pat an angle e to the horizontal.
One of them pushes the block; the other one pulls. Which,
one would require less effort to cause impending motion·
of the block?
,Cb) What is. the minimum force required to move the block byj
pulling it?
i
(c) Show that if the block is pushed at a certain angle 0o, it':
_ . canJ!pLb.e _rr,Qved_wh<I_teyq _tltf _yalu~ of_l' be. _
The vector triangle of forces is shown in Fig. 2E.65. The
minimum value of P will occur when the lines of action of P
and Rare perpendicular to each other, as shown by the force
P".
and
µ,mg
P=------cos0m +µ, sin0m ·
1
cos<j>, = - - - J1 +µ;
sin <I>,
and
Solution:
(a), _Equation~_ f~r_ pullin; _!o~ce :
Therefore
! aI
_...,__.,_p·ro:
YLi
r
mg
x
is
p = -~µ~,'-m-'g'-cos 0 +µ,·sin 0
which shows that latger force is_ required to push the
block. Note that normal reaction in case of pushing is
greater than that in case of pulling force. Consequently
friction force is increased.
(b) The body is in equilibrium under the action of three
forces: applied force P, total reaction R and weight mg.
~
I
Pmin
',~
µ, mg
P=-~~~-cos0-µ, sin0
When cos 0 - µ, sin 0 = 0 or cot 0 = µ,, the force P
tends to infinity, i.e., the block cannot be moved. Secondly,
force P must be positive so that it remains the pushing force;
therefore cot 0 :c, Ois the required condition.
~.J;,~91E:(3;1g~
Equations for pushing force : From conditions of
equilibrium,
Ux=Pcos0-µ,N=O
... (3)
UY = N - P sin 0 - mg= 0
... (4)
On eliminating N from eqns. (3) and (4), we obtain
p = -~µ-','-m-'g'-_
cos0-µ, sin0
mg
µ,mg
(l+µ;)/J1+µ;
µ,mg
J1+µ;
(c) From (a), pushing forcePrequired to move the block
I
From conditions of equilibrium,
Ux = P cos 0 - µ, N = 0
... (1)
Uy = P sin 0 + N - mg = 0
... (2)
On eliminatingN from equations (1) and (2), we obtain
µsN
P=·
= -"======
Fig. 2E.65 (a)
_fil
µ,
= -~'-Ji+µ;
P sin Bp
N ...
mg
µ
I
A block of mass m rests on a b;~~ke; ~J-mass ~ -Th~;
coefficients of friction between block and bracket are µ, and
·µ k . The bracket rests on a frictionless surface. What is the
maximum force F that can be applied if the block is not to
slide on the bracket?.
1
F!~ 2E.66_ (_a)_ _ _
Solution: Block and bracket must have common
acceleration in order to move in combined form. For the
system shown in Fig. 2E.66(b) , Fis external force; therefore
block and bracket will have acceleration in the direction of E
~-~:·:
I
" P><<!,,mg I,~sR
,,
Direction of R
-+
mg
'
I
·~
Fig. 2E.65 (b) -
mg
-
,
\
N mg
'
m
'
\. I
M ,,
'~----·--
..,,"II,
www.puucho.com
I
------
_ Fig.2E.66(b) ________ ___ --·-·-
Anurag Mishra Mechanics 1 with www.puucho.com
From Newton's second law,
Equations for block:
LP,·= µ,N -F =ma
LPy = N - mg= 0
.
Equations for bracket:
LP, = 2F - µ,N =Ma
or P
... (1)
... (3)
... (4)
On solving eqns. (1) and (3), we obtain
F
a=---
M+m
N = mg ,
Therefore the block will be in static equilibrium for
mg
,;;p,;;
mg
sine-µ cose
sine+µ case .
From eqn. (2),
On substituting a and Nin eqn. (1), we obtain
F = (µ, mg)(M + m)
(M + 2m)
a
l:,E,x(l:_~m:J e~ 67 ~d>
·
~
~lA block of mass 'm' Is supported oit arough wall by aj,p(yjrig a
l..<1 blockis ke'pfiln rough mcline \.I/hose a~1e ofinclinatTa;;
lg,:eater· thcin;ang;le of repose'.
. '
·. · ·
'(a) Find the minimum and m<Vf mum fore~ F applieq pc1rnllel
to incline. "t.hat will keep. if i!} ~quilibrium. . . ·. ;) ,,
fb) What is the ,required force if it is applied 'non71alto_<the
.....:::.:.::~ > )~£?~~!
r
. ,~~~im::~•' · . •.
!,ore~ P asshov,m inFig. 2E:67 (a/Coeff!.dentof static.fri,ctioTJ,
!between blocJs:,and iyall isµ,. Eopyhµtrange ofvables,of:P;
incline~:";,.
".·1,
.(c)
p
Fig. 2E.67 (a)
--------·- ""'-~:..-~==--
-,.··1,,J
Solution: Impending state of motion is a critical
border line between static and dynamic states of body. The
block under the influence of P sin e (component of P) may
have a tendency to move upward or it may be assumed that
P sin 8 just prevents downward fall of the block. Therefore
there are two possibilities:
Case (i) Impending motion upwards : In this i:ase
force of friction is downward.
. . . . • ,.'
I
~'
l
IP
· _.··•~ N
.
cos o
l.·
~Y-
x .
-P-c~qs°"·~··.
·-
,,.·
'}
~
i
(•~~e_ _ ___,
µN
F/4,s
. /4~f
.
-~e
mgcose
iLl-.-~--Fi_g._2_E--6~----L
I
,
.
Solution: According to condition of problem, the angle
of incline is greater than the angle of repose; therefore the
block will slide downwards. An external force can keep it in
equilibrium. We will consider two cases:
(a) Case (i) Impending motion downwards : In
this case force F first prevents the· block
y ~ . 'Jfx_-., -.
from slipping downwards. This is the
N ·V ·
minimum value of required force ·R
Friction force acts · upwards. From
conditions of equilibrium,
LP, =F+µN-mgsin8=0
c--::--
!
~
'\
~
,
----------··
·
(c) vyhat !s,the ~ange ofF if zt. is~applfed horizontally 9/Jc/he
· · block?:cL:. ..
.
·· · '·' · ____ , '\' ·
·in
i
mg
sin e - µ cos e
Case (ii) Impending motion downward : In
this case friction force acts upward.
:r.F, =N-P cos8= 0 or N =P cos8
LPy = P sin·e + µN - mg= o
or
P sine+ µP cos e - mg= o
p. =
mg
or
mm.
sine+µ cose
... (2)
. , LPy=N'-N~Mg=O
,;
=
max.
l.
mg
I
, -mg,
I
, µN
I
P sih,{)
l ---~------~~~---··-Fig. 2E:sr (bl
..~
From conditions of equilibrium,
:r.F, = N - P cos 8 = 0
or N = P cos 8·
LPY = P sin e - µN - mg = o
or Psin8-µPcos8-mg=0
... (1)
LPY = N - mg cos e =
o ... (2)
From eqns. (1) and (2), we obtain
Fmm. = mg (sine - µ cos 8)
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
.. - . ·1
_-- _,.177 _,
Case (ii) Impending motion r· -- - - - -· - · I
upwards : In this case, force F is large
enough to just push the block upwards.
':
This is the maximum value of required
I
c}<:.-~
.1'
force F. Note that friction force will
'I, '
reverse its direction. From conditions iI' "'"'
. ~ ':
of equilibrium,
l_ Fig. ~-68 (c) I
Ll'x = Fcos e + µ,N- mg sin 8= 0
. ~YVf:
: ~
... (3)
Ll'x =F-µN-mgsin8=.0
Ll'y = N - mg cos 8 = 0
... (4)
From eqns. (3) and (4), we obtain,
F"""' = mg(sin e +µcos 8)
Therefore the block will not slip if
mg(sin e - µ cos 8) ~ F ~ mg(sin 8 + µ cos 8)
Note that when force F is increased from its minimum
value the friction force is reduced from its maximum value
µ,N .. When F equals mg sin 8, friction force is zero. Block
will have a tendency to move upward only when F equals its
maximum -value. Static friction is a variable force; its
magnitude can change and, as the example illustrates, even
its direction can reverse.
(b) From condition of equilibrium,
Ll'x=mgsin8-f'."0
... (1)
Ll'y=N-F-mgcos8=0
... (2)
Note that, due to external normal force the normal
reaction increases, thereby increasing friction force.
Therefore for minimum force F we must have maximum
friction force µ,N.
Thus
mg sine= µ,N
or
mg sin 8 = µ, (F + mg cos 8)
=N
+ F sin e - mg cos e
From eqns. (1) arid (2),
(sine-µ, cos.8)
F . = mg
mm.
(cos 8 + µ, sin 8)
LFY
=o
... (1)
... (2)
Case (ii) Impending motion upwards : When
force F is increased the block has a tendency to move
upwards. Therefore friction force changes its direction to
downward. From conditions of equilibrium,
Ll'x =Fcos8-µ,N-mgsin8=0
... (1)
Ll'y = F sin e - mg cos e + N = o
... (2)
From eqns. (1) and (2),
F=mg (sine+µ, c~se)
case-µ, sme
Thus the range of force P for which the block remains in
equilibrium is
mg (sine-µ, cos8) ~F~ mg (sine+µ, cos8)
cos8+µ, sine
case-µ, sine
~~~E,Kq.ta~C:>
-;--·-····------"·
'
'
'A wooden block" slides down the right angle channel as shofvn
)in Fig. 2E.69 (a). The channel is inclined at an angle 8 w.r.t.
:the horizontal. The,angle a is 45°, i.e., the channel is oriented
•symmetrically with the vertical If the coefficient of friction.
,between the block and the channel is µ k, find the acceleration
;of the block.
F n,;~ = : (sin 8 - µ cos 8)
or
(c) Case (i) Impending motion downwards:
Block has a tendency to slip downwards and external force
just prevents it from sliding. In this case Fis minimum. From
_ _
conditions of equilibrium, ____ .
'-
~
I
I
I
I
I
1·
I
Fnction force
-<'.J';·
"(".)..., N
F~
'< ,},,."'
is upwards
for i~pending l
motion down ,
~
,,p,
N
µ
mgcos0
µN
;
I
0
Frtctioll force is dowOwards
for impendiilg motion up
'I
Fig. 2E.69 (a)
· Solution: The block is kept symmetrically in the
channel, therefore normal reactions on both its surfaces are
equal in magnitude. If the channel had been on a horizontal
surface, the reaction would be vertical_ [see Fig. 2E.69
(b)(iii)J, since the charmel is inclined to the horizontal
surface. Net reaction is normal to length of channel AB.
Since the charmel is symmetrical,
Net reaction
Fig. 2E.68 (d)
www.puucho.com
N 1 =N 2 =N
2N cos 45° = mg cos 8
Anurag Mishra Mechanics 1 with www.puucho.com
-MECHANICS~ij
From conditions of equilibrium,
y
x.J
mg
(I)
Fig. 2E.70 (b)
- (Ii)
For blockM:
Il'x = T - Mg sin e - µ ,N = O
ITy = N - mg cos e = o
,-·-N.1 cos'45°t N2 cos 45°:
N2
N1
j
N
... (1)
... (2)
For block m:
B
A
mg
hlg cos B
(iii)
(iv)
Fig. 2E.69 {b)
N=mgcose
-/2
From Newton's law,
mgsin0-2,!N=ma
.
(mg cos 0)
mg sm0-2µ----=ma
-/2
a= g[sin e-:- -/2µ cos 0]
or
-E-xa -:. ·.-e.- r-;;;1--.,
b~,__~--,_,-i':,@J~-±c::: "; - ~
~
~:~~:r~;ti~y.-~-~--
1
-,,a-l~~cd~~:e~~~!hi:~u~~~~
an angle e ii,tth the hortzontql and.'l(l' ~ luinging vertical()> asi
shown tn .Eig. 2E.70 (a).
co_effic/ent'of static friction'
between 'M' and _the tncline isµ,. 'Find the minimurn an<i
_maximu_myqlues of/m' so thatt_hef____sys(e_'f'"is at rest.
·
11,e
I
m
I
/
e
L_ --·----·- .__
Fi~~~-~--
·!
I
ITY = T - mg = 0
... (3)
From eqns. (2) and (3) we substitute values of N and T
in eqn. (1) to obtain
mg=Mgsin0+µ,Mg case
Therefore maximum value of m = M(sin0+µ, cos0)
Case (ii) Impending motion downwards of
block M : In this case friction force acts up the incline.
From conditions of equilibrium,
For blockM:
... (1)
Il'x =-Mgsin0+T+µ,N=O
... (2)
ITy =N -Mgcose = 0
For blockm:
... (3)
ITY = T - mg = O
Now we substitute N and T from eqns. (2) and (3)
respectively in eqn. (1), to obtain
mg+µ, Mg cos e = Mg sine
or
m = M(sin 0-µ, cos0)
Therefore the blocks are at rest if
M(sin0-µ, cos0),-;m,-;M(sin0+µ, cos0)
··- - --·
,_
.
~?-,.._
~J=~.2€5i!,~P,_\~_ ,:
Ji,-->
'
71
0
!r;;,o bl~cks-arekept
0~ ~n incline in contact with ea;;, othe;.:
,Masses of blocks are m1 and m 2 and coefficients offriction are,
'µ 1 and µ 2 respectively. The angle of inclination is e.;
"Determine:
'(a) acceleration of blocks, and
'(b) force F w1th which the blocks press against each other.
_L ·---- ~-----j
Solution: The block of mass M can have a tendency to
move downwards as well as upwards. It depends on relative
values of masses m and M. If M is heavier it tends to slide
down, and if m is heavier it tends to move down.
Case (i) Impending motion upwards of mass
M: In this case friction force µ,N is down the plane.
Fig. 2E.71 (a)
Solution : It is not clear whether the blocks slip or not.
So we arbitrarily assume that both the blocks accelerate
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
-- - --.
FORCE ANALYSIS
downwards. Contact force between m1 and m 2 is R; it should
not be negative or zero. Contact force between two bodies
reduces to zero when the bodies are separated.
1··-----·-
·y
-
-
----------
µ,N1 \
µ,½_I
!Trµ,mg
·- ---
-- - -- -
-
°
C ••
mg
i-·-1 ·--
;_~~?1'~~pJ~.;, 72 ~
----- - - --- - -- -- ------·1
as!
I
Four blocks are arranged on a smooth horizontal surface
1shown. The masses of the blocks are given (see the diagram).
_,The coefficient of static friction between the top and the,
·bottom blocks i§ µ ,. What is the maximum value of the',
/horizontal force -F, applied to one of the bottom blocks .as;
:show!', thilt makes all four blocks move with the same!
\acceleration ? __ _
___ _, ____ · ____ -,
i
/
:MBJ MB=r:
Fig. 2E.72 (a)
,
1
!
-
Lf--J
.
-F,
Mg
!
.
From Newton's second law:
Block 1 :
Ll'x =m1gsin8-R-µ 1 N 1 =m 1a
... (1)
Ll'y = N 1 - m1g cos 8 = 0
... (2)
Block 2:
Ll'x =m 2 gsin8+R-µ 2 N 2 =m 2 a
... (3)
Ll'y = N 2 - 11 2g cos8 = 0
... ( 4)
Now we substitute N 1 and N 2 from eqns. (2) and (3) in
eqns. (1) and (2) respectively. Now add eqns. (I) and (2) to
obtain
a= (m 1 + m2 )g sin 8- (µ 1 m1 + µ 2 m2 )g cos8
m1 + m2
From eqn. (1) we obtain R.
R = (µ 2 -µ 1 )m1 m 2g cos8
m1 +m2
which shows that ifµ 1 > µ 2 then reaction R comes out
to be negative, which is impossible. It also implies that
blocks have separated.
[
I
-
-
--~
·--- -
179
Solution : Step 1: Draw free body diagram of all the
blocks.
Step 2: 1iy to identify the cause of motion of blocks on
which force is not applied.
Block A moves due to static friction. When slipping
starts it is f, mu = µ,mg. This force must be greater than
tension T, only then it accelerate forward block C moves due
to tension, Twhich must be greater than/the static friction
between C and D. Block D moves due to f
-
Fig. 2E.71,(b)
- ---
-----~
~--------------------------------- - -------- - - " - - -
i
•• _J
Mg
Fig. 2E.72 (b)
From FBD of block B
F-,t,mg =Ma
From FBD of block A
.. '(1)
... (2)
µ,mg-T=ma
From FBD of block C
T-f=ma
From FBD of block D,
f=Ma
... (3)
... (4)
from eqns. (3) and (4),
... (5)
T=(m+M)a
a=(m:M)
putting T in eqns. (2) from (5)
µ,mg-(m+M)a=ma
µ,mg
=a
(M+2m)
putting'a' in (1) F-pmg = µ,Mmg
(M +2m)
F
= 2µ,mg( m+M)
2m+M
(,E,Xei'?'c',l".".f;e•t_7_3 ', ,.__-_
~,=-,-~'!3!,le;,,c,,.s,
,- - -- -,
-
~
--
------,
.
.
. .
. ..
'A car begins from rest at time t = 0 and then accelerates cilongl
la straight track during the interval O < t ,:; 2s and thereafter
'.with constant velocity as shown in the graph. A coin is;
'initially at rest on the floor of the car. At t =1 s, the coin I
'·begins w slip and it stops slipping at t = 3s. Find the!
.s-.oefficient of static friction bctwren the floorpnd tlze cqi11•. J
www.puucho.com
-
1
•
Anurag Mishra Mechanics 1 with www.puucho.com
11ao
~-~··-------
l1s~---
,.
j~
N
Parabo
. ' z.
lj
co."'o'-J-;;--2;--<3i---'4-+ icsJ
I
Fig. 2E.73 (a)
~ - - - ~---···--'<.'--·""'···- -- -,.,-·
(c)
(b)
Fig. 2E.74
Solution:
r·-·-. ·-· ---------·
i Concept: What IS ':~11_se_ of acceleration of coin?
w-
= ma;;
ay = 1mJs2
(B0-72)g
Friction force accelerates it, when slipping starts.
µ,mg=maora=µ,g
Given th~t graph is parabola having
vertex at origin then function of velocity
is.
now
2
or
a= 5/3m/s
Now apply Newton's second law on man is direction of
acceleration. Note that x component of acceleration of man
is due to friction.
mg sin37°-µmg cos37°= m x (5/3)
6-8µ = 5/3
6-(5/3) = 8µ
µ = 13/24
. . . . . . . ,---:::::7
a'
Fig. 2E.73 (b)
V
= ~t 2
at t = 2sec.; v = Bm/s we have
B=k-4
:;
V
= 2t 2
=}
ag = asin37°= lm/s 2
k=2
dv
-=4t
dt
the coin slips over floor if
ao = µg
Thus,
µ = ao = 4 x 1 = 0.4
g
10
~-.'.fil~~!I'PJ~ j 75
1>
-In the figure shown, the static friction .coeffici;n; betwe~;~u]
contact surfaces is 1/2. What minimum force applied leftward;
'on block 1 will move the system ? Repeat problem if tbe force)
is now applied on block 2;
_______ •
~~~?~fr
i - - - - - - . - - - ·', - - · . - - - - - -
iA man:;of m~§ •eyo kg, stands on a hqrizontal weighing.
machine, ofi'!eg{lgible mass, attached to a massless platform P.
that slides do111ri ·at 37° incline. The weighing machine read
72 kg. ,:nan is ci/w;;ys~_
a£. r_esP_t _w.r. t. ·weighing machine.
\
I
'
l'
!l
;:, _.
. "'
•
Fig. 2E.74 (a)
!Calculat,e :
, .
( a) The vertical- acceleration of the man
,
'(b) The coefficient of kinetic frictionµ between the platform
l_____q_n(IJnriline,_:__ . _ ___ _
~
Fig. 2E.75 (af
Solution:
Step 1: Calculate maximum friction force that acts on
all the rough surfaces.
Step 2: Check the tendency of motion of each block,
static friction opposes that. When slipping just begins f, is
,
·1
. l
-~i
·
Solution: Weighing machine measures normal
reaction. Draw FBD of man. System of man and platform
h& e acceleration at an angle of 37° ax and ay are x and y
components of ·acceleration. What is cause of vertical
acceleration and horizontal acceleration.
maximum.
Note that due to string constraint both the blocks'will b~
at the verge of slipping simultaneously.
Case I: f 1,- =0.5x3xl0=15N
'
f 2smax = 0.5x2xl0= ION
from FBD of 1 kg
F = f1 smax + T + f2 smax = 15 + 5 + 10 = 30 N
from FBD of 2 kg
zr ". f2,m~ = 10
T=5N
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
I FOiCE ANALYSIS
-- -- .. --- . - '" .
181]
..:·===========
Note: _ _:__:__ _:_--_:_·-:..:-:.:·:::.-
First decide whether there is slipping between blocks or not.
If blocks have same acceleration then friction force
between blocks must be less than µ,N.
2kg
f2,
1kg
f1,
(b)
N1
f2,
2 kg
T
f2,
f1,
20N
10N
(c)
Fig. 2E.75
Case II: From FBDof 1 kg T =
Fmin
f 2 smax + f 1smax = 10+ 15
= ZT+ f, ,m~ = 2x 25+10= 60N
N1
~-~T-;;,
F
Let blocks move together with common acceleration a
6mg - f 2 6mg - 3mg
a=
=-~-~=g
3m
3m
then for upper block
T-f1 =ma
=>
3mg-f1 =·mg
=>
f, = 2mg
but f 1 ,;; limiting static friction but here f, is coming out
·
to be greater than µ,N.
Assumption of no relative motion between blocks is
incorrect that means there is relative motion. Therefore f 1 is
kinetic friction.
3mg-µmg --2g
(towards right)
m
2T
12,
3mg+µmg-3µmg
.
-~~~~~~ = g/2 (towards nght)
2m
Applying pulley constraint to get acceleration of hand
ap
A block of mass m rests on top of a block of mass 2m which ls
·kept on a table. The coefficient of kinetic friction between all
,surfaces ls µ = 1 A massless string ls connected to each mass
.and wraps halfway around a massless pulley, as shown.
'Assume that you pull on the pulley with a force of 6 mg.
What ls the acceleration of your hand ?
F=6mg
µ=1[mJ
'µ=11
2m
-
~
Fig. 2E.76 (a)
Solution : The free body diagrams both the blocks are:
~~.
f11
N2~
mg
a2
2
= 5g / 4 (towards right)
r- ~.~p~e} .!:: ..:-1?!;.>
A 4 kg block ls placed on top of a. long .12 kg block, which is
accelerating along a smooth horizontal table at a= 5.2 m/si:
under application of an external constant force. Let minimum;
coefficient of friction between the two blocks which Willi
prevent the 4 kg block from sliding ls µ, and coefficient of
friction between blocks ls only half of this minimum value. of,
(i.e., µ/2).Find the amount of heat (in joules) generated due•
to sliding between the two blocks during the time in which. 121
kg block moves 10 m starting from reg,
~
= 5.2m/s
12kg 1-I
smooth.___,__- - - - ' · - r"l'.'.":J'---'-, a
T
N 1 2mg
(b)
+
acceleration of pulley= acceleration of hand
am +a2m 2g+g/2
ap =~-2~=
2
11
12
= a1
2
Fig. 2E.77 (a)
(c)
Fig. 2E.76
f 1 is force of friction between blocks
f 2 is force of friction between block and
___ j
Solution: First assume that blocks have common
acceleration, for both block to move together acceleration of
4kg block must be 5.2 m/ s 2
ground.
4kg 1
·
I' ,'-"=1--1-2..::kg....=ll:;-b+f a = 5.2 mis~4
from FBD of pulley, we get
T = F/2 = 3mg
Fig. 2E.77 (b)
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
-
rt:;~·:·~~? b··:;· ·-,~. --.-- ·7 ·
I
a= 5.2m s 2
f= 4x5.2
µ 0 mg = m(S.2m/s 2 )
I.~--. .:--r·, .,.., '· · ,.
· µ 0 = 0.52
Ifµ= .!co.52)
= 0.26 the
acceleration of 4kg block is
2
.
.
due to friction
.
'
' 2
a 1 = µg = 2.6m/s
As there is relative motion.between blocks we apply s,,1
Sre1
L
= - 2,6 m/s 2 ,
1
2
= -(-2.6)t
2
Time of motion can be determined from motion oflower
block·
~ = .! (52)t 2 = 10 (given)
For 12 kg.block
2
sre,
=-Sm
work done by friction is given by.
w1 =µmgS,., 1 = 0.26x4x lOx (-5) = -52J
Heat generated = 52 J ·
Ii+ 12 = 0
b+ /4 = 0
From which we get .a = b = c
Applying Newton's Law on.block A
Mg-T=Ma
on block B
T-T1 -µmg= ma
on blockC
T1 ~µmg= ma
solving eqns. (1), (2) and (3), we get
~,,.,~..,--.-rat·-,.,
·, •
·. ,·.:]
.., ~{~·
,
,'
•:,_ \·
• . .: 1 .
w
eC
mC
m
. . •
r:'
an{c\~;;J
B
.
;,
j
!::,::;~::::.: '.
::···,;
M A I ·. : '
:J
ta) ~:~:io·~.$.,·~.t~:!~~.~t?E.i~.-.·.·~~.-.i·.:t~::~~~i.!.·.·~.R.~.~·;~
I ..
!,
i. 'small: , J . : ·
,,; ' , ·
·. · , ,;;,,· ,
'(~) if t/zci Fri~$ of blq~k ;t';:, z;;;ss th~,.; some critical vi:zltii!, the
I ..blocks will not.'accelerate1whefr·relea.sed from" rest.,Write
t__.<;/9.wJ1.q ezyr_i§~jqn_for_thaf41nca,l_111C1§~.~ ·. ,~ ~- ..2:"·
··
Solution: AppJy·constraint equation on strings, length
-~~-o,
a•T
,..,__
, ~ · T ~ , ,"
,,T, , ~ , ~ µ m g
,·, '
'"• a__ J
•
,
' .
'
;
.
Mg-T =M(M -2µm)g
M+2m
T
= 2.mMg(l +µ)
(M+2m)
(b) As there is relative motion between blocks we
apply
v~l = v:el + 2arel srel
If system is released from rest, u,.1 = 0
'(b) SupJJOS~ ~he syste(n 1..< relea.sedfror1gestwith b!ockC heajJ
_the/,g.ht end of blockB,.as·s1ibik,rin the above'figjlre. Ifj
the.le,ng~.1.1'+ of. blb.ckB. is give·n·; w·h·.··a.tis tl.,e· sp_.jed. of·,.·. .b!iJc.k
· · C. 1;1.s' i( '!'~aches th~ (eft end.pf-block B? Treat,,siJ!~,oj\C
... (3)
putting 'a' in eqn. (1), we get
_·: ' :·
•
.•. -:.,, ·:.
... (2)
a=(M-2µm)g
M+2m
;j ··,: -,.--.\q,:-··,1·
';:: .. , ,''
r
~~
... (1)
L.._.~.:.-~-· fl~:.;~E-?8_\':l, ~- --~-.:.. .
0
--- •
[: t,19, '
I . ,
figU;,/
•
µ~ ",.
*A
Given t/J7.Ia;·sho;v;, .in °the
Bliicks~A, B
masses m,1. "'M·& mB = rnc.1" ni. TI1e strings are q5symedi
massless and 11v.itretchable, 'iJ.11il t/iii pulleys frjctionless:)'11.efe ·
is no frict;.on l>'.etween blocks B qri:d (he support table, bu"ttherel
is frict/o~ /zetivee,; qlocks -p:~ncl,\i/denoted; bY,,<1::giveni
coei.;
.· • . ; , ·- . •: «,. .·- • ~ . ;·. ,·:i.
'.JJ ienfµ. • . . ,.
\: , , , ··1
_w/-;~ - - -·~
~~:,'i ,~
l
:r
e-e><.~~:~e ~ 78 ;;;'1>
!:Er"'=~,.::.,_,,;_ ·;;;_~·,".....-""'"-V
,
. :
z;+z~=o
2
a;<I = a 1 -a= 2.6-5.2
0
U.,
a
_:c!''!l· ~~,,?~_!b!
11 + 12 = constant
=u,,it + -1 a,,1t 2
sre,
'·1 !I
v;el- =
Of,
2arel.srel
Vrel
= ~2arel Srel
arel,
·=·.2a·
= L- - - - v = , I,4g--"-L('-M_-___,2µm--")
S,.1
.
(M +2m)
(c) If blocks will not accelerate, then
of strings is canst. Differentiate. twice to get relation
a = 0 in express in of a to get
Put
between acceleration. Let acceleration of blocks A, B and C
M=2µm
be a, b and c respectively.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
FORCE ANALYSIS
183'
l2c.S-=K9.~R.1.C?-
, 79 :__.;-
Board A is placed on board B as shown. Both boards slide,
without moving with respect to each other, along a frictionless
horizontal surface at a speed 6 m/ s. Board B hits a resulting
board C "head·on". After the collision, board B and C stick
together and board A slides on top of board C and stops its
motion relative to C in the position shown on the diagram.
What is the length (in m) of each board? All three boards
have the same mass, size and shape. The coefficient of kinetic
friction between boards A and C and between board A & B is
0.3.
v=O
Before
--+-
Fig. 2E.80 (a)
Solution: For equilibrium of block B
ftl:gN
L . ~ 4 s · Lx ... ··•
·~"71,Y
-
µN
and
IFy = 0
= .!!_
..fz ..fz
N = ..fzmg
1-µ
mg+ µN
Solution :
Concept: Initially block A slips on block B and C.
Finally A
Lx
Fig. 2E.BO (ti)
After
Fig. 2E.79 (a)
y
and
C have
common
acceleration
apply
For equilibrium of block A
IF = F-.!!_- µN = 0
..f2 ..f2
X
F=!!..[l+µ]=mg(l+µ)
or
Block A:
µkNk
Block B: µkN A -µkNC
Thus
= mAaA
= mag,
are1 = 4.5
VreJ
m/ s2
0=~-4.5t
= O;
A carriage of mass M and length l is joined to the end of a
slope as shown in the Fig 2£.81 (a). A block of mass m is
released from the slope from height h. It slides till end of the
carriage (The friction between the body and the slope and also
friction between carriage and horizontal floor is negligible)
Coefficient of friction between block and carriage is µ. Find
,minimum h in the given terms.
=u!1
+ZaretSret;
2
For
=6xl.4=14N
0.6
2
v 1 = 6-2= 4 m/sec
V~t
(l-0.4)
Fig. 2E.79 (b)
2
6
(1-µ)
= 0.6 X 10 (1 + 0.4)
~~~f-+v/2
t=--=-sec.
2x 4.5 3
For block A, aA = -3m/s 2 ;
Apply
..fz
o-v
0 2 = ~ - 2 ( + 0.45) (L)
4
v = 6 m/sec
m
L=lOm
r:..:·l;;~gmp)~i
-·-· ·. --.
BOV
A side view of a simplified form of vertical latch Bis as shown.
The lower member A can be pushed forward in its horizontal
channel. The sides of the channels are smooth, but at the
inteifaces of A and B, which are at 45° with the horizontal,,
there exists a static coefficient of frictionµ = 0.4. What is the'
minimum force F (in NJ that must be applied horizontally to
A to start motion of the latch B if it has a mass m = 0.6 kg;
question
·
:F. . . .
"'=,---,
M
,...,_~~..-, smooth
Fig. 2E.81 (a)
(a)
_ (c)
www.puucho.com
2µ( l+ : }
µ(1+:)z
(b)
µ(2+: )z
(dJ.µ(1+~J1
Anurag Mishra Mechanics 1 with www.puucho.com
MECf,f/\Nl(Ssl
·L184
----..---~·--· ----· -------'-----~
Solution:
'
Concept: Block slips relative to carriage, use relative,
motion equations of kinematics.
+-·,
'
Solution: Most important concept here is that man
moves slowly. Slowly means, always in equilibrium
For the man,
(vertical)
N+Tsin8-Mg=0
(horizontal)
F-Tcose = O
~
mg
Fig. 2E.81 (b)
velocity of block, just before reaching carriage
j
Vo =~2gh
Now acceleration of block
µmg
a, =---=-µg
m
acceleration of carriage
µmg
a2;;::-M
considering this moment as t = 0, motion of block as
seen from carriage
U,el = Vo = ~2gh
Relative velocity of block when block moves through
distance x with respect to carriage
2
2
2
Vrel = Vrel + arel X
when
x=l,vrel =0
= 2µg( i
+: )z
h=µ(1+:)z
~
---,
A man with mass M has its string'.attached to one end. of a:
'spring which can move without friction along a horizontal,
overhead fixed rod. The other end of the spring is fixed to a,
wall. The spring constant is k The string is massless. and:
'inextensible and it maintains a constant angle 8 with the,
;overhead rod, even when the man moves. There is friction,
iwith coefficientµ between the man and the ground. What isl
,the maximum. distance (in m ) that the man moving slow[y;
;can stretch the spring /?eyqnf!Jts_J!atyral length?
k
'
I
Maximum extension is obtained when static friction on
man is maximum
For maximum extension, f = µN
For spring,
T cos8 - kx = 0
T case = kx
~
T = kx/cos8
Substitute for F and solve for N
=0
N = T cos8/µ = kx/µ
kx/µ + kx sin 8/ cos8 = Mg
µN-Tcos8
or
kx(l +µtan 8) = µMg
µMg
x=-~~~k(l+µ tan8)
-'...
- - --· ·-·------- --------"-~~--~.-----;
!Find minimum normal force to be. applied by each hand toj
;hold three identical books in vertical position. Each booJi hasi
:mass' m' and value of coefficient offriction between the l,ooks1
.as well as between hand qnd the bpok is µ.
!
I
Fig. 2E.83 (a)
----~"'~"',--
'--------
Solution:
From FBD, for center book
2f, = mg
M
µ
(friction coeff,)
Fig. 2E,82 (a)
(horizontal)
Solve for T :
or
So,
a" 1 =a1 -a 2 =-µg(l+ : )
2gh
_Fig, 2E.~3_ (bL__,
f ,,; µN
~ 1,1N > mg
-
•
2
N>mg
2µ
... (1)
For side book f - f,
f= 3mg :o;µN
= mg
www.puucho.com
2
i
iN
j'
.
- - ------ -·-=~
--:~.M
1J, '
1,
f,
__ Fig.2E.83(b)
.,,_,J
Anurag Mishra Mechanics 1 with www.puucho.com
f-FORCE ANALYSIS
N;;, 3mg
2µ
By eqns. (1) and (2)
N
. .
Thus,
M1mmum
~E~a"~PI':-
... (2)
B
:
--A--/~~i
3mg
=2µ
Fig. 2E.85 (a)
~>
.
..:--
_
.....
''
- -----·--··'
- ........ -.-~~,
Solution: During upward acceleration
. a1
L0.!.:.
µN
I
4a !
a 1 cos 37q =~
mg
N - mg
= m(
3
;
µN = m( ~
a1
150N/m
1
\
.I
Fig. 21:.~5 j~).
450N/m
)
1
on solving we get
}
= 15g m/s2
31
Fig. 2E.84
.
Solution : Suppose origin is at the equilibrium position
and the direction of increasing x is towards the right. If the
blocks are at the origin, the net force on them is zero. If the
blocks are a small distance x to the right of the origin, value
of the net force on them is -4kx. Applying Newton's second
law to the two-block system gives
-4kx= 2ma
Applying Newton's second law to the lower block gives
k(x1 - x)- f = ma
where x1 = initial stretch and f is the magnitude of the
frictional force.
f=k(x 1 +x)
The maximum value of x is the amplitude A and the
maximum value for f is µ,mg. Thus, µ,mg= k(x 1 + Amrucl·
Solving Amax gives
A
=µ,mg -x =3
k
= a.1 .sin 37' •
N ,
,When the system shown in the diagram is in equilibrium, the
,right spring is stretched by 1 cm. The coefficient of static:
'
I
::Jriction between the blocks is 0.3. There is no friction between[
\the bottom block and the supporting surface. The force)
,constants of the springs are lS0N/mand 450N/m (refer Fig,;
2E.84). The blocks have equal mass of 2 kg each.
:
Find the maximum amplitude (in cm) of the oscillations of1
·the system shown in the figure that does not allow the top:
'block to slide on the botto111.
;
max
~a
/
Concept: When lifting arms accelerate up, caus.e ofj
,acceleration a1 cos37° is friction µN. And resultant force up1
:is .(N -c 111g):,1chich causes acce!er;it!~n,J{i.sj.n}J_0,. _.
-
-- - -
- - --·· ''f
4a~
a2 cos37°=5
··71
37°
82
+
a2 sin
37°=,
1
·
I
Fig. 2E.85 (c)
FBD when Arm is in Deceleration
.
Concept: During deceleration direction of friction force
,is towards left. Student is advised to ponder over a simple
question.
I.-----
- .
''Which force is cause of component of acceleration
a 2 cos37° parallel to surface."
1
mg -N =
'In the manufacturing process disks are moved from level A to:
B by the ·lifting arms shown. The arms start from level A withi
lno initial velocity, moves first with a constant acceleration a,!
'as shown and then with a constant deceleration a 2 and comes:
,to step level B. Knowing the coefficient friction between disks
'.and the arm is 0.30, determine the largest allawable.
:acceleration a 1 and the largest allawable deceleration a 2 of
·the disks are not to slide.
'
Which on solving given
m( 3;
2
)
a2 = lSg m/ s 2
4a
'·-------..---···-···-~ ~
6-,~~~.~~~-~-·~
"-
- ---
--- - --- ·- - --
- --- -------·--- -·
- 1
;In the Fig. 2E.86 (a) shown a constant force Fis applied on,
:lower /:,lock, just large enough to make this block sliding.outi
from between the upper block and the table. Determine the i
'force F at this instant and acceleration of each block. Take:
g_= l()_m/s 2_. • • •.. ______ - - - - - - - - · - · · _______ ...
www.puucho.com
.!
Anurag Mishra Mechanics 1 with www.puucho.com
11as
, / , f,
>'
r:;1·;;,d ~2 (11)(~2>
<,,
II ~.s7 r·- - - - {
'
"
I
'
I
•
_:_S_o-lu-tio~-lt i-n-s;;:~::~~:~:::;:e_F_w_e-ap;y
I
i
time force F = (2Ot) Newton. Plot a graph between
.
acceleration of both the blocks .and time.
Let f1 =·force of friction between 5 kg and 15 kg block
and h = force of friction between 15 kg block and
gronnd.
Then, (f1 lmax = maximum static friction
= (O.3)(5)(1O) = 15 Newton
(f1 ) k = kinetic friction
= (O.1)(5)(10) = 5 Newton
similarly (f2 lmax = (0.5)(15 + 5)(10) = 100 Newton
and
(f2 h = (O.4)(15 + 5)(10) = 80 Newton
Now when F $ lOON, the -,. ------------····""·system of block will not move. i, : ,' : @5) , 15 Newton
In this case f 1 = o," i.e., f 1 [ 1 5 N e ~ _ ~ •
starts acting for F > 100 I
~F
Newton. At the time of lso Newton
,....... a
slipping between 5 kg and· 15 l' "
Fig. ze.as (b)
kg block f1 will be Cf1 lmax and - - - - - - · ........- .
f 2 will (f2 )k and obviously F > 100 N.
l
l
.•
'
'
3
!
0
I,
I.(. ,
5
t (sec)
7
Flg::1E:as (c)
,.. , . ______ -··---·-----~
CIRCULAR MOTION
· Consider a string of beads whirled in a circle as shown in
Fig. 2.44. Each bead moves along a different arc but sweeps
the same angle. If the arc length traced by a bead at a radial
distance r is I, then we define 0 as
0 = 1/r
360°
I = r, 0 = 1 radian, 1 rad= - 21t
= 3600 = 57,30
6.283...
Any angle 0 can be transformed in_to degrees by ·
·e (radian) = --'---"----'e(degree)
--,--,--,2it (radian)
36O(degree)
When
lllustration-7
Diameter of moon, D ~ 3.4 x 10 6• m
Distance from earth, r = 3.8 x 10 8 m
At this instant both the blocks will have the same
accelerati,on. ,
~quations of motion are as u11der :
15= Sa
a=3m/s 2
F-95 =15xa = 45
F = 140 Newton
We saw that· a1 = a 2 = 0 upto the instant when
F = 10ON or t = 5sec.
Both the blocks move· with same acceleration,
a = F - 80 = 20t - 80 = t _ 4
. 20
20
till force becomes 140 Newton or -.7 second. After 7
seconds acceleration of upper block a 1 becomes constant i.e., ·
1 m/ s2 while that of lower block
F-8O- 5
a2 =
15
.
2O 85
=
t= l,33t - 5,67
15
The corresponding graph is as shown in figure.
If we approximate its straight line diameter as an arc
length, then the angle 0 subtended at the earth l,y the-moon
is
0=i=D
r
r
6
= 3.4 x 108 m = 0.009 rad.
3.8 X 10 m
Diameter of sun
www.puucho.com
(l)=l.4x10 9 m
Anurag Mishra Mechanics 1 with www.puucho.com
IFORCE ANALYSIS- Distance of sun from earth (r}= 1.5 x 1011 m
Angle subtended at the earth by sun is
S = _! = E_ = 1.4 X 10
r
9
1.5 'X 1011
r
=- 0;009 rad.
X
That is why the sun and moon seem to be of same size.
Average and Instantaneous Angular Speed
When the beads move in a circle of radius r, the radius
sweeps angle 8, we refer to it as' angular displacement. After
taking 3 complete counter clockwise turns, 8 = 3 x (21t) rad
rather than 8 = 0. Arc length 1 is different from vector
displacement, but we can take counter clockwise l positive
and clockwise negative. We can call it curvilinear
displacement.
'-------F_l=.g;__2:~6 '________
J
The angle 8 is measured w.r. t. the x-axis ..
Acceleration in Circular Motion
The position vector of velocity and angular
velocity for circular motion : Position vector i(t) is
---- ---y -- -- - --- '
'
,
-
'
.
v(t)
----·1
I
-
I
[
I
;, ~ l
Particle
y(t)J r sin B(t)_ I
'
B(t)
x(t) i
From figure,
where
111 = 11 - l; and ,118 = 8 f
If time duration is /J.t, then
111
!J.t
... (1)
-
118
CO avg.
= ~t
We may call 118 as average angular speed, angular speed
=
!J.t
of 1 rev/s 21trad/s
Instantaneous angular speed
00 = Jim 110 = _de
M...;O
and eqn. (2) becomes
/J.t
dt
----~:::_ J
(t) = [r cos 8(t)]
i + [r sin 8(t)] j
dt
dt
The velocity vector is tangent to the circular trajectory.
Velocity vector ,I (t) is perpendicular to the position vector
i/(t) at all times. Students can verify it by scalar product
... (3)
dl
d8
----,-=rdt
dt
v
I
,I (t}= r ~ [cos 8(t)]i + r ~ [sin 8(t)]j
'
dt
dt
= r[- sin 8(t)] dS(t) i + r[cos 8(t)] dS(t) j
... (2)
where
_ ·x
Velocity of particle is
/18
!J.t
-=rv avg. =· rco avg.
r
8;,
·
1(t) - ,l(t), which is zero, independent of time t .
In circular motion the three vectors ,l(t),
r (t) are related to each via the vector product
= rm
,I (t) = 00 (t) X t (t)
Angular Velocity Vector
· Angular speed ro is the magnitude of vector called the
angular velocity oo of the particle. Direction of oo can be
determined from circular motion right hand rule.
Curl your fingers of right hand in the sense of rotation of
particle, then the extended thumb points in the direction of
In magnitude, v(t) = rro(t)
Acceleration of particle at any
instant of time t,
...;
ro.
www.puucho.com
oo(t)
and
Anurag Mishra Mechanics 1 with www.puucho.com
·. MECHAN'iS~lB
Since the particle is in circular motion, the radius r is
constant. If the particle is undergoing uniform circular
motion,
de
dt
-- = OJ = constant
' ~_~;:?/. '' .
.
li(t)=-{· rdO[cos0(t)]
dt
=-(d
8
dt
act)= -
00
de}i+
dt
{r dOdt [-sinO(t)]- dO}J
dt
2
) {[r cos ·e(t)] i + [r sin O(t)] J}
.
2
1ct)
Note.that term in { } is position vector i(t). Negative
sign indicates that the acceleration is antiparallel to the
position vector i(t). That is a(t) is directed towards the
centre of the circle. This is called centripetal acceleration.
The 'magnitude of centripetal acceleration
.
a, =ro 2 r=(v/r) 2 r=v 2 /r
Alternatively, centripetal acceleration can be obtained
by differentiating the expression
v(t)
=0) X i(t)
and
1a,Ctl I=~= ro 2 r
,-
Since
Hence
where
·. A )' c'"tJ;:
~
1
c
dt
.,_v_
4
·-
.. _ ·- l
_
~
, at
a
~ For a particle slowing
down in circular motion
(b)
____ Fig. 2.50
a, ,;,resultant acceleration • ·
~-
I
.
.a, =' taT1gential acceleration
~
:
''
.
-
,,
,.'
'
_ a, ;"' centripetal accelemticin
,r-·-
,1
I
d1 Ct)= v(t).
dt
dt
4
(a)
3.
(t)
_
For a particle speeding
up in circular motion
,..,
a (t) = d OO (t) X r (t) + 0) (t) X d t
d ol (t)
I
,
a,=a(t)xr(t)
Fig. 2.49
·-c . >"'·,~----,
1l~ r;;}~
ac...___
_,
v(t) =ro (t) x r (t)
In the first term
I
~p
a, =oict) x vet)
\
:' ·-· • .
!speeding dl~ng t~e circle; or antiparallel to c6(t) w,he~;the-1
fparti_cl_e _isc5lp}ll_ing"'.'-:______ •.____.____ -----..-----, .
I
dt
Note that second term in this equation is
". ,. .
7
dv Ct)
_,
:·
ir ' !
The acceleration of a particle is rate of change of
velocity.
_,
~
.
i6(t). If speed of particle increases or de~reases, the angllfor
1
~locity vectqt-falso increases or decreases. ·.
[ Dfrection 'of angular velocity vector is always normal tc;,
lane of rota'i;ion. Therefore the -angular acceleration vecfor
lot
ct) is.- direct~/i
either' parallel to :ol(t)
when the part(~le
is.
I
'
, <:
"
,
. r
Non-uniform Circular Motion and Angular
Acceleration:
dt
;~~:~.,]
is speeding 'up and_ antiparallel to 'v(t) if the partic[; is
slowing dowij; .... ______ ..____ ----·--"··--·
___ ..
2
act)=
'--+
tdnl!;ent to t/tecircular path andpa\dllel to v(t) ifthepdr,ticle
ol xv(t) = -00 2 1 (t)
.
'
2. At" ~n( instant tangential: _acceleration _is. ~-lwaysl
d--+--+
--+--+
(OJ x r(t)) = OJ x v(t)
'it,(t) =
' ·;.,~
Concept: 1. The angularucceleration vector a(t)pointsl
in the direci{on of the change in tile angular ve(ocit;yye~torl
'-----:~---.
a(t) = dt
or
Hence first term is cit(t) x i1 (t ), it is termed tangential
acceleration.
is defined as angular
acceleration.
,
Angular acceleration is rate of change of angular
· velocity.
o1(t) = di6 (t)
dt
www.puucho.com
- ..::; ...- ........ ..
~
V
.,
~,
- - - • a,
"
I
Anurag Mishra Mechanics 1 with www.puucho.com
A1{ALYSI$ . ·.
·
:
: • : . ci .
Ir FORCE
::::::::-¢=-;:::::;:~~~;;:;.;:;;~~~:-:::::'':::-'::::'
·-=~~-_;.•"..:'..=·
. L....
,,1.,
'-~-'---.;_:_c
--
e, = (cose) i + (sine)j
e, = (-sin0)i+ (cose)j
and
Radius vector of the particle at time 't'.
->
•
•
r = r[(cos0) i + (sin0)j ]
Differentiating both sides w.r. t. time, we get
II'
.
->
a,
dt= r [ -sm
. 0 -1+cos-J
de •
de "]
;
, Fig. 2.51 (b)
----~--·--
-2+ -- .
dt
dt
di
i
= roo[ (- sin0) i + (cose)j]
... (1)
[,
When a,. is )n direction of motiol! i.e. parallel to velocibvector speed of object i'!creases.
:,
Centripetal acceleration chrmges direction of ,;elocity
vector.
,_
1_,:
'
'
'
'
When ta11gential acceleration is opposite ·to veJoi:ilX
vector speed ·of_ o_bject decreases.
·
·
· .. :
· Note that aligular velocity> vector, position vecton: cmd
tangential ac;celeratio/1 vector are -rtormal to each other. . ··
Total acceleration of particle a(t) is
;
j
<
~
4
'~
'
~
a (t) = ex (t)Xl'.. .(t) + OJ
->
-:+
(t) xv (t).
->
= a,(t)+a,
Again differentiating (i} w.r. t. time, we get
(t)
di
Total acceleration is vector slim of the two mlltu~llyj
P!,yen_d_icu_la!L!'!E.g~ntiq1: '!'!_d centripetal acceler'!.U.~:-~-~-J
dt
dt
To Find the Angular Velocity of a Particle
with Respect to the Other as Shown in the
Fig.
From the Fig. 2.52, angular velocity of B with respect to
particle A is:
r· '"'JL'·
j
½·ll
A
. / :_
.
.
; ____.
_ Fig.2.5~
I
..
B
=ro>~{(,-sin0)i+(cose)j}
dt
doo
•
. •.
I
+r-{(-sin0)i + (cos0)j} . ,
do>
=--{J) r{(cos0)i+(sin0)i}+rdte,
2
i
:I
I
'
do> •
=- ( co 2 r)er+r-e,
.
i
dt
i,
anet
Hence
= -(oo 2 r)e, + (exr)e,
where ex r is the tangential acceleration and
radial or centripetal acceleration.
OJ
2
r is the · I
·
lllustration-8
,
I
_______,'
linear velocity of B w. r. t.
A .l to the line joining them
separation distance between them
(v 2 sin0 2 -v 1 sln0 1 )
l
Consider a particle moving in the x-y plane according to
r = r(cosooti + sinooti), where rand OJ are constants. Find the
tra~ectory; the velocity, and the acceleration.
Unit Vectors along the Radius and the Tangent
Let us consider that a particle P is moving in a circle of
radius 'r', at any time 't' the particle's angular_position is 0. •.
Let
and. e, denote the unit vectors along the ,a<!ical
and tangential directions then from the Fi.g 2.53.
e,
y
--
',,.,,
x=.rcoswf
:I;
'1.v_
----~-~- '· \•,
'
oot
' ·l
L..l:'_
. .y~ r sin,oot
I \'
)
i •
'
..
I
x)
!
~----F_lg_._c·~~:~_4_ _ _~
www.puucho.com
If
I
Anurag Mishra Mechanics 1 with www.puucho.com
I190
MECHANICS-I
Irj = [r
2
2
2
2
cos rot+ r sin rot]1/
2
Using the familiar identity .sin e + cos e = 1,
Irl= [r 2 (cos 2 rot+ sin 2 rot)J1/2
2
2
= r = constant.
The trajectory is circle.
The particle moves counterclockwise around the ~ircle,
starting from (r, OJ at t = 0. It traverses the circle in a time T
such that roT = 21t, ro is called the angular velocity of the
motion and is measured in radians per second. T, the time
required to execute one complete cycle, is called the
period.
,-··--···- ·--
ro I
-- --
I
I
·x
.,..,/
ro A/B =
I
I
\
\
....... __
ro
(ii)
' ----
,./
I
a,
-
ro A/B =
' .I
,• I
I
X
, i
1,
I
r sinB/2
1
I
\
,.._
I
......
!
_:_
1
-- --
I
I
V
.
.A/B
A/B - 2r sin(0/2)
I
1_.[ _-.~-_ _F_lg_.2_,s_s_(a_J_ _~J
1
·y
. . ,. ,,,...--
I
I
:
i
r sin0/2 ;
ro2 r ~-
,.
. I
I
1
r··.. :
oo 2 r sin012·· •• :
•'B
I!
} ·
"
., t
I
}
I ________F_:lg-J·~'!__ _______ __
L
•
,
.
·/J,i+-- ' . '
J
.r
1
•A
ro 1r sin0!2/.:
r/ :
J
I:
I
'\\
•
It
dt
I
separation distance between A and B
-------., , ~--.·-·--·--7
= rro 2 [ - cosroti- sinrotJ]
I
VA/B
(anticlockwise)
dv
a=-
I
I
line joining them.
ro 1 r sin(0/2) + ro 2 r sin(B/2)
ro A/B =
2r sin0/2
Iv I= rro = constant.
--
. _
r sine12j
I
v A/B => Velocity of'A' relative to 'B' perpendicular to the
and
,/
r
2
Angular velocity of 'A' .relative to B (ro A/B)
(i)
v · r = r 2ro (-sin rot cosrot + cosrot sin rot)
=0.
Since v is perpendicular to r, it is tangent to the circle
.
oj
':~
'•.'
~- dr
· ~-- ~ - - ~
v=dt
= rro (- sinroti + cosrotJ)
We can show that v is tangent to the trajectory by
calculating v · r : .
y
}rsin0t2!
...____...- iB ro,r sine/2 J
.
I
___, ___ _ Fig:2.57 ______ !
, Fig: 2.55. ·
= --co2r
!,·/\ .
.~0/2 :
~-..
1
I
......
:
ro 1r sine/
•:\J,e/2·r·
I
1
'~, A
:
:
'\\
I
.,
"Iwo particle 'A' and 'B' are moving on the same circle
with angular velocities ro 1 and ro 2 respectively w.r.t. the
centre of circle. Find the angular velocity of 'A' w.r.t. 'B'
when,
·
(i) their sense of rotation is same,
'
.
(ii) and their sense of rotation is opposite.
.,,, .
;
---....,r
,
,.._
lllustration-9
'
' ,/
I
\
The acceleration is directed radically inward and is
known as the centripetal acc~leration.
' 'y
,.,., ... -
I
ro 1 r sin(0/2)- ro 2 r sin(B/2)
.
2r sin(B/2)
,x
_,./
If ro 1 > ro 2 , then ro A/B is in anticlockwise.
If ro 1 < ro 2 then ro A/B is in clockwise.
If ro 1 = ro 2 ,then ro A/B = 0.
Fig. 2.56 (b)
,..
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
(!oRCEANALYS~
191'
-------~
Equations of Motion
tan0=vx
(u sin 0 - gt)
= - - ~0~ ~
u cos 0 0
Similarlyvand0can be determined in terms of0 0 andy.
2
Vy= (u sin 0 0 ) - 2gy
Case I : Constant angular acceleration
ro = ro 0 ±at
1 2
0 = ro 0t ±-at
2
= (J)~ ± 2a0
(J)2
vx=ucos0 0
ro 0 is initial angular velocity
ro is final angular velocity
a is constant angular acceleration
0 is angular displacement in time t
Case II : When angular acceleration a is variable
(a) if
a= f(t)
(function of time)
dro
a=dt
or
J dro= Ja.dt= Jf(t)dt
(b) if
a= f(0) 0 or f(ro)(function of 0 or ro)
rodro
a=-d0
Jrodro = Jade
I; I =Jv; +v;
= ~~(u-s1-·n_0__)_2___2gy
__+_(_u_c_o_s_0_)20
~(u sin 0 ) - 2gy
= ~ - -0- - - vx
u cos8 0
Vy
,-
I
gsin8
Thus
a= g = ~a; + a;
From Fig. 2E.92 we can see that
- cos e;:::: vx
V
= an ;:::: an
a
g
,.y
I ' 8
'- l \,
I \,
Vy. \
',,
a::
~/
' ', , , ,
.
,, '
____ Fi~_-~E-~!- __
a =g!!.!_=
n
.
''
V
a
\igcosO
I
.
-~
and
and
1
.
X'
L__ -- -- ___ Fi?:.2~~9. _________ !
=P
where p is radius of curvature of the trajectory at the
instant under consideration. Thus,
v2
v2
p=-=--.
a, g cose
where and 0 can be· determined in terms of
(velocity and angle of projection) and time t.
vx =u cos8 0
vy=usin0 0 -gt.
vand 0
v ~v; +g2t2
0
~~2$...,.~~J 88 ~
ltt~lloon -s~ar~ risi~g fro,;_ th~ ;~rface of th~- ,"~,;h~itiil
!vertical component of velocity v 0 • The balloon gathers ai
ihorizontal velocityvx = ay, where a is a constant andy is the!
;height from the swface df the earth, due to a horizontal wind. i
:netennine
(a) the equation of trajectory of the balloon.
I
(b) the tangential, normal and ,tqtal accelercation of the:' ·
[_ l?ailg_O_I! ~fun~tion ofy.
· "': __________ -· ____:
l;l=Jv;+v;
= ~(u cos 8 0 ) 2 + (u sin 0 0
gvx
On substituting numerical values, vx = 15 m/s, g = 9.8
m/s 2 , we get a, = 5.4 m/s 2 and a,, = 8.2m/s 2 •
v2
a,
X
~Ir--~
a,
9
--
Solution : The horizontal component of acceleration is
zero. The net acceleration of the stone is directed vertically
downward and is equal to the acceleration due to gravity, g.
t-axis
v/
',
- --
-
Hence
\
I ..-;
v
--
is thrown horizontally with the velocity v x = 15 m/s. ,
!Determine the normal and tangential accelerations of the i
~t9_n!l_ (11_1 s_ecol)_d_ after it 1,_egbls to_ mo_ve._ ___ __ ___
'
n-axis
j,
--
;A stone
Consider a projectile at any instant t with its velocity
vector v at an angle 0 with the horizontal. We choose
tangential and normal axis as shown in Fig. 2.59.
Component of g towards normal axis provides centripetal
acceleration.
\
0
2
tan e = -
Radius of Curvature at any Point on the Path
of a Projectile
\
Vy
and
-
gt) 2
Solution : (a) Balloon's vertical velocity is constant
and horizontal velocity is variable w.r. t. height y. So we have
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
.. ,
'
·:·-v ::,. :f>
= ay;
Vx
Vy =Vo
dx
dy
-=ay·
-=·vo
dt
'
dt
·:dy dy/dt '!o
As
-=---·=... (1)
dx dx/dt ay
On rearranging eqn. (1), we get
ay dy =v 0 dx
... (2)
On integrating eqn. (2), we get trajectory as
ay2
--=VoX
2
or
y
2
-(2v
i'
Concept of Pseudo Force
Newton's laws of motion are applicable in inertial
reference frame but not non-inertial reference frames. In
this section we will see show Newton's law can be modified
so that they work in non-inertial reference frame too! !I
In the Fig. 2.60 two observers, one on ground and the
other in a balloon moving with constant velocity, observe an
airplane.
1 : Position vector of plane in gro~nil reference frame.
1• : Position vector of plane in balloon reference frame.
0) X
- -
.
::.•'
I
I
l
a
(b) ·x-component of acceleration,
dvx
dy
ax =~=a dt =avy =av 0
.
dvy
y-component of acceleration, ay = - - = 0
dt
->
Resultant acceleration a = a)+ ayj = av 0 i
, From Fig 2E.88,
= tan 8; therefore
:
Flg.2.60
,,
->
R : Position vector of balloon.
,-+
-+
-+
f =R+ r'
->
',
.
',
',
'.,.,a~'.
·~ .....
e·
->
->
dr dR dr'
or
-=-+-... (1)
dt
dt
dt
We- assign letters to each body: P, airplane; B, balloon; G,
ground.
L,
~
......~..-n-axls
->
Vpa
->
->
= Vpa + VaG
... (2)
If we differentiate above eqn. (2) again,
...
cos8=
~
->
1+(:r
1
2
=
dy/dx
2
~1 + (dy /dx)
=
->
since
v BG
= constant .
y Non-inertial
Y Inertial
reference
!rams S •
v 0 /ay
~1 + (v 0 /ay) 2 ·
Vo
~(ay)2
->
= ap8 ,
Therefore accelerations of a particle with respect to two
coordinate systems that are moving at constant velocity with
respect to each other are same .
. Now consider two reference frames as shown in Fig.
2.61.
.
ay
=-;========
~(ay) + v~
sin8 =
a PG
1
r'efereflce
frames·
Origins coincide
when r= Os
+ v~
o,,,--~-o.x
2
Tangential acceleration, a, = a cos 8 =
a VoY
, .
~(ay) 2 + v~
-
Normal acceleration, a =asin8=
n
.
av 20
~(ay)2 + v~ . ·,.,~.
z
·zi
A non-inertial refereflce fr8me' S' a~elerating with '
respect to !ne_i;tlal frame S.
'
•'
Flg.2.61
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
1' 1~~'":;'.",~;[t~,-. - ; , . .
F'·FO~CEANALYSiS
-~-::::.fr, ~--::1 ''" --·: -~-"'.-~~~: "": ~ ~ ;.-;_.
-
. , ..
-;1,,-"' ';;,
The tr~nsformati~n equati~ns .relating :the coorc:iinat~s
of the particle in each reference frame are
· ·
1
2
x = x' + vxut + - a,,,t
According to the observer (non-inertial) riding in the car
. the pendulum bob is at rest. The thinking point for him is:
...,
which force has balanced horizontal component of T.
2
Equations for non-inertial observer:
y =y"
E~--~ (:~"~~~
z =.z'
I·'
Acceleration of particle in inertial refe~ence fr~me Sis
_,
d 2 X • . d 2 y o d 2z i'
a =--i+·--J+--K
dt 2
dt 2 . di 2
. d2
·
r
2 ,
d 2 y' , d 2:z'
=(x'+vx 0 t+-ax 0 t )1+-2-J+--.k
2
dt . .
2.
dt
dt 2 '
2
2
2
d x', . · , d y' • d z'
= --.
1 + a, 1+--j
+--k
dt 2
O . dt 2
dt 2
--+ . --+.
a
Therefore
=.
lnerllal .
·obsel'VSri
L.
[f ,•, .
(al,
Non-iriertial': 0bS8~er:1 ,
"
-t .
For inertial observer,
~xpression
for.Newton's
_,
_,
_,
. . . .law
. is .
= ma + ma 0
\
'- r a:
··: f/ftictiiioUs:, ·:• ·
', .::
.
_c,:;'.-f:! ..
;; · • . JnQ,
... (4)
~ .....
which is wrong, There is an additional term 'mao iri .
equation (3). ·
· The additional term on the left hand side {-m
called a pseudoforce.
Ho;_.,ever, if we rearrange c (3) in the form ·
--+
F1o1a1
.__.-,
.-t
a:i) is·
-t•
+ (-ma 0 ) = in a
Real force
Pseudo force
therefore in a non'inertial reference· frame, Newton's
law can be written as
_,
_,
F real-+ pseudo = m Ji'
i.e., vector· sum of real forces and, pseudo forces on the
system is m °ii' where. a' is observed acceleration of mass in
non'.inertial reference frame.
Newton's second law_ can be applied ·by considering an
pseudo (imaginary) force -m ~ o on the left hand side- oflaw.
Non-Inertial Reference Fram~
Illustration 10.: Consider a pendulum bob in an
accelerated train car. Pendulum is inclined co vertical at an
angle 0. According to an observer on_ ground -(inertial
observer) the forces acting on the bob are: tension of string
'i and weight of the bob mg. ·The :icce1eration a is provided
to the bob by horizontal· component of T, and vertical
component of tension balances weight.;.
l:F'x = T sin 0 = ma
... (1)
l:Fy = T cos 0 - mg = 0
... (2)
On solving eqns. (1) and (2) simultaneously, we obtain
a=gtan0
:.'
. ·:0,",t,;,.:_ i-J't~
... (3)
= ma'
~
. 'l._' ,-
For non-inertial_, observer
. _,expression will be_·
Ftotat
.
•s
a' + a_0
F,0 ta1
~ '
I .
;;-·:, C!>i
,. : d
.'
T sin 0 - Fpso.,,io = 0
ITy = T cos 0 - mg = 0
The non-inertial observer must obtain
same
mathematical result as the inertial observer does, whtch is
possible if
. l:Fx ".'
·~pseudo = _mainertial ;;;;; ma.
Illustration 11 : Consider a block kept on a
frictionless turntable, connected to centre by a string.
According to ground. observer (inertial. reference frame)
block moves ;,long a circular path. Therefore it must have.a
centripetal acceleration provided by tension of the string.
From Newton's second law,
·
Non-inertial i
h
!:·,· . / "-n,·':· 'r-~1:t"'..-'", i'..:.: ,;··~:.,.~::
obseiv6r' -'. ' '
r~
,,,,--,
i r
...
! ;
·,
··'.",9
·. ~ ~5-'
' " " '
. •
' .
(·::~
."1t ' • ,, ,:, , Inertial observer, ,,.~ •
I,"·
~ (a) .., ' '
·
,
L
,, \.-'.·:,":. ,_
l
!
'(b)
-Fig.'2,63:· ___ -,.,,.,;~~ ------~-~mv2
T=-
!
r
According to observer ·(non-inertial) on the block the
· block is at rest. Since observer and block turn through same
angle, the observer will always see the block in front of him.
In order to explain equilibr(um of- the block, ·the observer
must imagine an outward force to balance tension, i.e.,
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
1194
mv 2
r.F=T--=0
r
This outward acting pseudo force (imaginary force) is
termed centrifugal force.
(b) Minimum contact force between two bodies is zero.;
at this point contact between two bodies breaks.
mv 2
From eqn. (1)
0 = -·-' - mg
r
or
v, =-..fir'
(c) At the topmost point, IFy = may
·
mv 2
l~Exi&.~\~~~
E=.=,E__
' -~.....-
I
A pall of }'later 1s whirled ''L.i: circle of r~di~f,,,;r1:.i&J
1topmosfpoirit the speed of the pqi[is·v, : .,· , 'L, .i{,,.i/·I
'(a) I)eterrnine•the force exerted,on:\vater by the'.pa/Uq(the1
top of:the'drcle
. .• " : .
. . ' ·• "' :· .
(b). Pihgtke•;mini111~m value ofv, jqr the .,;at,?ta
in
l
r~#taj~
I
Fp -mg = - - b
r
Fp
or
= mvi + mg
r
Remark:
When a particle moves along a curved path, no
particular force can be said to be centripetal force, It is
the name for resultant force that must be directed ·
towards of circular path,
'the.pair .• ·. . · · · . , _ . . · . . :,:~ .. _'.
(c) Find th.e force· exerted by th¢ pail at: the ·gottomtoj;'the
l_ circle ,,/;hiJreCspeed is Vi, . .£:.:"' . , _' .. ::; >, ; ',
Solution : Forces acting on water are weight mg and
........
'
the force of pail on water Fp . Fp may be termed reaction of
pail on water. Same force will be exerted on pail by·water.
~·,,,...,.··---- .
.~·--c,:- .·- i
·Y
t-=-~~
r,:-"1
~-t
:<'"'"' _ _ _ --~---·--,,...,,_.,
'
".
····-~·- ··,· -
l'd~rib.ei 1'hcrriiontal ~irele of raqius r with speed v. T(t(ropel
!makes aniingle8.withverticC1lgivenby sine= rf 2,Determine'1
'(a) 'the, (ension}(I the rope,.,and (b) the speed .of the'balZ: j
1
(c):Ji,111e,per;ip_cl,_of b_gl[._ , .. ...: .. •
. ...
.::___, __ .. J·
.··: ''",
L.:,;'
•
---,
Solution : Forces acting on the ball are: weight mg .
and the tension in the string.
Note that component of tension T cos 8 towards centre
of the.horizontal circle is the required centripetal force.
r.FY = T cos 8 - mg = may = 0
... (1)
2
.
mv
r.Fx =T sm 8 = max =-r... (2)
t
L·
(a) At top of the circle,
r.Fy = may = m (- v!)
( '2)
-Fp -mg=m -~,.
2
or
Fp
mv
=--'
-mg
... (1)
r
Note that there are two ways to write a force equation:
(1) Assign positive and negative x, y axes; e._g,,
centripetal acceleration is towards centre of circle. At the
topmost point it points in negative y direction; F; and mg
also point in negative y direction. ·
mv 2
·
(2) Set net force towards centre equal to - - ,
r
i.e.,
mv 2
Fp +mg=---'
J
From eqns. (1) and (2),
V
= .Jgr tan 8
(c)
v
= rro
(J) -
r
Fp
mv
=-.-'
-mg
r'
www.puucho.com
v2
tan0=-
or
2
or
--~~--
0 ball of'rnas~ m is suspended from· a rope oflength r . rtl
~
v~
T = 21t = 21tpcos8
(J)
g.
rg
Anurag Mishra Mechanics 1 with www.puucho.com
'
FORCE ANALYSIS
195 I
... -1
WHIRLING ROPE
Car Negotiating a Circular Bend:
A uniform rope of mass M and length Lis pivoted at one
end and whirls with uniform angular velocity ro. What is the
tension in the rope at
distance r from the pivot?
(l)
- - - -....
Neglect gravity.
.________
Ii ii ii // I :7n,
Consider the small ~ I T/1 rc.)'1
L
'1
section of rope between r
_/
,
and r + l!.r.The length of
the section is l!.r and its
ii/Iii/I
mass is t.m = M l!.r/ L.
Because of its circular
r+M ___.
motion, the section has a
radial
acceleration.
Therefore, the forces
T(r+l>r)
pulling either end of the T(r)
section cannot be equal,
Fig. 2.64
and we conclude that the
tension must vary with r.
The inward force on the section is T(r), the tension at r,
and the outward force is T(r + l!.r ). Treating the section as a
particle, its inward radial acceleration is rro 2 •
The equation of motion for the section is
T(r + l!.r)-T(r) = -(/!,.m)rro 2
t
=
,_j
Mrro 2 !!.r
L
However, by dividing the last equation by l!.r and taking
the limit l!.r ~ 0, we can find an exact expression for dT/ dr.
dT = lim T(r + 1!.r)-T(r)
dr or-,o
l!.r
2
Mrro
A car, travelling along a level road, enters a tum with a
radius of curvature R.The coefficient of friction between the
road and the tires is µ.What is the maximum speed at which
the car can negotiate the turn ?
Concept: When a car turns a comer on a level road,
friction is the only force acting horizontally on the car. It is
therefore the friction exerted by the road on the car that
accelerates it around the turn that is provides necessary
centripetal force required for circular motion. Because the
tires roll without slipping, friction force involved is static
friction, and it is the limit on static friction that sets a
maximum speed for rounding the turn.
Figure shows two views of the car. Since the car is not
accelerating vertically :
0=IFy =N-W
=> N-W= 0 or N=Mg
... (i)
...
The top view shows the horizontal force f, acting on the
car. Since friction is the only unbalanced force acting, it
equals the ca(s mass times its acceleration :
f, = IFx = Max = Mv 2 /R
... (ii)
The maximum speed is that which requires maximum
possible friction f max = µ ,N. Combining this result with
eqns. (i) and (ii), we have: Mv~,,jR = µ,N = µ,Mg
y
-,
N
=--L
...f
To find the tension, we integrate.
Mro 2
dT=---rdr
X
L
2
dT=-J'Mro rdr
To
o L
'
where T0 is the tension at r = 0.
Mro 2 r 2
T(r)-T0 = - - - -
J
T(c)
L
(a) end view
(b) Free-body diagram
2
Mro 2 2
T(r)=T0 - U r
or
To evaluate T0 we need one additional piece of
information. Since the end of the rope at r = L is free, the
tension there must be zero.
We have
1
(c) Top view
Fig. 2.65
2
T(L)=0=T0 --Mro L
2
1
.
Hence, T 0 = -Mro 2L, and the final result can be written
2
2
T(r)
= -Mro
- ( L2 2L
r 2 ).
The mass of the car cancels out,
and
www.puucho.com
V max
= ~µ
5
gR
Anurag Mishra Mechanics 1 with www.puucho.com
'"''""·----·
119s
"----··-··- .........
The maximum speed depends on the. road conditions via
the. coefficient of friction. On a wet road, the coefficient of
friction between the tires and the road is reduced, and the
car cannot turn as rapidly as on dry pavement.
Roads designed
for high-speed traffic
have banked turns
(Fig 2.66). Then both
the
friction
and
normal forces exerted
by the road on the car
--,_
have
horizontal
w
components
that
.,.__to center of turn R
together ~ause the
Fig. 2.66
necessary
acceleration:
No
friction is necessary, a11d you can round the turn even on an
icy road at the proper speed for a given bank angle. Example
96 for motion along banked road.
Lift Fore~ on an Airplane
Airplanes also make turns by banking. The lift force, due
to contact forces of moving air on the wing, acts at right
· angles to the wing chord when the aircraft banks, the pilot
maneuvers to obtain greater lift than necessary for level
flight the vertical component of lift balances the airplane's
· weight, and the horizontal component accelerates the plane.
Concept': What does it mean t~ feel heavier? In level
'flight, ea~h pass.enge~'s weight is balanced by the normalf~rce
!exerted by tlie"se(lr; -exactly as if the person were at rest on the
,ground.' fhe person's muscles tense to maintain an upright
,posture. This 1m1scle tension· and the pressure on our l:>ottoms
lis what ~e ·se!1s~ )Vhen we speak offeeling our weight. Wlien
!the airpla11e:biinks, the seat has. to exert enough normal force
ito·balance weiglit and to accelerate· the plane.
•
-·--
>•·
---·
-
•
!
-+
L,
!
,.
0
0
-+
w
• (aj.
(b)
~1
I
'
L__C_. - - - - --- ' - · · · ·- · - -
8b:;;
:-7
MECHANICS-I
s·
'
1
___ }
Motorcycle Stunt
~·
f
_,
_,
N
_,
N
w
--; I
w:
Fig. 2.68
Fig 2.68 is a free-body diagram for the motorcycle and
rider, modeled as a single particle.
Concept: Static friction, exerted by the cylinder walls·
on the motorcycle tires, balances-the weight of cycle and rider.
'The normalforce acting on the tires causes the centripetal·
iaccel_':ration of cycle and rider.
If the rider tries the stunt at too low a speed, the normal
force will be correspondingly small, and the maximum
possible friction will be too small to balance the weight. (On
a straight wall, there is no horizontal acceleration, no
normal force arises no matter what the speed, and the stunt
cannot be done.) The minimum speed for the stunt is that
for which maximum friction can just balance the weight.
Vertical Components Horizontal Components
'I.Fy = 0
'I.Fx = Max
f~Mg=O
N=Mv 2 /R
At the minimum speed, friction is at its limit;
µ,N = fmax =Mg.Thusµ, Mv~;n/R = Mg; so:
Non-uniform Circular Motion on Horizontal
Plane
Let us consider that a particle of mass 'm' is moving in a
horizontal circle of radius 'r' with velocity' v' and tangenti;tl
acceleration
We will solve problem in reference frame of
car. To oppose the tendency of skiilding of the particle
(body) in the direction of net force F,er, a static frictional
force F, is developed as shown in the Fig. 2.69 .
To avoid skidding,
a,.
rnv 2
\
-·-~~ ' · -
· - - - - ._ - - · · ~.,u,.
r
bank angle
-+
w·
Fig. 2.69'
·(c) ,
Fig. 2_.67 , --
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
FORCE ANALYSIS
i.e.,
- - - - - - - ' - ' - - · - - - ____________ ------~-'-----'-'---'---'1~9.:-111
F,
=
(m;2r+(ma,)2
.
mv 2
LFx = N sm 0 = max = - -
... (1)
r
Since
F, <; µ ,N
... (2)
Where N is the normal reaction (N = mg) and µ, is the
coefficient of static friction between the body and the
ground.
From eqns. (1) and (2), we get
2
mv ) ·
2
( -r- +(ma,) <;µ,mg
( vr2
r
+
And
• LFY = N cos 0 - mg = may = 0
... (2)
Thus from eqns. (1) and (2),
v2
tan0=rg
(b) If driver goes faster than designed speed v m/s, a
frictional force must act parallel to road and inward toward
the centre of road.
mv 2
Il'x =N sin0+Ffri,. -cos0=max = - - ·
r
a; <; µ ,g
v <; [r2(µ;g2 -a;)J1/4
Hence the maximum velocity, so that the body with
tangential acceleration a, in a horizontal circle of radius 'r'
can move safely without skidding is given by,
vmax = [r2(µ;g2 - a;)J1/4
For uniform circular motion (a, = 0), the maximum
velocity with which a body can perform·a horizontal circular
motion safely without skidding is given by
LFY = N cos 0 - Ffri,. · sin 0 - mg = may = O
When a rolls without slipping, there is no slipping
between the road and point of contact with road. Therefore
static frictional force comes into play. Since we require
maximum speed with which the curved road may be
negotiated, we will require maximum frictional force. ·
Ffri,. = µ,N
Thus our equations are
mv 2
N[sin e + µ, cos 0] = - ... (3)
r
N[cose -µ, sin 0] = mg
... (4)
We can eliminate N by dividing eqn. (3) by (4).
sine+µ, case v 2
case-µ, sine . rg
·-
= ~µsgr
Vmax
v =
or
r :Etxcii,;:.;~,ef9iT---,,
f~
[~-,.~~::;,,,,~~-iL-i~~::,i.~
A section of a hilly highway is a circle with ·radius r.
:
(a) What should be the banking angle e of the roadbed sol
that cars travelling at v mis need n_o frictional force from 1
the tyres to negotiate the tum?
(b) The coefficients of friction are µ, and µ k • At what'
maximum speed can a car enter the curve without sliding:
towards the top edge of the banked;curve?
'
•
:-~·
·--~...
--
,
-
N
-··
._,.
-
..
1-
Ncose
X
e
mg
mg
N 81§'Ny~
cos
.· I
(sine+µ, c~s e)gr
cos0-µ, sU18
.
- .
-
.
- -- -·
- .
--- ---·;" - · - 7
A small block B is supported by a tum-table. The friction
:
coefficient between block and'suiface'is µ: '. ,-: .'" .
(a) If tum-table rotates at constant_'angular'speed_ OJ, what,
can the. maximum angular speed OJ be fo·r w/tich the block
doesnprslip?' •. :
-- .. , .-.
- .. (b) ~f the ang,;la,· speed is increased.uniformly from rest'with
an angular acceleration a, at ivhat 'speed will' th'e block
slip?
'
· · ".:
. · ·' '· ' " · i
(c) Of the tum-table rotates in such a way that the block'
undergoes a constant tangential acceleration,· what is. the
smallest interval of time in which the block can ·reach the
speedv?
'
·
- I
--- • j '
I
1
~-------
~,;~~H~JIJ.~J92!p
I
• 'i
... (1)
~r-+
-1
II
I
N sin 8
Fmcsin8l
I
cCOS9
!
mg
.·',':'·.
'
--------
Fig. 2E.91
Solution : (a) Fig. 2E.91 shows front view of car. We
assume no friction, hence the only forces· acting on the car
are normal reaction and weight. From Newton's second law,
www.puucho.com
i
Anurag Mishra Mechanics 1 with www.puucho.com
MECHANICS-I
1198
Solution : (a) In Fig. 2E.92(a), the circular path of the
block is shown. The only force directed towards centre is
Ffriction•
Since angular speed is · constant, the bloc!< has
centripetal acceleration only.
From Newton's second law;
mv 2
,
LFx = Ffri,tion = max = - r -
µ,mg=--
r
Thus
V nnx. = ~µ ,gr = r 0) nnx.
or
ffimax. = ~µ,g/r
(b) When the tum-table rotates with angular
acceleration, the block has centripetal as well as tangential
acceleration.
r-, - -- - ---
Frriction
- Tapvi8w
Ffric:tion
..
·side view
mg
r
Fig. 2E.92 (c)
(c) From equation of kinematics,
v =v 0 +at·
if block ~tarts from rest, v O = 0.
So
· t = !:_
a
where
a = a, = ra
When the block is on the verge of slipping,
V
= ~(µ,g) 2 -
t
=- - - - - - -
(ra)
2
~(µ,g)' - (ra)'
(ra)
k,_g~~~~
mg
r-·
,
· .
-~---"'~---~w-:·-7
IA.50 kg wo'man.is on a large swing (generally seen inft;Iirs)of1
radius 9 m that rotates in a· vertical circle at 6 reve.1/min. i
Side view
Fig. 2E.92 (b)
aR
. .
Therefore
:
Therefore
I
... (1)
r.Fy = N - mg = may = 0
... (2)
Since F motion, ""'- =µ ,N from eqn. (2) we substitute N
. into eqn. (1).
mv 2
...,
Toµ,vlew~··
.
N
1
What is ·the magnitude of her weight when she has movedi
~~~,
..., ...,
= a,+ a,
~
.
1
I~aR I = '\/Ia,2 + a,2
= ~(ro 2 r) 2 + (ra) 2
Resultant acceleration of block is parallel to surface of
tum-table. The only force that is parallel to surface is force
of friction.
So
LF'r
= m~(ro 2 r) 2
+ (ra) 2
µ,mg= m~(ro 2 r) 2 + (ra) 2
or
ro
= (µ,g) 2
-
jy
Flg.2E.93
and
LFY = N - mg = O
From eqns. (3) and ( 4)
(ro 2 r) 2
N,
mgX,
= Ffriction, m:ix.
=µ,N = max
-- mVla2C + a2l
or
N,-f
., I.
(ra) 2
2 ]1/4
= [(µ;g) -a•
... (3)
.... (4)
;l
I
Solution· : The woman experiences three forces: mg,
her weight acting vertically downwards; N 1 , reaction due to
her ·weight; N 2 , horizontal reaction whlch provides the
centripetal acceleration.
From Newton's secon(i la.v,
mi,2
LFX =N, = - r
r.Fy=N 1 -mg=O
v = (21tr)v
(wh'erevis·frequency)
;. (2!t X 9)(6)
= l.81t m/s
Therefore,
N
www.puucho.com
2
= (SO)(l.Bit)' = 178 N
9
Anurag Mishra Mechanics 1 with www.puucho.com
i--FORCE ANALYSIS. . .
1-- --
199:
-
N 1 =mg= 490N
The magnitude of her weight is the magnitude of the
resultant force exerted on her by the chair.
/
2
Solution:
Concept: Detennine tangential and nonnal component
of force E Apply
IFn-mRro 2
2
N=-yN 1 +N 2
=~490 2 + 178 2
LF, =mRt.
= 521 N
,
F case = mro 2R
r--·.
i s.-"t:'SJ?·~P r.e i 9_4 _i>
F sine= _ma,
Angular velocity ro of Line joining P and C is
ro = d(20l = 2 de
dt
dt
:In amusement parks there is a device called rotor where
people stand on a platfonn inside a large cylinder that rotates.
about a vertical axis. When the rotor reaches a certain
angular velocity, the platfonn drops away. Find the minimum·
coefficient offriction for the people not to slide down. Take the,
radius to be 2 m and the period to be 2 s.
'
...:>
'
: N
;
......
... (2)
de = (~) and tangential acc. of particle about
dt
2
Ca =Fsine
'
m
ar=Ra=(F:::'e}
y
f
... (1)
a=F:e
t
~
a=dro= d[2~]=2d2e d2e=~
dt
dt
dt 2 ' dt 2 2
1---1mg
'
Fig. 2E.94
Solution : In this case normal reaction of surface
provides centripetal force and friction force prevents the
man from sliding vertically.
From Newton's second law,
mv 2
Lf'x =N = - -
... (1)
r
Lf'y
where
=f
- mg = 0
Fig. 2E.95 (b)
... (2)
d 2e Fsine
-=-dt2
2mR
2
f = µN = µmv
r
µmv 2
From eqns. (2) and (3), - - = mg
r
or
C
µ = rg =
v2
gr
(2nr/T) 2
,.,(3)
From eqn. (1)
de = ~ = _! x (F cose)1/
dt
2 2
mR
= 0.5
(:~r ::(~r
(:~r
la~~R~P Ie i_~5,:y
A particle Pis moving on a circle under the action of only one,
force acting always towards fixed faint O on the:
p
=¼F:e
... (4)
~ = Fsine x 4mR = 2 tane.
2mR Fcos8
1···-- . . . . .
r-c.
ks~q_!TI__p!~J -~:.---96 l >
'
.
IA car is moving in a circular path of radius 50 m, on a flat.,
!rough horizontal ground. The mass of the car is 1000 kg. Ata•
;certain moment, when the speed of the car is 5 m/s, the driver'
s 2• Find the value of
,is increasing speed at the rate of 1
•sta_ti,frictio!J.. on tyres at this moment, in Newtons,
m/
Fig. 2E.95 (a)
2
d 2e
- ·:
.
,,
F'md ratw
. oif -d2e & (de)
czrcumJerence.
.
dt 2
dt
... (3)
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
- -- .-- - -- -- --- -MECHAN·1cs:1 i
--·-----------··- ·---- __.
1202
~-------------·-·--,.-··
. 5m
···IIJ
:......._____~-----··.-·
B
'
Fig. 2E.103 (a)
'(a) Tangential acceleration of the block.
(b) Speed of the block at time t.
(c) Time when tension in _rope becomes zero.
Fig. 2E.102 (a)
Solution: (a) Tangential acceleration is the retardation
produced by the friction
a= -Jim= -µmg/m
a, = -0.2x 10 = -2m/s 2
dv
(b)
- = a, =-2
1 ••.•------·:· ......
dt
Solution: Radial direction:
T1 sin60°+T2 sin60°= mco 2 r
11:......... .
(T1 +T2 )sin600= mco 2 Lsin60°
= mco 2 L
Vertical direction :
T1 cos60°-T2 cos60°= mg
T1 -T2 = 2mg
adding eqns. (1) and (2),
2T1 = 2mg + mco 2 L
mco 2L
T1 =mg+--
... (1)
... (2)
60°
10
0
•/
5m
p1Qm/s:
.
·'·
: ·~········--------------·· ·,
·
I
'.
Fig. 2E.103 (b)_
v-10=-2t
V = 10-2t
(c) Tension in the rope will become zero when
centripetal acceleration becomes zero i.e., when speed
becomes zero
v=0
=>
10-2t=O =>
t=Ssec.
L.s,dq~p,,e;__[W41~
·A ball of mass M is swing around in a circle around on a lighti
·spring which has spring constant k The ball describes a
:horizontal circle a distance h above the floor. The stretched
spring has a length I and makes an angle ewith the vertical as,
,shown in Fig. 2E.104 (a). Neglect air resistance.
:
. 2
_____
C
f dv = -2 f dt
Fig. 2E.102 {b)
T1 + T2
V
60°!
,::·..··.·.·.::·"·..--+-.C...............
.......... .
<:~.·-·.·...........
Fig. 2E.102 (c)
M
(b) Tension in lower string= zero
... (1)
Tcos60°= mg
T sin 60° < mco 2 r
T sin 60° < mco 2L sin 60°
T < mco 2L
substituting eqn. (2) in (1)
mco 2L cos 60° > mg
(02
> 2g
L
=>
/illlllllll/111/JIIIIIIUIIIIJ/J
Fig. 2E.104 (a)
... (2)
co>Ff
~~mpJ~f 103 )>
25
~-block of~ass
~-res~ on a hor~o~taiflo;r (~- = 0.2). It;
lis attached by a 5 m long horizontal rope to a peg fixed on,
,floor. The block is pushed along the ground with an initial.
\;L~f~
(a) In terms of only the given quantities, what is the·
magnitude of the force F that the spring exerts on the
mass M?
,Cb) In terms of F, k and i what is the natural length 10 of the:
spring, i.e., the length of the spring when it is not:
stretched?
'(c) In terms ofF, l,M and 0, what is the speed v of the ball?
(d) At same instant aftime, the spring breaks. The ball moves'
a horizontal distance x before it hits the floor. In terms al,
l .v, Ii, ,ind g, what is x_?
of 10 ~/s so that it mov~ in a circle_ aroun_d ~he_ ~eg. j
www.puucho.com
I
Anurag Mishra Mechanics 1 with www.puucho.com
l_Fo~c~_~N_A~s~ -- · ·--- · ·__ .. __ -·--~ .::_:_ ~~-
-- ------
Solution:
Concept: When a particle moves in a circle,
,perpendicular forces along y-axis balance out. Towards of
horizontal circle centripetal acceleration acts therefore that
must be a resultant force. Note that along the length of spring
forces are not balanced because this direction has component
of acceleration.
~~
friction between the shoes and the drum is µ, find the power
required in watt to tum the governor shaft.
Solution
Centripetal
force for rotation of brake shoe
comes from normal reaction
between brake shoe and drum.
N = mrro 2 = mr(2rrf) 2
! \._
+.
mg
Friction force
(c)
(b)
(F)
Fig. 2E.104
(b)
F,p
=>
=
F
sp
= µmr(2rrf)2
10 = 1 - ~
Kcose
2
.
mv
F,p sme = - r -
Mg
case
P=2Fv
= 2 x [µmr(21tf) 2 ] x r(21tf)
= l6mµ1t3 f3r2
r
~
,--~,.
'-................... ;r=l~_;_~~---·
,
Fig. 2E.104 (d)
mg .
mv 2
--sme=-cose
I sine
--~
---
- · - - ----
A particle suspended from the ceiling by inextensible light
string is moving along a horizontal circle of radius 1.5 m as
shown. The string traces a cone of height 2 m. The string
breaks and the particle finally hits the floor (which is zy plane
5.76 m below the circle) at point P. Find the distance OP.
I _
glsin 2 e
cose
h
X
1 2
= -gt
2
=>
t=f!
~
1.5m
.-, ,./
/
= Vt
x=vf!
l.,S:~R\J\.i:?J?
--r--,..
~~gm~,!,~ }106 I >
v=.1=--(d)
Fig. 2E.105 (b)
Power
required
to
overcome friction force on both the brake shoes
= K!il
Iii = F,p = ~
K
Kcose
10 = l - Iii
(c)
--
Fig. 2E.105 (a)
(~·.·.·.·-.-,
-~~---5t
--~···· ,.,
F,p cos0 = Mg
- - - - - - - ~------ -- ---
acceleration
"\:T··.0,
(a)
·---- ..
component of
+y-axis
tTcosB
8
-
y
!1057>
Q
112m
5.76m
X
.! -------p[_..-·····
Fig. 2E.106 (a)
The essential elements of one form of simple speed governor
are as shown : to a vertical shaft a horizontal rod is mounted
symmetrically and on the horizontal rod are freely sliding'
brake shoes,
'
.
When the shaft turns at a frequency of rotation f the brakel
shoes press against the inner surface of a stationary,
cylindrical brake drum. If the brake shoes are each of mass m!
and their thickness dimension is negligible compared to the
inner radius of the brake drum rand the coefficien~ of sliding'
Solution : Let the string breaks when the particle is 1.5
m right of point O and direction of its velocity v is along
y-axis.
.
mv 2
Tsme=--
r
and
Tcos0=mg
--v = .Jgrtan0
Now time to reach the floor,
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
f 204
. ·-
. -. - --
______ M_ECHANICS-1_i
I _____ -- ------- ---- . - - - -
t =
~
=;
Before it hits the floor,
l'>.y = vc:t,---c:
= ~2h 2 rtane
r
where cane= h1
T
0
--,-
v=~=Sm/s
(b) Tangential component of
force
=k(3:)sine
dv
9kR
m-=dt
25
Rate of change in speed
dv
9kR
-=-dt
25m
Fig. 2E.106 (b) ,
l'>.y=~
=
2xl44x(l.5)2
25 . 2
18
=-m=3.6m
.
Fig. 2E.107 (b)
·=;
24m/s 2
5
Its position from 0, when it hits the floor = LS i + 3.6]
OP= ~(1.5) 2 + (3.6) 2 = 3.9 m
A bead of mass m = 300 gm moves in gravity free region.
.along a smooth fixed ring of radius R = 2 m. The bead is
'attached to a spring having natural length R and.spring,
constant k = 10 N / m The other end of spring is connected to;
An inclined plane of angle a is fixed onto a horb:c,ntal.
tum-table, with its line of greatest slope in same plane as a
'diameter of tum-table. A small block is placed on the inclined
plane a distance r from the axis of rotation of the tum-table
and the coefficient of friction between the block and the
inclined plane is µ. The tum-table along with incline plane,
spins about its axis with constant minimum angular velocity
~-
'
:
6
r
:
,.__,.;
a fixed point O on the ring. AB= R_ Line OB is diameter of
!
-5
!
1ring:
Fig. 2E.108 (a)
Fig. 2E.107 (a)
_Find (a) Speed of bead at A if normal reaction on bead due to
,ring at A is zero.
i(b) The rate of chqnge_ in SJJ.eec:( qt this irJ5.tanf.
Solution:
Concept: Spring force has component in radial as well
_as tangential dir_ection.
(a)
Elongation in spring= (2R) 2
-(
6
:
r
-R
(a) Draw a free body diagramfor the block from reference of.
ground, showing the force that act on it.
(b) Find an expression for the minimum angular velocity, oo,,
to prevent the block from sliding down the plane, in terms
of g, r, µ and the angle of the plane a.
(c) Now a block of same mass but having coefficient of
friction (with inclined plane) 2µ is kept instead of the
original block. Find ratio of friction force acting between
block. and incline now to the friction force acting in part
(b).
Solution: (a)
N
k
=SR_R=3R
5
5
Radial component of spring force
=
k(3R)
case= 12kR
5
25
As normal reaction is zero
12kR mv 2
--=-25
R
.
Fig. 2E.10B_(b),•
Concept: Not force along vertical axis is zero and along
radial axis provides centripetal acceleration vertical axis.
. .
www.puucho.com
.
Anurag Mishra Mechanics 1 with www.puucho.com
[ioRCE ANALYSIS .
~~~~
Ct. stone is ia~~ched upward at '45° with speed v~. -fl beej
(b)
0
[follows the tra;ecto,y of the stone a_t a constant speed equal to/
\the initial speed of the stone.
··
.
;
(a)
Find
the
mdius
of
curvature
at
the
top
point
of
thei
•
•.
.
.·
'I
I
traJectory. · ·
. ·
. . i
'.Cb) What is t/ie acceleration of the bee atthe top po/nt of, the1
L_. trajectorxLf.QrJ/Je store,J1_egle_c_t.thLair.J:.<J,~tg.1J<;.e; __ ... .:
Solution: (a) At the topmost point of trajectory weight
mg acts as centripetal force. Thus,
Radius of curvature,
µN sina+N cosa- mg= 0
N sina-µN cosa = mro 2 r
mg (sin a-µ cosa) mro•r
(µ sin a+ cosa)
(J)
g(sina -µ cosa)
r(µ sin a+ cosa)
=
.I
1~-~-
kif ...., .. '.. ··--r-t·-
1A circular r~ce. track is banked at
4s
v,2
an
2g
(b) For bee speed is u 0 and radius of curvature
trajectory is same as that in part (a) :
and has a radius ~},
:40 m At what speed does a car have no tendency to slip ?If tlie !
;coefficient of friction between the wheels and the track is .!.
0
.
a =-=-o_
n
Re
Vo2 /2g
. an= 2g
~:::fml:>:;:,,,,.
t:Exam,.,; I
~CE===~tl!?=s·....~
---·----- - --- -----~---··- -·-
Solution : (a) Banking angle is given by
v•
tan8=rg
v 2 = -Jgrtan8 = .J400 = 20m/s
mv 2
(b) Normal to plane N = mg cos45°+--cos45°
'.
'"
.
"
.
.
r
)
= ½;{g+ vr2)
I
Along the plane
2
friction + mg sin 45° = mv cos 45°
r
m (
2.Jz
v
2
)
g+7 +
·g
v2
2
2r
mg
mv
!I
2
c
.Jz = .Jzr
.,.__.
'
· Fig. 2E.111
\
--
g
v2
r
2
v
3g
-=2r
2
2
v =3gr=3xl0x40
V maxc
j
'
(a)
~-----------~----·-- -- .__
-+-+g=-
= 1200
= .J1200
l
;A .rock .is· launched upward at 45°; 'A bee moves along the;
itrajecto,y of,,the rock at a i;onstant speed equal to the_ initial!
!speed of thejoi:k. What is the magnitude of acceleration (in!
!m/s 2) ofthe bee at.the top point of the trajectory? For the'
kos(<, _nfg/iftth_e q_ir..resistaric;e.... ., _ __ ____ _ _ .
Solution : From previous problem we have at highest
point a,= g
i
. - ••
2
= ;(g+ vr
v.2
vz
,!
2 I
find the maximum speed at which the car can 'travel round theJ
'trg.ck.without ~l,idding.
, · • . . . __ . . ........ J
f,maxµN
vz
R=-=_o_
..::_~~~~~,~t::1,10~~
.
.,."-.
·an =g
Force of friction will remain unchanged.Hence ratio is 1.
'
~,
=.p..
i__ ___ .. ~i_!!::e.~10_ _
(b).
·
45°
iv~al=g"""
(c) As block remains static at same height and
radial distance, requirement of friction is same as in pact
:
'IJ
_ 0cos
.
[R, = Radius· of curvature]
u2
-=g
2R,
.:,",
www.puucho.com
... (1)
Anurag Mishra Mechanics 1 with www.puucho.com
--- ------·- - ---------
___________·:.~.- _. _- _ MEfiiAN!CS-i
-------...·-------·---
Now when bee moves along the same path with
constant speed u, then at top point, since radius of curvature
(R, ) remains same
u2
R=a,
... c2i
:i:.Fx = mrm
2
cos 0 + µN - mg sine= 0
i
... (3)
:i:.Fy = N - mg sine - (7lrro 2 sine= O
... (4)
Substitute N from eqn. (4) in eqn. (3) to obtain
'
g(sin0-µcos0)
= [ Rsin0(cos0+µsin0)
Olmin
]1/2
Therefore the block will remain stationary relative to
. < ro < ro max.
bowl if its angular speed lies in the range ro mm.
Students are advised to write the equations for block in
ground reference frame also and verify the similarity of
results in both the approaches.
Fig. 2E.111 (b)
From eqns. (1) and (2), we have
1 g
-2==>
a,=2g
a,
a,= 20m/s 2
-- -----· ---- -- -- -.
kF.~P-ID;PJ:.c~,! 112
~-~
j~
•
'A wedge with mass M rests on a frictionless horizontal
;A
block is kept inside a hemispherical bowl rotating with;
,angular velocity Ol. Inner surface of bowl is rough, coefficient
'of friction is µ. The block is kept at a position where radius
,(Jlakes an angle 0 with the vertical. What is the range of the'
'angular speed for which the block will stay at the given.
position?_
·
,w:
;_
I8 ,
'
I
I
. I
,surface. A block with mqss (1l is placed on the wedge. 111ere is,
:no friction lietween the block and the wedge. A horizontal·
force F is applied to the wedge. What (Jlagnitude F must have
.if the block is to re(Jlain at constant height above the table
top?
vvx
Observer N
jf mrro2 cos 8
18
1
----1--~--
I
/
mrro2
------/
'll
,t
mg sine
Fig. 2E.113 (a)
(Pseudo force)
mrw2 sin 0
r = R sin B
mg cos 8
mg
i
I
(a)
'I'
(b)
I
Fig. 2E.112
.I
Solution: We analyse this problem in the reference
frame of bpwl. As angular ·velocity is increased the
centrifugal force will increase. When the component of
centrifugal force, tangential to surface, (7lrro 2 cos e will
increase, the block will have a tendency to slip upwards. In
this case friction force will a~t downwards. Similarly at low
angular speed the block ha_s a tendency to slip downwards
and friction force will act upwards.
Impending motion upwards:
2
:i:.Fx = mr ro cos 0 - (Jlg sin 0 - µN = 0
... (1)
:i:.FY = N - mg sine - (7lrro 2 sine=
o
Solution : This problem can be solved very easily if we
analyse the block in the reference frame of wedge instead of
analysing it in ground reference frame. Reference frame of
wedge is non-inertial, therefore we must apply a pseudo
force on block m.
According to condition of problem the block m remains
at constant height h, i.e., it does not slip downwards along
the incline. For an observer on the wedge the block will be
stationary.
System,,. ...
"'-../
-"?,-.. . ,,
.,
N
"
' ,
Pseudo force 8~ ,
\f
m
\
mA+-cc'-i'...;,;.
\
M
~A
~c::,.~I /a
',..__ _8~ I
,I"'
I\
~'<'/
.......
.,,'
... (2)
0~
l'0
----
mg
Fig. 2E.113 (b)
Substitute N from eqn. (2) in (1) to obtain
=[
Ol
max
g(sin0 -µ cos0)
R sin0(cos0-µ sin0)
Impending motion downwards:
]1/2
:i:.F, = (Jlg sin e - mA cos e = o
:i:.FY = N - mA sine+ (Jlg cos e =
From eqn. (1), A = g tan e
From eqn. (2), we may obtain
www.puucho.com
o
... (1)
... (2)
Anurag Mishra Mechanics 1 with www.puucho.com
!
·[FORCEANALYSIS _ _ _ _ _h--~----'------'-------'---------·2~0__.7
N = mg/cos0
If the block is kept on a scale its reading will be
N = mg/ cos 0. We may consider block and wedge as a single
· body (because block does not slip).
Therefore
F = (M + m)A
= (M + m)g tan e
~~g~fiJ~~
;;,; ·smo~~;;· se~icir;u/~;-~if~-;;;,,;, of radi;,; ;_· is fix;d-in a\
/vertical plane (Fig. 2£.114). One end of a massless spring ofi
·natural length 3R/4 is attached to the lowest point O of the,·
1wire track. A small ring of mass m, which can slide on the
:track, is attached to the other end of the spring, The ring isj
iheld stationary at point P S[!Ch thdt the spring makes an angle;
'of 60° with the vertical. The spting constant .K = mg/R J
!Consider the insta7.1t when the ring is released, and (i) drawi
:the free body diagram of the ring, (ii) determine thei
tangential acceleration of the ring and the normal reaction. '
,
- - · -·--- -·- --
- ---------- ·-1
.0
R'
,·
....................
'
''
'
l
~
small-bead-of-,,;~;·;;; ~.-g-i-ve_n_. ;~i;id~l-velo;iryefi
jmagnitude· v O on a horizontal circular wire. If the coefficienti
of kinetic friction is µ k , determine the distance travelled I
1
[QefQ[e.£he_collfl[_COm~ tq J:est,_ ____ __ ..... _______ , ___ •.l
Solution : Reaction of the wire on the bead is
unknown. We assumes it to be N at a!' angle 0.
From Newton's law,
mv 2
LFx =N sin0=--
,,,
........'IJ
.Sil]
... (lJ
r
... (2)
LF'y = N cos 0 - mg = may = 0
Eliminating 0 from eqns. (1) t ~ · . l
'
I
'
and (2), we get
r
!I
----~-
N
= (-m-;_2
+ (mg )2
r
_ _Fig. 2E.115 (a) ___ _
From Newton's law,
------
I
dv
=-mv-
ds
or
....... Joo'
N
N tNcos0
81
Nslnl
+--(a)
XJ i
mg
mg
' '
4
4
.or
or
or
From eqn. (2),
ma = (mg · ~) ../3 + mg ../3 = 5-,/3 mg
'
R
4 2
.2
8
a,
I
.I
Fig. 2E.115 (b)
_Jo
2
d(v )
vo ~r2g2 + (v2)2
2µk
r'
r Jo
as
[1n(v2 +~g2r2 +v2lJ:, .= _2µ_/_s
4
spring=R- 3R=~
or
y
(b) .
Solution : Note that the ring slides along a circular
wire. It starts from rest, hence centripetal force is zero. From
Newton's second law,
... (1)
LF'n = N + kx sin 30° - mg sin30° = 0
... (2)
LF, = kx cos 30° + mg cos 30° = ma,
3
Natural length of spring is R ; therefore stretch in
5,/3
=8
Thus
g
www.puucho.com
v2 + /r2g2 + v•
In o 'I/
o
rg
5= _r_ 1n
2µk
[v~ +
l
'
'
l
I1-
m
,.,_
I Vo
I
,
'
Friction force is tangential force
on bead .
'
''
----+-'~..............
Ct----+--
!
l!
fl>',
~,,
n-axis
l
[E.1!S:am,mte·l
115 ~:~
i§==--- -·· -,~~~=-~:.:~l-.J~
2µks
r.
2 2
~r g
rg
+ Vci]
Anurag Mishra Mechanics 1 with www.puucho.com
v-:-, . ' ,;
' ' ,-,--,~
,''
;-1 -)1-
, ,
qtl
.,rert.angu.la.,rb,lpck ofm.ass,M
... r~ts.. 011.· .an. inclinedplanel. .h.
miakes an angle a.with the,horizontal as shown in'Fig. 2E. 116
fca). Find th~ m~itude of a ?O~Ontal foq:e Papplied io thej
kentre of _the 1:i/q.ck and acting, m: a plane parallel to ;t~el
!inclined :plCf1!e,:thb;t "'.ill cause th,e· riwtipn of t~e. q]oc~ ;toi
1irnpenci
·, /·· ·. . . . .
., .:
.
,i
iAssume that tM dngle offtictioh ~/or the surface of contact is 1
:,A·
,'""r·"" ..i,,, "''"" """ ~'""";'--
i
I
! "''..l'
re-·-~ -~···· ···-:--~'~le~~,~~~~
. : :.
rI
X
For the limiting case when a = ,i,, eqn. (3) gives P = 0.
Also for a = 0, the inclined plane becomes a horizontal
plane. Eqn. (3) gives P = µMg
In the first limiting case all the available friction is used
to resist sliding of the block down the plane; then there is no
resistance to lateral slipping. That's why a rear wheel drive
automobile can skid so freely from side to side when
climbing a wet or icy pavement. For the same reason a car
loses lateral stability if the brakes are too suddenly applied
so as to cause the tyres to slip.
~ ;~n hangs ;;;~~h~·-;;d;;;~~ ~j a'.);p~·-1;;,· l~~ih~ e~ll
0
I
'. !
;,. ,
i,
j'
6fwhich are tied to two light rings which are free to.m6v{!J,ni
a horizontal rod (see Fig. 2E.117); ,Wluit is the maximttm i .
possible sep4raticin d of the rings y;hen the man isHapstngiiiJ
!equilibrium,. if the relevant coefficient of static. ft'fcti9r't}isi
!o.335?
i
·J
..·. I
L~~L.._.._
Fig. 2E.116 (a)
I'!, '.
I
·
· -
·,
-, ·-:t
..,,, ,_
I
1----'-=--~
1 . '
Solution : At .the. instant of impending slipping the
block is in equilibrium under the action of three forces:
weight, Mg; the external force P; and a reaction, R, exerted
by the inclined plane. These three forces mu.st intersect in
one point and also lie in one plane. When sliding impends,
the.reaction Ris inclined to the normal to the inclined plane
by angle of friction qi.
For the equilibrium of the block
LF2 = R cos qi - Mg sin a = 0
... (l)
Fig. 2E.116 (b) shows the system of coplanar forces in
equilibrium in the plane of incline. So we have
,,~-----:--~--
'
h-~-'-"-----,,
,
--- -!
I
!
'
'I
ot:-+-_r-r_.. ,
v;
_,
''. .I
p
'i
X
Solution : Since the man hangs from the midpoint of
the rope, by symmetry the tensions in the two portions ofthe
rope must be equal and have _magnitude T, and each portion
will be inclined at the same angle 0 to the vertical. Thus.the
system of forces .acting on each ring will be the same.
Now consider one of the rings. Three forces are acting
on it: the tensional pull on the ring due to the rope, the
normal force exerted upward by the rod, and the frictional
force attempting. to prevent motion of the ring toward its
fellow. Since the ring is light, its weight may be ignored. If
the ring is too far out, slipping will occur. At ·the maximum
distance apart, each ring is just on the point of slipping.
Hence F = µ,N.
·: ''
,'
'•/
i
,fig; i!E._1_1~_(_!1~)--~'
R 2 sin 2 qi= P 2 + (Mg.sin a)2
· ... (2)
Eliminating R from eqns. (1) and (2), we get
·
P=MgJµ 2 cos 2 a-sin 2 a
where
µ = tan qi = coefficient of friction.
, .. (3)
When we resolve T into its horizontal and vertical
components, the equations for equilibrium become
· Uy = N - T cos 0 = 0
Lf'x =Tsin0-F=0
where we take the positive perpendicular direction as
pointing upward and the positive parallel direction as
pointing to the right. Then
·
N=T.cosa and F=µ,N=Tsin0.
· _ ·µ ,N _ T sin 0 _
_
µ, - - - - - - - - t a n 0 - 0.35
N
T cos a
or
0 = 19.6°
Finally, we solve for d :
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
[_FORGEANALY.,$~_-·-_ __
---'.,__:.···=20!]
--~-~--"---'----~--k·~-------''~~
'~'
sin0 = sinl9.6°= ~ = dm-1
1/2m
0.33 = d m·'
-----1'-
d = O.33m
which is the maximum separation permissible.
Note that 0 and d do not depend on T and therefore the
ring separation is not dependent on, mass hanging from the
midpoint of the rope.
'
---
In Fig. 2E.119 (a) shown two,b}ocks are kept on a. rough
table, where nfa = 0.9 kg, m 8 =I.'i"kg, r = 13 cm,µ, = 0,').
Consider fticti.on between all the contact swfaces, pulley is
(frictionless. JJetennine the angularspeed of the turri-tab/efor
· ---'-~
!which the blo~15sJust begin to slide_._._ _ _
Iiii!(·Exa-tn.t:r!e
~118,....__,__
' - ••. - ·.·· ~'LWT. ~ : . ' : . . ~
C
- ·- ... -··. ··-------··-··--··---·-----~
B
;Figure shows top view of a circular rotating table, rotating
,with speed o). Thto particles connected by string are kept,(in
two mutually petpendicular radii. Coefficient of friction. i~ i!,,
•What can be the maximum angular speed of the table so that
/the.particles do not slip on it?
.
a
i
.
--·· --------- ---
l ~-Var
J{~ f
1
'"
mrco2
I
(a) Side view
-
(Pseudo
force)
[·
(b)
(Pseudo force)
msrw' ~
I
I
2
mArw~
· ·.... ' .' .. --"'·~,'!!,..__~J
Solution: We will solve this problem in the reference
frame of table. Friction force is static, therefore it is variable.
Letfrictionf act at an angle0as shown in Fig. 2E.118 (b).
fmax. = µN
In the impending state of motion,
:l:F,
... (1).
= f sin 0 -Tsin45°
... (2)
From eqns. (1) and (2) we eliminate T, to obtain
2
~rro = f(sin 0 + cos 0}=
f.Jz ( J'z sin 0 + J'z cos 0)
= µmg.Jz sin (45° + 0)
ro 2 = ,Jzµg sin(45° + 0)
r
Since.maximum value of sin(45° + 0)_= 1,
or
therefore
ro max. =
~ .Jz:
0......__..,: TYL
X
fmax = µs (mA + ma)9
(Pseudo force)
j·
:i::FY = mrro 2 - (J cos 0 + Tcos45°) = 0
(b)Top view
(a)
•
f X
~-~
:r=13cm
1
T
: 84~• e
!
m _r_J
. -----·-1.
w
B
[Al......__.. ,_ T
YL
fmax = JlsmAg
X
Solution : We will solve this problem in the reference
of tum- table. Due to larger pseudo force on B it will move
outward, and A will move.inward. When blocks just begin to
slide, the static friction force is maximum.
Equation for block B:
2
:l:F, =T+µ,(mA+m 8 )g-mnrro =.O
... (1)
Equation for block A:
:i::F, = T - µ,mAg - mArro 2 = 0
... (2)
From eqns. (1) and (2), we eliminate T to obtain
2µ,mAg +µ,(mA + mnlg
= Cmn - mA)rro 2
or
ro
= [µ,g(3mA
+ mB
)]1/
2
r(mn '-mA)
which on substituting numerical values yields ro = 6.4
rad/s.
g
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
-~~::~~-~--,-~~-;;µr:·::; _
,t:~:::-~1
On(!iOne Alternative· is Cortept ; ;
. ·· <Y ·i
L:.:.:i~ -~
.·
~
~-2~~·0: · · 1
1:-••
___
·'"
,
'i • ,,".
-~----~---_ _ _
•
7
1, ,A cyclist move~.with uniform velocity down a. rough
-.inclinecJ; plane of inclination a. Total mass of cycle &
cyclist is ~.Then the magnitude and direction of force
acting on .the :cycle from inclined plane is :
(a) mg cosa perpendicularly into the inclined plane ·
(b) mgcosa perpendlcu\arly outward·ofthe i~cjined
plane .
·
(c) mg perpendicularly outward of the inclined pla11e
(d) mg vertical upwards
2. A block of in:ass s.kg is dropped from top of a building.
Then the' mamiitude -of force applied by the block on
the earth whne falling is :
· (a) SgN .upwa,ds. . (b} Sg N downwards
N down~ards (d) None of these
0
(~).
Sg
3 .. .In•a.vertical _disc two grooves-are made
as shown in figure. AB is a diameter.
Two balls.are dropped at A one in each-
--·A,;· .
'.@
,
.' ·. .
~oove;·;in'i.ultaneously. Then:
·
·:, ·
c '· • :
(a) ,Time to. each.at. C is less than t:µat
to reach at B
·6_,_.l
(b). Time.to reach atC_:is greater than that to reach at
B
. (c) :'Dille to r_each at c; _is equal to that to reaclT a! B
· (d) The difference in time to reach at C and to reach
, . . at B may be positive, negative or zero depending
•'
I•.,
.•
•"
,
ona,,
.. ·
4, With ~h~t ·f;~ce mu~t.a man. pull
on the rope to hold the plank in
position if the man weights 60 kg ?
Neglec_t the wt. of ~e plank? rope
and pulley. [Take g ~ 10 m s 2]
.
-
•
5. In the•.situation shown
in· · figure
the
magnitude · of total
external . force acting'
on the block_A is (all the surfaces are smooth/::,
(a) 21 N
(b) 1_4 N
(t) TN
(d) Zero
.
··--1-
6. In the figure_ a _ro.pe o.f m_.a_ ss m !. · .._ ;. , ._, ·._:_:_:_ .;::,_?_;_,,J
. and length Z1s such that its one I
.... ,;,•,t;, ;:''·: •
end is fixed_ to a ri~d wal_l_ and
fixed to. .ih.·e_J:_1g[ti'w_"_I_I. ;
the. other IS applied With a.
honzontal force F as shown · ,
·
. below, then tension at the mi<ldle of the string.is.:
(a) F
(b) 2 F
(c) Zero
(d) F/2
7. the sum of all electromagnetic force between diffe~ent.
particles of a system of charged particles is_ zero :
(a) Only if ~ the particles ar~ n~gatively charged
(b) Only if half the partides are positiv~ly _charged &
half are· negatively charged
· ·
·,
(c) Only if all the particles are positively charged
(d). ,Irie_spective of the signs of the charges
8. Figure shows a light spring ' ['..,, ,_ -,-;~--:-1
balance connected to two
. ·
.,
blocks of mass 20 kg each.
~·. ,• a 'I
The graduations in the f,-J, 20 kg · • • -·
balance measure the tension· I!:::!:-~~~ • ---1
in the spring. The reading- of the balance is :
(a) 40kg .. ,
'
(b) Zero kg
(c) 20 kg
(d) Depends on mass of spring balance .
!'·
U-'·--·-··:.Jrfuj ·
db t;· '1
',
(a) 100 N
'
(b) 150 N
(c) 125 N
(d) None _of these
,.
'
(
. ' '
'
www.puucho.com
. ' . '.'. :, ,,
.'
Anurag Mishra Mechanics 1 with www.puucho.com
9. A block of mass 10 kg is suspended
through two light spring .balances as.
shown below :
(a) Both the· scales will read 5 kg
(b) Theupperscalewillread lOkg&the
lower zero
(c) Both the scales will read 10 kg
(d) The readings may be anything but
their sum will be 10 kg.
10. A force F1 acts on a particle so as to accelerate it from
rest to a velocity v. The force F1 is then replaced by F2
which decelerates it to rest:
(a) F1 must be unequal to F2
(b) F1 may be equal to F2
(c) F1 must be equal to F2
(d) None .of these'
11. 1\vo objects A and B are thrown upward
simultaneously with the same speed. The mass of A is
greater than the mass of B. Suppose the air exerts a
constant and equal force of resistance on the twci
bodies :
(a) A will _go higher than B
(b) B will go higher than A
(c) The two bodies will reach the same height
(d) Any of the above three may happen depending on
the speed with whiclt the objects are thrown
12, A smooth wedge A-is fitted in a chamber hanging from
a fixed ceiling near the earth's surface. A block B
placed at the top of the wedge takes a time T to slide
down the length of the wedge. If the block is placed at
the top of the wedge and the cable supporting the
chamber is .broken at the same instant, the block will :
(a) Take a time shorter than T to slide down the
wedge
(b) Remain at the top cif the wedge
(c) Take a time longer than T to slide down the wedge,
(d) Jump off the wedge
13. In an imaginary atmosphere, the·air exerts a small
force;'=.'."'~any_particle in the direction.of.the particle's
motion, A p~rticle of mass m projected upward takes a
time t1 in reaching the maximum height and t 2 in the
, .return journey to the original point. Then:
(a) t 1 > t 2
(b) t, = t2 .
(c) t 1 < t 2
(d) The relation betwee\1 t 1 &t 2 depends on the mass
of the particle.
14. A person standing on the floor of an elevator drops a
coin. The coin reaches the floor of the elevator in a
time t 1 if the elevator is stationary an~c,time t 2 if it is
moving uniformly. Then:
(a) t 1 <t 2
(b) t, > t 2
(c) t 1 = t 2
(d) t 1 < t 2 ort1 >-t 2 depending on· whether the lifr is
going up or down.·
_·
·
15. Three blocks A, Band Care suspe~dedl- -·---·--,,i-:·
,,eh
as sh<>~
....
ofhloek
A
and Bbelow
is .m.
If ofsyst~m··
is in
m
',. '.\_ ....
,_·-._.- ·: ·,,
.._ ,.,
equilibrium, and m.ass ofC is"'!' then:
-A .•/,c::
'
',
':
.-:f-__,~;,A
(a) M<2m
(b)"M>2m
(c) M = 2m
(d) M !> 2m
16. A light spring is compressed and - - - r '""' "l
-e~~:~1 ~~~z;~~!d
l~~
:~:;,nfre!
to slide over a smooth horizontal
table tcip as shown in the figure. If the system is
released ,from rest, which of the graphs· · below
represents the relation between the acceleration' a' of
the block and the distance 'x' traveled by it ?,
·
(b)
~K _-_7-
~
·x~
o_
,~~
r- 7 t1 ·1,j
r~-------.
(c}
[d)
__ - __ · __ -..,"~.!
17. A steel ball is placed on the surface of water in a deep ·
tank. Water exerts a · resistive force which is
proportional to the velocity of the ball. The steel sinks
· '
into the water : ·
(a) with decreasing acceleration and finally attains _a
constant velocity ·
,i,
(b) with constant acceleration equal, to the
gravitational acceleration ·..
(c) with constant acceleration less than the
gravitational acceleration
(d) with acceleration detreasing _initially and
reversing _after a finite )ime. ' '' ' '
18. In the arrangement, slj.ownbeic\w,p,ulieys are'massless
' and m_monleSS
threads
II~9:1'illlf..bJock Of
mass m1'win remam at rest 1f :,.
, _
·
4
1
1
.
an~
(a)-=-+m1
m2
m3
(b) m 1 = m 2 = m 3
1
1
1
1
2
3'
(c) - = - + m1
m2 m3
(d) - = - + - ·
m3' m2 _mi
www.puucho.com
~re
1.
Anurag Mishra Mechanics 1 with www.puucho.com
I 212 ~
,MECH,\Nl~S-1
19. A fireman want to slide down a rope. The breaking
load the rope is 3/4 th of the-weight of the man. With
what minimum acceleration should the fireman slide
-down?,
' ·
_(a) g/6
(b) g/4
J/43
(c}
(d) g/2
20. An einpty pl1!5tic.box of mass Mis found to accelerate
UIJ _at the r~te of g/6 when placed deep inside water.
, How much ~and should be put inside the box so that it
may accelerate down at the rate of g/6?
(a) , '2M/5
(b) M/5
(c) ZNf/3
· ·
(d) 6M/7
,
21.
A
m"'i ;>i.~
t
a-
·ji
is hrn,g by
fu<ol W
~ T,
aw.al): ..Tjie fo.rces acting on the sphere are '~ .
shown ·in figure. Which of the following N .
statement 'is/are wrong ?
.' . · · ··w ·
(a) T 2 =N 2 ·+ W 2
(b) T = N + W
·-·--·---+,
-t
--+
(c) N+T+W=0
--+
(d)
N =Wtan8
:.+
--+
22. A force F = vx·A is exerted on a particle in addition to
.
....
the force of gravity, where v is the velocity of the
....
particle and A is a constant vector in the horizontal
direction. The minimum -speed of projection for a
particle of inass m so that it continues to move with a
constant velocity is given by :
(a) mg
(b) mg
3A
(c)
A
mg
(d) mg
2A
23. ,A pa,~icle of small mis joined to a very heavy body by
a lig~t string passing over a light pulley. Both bodies
are'·f'ree to move. The' total· downward force on the
Pl!lley is : .
.. ,-1 ' .
.,
(a) 2 mg
(b), 4.mg
.. . i: '
Jc) ,ng
(d) ·>>mg
24. Blocks A & C starts from rest & inoves to the right with
acceleration aA =12tm/s 2 & ac = 3m/s 2 • Here''t' is
in seconds. the time when block B again comes to rest
is :
.I . A.
,.....+
I,
,_ -:
25. In order to raise a mass of 100 kg a man 60 kg fasterts
a rope to it passed the rope.over a smooth pulley. He
climbs the rope with acceleration Sg/4relative to rope.
The tension in the rope is: (g =.10m/s 2 ) ·. ,
(a) 928 N
(b) 1218 N
(c) 1432 N
(d) 642 N .
26. A ball is held at rest in
position A by two light
cords. The horizontal cord is
now cut and the ball swings
to the position B. What is the
ratio of the tension in the
cord in position B to that in
position A?
. (b), 1/2
(a) 3/4
(c) 3
(d) 1
27. In the shown figure two beads
slide along a smooth horizontal
rod as shown in figure. The
relation- between v and v O in
the shown position will be :
(a) v = v 0 cote
(b) v,;, v 0 sine
(c) v = v 0 ,tan8
(d) v = v 0 case
28. Two masses each equal to m· ,~,--·--.-;..,+;
are constrained to move only
· >,..f · ' ., · ·
along x-axis. Initially they
m
' '· m x
are at (-a, 0) and (+a, 0).
(-a, O) i.
(~. of .
They are connected by a light
string. A force F is applied at the origin along y-axis
resulting into motion of. masses towards each other.
The accel~ration of each mass when position of masses
at any instant becomes (-x,0)_and (+x, 0)is given by:
F.Ja 2 -x 2 .
Fx
y~d.
(a)
(c)
m
---
(c) 2·s
2m.Ja2 -x2
(d)
_!_
~
,2mV~
29 . All surfaces shown in figure are smooth. System is
released with the spring unstretched. In equilibrium,
compression in the spring'will be :
I
3
(b) -s
2
1
(d) -s
2
(b)
X
F
x
m:J 0 2_x2
[---'---·--·-~II
(a) 1 s
j
~!
·I
1
(a)
2mg
k
mg
(c)
www.puucho.com
..J2k
(b) (M +m)g
..J2k
(d) mg
k
Anurag Mishra Mechanics 1 with www.puucho.com
I FORCEANA~iL:s.L'..;..,.---'-"~30. Find
the ' maximum · ('."".
/ ~;~fi~!~~~ l~~~~~:
.
-- ""-
kufu!::::e~~
with acceleration 'a'. All
the surfaces ate smooth :
ma
2ma
(a)
(b)
2k .
k
ma
4ma
(c)
(d)
k
k
31. A block of mass M is sliding down
the plane. Coefficient of ·static
friction is µ, and kinetic friction is
0
-- ,
µ k. Then friction force acting on the ,
=block is :
(a) (F+Mg)sin8
. (b) µk{F +Mg)cos8
(c) µ,Mg cos0
(d) (Mg +F)tan8
32. The displacement time curve
of a particle is shown in the
figure. The external force
acting on the particle is :
' a.
I-~
,
,o~---~--..
(a) Acting at the beginning
0
Tim~~,
part of motion
(b) Zero
(c) Not .zero
(d) None of these
33. A block of mass 'M' is slipping down on a rough
inclined of inclination a with horizontal with a
constant velocity. The magnitude and direction of total
reaction from the inclined plane on the block is :
(a) Mg sin a down the inclined
(b) less than Mg sin a down the inclined
(c) Mg upwards·
(d) Mg down wards
34. A block of mass 0.1 kg is held against a wall by·
applying a horizontal force of SN on the block. If the
coefficient of friction between the block and the wall is
0.5, the magnitude of the frictional force acting on the
block is :
(a) 2.5 N
(b) 0.98 N
(c) 4.9 N
(d) 0.49 N
35. A body of mass Mis kept on a rough horizontal surface
(friction coefficient= µ). A person is trying to pull the
body by.applying a horizontal force but the body is not
moving. The force by the surface on the body is F
where:
·
2
(b) Mg ,,;p,,;Mg~l+µ
(a) F = mg
·. _•.-·.
[E
lI . lj
(c) F =µMg
(d) Mg?. F?. Mg~l-µ 2
36. A spring of force-constant kis cut into•two pieces such
,,
that one piece is double.the length of the. other. Then
the long piece will have a force-constant· of :
00 ~k
(b) ~k
3
2
(c) 3 k
(d) 6k
' . '
37. In the arrangement shown in figure -----,
tlie wall is smooth and friction
coefficient between the blocks is
µ =0.1. A horizontal force F =1000
N is applied on the 2 kg block.The
wrong statement is :
(a) The normal interaction force
1:>etween the blocks· i~"lOOON.
(b) The friction force between the blocks is zero.
(c) Both the blocks accelerate -downward with
acceleration g m/ s 2
(d) Both the blocks remain at rest
-r--,~38. 1\vo blocks are kept on an
inclined plane and tied to
each other with a mass-less
string. Coefficient of friction
between m1 and inclined
plane is µ 1 & that between
m 2 & the inclined is µ 2 .
Then:
(a) The tension in the string is zero if µ 1 > µ 2
(b) The tension in the string is zero ifµ 1 < µ 2
(c) Tension in the string is always zero irrespective of
µ, &µ2
(d) None of these
39. A block kept on an inclined surface, just begins to slide
if the inclination is 30°. The block is replaced by
another block B and it is just begins to slide if the
inclination is 40°, then :
(a) Mass of A > mass of B
(b) Mass of A< mass of B
(c) Mass of A =mass of B
(d) All the three are possible
40. A force of 100 N is applied on
a block of mass 3kg as shown
below. ·The coefficient of .
friction between wall and the I .. 1
F = 100N
block is 1/ 4. The friction force
Fixed vertical'wan: __ _
acting on the block is :
,:--
"d
.,' '~-~·-·_.
h..,,,, .
0
l
(a) 15 N downwards
(c) 20 N downwards
www.puucho.com
...
_
(b) 25 N upwards
(d) 20 N upwards
Anurag Mishra Mechanics 1 with www.puucho.com
:,., ..
- ' ......
41.
An
insect surface
crawlsveryup
hemispherical
slowlya
1.• ME~HANICS-1
·
iw·
_ ·-.
I
(X
·• _
•
.(see the figure). The coefficient of
f'r!cti~n betwe_en the insect and
· ):lie surface is 1/3: If the line
joi1Jing the :centre of the hemispherical surface to the
insect · -makes· an" angle a with the vertical, the
· maximum possible value of a is given by :
(b) tan a= 3
(a) cot a=' 3
(c) seca = 3 ·
(d) coseca = 3
;ri,1ocl(of mass 2 kg is held. at rest. against a rough
vertical wall by passing a horizontal (normal) force of
45 N, Coefficient of friction between wall and the block
is equal to 0.5 .. ·Now a horizontal force of 15 N
(tangential to wall) is also applied on the .block. Then
the block will :.
. .
M~;e horizontally with acceleration of 5,m/s2
.
'
(b) ·.. Start to move with an acceleration of magnitude
·
. •
:· .L25 '!'( s2
(c) .Remain stationary
(d) 'Start to .move horizontally with acceleration
.,• .gte~t~r than 5 m/ s2
46. A stationary bcidy of mass m is slowly lowered onto a
rough massive platform moving at a constant velocity
v O = 4 m/s. The distance the body will slide with
is :
·
respect to the platform µ· =
:o.~
(a)
(b)
(c)
(d)
(a)
(b)
(c)
J.'i. Mg.
42 mg
0CM + m) + m )g
(d) (~(M+m) 2 +M 2 )g
~:~.
~
'
'
(b) 30°
(a) ·_Zerp ·.
(d) 60°
45~· ·
45. The',force· F1 required to just moving a body up an
incljned plane is double the force F2 · required to just
preyertt 'the body from sliding down the plane. The
coefficient of friction isµ. The inclination 0 of the plane
.
is·: '
'
ca)
'
truJ..:1
·
µ
(c) tan~' 2µ
,.
•'
(b) t an -1 -µ
(d)
tan-1
2
3µ
''
•
Smooth surface
--;:=-=·
::;;=;-::-::------7
m =10kg
F ,
.
·
·
1
m2:::::15kg
, 1_ µ·::o 0.1 between the blocks
,
(µ:-coefficientoffrl,ction)
F'·' .. (Smooth ground) '
WeSI
Eas\
. (a) m1 experiences frictional force towards west only
iJ; F1 > F2
(b) If F1 '# F2 then it is possible to keep the system in
equilibrium
certain suitable values of F1 &F2
for
(c) · If the system is to remain in equilibrium then F1
must be equal to F2 & F2 :s-10 N ·
.!i = !-!,_,
m1
m2
then frictional force betwe'e~ the
blocks is zero
48. Consider the system as shown.
The wall is smooth, but the ·
surface of block A & B in contact is
rough. the friction force on B due
...
to A is equilibrium is:
(a) Zero
(b) Upwards
(c) Downwards
(d) The system cannot remain in equilibrium
49. Given mA = 30 kg, mB = 10 kg, .
m, = 20 kg. Between A&B µ 1 = 0.3, ~
A ,. F
between B&C µ 2 = 0.2 & between
B · •
C & gronnd µ 3 = O.L The least
.
c ·
horizontal force F to start motion of
·
·· ·
any part of the system cif three blocks resting upon one
another as shown below is:
(Take g = 10m/s 2 )
(a) 90 N
(b) 80 N
(c)' 60 N
(d) 150 N
SO. The coefficient of friction
between the block A of mass m
& block B of mass 2m is µ.
.
There is no friction between
blockB & the inclined plane. If
I ~
·
www.puucho.com
' ',.,
I
v0 =-4m/s1
rnw·.
m
cc1
',.,
Es]
2
44. The pulleys -and strings shown in
tit!! ,figure are smooth and of
negligible mass. For the system to
remain in equilibrium, the angle e
should be :
i
I\ " ''.. Platform
1,m,J,,,;;;;;,t;,,, I
r-·
i<'·. ,
fF
(d) If
,, ..
' .
2
4m
6m
12 m
8m
47. In the diagram shown the ground is smooth and F1 &
F2 are both.horizontal forces. The mass of the upper
block is 10 kg while that of lower block is 15 kg. •The .
correct statement is :
·c;)
, 43. Astring;?f h¢gllgible ~a~s going over a
clamped'-'p~lley.' of mass m supports ·a
block of mass M as shown in the figure.
The force on the pulley by the clamp is
givel},,by : , · · ,
j
B
8
.
Fixed'
Anurag Mishra Mechanics 1 with www.puucho.com
~1-_FO_RC~~-·AN_A_LYS_l_s'_-~:_._ _ ~;~---'_·,~-----------~------'-----~~-"''~~15....JI
the system of blocks A &Bis released from rest & there
is no slipping between A ~ B then :
(b) 0 ~ tan-1 (µ)
(a) 28~ sin- 1 (2µ)
(c)
20~ cos-1 (2µ)
· (d) "28 ~ tan-1 (µ/2)
51. The system is pushed by the
force F as shown. All surfaces
are smooth expect between
B&C.
Friction
coefficient
between B&C is µ. Minimum
value of F to prevent bloc!< B from downward slipping
is :
Cal
(c)
(2:)mg
·(..!.)mg
2µ
.(b)
(d)
(!)µmg
(;)µmg
,-,---,
jvr&'v2.
·v,
I
'
-+-------;;,. 11
V2
L
r·1 & V2
F"
,m,;;m,D~
.
-~-~
. ·-
_.,
~
(d) 15 N
54. Two beads A &B of equal mass m
A
are connected by a· light
·inextensible cord. They are ' ·
connected to move on a
frictionless ring in. vertical pla1?:e. · _,8 . ·----····· ·······•··••
The beads are released from rest ' · ·
1
_...::::::-d:::::::::___.J
as shown. The tension in the
cord just after the release is :
(a) ../2.mg
(b) mg.
2
(c) mg
4
Cd) mg:_
,./2
55. A bead of mass 'm' is attached
to one end of a spring of
natural length R & spring
(-fl+
constant k = ---~. The
l)mg
other end of the sp!ng is fixed \
1 _
at point A on a smooth vertical L
_:::::::=C:::::..-_J
ring of radius R as shown. The normai reaction at B
just after it is released to move is : '
(a) .fl mg
(b) 3,.J3 mg
(d)·,3,.J3mg.
2
2 ...
56. In the above question 55 tangential acceleration of the
bead just after it is released is .?,
·
I
U2
I-_
V1
. (a)
!
2
& U2
(c)
v,
v1&v2
·(d)
1
(a) 20 N
(b) 10 N
(c) 12 N
(c) mg
(b)
-
~ m-"3k1/µ =~
LI
52. A block A is placed over a long
rough plank Bsame mass as shown
below. The. plank is placed over a
smooth horizontal surface. At time
t = 0, block A is given a velocity v O in horizontal
dqection. Let v 1 and v 2 be the velocity of A &Bat time
't' . _Then choose the correct graph between v 1 or v 2
'l"d t:
(a)
53. What is the maximum
value of the force F such ,.
that the block shown in
the arrangement, does not
move question:
(c)
!
(b) ~ g· .
4
ig
(d)
4
57. If you want to pile up sand onto a circular area of
radius R.The greatest height of the sand. pile that can
be created without spilling the sand onto the
surrounding. area, if µ .i's the coefficient of friction
between sand particle is :
(a) µ 2R
(b) µR
(c) R
www.puucho.com
(d) R
µ
Anurag Mishra Mechanics 1 with www.puucho.com
58. A man of mass 60 kg is pulling a
mass 'M, by an inextensible light
rope passing througlf · a smooth
& mass'.less pulley as shown.
The coefficient of . friction
between the man & the ground
isµ = 1/2. 'Ihe maximum value
. of M that can be pulled by the man without slipping on
the ground is approximately :
(a) 26 kg
· (b) . 46 kg ·
(c) 51 kg
(d) 32 kg
59. A weightless string passes through a slit
over a pulley. The slit offers frictional
force 'f' to the string. The sp-ing carries
two weights having masses m 1 and m 2
where in 2 . > m1 , then acceleration of
the weights i~ :
hanging icleal string. The
maximum
possible
tension in the string is
1000N. --The minimum
time taken by the man to
reach -upto the pulley :
(a)
m
!~-~
\
60. A plank of mass 3 m is
.placed on a rough inclined
plane and a man of mass m
walks down the board. 1f
the coefficient of friction
between the board and
inclined plane isµ = 0.5, the minimum .acceleration of
does not slide is :
(a) 8 m/s 2
(b) 4m/s 2
*
·.
0.2
l ; l;;;;, ~kgl;?
(5 - 2t)N
di/II 1/IIJJJJ
(a) mg
,_--~---.
---
fig
t
' ~-
-
.•
l
I
30°
;
~-~- -· ....,.._ __ I
(b) 2n + 1
2n
2n-l
(c) 2n,- l
(d)
2n
2n+l
· 2n+l
'•
62., A· wedge of mass 2 m and a cube of
:1.'
mass m are shown in figure. Between ":'
'
cube and. wedge, there is -no friction.
.
The minimum coefficient of friction
45°
between. wedge and ground ·so that
~
wedge does not move 'is :
caJ 0.20
CbJ 0.25
(c) 0.10
(d} 0.50
63. The figure shows a block 'A' resting on a rough
horizontal surface with µ = 0.2 A man of mass 50 kg
standing on the ground surface starts climbing the
- J
. (b) 3 N
!
:
sn+l
il
67. In the above question 66, if the same acceleration is
towards right the frictional force exerted by wedge on
th,;, block will be : (Coefficient of friction between
wedge & block = ../3/2}
61. A small block slides without friction down an inclined
plane starting . from rest. Let Sn be the distance
~ n -.1 to .t =·n. Then ..!!.E_ is
4
· (d) Zero
66. The acceleration of small block m with respect to
ground is (all the surface are ·smooth) :
(a) g
. ,(b} g/2
(c) · Zero
· (d) .fig
(d) 3 m/s 2
.
'jto~-J ·
. r,L-----~
. - _,,_ '"--··
,. ...·___I,
--·---·---· ~ ~ - - · · - · - - - - - ~ -
m1 +m2 ,
. :
¢•··'"· - · - - · - --.-
·""r
(a) 2 N
(c) 1 N
(c) .Cm2 - m1Jg- f ·
(a) 2n - 1
20m
(b) 1
,------.
m1 +m2
traveled from time t
i
(c) .Ji
(d) none of these
64. In the above question 63 distance between the man ·
and the block' A', when man reaches the pulley is :(a) 10 m ·
(b) 2 m
(c) 20 m
(d) None bf these
65. The force acting on the block is give1_1 by F = 5 - 2t.
The frictional force acting:1m the block after time t = 2
seconds will be : (µ = 0.2)
(a) f :- (m2 - -m1) g
(c} 6m/s 2
,;r-~·
: 50kg
(c) 2mg
(b) 3mg
2
(d) mg
2
68, A· block of mass 'm' is held
stationary against a rough wall by
applying a force F as shown. Which
one of the following statement is
incorrect ?
'
(a) Friction force f = mg
(b) Normal reaction N = F
(c) F will not produce a torque
(d) N will not produce any torque
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
[}011.cEA_NA_-LY_s1s______________~ - - - - - - - - - - ______ 31?]
2m
69. Two blocks A and B of masses
and m,
respectively, are connected by a massless
inextensive string. The whole system is
suspended by ·a massless spring as shown
in the figure. The magnitude of
acceleration of A and B, immediately
after the string is cut, are respectively :
(a) g,g/2
(b) g/2,g
(c) g,g.
(d) g/2,g/2
70. Two particles of mass m each
are tied at the ends of a light 1
string of length 2a. The whole
O
system is kept on a frictionless
nJ
p '',m
horizontal surface with the
string held tight so that each : Jc a >Jc ., a >j
mass is at a distance 'a' from
the center P (as shown in the figure). Now, the
mid-point of the string is pulled vertically upwards
with a small but constant force F. As a result, the
particles move towards each other on the surface. The
magnitude of acceleration, when the separation
between them becomes 2x, is :
o-
(a)
a
F
2m )
(c)
(b)
2m )
0 2 -x2
F x
2m a
F
(d)
r
00
0 2 -x2
F )a 2 -x 2
2m
X
(b). mro~a
+ 000 ) \
2
,., :I
+=-.er
i
(b)_
.
r-.·- ---- ~--1
, , h-..!
. Jc· I
l
[
.
. i
(d) mroro 0 a
73. A particle of mass m1 is fastened to one end of a
massless string and another particle of mass m2 is
,fastened to the middle point of the same string. The
other end of the string being fastened to a fixed point
on a smooth horizontal table. The particles are then
projected, so that the two particles and the string are
always in t:lie same straight line and describe
-- ~I
r----·
{d) : ~
.J
____t
----------'
particle moves
along on a road with
constant speed at all
points as shown in
figure. The normal
reaction of the road on the particle is :
(a) Same at all points
(b) Maximum at point B
(c) Maximum at point C
(d) Maximum at point E
72. A particle of mass m rotates about Z-axis in a circle of
radius a with a uniform angular speed ro. It is viewed
from a frame rotating about the same Z-axis with a
uniform angular speed ro O• The centrifugal force on
the particle is :
(c) m(
released
from rest
from point
A
ii~-A~
:_:·::~--~::_:~-~----·~;;-.
inside a· smooth
hemisphere
bowl
\ _,.,___, ,
8
as shown. The ratio (x) of
magnitude of centripetal force &
normal reaction on the particle at any point B varies
withe as:
X
71. A
(a) mro 2a
horizontal circles. Then, the ratio of tensions in the
two parts of the string is :
(a) m,/(m1 + m 2 )
(b) (m,. + m 2 )/m1
(c) (2m 1 + m 2 )/2m1
(d) 2m 1 /(m 1 + m 2 )
74. A small particle of mass 'm' is
75. A particle of mass' m' oscillates
along the horizontal diameter
AB inside a smooth spherical
shell of radius R. At any instant
KE. of the particle is K. Then
force applied by particle on the
shell at this instant is :
K
(b) 2K
~@i1··-~-
A--- --------.;......... ••· B
i
_____ ·;_ _ _ _ _j
(a)
R
(c)
R
3K
R
K
2R
76. A particle of mass m is moving in a circular path of
constant radius r such that its centripetal acceleration
a, is varying with time t as a, = k 2 rt~ where k is a
constant. The power delivered to the 'particle by the
forces acting on it is:
2 2
2 2
(a) 21tink r t
(b) mk r t
(c) (rrik 4 r 2ts)/3
(d) Zero
(c)
77. A long ·horizontal rod has a bead which can slide along
its length and is initially pl~ced at a distance L from
one end A ofthe rod. The rod·is set in angular motion
about A with a constant angular acceleration, a. If the
coefficient of friction between the rod arid bead is µ,
and gravity is neglected, then the time after which the
bead starts slipping is :
(a)
fa
(c)
../µa.
www.puucho.com
1
(b)
µ
../ri.
(d) infinitesimal
Anurag Mishra Mechanics 1 with www.puucho.com
,·,
-',"
, _1, ..
Cl",
-.
a
78. · In gravity-free space, a particle is in constant with the
. -, ', ', ME~f!{INICS-1
:I
·1
witho1;1t fri_ction thr~ugh it. B is . r.;· .. .
hii,er- surface• of. a hallow cylinder and ~aves in a
circular path aJo!'g the surface .. There is some friction
Dunng. •CA
the moaon
from. A, to
C if
. · ·+::: .B .•..
'"""
. •,d ru,,i.""1
""'""'
·between the particle· and the -surface, The.retardation
will:
•'
.
: '' ·::·
of the particle is: ... : .:·
·
·
.
(a} Alw~ys be in contact witlr the ... ' .,:,.~:~:-'.· . 9,
· Ca) Zero · .
.'
inner wall of the tube
(b) Independent of_the velocity
.(b) Always be in contact with the outer wall of the
Cc) Proportional to
its velocity
, .tube : · _
·
.
. -.
,.
'.
Cd) .Proportional tQ'the square of its vel99ty
Ci)- Initially be in contact w:it:li. the inner wall and later
'with the outer wall
79. A curved ·section of a road is banked for a speed v. If
there is no friction Between· the road and the tyres
(d) Initially be in contact with the outer wall and later
·
then:
·
with the inner Wall ,
85. A particle is ~i>ving iri. ·a: 'circle ~t radius R in such a
Ca) .a car. moving with speed v ·does noFslip on the
.road
way .thilt
any instant 'the normal. and tangential
components ·a( the acceleration '!i:e· equal. If its speed
(b) a caris more likely to.slip on the road at speeds
at t ·=· 0 is ·u 0 ,. :the tim_e taken to complete the first
higher than ·v, than iit speeds lower .than v
revolution
is :
Cc) a car is mqre likely to slip on the road at speeds
Ca)
R/u
(b) u 0 /R
0
,, _lower thaji v, t!i~. at _speeds. hlgher than v ·,- ,
Cd} ,a ·car can remain stationary on. the road without
Cc) ~(1--e"2lt)
Cd) ~e-2'
slipping .
-. ' .' _
·
Uo
'Uo
80. In' a, circular :mbti~n- of ~ particle the tangential
86. A ·particle P is inoving in a circli ~f radius r with a
acce)eration of the particle is given by
2t m/s 2 •
uniform speed u. C is the center of the circle and AB is
diameter. The angular velocity of P about A and C are
The radius of the circle described is 4 m. The particle is
·
in
the ratio·:
initially at rest. Til)l.e• after Which-total acceleration of
(a) ~ : .2
the' particle makes,45° with'radial acceieration is :
(b) 2 : 1
Ca): sec · ... -· .r,'. ,_ (b),'2 sec
(c} 1: 3
Cd) 3 : i
87. A small body of mass m can
Cc}"3 sec
-Cd)' '4 sec
81 .. A partide travels along the arc ofa circle bfradius Its
s_lide without friction along 1
'
]
a trough bentwhlch is iri the . ·1
_·
•
•
.
;
''..
,j'
·
· speed depends on the distance. travelled l as v = a.ff.,
7
where 'a' is a -constant. The angle a; between the
form of a semi-circular arc .
of radius R At what height h i · ~ h ; ·
vectors of total, accejeration. and the velocity of the
particle is : , ,_
will the body be at rest with " · ~----1
1
respect
to
the
trough;
jf
the
trough
rotates
with
(a)_.g = tan- C2l/r) - . Cb),. a= ~os- (2Zjr) ·.,
angular
velocity
OJ
about
a
vertical
axis.:
uniform
Cc} a; =·sin-1 C2Z/r) · ·(d), ·a= cot" 1{2Z/r)
Ca) R ·
(D) R -· 2,g
82.
p~~le of ·mas~ m is atta~~ed to· ~ne end of a string
OJ2.
oflength 1while the othe'r end' is fixed to point Ii (h < l)
'
(c)
2,g· -Cd)· R-.L ·
· above a horizontal table. The particle is made to
2
. QJ2
OJ .
revolve ·in a circle ion the table so as to make p
88. A car moves, along a horizontal circular road of radius r
revolutions per second. The maximum value of p, if
with constant speed v. The coefficient of friction
the particle is to be in i:ontaciwith the table, is : CZ > h)
between the wheels and the road is µ. Which .of the
(a}-• 2rc.Jifi
· Cb)' ..jg/h
following
statement is not true ?
; -. '. '
. - '.I:
'
Cc) 2rc..jh/g
Cd) -, ..jh/g
Ca) The car slips if v > .Jµii
· 2rc
Cb)
Th~ car slips ifµ < (v 2 /rg~
83. A. sto~e is thrown horizontally with a velocity of 10 mfs
1~_· .- .·.-
·
at
a,·=
.i
r.
W·
0~ ·_· ·. .
i
R_;
Cc) The car slips ifµ_>(v 2 /rg)
at t = 0. · The radius · of curvature of the stone's
trajectory at t = 3 s is : [Take g = 10 m/ s 2 J
Cd) _The- car slips at a lower speed if it moves with
·cal 1oJio m
.
Cbl 100 m some tangential acceleration, than if it moves at
constant speed Cc) 10oJio ~
Cd) 1000 m
89. A smi>otli liollow cone whose vertical angle is 2a; with
84. T)Ie narrow tube· AC forms a quarter circles in a
its axis vertical and vertex downwards revolves about
vertical plane. A ball B has an area of cross-section
i~
axis 11 tl!ne p~r seconds. A Particle is placed on the
can _move
slightly smaller than that of the tube and www.puucho.com
;·
.
''
·, . .
,,
Anurag Mishra Mechanics 1 with www.puucho.com
',
,·· '.,
•• ..
'
.
•.
inner surface of cone so that· it rotates with same
speed. The ,radius of rotation for the particle is :
(a)gcota/4rr 2 T] 2 '·, (b) ·gshl'a/4rr 2 Tj 2
(c) 4rr 2 TJ~/g
(d)
g/ 4rr 2 TJ,2 sin a
90. A particle is kept fixed ~n a turntable rotating
uniformly.As seen frpm the. ground the partjcle goes in
a circle, its speed is 20 cnys & acceleration is 20 cm/s 2 •
The particle is now shifted to a new po~itiol) to make
the radius half of the original.yalue. The new values of
·
the speed & acceleration will. be :
(a)
cm/s,·'10 cin/s 2 , (b) '10, ~/s, 80 cm/s 2
'10
(c) 40 emfs, 10
cm/s
2
(d) 40 cm/s, 40. cm/s\::,
'91. A particle· of mass in is suspended
from a fixed point O by a: string of
length' l. At t ~ 0, it is displaced from .
its equilibrium position and released. .
The graph which shows the variation
of the tension T in the string with
time t is :
(a)
.
(b)
~o
· ~. . . .1.. ,:,-,:·
rn
e
I
1-.
:.
' ,
f,~r, ,; ·~.·
:;• .· .•..·._·J·
:j
,I_,_
··'
1·
---·-.
:,,,
-·-L.:J
~~-1·;r~,·
r~I .,.
(c) l : , . ~ ( d )
It
(b) 10 se{
(d)'-'s tee'·,
(a) 20_,sec
(c) 40 sec
~
96.
'~;;7
!1~:S~~cl~o:ti;ll~s :~::_
•
figure'.
.The , · · approximate·
~
· ./" = 01
variatio~ · of· direction . ·of · · t :· . ·· ·.::'.,;,·{. •, i' j
resultant
acceleration . as·· B ·····;,; >\J:~:.J
particle µioves 'from A to B is : .. ' .
.
(a) clo~se
,.
(b) anticiockwise
'. ,
(c) · direction· does not changes.'
(d). no~e o~ the~e ,.,
.· .
. .
.
97. In the above question 96 the net acceleration of
particl~ is h«;>rizont'al only at 8 (8 is acute angle made
by string.~th liµe OB)':. '
··
=
•., .• ('1)·
'(1)
./3
· .·'
.'J3 .
·c· )..
.i)
~;(·.1r,;:J.
r,; . . , .
' ·. (a) cps-' .
(b{si~~' .
. ~'(
(~)_:rr:
-rr:. -.sm
· ..
;c-.,-cos_.
2.
.
'!3: . . ,_.,,.,.2 ',,: . ....,3 '·
98. Two similar trains are moving along the equatorial line
with same.speep_but in·«;>pposite dire~tion. Then:
(a) they'.will exert 'equaH9rce ,6n rails
,: :-,
(b) they ~ :not exert 'any f~rce. as they are on
equ!'ltorial line · · ·. ,
(c) on~. of them will exert.'zero ·force.
(d) both exert.different forces _... ' . . ·. •.
99. Two b~· of'mass m and 2111 are attached with strings
of length 2L and L respectively They ,are released from
horizontal position. Find ratio tensions in the.string
when the accelerati9n ·of b9th, is only ,in vertical
direction: · . · ·
· ·. _. ' ·.
··
(a)
5 · . ,
·
'(b).
5
c
92. A rod of length Lis pivoted at one end is rotated with a
uniform angular velocity in a horizontal plane. Let
T1 &T2 be the tensions atthe pointL/4and 3L/4away
from the pivoted ends.
(a) T1 > T2
(b) T2 > T1
(c) T, ~ T2
(d) The relation between T1 &T2 depends on whether
the rod rotates clockwise or anticlockwise
The
driver
of a car-travelling at speed V suddenly sees
93.
ci:). 2 ,.s,·:·_
. ca)
a wall at a distance r directly infront of him. To avoid
collision. He should : ·
100. Indicate.the direction offrictional f~r~e·6l' a car which
. is movhlg along. ,the.· ctJrv~d. path- with .. ,non-zerf>
(a) apply the brakes
tangential acceleration; ih a,nti-clock' directioJi': . . .
(b) tum the car simply away from the wa:11
(c) do any of the above options
. (a)
·
(d) none of these
.·,.-< _:I · .
94. A body is undergoing uniform. circular motion then
which of the following quantity is constant :
(d) .
(a) velocity
(b) acceleration •
(c) force
(d) kinetic energy
'il
A
particle
is
resting
on
an
inverted cone as shown. It
95.
. ·'
is attached to cone by a thread of length 20
String
is_ given
remains parallel to slope of cone. The cone
www.puucho.com
of
,t
3·:
-i.i~- . ...
,~~r~. . .. _: ~) -F~,,J· :,: . .;.-
[0_·.LL,
_,
['\I ,.
. lil$.J
cm.
Anurag Mishra Mechanics 1 with www.puucho.com
I 220
j
M_ECH~N.lCS-1 ]
---~-Cc-------~------ -------c:,-------:.....-....J.
•
,,
•••
•
,.
•
10 I. If a particle starts from A along the curved circular
path shown in figure with tangential acceleration 'a'.
Then acceleration at B in magnitude is :
r·--···-s-··_ . 7
l' C'\!.
.:f':.__ .•
103. A simple pendulum is oscillating without damping.
When the displacement of the bob is less 'than
maximum, its acceleration vector
in:
.c•
(a) 2a~1+1t 2
(b) a~l +1t 2
(c) a~1t 2 -1
(d) a1t~l + 1t 2
(a)
102. A small block is shot into each of the four tracks as
shown below. Each of the tracks rises to the same
height.The speed with which the block enters the
tracks is the same in all cases. At the highest point of
the track, the normal reaction is maximums in :
':
,:
: -
L-,--." -,- ,_ · - - - - ~
¾,~
~
: __
(b)
······et 'i
-·-1.
,''
I ~ ; (b)
l :
I
(c) r ·...: .~ .
'· . . ~. lI
, , G'. ~ ) "'
cai
!~i
-----1
':,:'
www.puucho.com
'~
a
........··
.
.. ··
(d)
'
!
a is correctly shown
Anurag Mishra Mechanics 1 with www.puucho.com
[ FORCE ANA~YSIS
···-
----
2 __ --~~~~ ,t~-~~-~~~-~ltern~ti~=-~~~~~~
1. A particle stays at rest as seen in a frame. We can
conclude that :
(a) Resultant force on the particle is zero
(b) The frame may be inertial but the resultant force
on the particle is zero
(c) The frame is inertial
(d) The frame may be non-inertial but there is a
non-zero resultant force
2. A particle is found to be at rest when seen from a
frame S1 and moving with a constant velocity when
seen from another frame S 2. Select the possible
options :
(a) Both the frames are non-inertial
(b) S 1 is inertial and S 2 is non-inertial
(c) Both the frames are inertial
(d) S1 is non-inertial and S 2 is inertial
3. Figure shows a heavy block kept on a frictionless
surfaces and being pulled by two
ropes of equal mass m. At t =0, , · ·· - - · ····-- - ·1
the force on the left rope is ' . ~
.i
.
1 m
m
Fj
withdrawn but the force on the 2@JN _ ! .
right end continues to act. Let F1
and F2 be the magnitudes of the forces acting on the
block by the right rope and the left rope on the block
respectively, then :
fort < 0
(a) F1 =F2 =F + mg
(b) F1 = F, F2 = F
fort > 0
for t < 0
(c) F1 = F2 = F
fort> 0
(d) F1 < F, F2 =F
4. The force exerted by the floor of an elevator on the
foot of a person standing there is more than the weight
of the person if the elevator is :
(a) going up and speeding up
(b) going down and slowing down
(c) going up and slowing down
(d) going down and speeding up
5. If the tension in the cable supporting an elevator is
equal to the weight of the elevator, the elevator may
be:
(a) going down with increasing speed
(b) going up with uniform speed
(c) going up with increasing speed
(d) going down with uniform speed
6. A particle is observed from two frames S1 and S 2. The
frame S 2 moves with respect to S1 with an acceleration
a. Let F1 and F2 be the pseudo forces on the particle
-~or~ect -~
when seen from S1 and S 2 respectively. Which of the
followings are not possible ?
(a) F1 ,;, 0, Fz = 0
(b) F,
0, F2 _o
(c) F1 = 0, F2 ,;, 0
(d) F1 = 0, F2 = 0
7. In the arrangement shown pulley r- ······
and thread are mass less. Mass of
plate is 20 kg and that of boy is 30 I
.
·:
kg.
'
:
'
'
'
Then: .
.
1
1~1a
_i
-.--- ---·-"~
~.
(a) If normal reaction on the boy is
equal to weight of the boy then the force applied
on the rope by the boy is (lS0g/7) newton
(b) If the boy applies no force on the string then the
normal reaction on him is 30 g.
(c) If the system is in equilibrium then the boy is
applying 125 newton force on the rope
(d) None of the above
8. A smooth ring of mass m can slide on
a fixed horizontal rod. A string tied
m
to the ring passes over a fixed pulley
B and carries a block C of mass 2m as
shown below. As the ring starts '
sliding:
*
*
:LI
-
f h . . 2g case
.
(a) The acce1eranon o t e nng 1s --"--'-1+2cos2 0
(b) The acceleration of the block is
2g
1+2cos 2 0
. m
'the stnng
. .1s - -2mg
-"-c
e tension
()Th
1 + 2cos 2 0
(d) If the block descends with velocity v then the ring
slides with velocity v cos8.
9. A block of mass mis kept on an inclined plane of mass
2m and inclination a to horizontal. If the whole system
is accelerated such that the block does not slip on the
wedge then:
(a) The normal reaction acting on 2m due to m is
mg sec8
(b) For the block m to remain at rest with respect to
wedge a force F = 3mg tan a must be applied on
2m
·
(c) The normal reaction acting on 2m due to m is
mg sece
(d) Pseudo force acting on m with respect to ground is
mg tan a towards west
www.puucho.com
.
Anurag Mishra Mechanics
1 with www.puucho.com
'
:,~:1\·>f·:
l--~-,:~22,.,:.'__·,f;~~~:~~~:>,~~4
. "~ J-!:=
! •
·.:.· ;~~~./: }'::~~~-1~~,f.
·i,, ~._:-~<;,.:, -.~: .°'.--,.3-::.1,:~S,..;;·0.:_ ..__. __ --'. ~.·: ._
.
_
1
io. pie ca-~i~ tile,glven-figure mov~s.
with ,co~gant vel9city v, When,'
x = 0 ends A and B ..were
.i6incident'.at c;· Then whicli ~f
'tlie foil~~tig -s~nterices · is/a~e ,'
corr~f~.: ·
'
·, ,-
·--··":::7··---:;:•·7
t .· ,·: ·- 2\_._~· .5
r-c .
l ·,
· ·"
~ t;:·:·
l,
_;,t.,J:1
i'+::c, , ,B" ~ ·!
L;:,:::.~~ ,.
,I
· -;-·· ·, '. :.·. '- , :_;· :, . :: .'i' ·. --:' :~
·:,-: (a)° ·The velocity of the block is
··
2
. :: _· . .,·:,',,/::.::.: :block
: ·_::.v~:t\
·:·
(b) Accelerattoh
, · , , ·, of the
, .
'
..- , . '_ v 2)' .3/2 : '
, ·is
- (H2
· --. ' . ._. ::; ' · - · ·_ ·.·
·· + ~· · · ·
(c) Ac'ce!~r~~oh'cif.block_A, is i:¢ro, ~ '.: ,_.,,
.(d)-Velocity'· ofthe'blockisti. . ···:,. .:!,· ·
'
'
.. ,
i1, .Two !Il.~n.'.01'f.:un~qual).~as,ses lio.ld'oo ~.di,
f Ii 'h ·
·
·
· !
·· J
0
tw.··
..
secti_o.n~.
-.c,.
,
a·
·g
t
~.~p
..
e.
;P·.
assip-·g·
,oyeta
[·.·f.
·•·.
smo~~. ljght pulley: Which ~£,othe (ollo'Vll1g f ;,, :•.·...'...•
., ..
- 'bl· ?". , -- . ·• , .. ,
- .
~~·;~
:~t:~oriaj,:
~~~
the.
1:
1
? ·.- ···" _.,,,,. ;-·· _·. · -· .... - ; 1 _ ·• ._, , • •• ,_ ·~·
": lighter miili-fui>ves wlth:some·acceleraticin
~) _th~)iiO:tet~~ sti;i ha;:y_ w!ill~--the h~~~er
.... man·moveswitlisciineaccelebition:'. ,
·
cd:'rhe light~r:in~n _ls stati\)hazy,~hile,the heavier
-_,: . •m·.an.·,m,,9.."e.'Yi,\li, some,,ac.~e\e,ra.t_io_h .. .-·-'.. , _
.
td)< The twii'
men move. wiih acceleration: cif the· same
.. ·(, \•: -~,amJf.U~e ll! O~po~it<s,4/i:ecti?nf . ':':·,
12. In, the situation' shown · iri '.figure · · : ·
F = soo ~evJton appJied, oft· t)le 'pulley.- : .m,· = s··J<g'and in~ 10 kg and' pulley ·. ·:.
and
, and , ·; • '·. •
. -. strings ', are·· '·massiess
.~
'
. frictionless. ,Then,. the- acceleration ·of ;rn
tJje p~lley is'.: fg =: 10.ir!./s2 ]
· .'.,~~"'"-3
I
I ,.., ,
:f 1i!\ :a~~~-
,:;;,,l]
ls
9
·
- ·--
"'~--.i:1('
. -·-"-"''----C·-:.u.:~Li~....:.::..-·,_
:: ,.,-~
-·-··.·MECt1AN1~s~:J. ·. -;>,-.
... _.~~w
f'
;,::.~:-,~-~,
, . (b).
The maximum force, which the man can exert on
the wall is the.maximum frictional force which
exists between his feet and the floor
Cc) 'rhe man can never exerts a force on ,the wall
°&hich exceeds his weight
,.(di .. The.man cannot be in equilibrium since, he is
· exerting a net force on the wall
·
15. A block of mass mis placed on a smooth wedge of
' ··
:~~;=~*re~~~~~!Jt:a:~:~e~s::~~~~~~~t::
from .the .grou.nd: ·. ' .
.
'
'
··· · · ··
Cb) ,(M + 171sin8)g -_
(d)• .('M_ +msin8)gcose
·
M+m·
'
(~) (M'.t- mig
(c) . Mg
16. A block of mass m is placed on ,a smooth Wedge of
inclination e m,·th the horizontal..J'he ~hole system is
accelerated so, that the_ block does not .slip on the
wedge. Theforce exerted by the· wedge on the block
has a magnitude:
Ca) mg/cose
. Cb) mgcose
(c)· mg
(d). mgtan0,
17. In ,arrangement .shown below, the
thre~d ,pulley :and spring, ~e .. all
massless and there is no friction
r,
:::t;~:~.
I~eadsi;ri:~ctin~ in4 ~ m.,·r,J'-;,-'--\_;"·1
cut then just after thread· is cuf:
·, '
=·
.
(~) 1s:1;1Js\0::',
(c) 40 rfi/s
2
. :/, :_': (b) 27:s:n;s
. ,... -~
' '.
7.S·m/s 2 : .: •
r - ::·. _; : (d)
', '.· i ·,;l : ,.
. .
2
' ' ,·, ' ' '• ''
4
' -
f -. .
l:-' ,/• , , '
13. !Ii.th~ figur,e; the puµey·P·mov:es· td
-7>~
the right-with 1i·const1!-'!t s~eed u. T_he ~ -"'
dOw,11\\'.ar~, speed M, 1ps VA, and tl}e'
-. speed of B to,the tjght is vB: , ,· L
, ,
AJ
(J)
(b)
VB
=h+~~'· <:, ' ' i '
VB
tu,{v:;,.
'
(c)
~
• '
. ,,, .
'
_,,
·,-,,,, •, ' ' - •
•
,, ,.
;. _,
•
VA "" VB : ,
.•
·.,, .., . I.
-
'
' ' , '. ·
•
'
,: '
~ '. •_-,
. ,
''-•
;.l . . . .
.
.,
~
.,•
.•. •:•::;;r;·. .·.•.••.:; :;:f.:.•·. ·
·-.·'
' ',, .
'.'
.
'
.,. '
,·,.
'
:-
=
m4
18. A,' trolley C can run on a smooth: ... :.. -~--"·;;--
1J[.,,,·····[!l·J
,· but
h~ri:i:ontal
ta.. ble. 1\vo,
much
smaller
equal masses
A .and
B- are
hung l
. ::::;: '; 8 J
by strings which pass over smooth A •.
,i' <j
•
•
'
r.
)
pulleys: The stnng are long enough --that when C is in equilibrium. A and B both are just on
the ground. The trolley is pulled slow to one. side and
released as shown below. The graph of its velocity' v'
· against ' t' ~II be as :.
f". •n-:'!"7'7
i.
<
0 - - • - • - ~ 0 , ... ,
'
,,
,_
(d)· The two·
blocks'.• have
'acceleration
of the same
< ,, .
.
i
.
, ' magnitude ·
·- .
,, ,·; · ·.
.
,
14. A man pusiles. against ~ rigid"fixed vei-iicaj wall'. Which.
of the folio~g .is · {!ll'e)', tjie most· accurate
,
· ,
statemerit(s) related 'to the·.siruation ?,
(a) Whaieve;· force the niari ·ex~it:s 'on th~ wall, the
. wa1La,i~o e'!'e,-u, an e<'J.ua1 and.opposite force on
<
(a) a<;cel~ration of m4 = 0
(b) acceleration of m1 = m2 = ~ 3 =, m4 = 0
(c) acceleration of m1 = m2 "' m3 = 0
(d) ,,acceleration ,of m
[(m, + m2 )-,(m 3 + m4 )lg
(a)
-~ t·'.:,:\l .
!
www.puucho.com
o , ::. ~
·,,, ·
i ·.•Is - , , " -.
..:.. .'. L.,.
- -. l< ~.
Anurag Mishra Mechanics 1 with www.puucho.com
"\'
..
I
,,
rl•· FORCE ANAr~""'
...,-5r,~"-''
-'"r
-' i"':~.:r_
;;;L : _," • _ " <1~, A:.~·"'"~=-··
-'-'-" . ,;;;.:,-.~~-'---"''-'--'----'-''-C.-·-'--~==
19. In the ,system showp. i~'. figu~~ '·r;:-½·-·---·-",
m1 >m2 ·,.System is held atrest.bj..'·j
,· · . '
thread BP. Just after the thread BP.is l ._
burnt :
'
(a) Magnitude of acceleration of
b th bl cks will be equal ·to . m1Lf'--1.,;1.....,
~m1 :_: ; 2 ) g:
(
+ 2m2 ,
· ,
(b) Acceleration
m 1 will be equal to zero . ,
(c). Accel,eration of m 2 'l½U be upwa!_ds '
(cl) Magnitudes of acceleration' of two blocks will be
non-zero and unequal ; .
, ·
·
m,.
of
20. A . particle .i,s resting over a · · smooth horizontal floor.. At t = 6,
' ·,.
a horizontal' force start's ·acting
on it. Magnitude · of the fore~
increases with time.according to
1
law '_F ~- at, 'Where · g.' ~1~ a 49-:,_.
:::z:::d~
constant. For figure which of the . ·· . ·. ·following statement is/are correct? · ·. ·
(a) Curve B indicates velocity against time
(bl Curve B-indicates velocity against acceleration
(c) Curve A indicates acceleration ag~inst time
(d) None of t:l\ese
.,
21. TwoparticlesA&Beachofmass '.[- ·
mare in equilibrium in a vertical ·r ,: a_:r'A. '. •
plane under action of a
·: . , : 2 B ' F~.;g\
horizontal force F = mg on
: :.
. _ I
particle B, as shown in figure.
Then:
(b)- T1 ./2 =-T245
(a) Zf1 = ST2
(d} None of these
(cl tane = 2tana
.=:~:=:==::·: : ': ·
·,::--·,1
, •· · - 2231
~ ','-· _.,_,~,"
(al a 1 > a 3 > a2 · · ,', · (bl. a,·= a2,a2 =.a,'
(c) a 1 =a2 =a 3 - , .. ', (d);:a1>a 2 ,a 2 :aa3
24. A man has falleli'into ~'ditch of i -_· ~
·· ·,
width d and two 'of liis· trienas' ,ar~ · ,
· '• "
- -1
~lowly' pulling him ciut using a light
rope and. ·two fixed pulleys as
.
·
shown in. figure .. Indicate ·the ' . · ·- correct statements : (assume, both the friends !ipply
equal forces of equal magn/l}lde)
·'
(a) The force exerted by both the friends·deqeases as
the man move up · ·, · _, .
·
,
mg
2
(b) The force iipplied by'each friend is , h ;-~ d~ + 4h
4
when the man is at depth of h
-~c) The force exerted by both the friends inc~~ases as ··
the man moves up ·
r· : . , ·. Jj
(df The force applied b~ ea~ fri~nd is
25.
7, .Jd
rd th~ figure shown m,.~ 1 'kg; m
(b) .~
.
.
(d) g/3i' ·
23. In the figure the block A, B and C of mass m each, have
accelerations a 1 , a 2 &a 3 respectively. F,.&Fi are
external. · forces of magnitude · '2 ·, mg - a_nd mg
. . ..
-
.
. .
E·I··;l
/
·
.
~
F J
: '..~-,
m : , -\ 1
• ,
m ,
/>.{ ['~ ~mg B zm_
~ 1
S
~
I'
.
r
""'"" ,.i
_•ram.
Afu= F - "
is applied to2pulley (t is in second) then
(g=1Dm/s
):
·
,
:
2~.
' ,
~~j• : . ;; .
.
iI ·
m
(c)
1 is lifted off the ground at t = l_O sec(d) both blocks are lifted off simultaneously
26. In the following figure all . surfaces are smooth. The
0
. . I -----'-..c
I
. ..
' ," .
~~- .,~"""''""
''. (a) acceleration ofwedgeis_greater then g sine
2
(c) acceleration of mis g '.
(d) acceleration· of wedg~ is g sine
27. In above question 26, the normal .fdrce acting,between:
(a) wedge and incline plan~ i~ Mg cose · . ,,
(b) m and wedge is mg cos~·
(c) m and wedge is•zero · ·. . ·
( d) m ~nd wedge is mg siri 8: :
28. In the figure shown .."'.i '." 5 f ~
kg, m 2 =10_kg & fnctt<_m f
J µ,=0.1·
coeffi_cient between ri11 &· m_2 i'n:·,~11_~-).1 Jlll?:tan
i,s µ = 0.1 and grou°:d. · ,is. l (Smoot_h ?rou~d) - ·
frictionless then: · _, .. :. ·-::: .. . .··• · ..<' .
www.puucho.com
"
.
I
(a) m~ is lifted off the ground at t = 20 ~
ro 2
1
sec
.
,
(b) acceleration of piillhwhen m 2 is about ,o lift off
is 5 m/s 2 ·
·
•
. . "T[~ ·:. '
. J,
·s:,
6tJ
i ·• -
,,..
~
>
.
i . "-.-.--1
j
\7okii.
' • •,
m , , .·.
.2,~..,.~~'.°
2
(b) acceleration of m is.g~i + 2cos e
.
(c) Zero
respect~iv~:
+h
2 = 2kg; pulley ls
ideaE At t = 0, both masses touches the
' ,
.---~~~J'
I
.
JlJ
22. The magnitude of difference _in ~ - accelerations of block of mass· {
.
m in both the cases shown
.
.
lielowis: ,·:,
·
·
m,
. m
·
· F-Zmg •. 2m
(a) g
.
.
~ •-
2
Anurag Mishra Mechanics 1 with www.puucho.com
I 224
.,,;,•, ·.•
· ;:. ·
' - - ' · - - ' - = " ' - - '- - ~ - - ' - ' ·;.•'
(a) r{ a horizontal force F ,i, 20 N is applied on mi
then the friction force ·11cting' .on m2 is· 5 N in the,
girection of F
(b) Maximum amount of horizontal force that can be
applied to m 2 s1,1ch that there is no relative motion
· between blocks is 15 N
(c), If a horizontal force F 20 N is applied on rri 2
then friction force acting on rrii is 20/3 N in the
,
direction of applied force
\ (d) Maximum amount of horizontal force that can be
applied fo mi such that there is no relative motion
between blocks is 8 N
·
·
,
29: Pi. block of mass 0.1 kg is kept on
=
an inclined plane whose angle of
inclination can be varied from
B.= 30° to B = 90°. The coefficient
of friction between the block &
the inclined plane is µ = 1. A force
.
+
1 .. .
of constant magnitude - mg
'
'
,-----.=,•-;:,:"'·\·t·.···'./l,.,...,•·I
,i<'"i )
·
8·· '
·.•·;:
µ =··1·J'".
} ;·
2
newton always acts on the . block directed up the ·
.inclined plane and parallel to it. Then :
(a)
1·f
~
"
d:
11.f
jt/6 .. lt/4
rr/2
(b) If0 < tan-i µ the block cannot be pushed forward
for any. value of F
(c) As B.decreases the magnitude. of force needed to
just push the block M forward increases ·
(d) None of these
31. .In
the
arrangement
. ~
.. . ·
.:
shown, coefficient of ,.,.
friction for all the [:' ,, ·.
A m . Ti ;, ' · .
l
.surfaces. isµ _and blocks f.t;;;;;;;;!~are movmgwith constant
t:.'t!'--
ii",...._._~,-="'·-
-J_
speeds, then :
·
·ca) Ti =µmg
(b) F= 3µmg
(c) Ti= 2µmg
(d) F = 5µmg
32. A triangular block of mass m rests on a fixed rough
inclined plane having friction coefficient µ with the
block. A horizontal forces F is
applied to it as shown in figure
below, then .the correct statement
is :.
(a) Friction force is zero when
F cos0 = mg sin0
.
(b) The value of limiting friction
is µ (mg sin B+ F cosB)
(c) Normal reaction on the block is F sinB + mg.·cosB
(d) The
value
of
limiting
friction
is
µ (mg sinB-F cosB)
33. A body is moving down .a long inclined plane of
inclination 45° with horizontal.The coefficierit ,of
friction between the body and the plane varies as
µ = x/2, where x is the distance moved down the
plane. Initially x = O&v = 0.
(a) When x =· 2 the velocity,of the body is
~g./2 m/s
(b) The velocity of the body increases all the time
(c). At an instant when. v ,;, 0 the instantaneous
acceleration of the body · down the plane is
g (2- x)
lt/2 8
.. 2./2
C.
,o
!ti
lit:'
.:
-··-!
.
(d) ~·--mli---.,,f;c.tsi.-.-cy"'.1r-r-+8
'
.
. . '- ,.- ~~;:~1::i-"'
30. · In, the .situa""t1'"·0-n-sh~o-wn__,
in the
'.'t'rgure.'
the
friction
coefficient
'
b~tween M and the horizontal
surface is µ. The force F is
applied at an angle B with
vertical. The cortect statements
~
'·
[flnnJ:n!nm mnlm»
are:
(a) If B > tan-i µ the block cannot be pushed forward
for any value of F
-~-
(d) The body first accelerates and then.decelerates·
34. , Suppose F, FN & f are the magnitudes of the contact
force, normal force and the frictional force exerted by
one surface on the other, kept in contact, if none of
these is zero :
(a) F > f
(b) FN > f
(c) F > FN
(d) (FN-f)<(FN+f)
35. Bl.ock A is placed on block B. ~ .
· ~-,.·,.···.,
There is friction between .the
.
P • : , ·\
blocks, while the ground. is
1
1l
smooth. A horizontal force P
· ':-""-"'"'...I
'
increasing linearly with time, begins
to act on A. The
accelerations ai & a 2 of A and B respectively are
plotted against time (t ). The correct graph is :
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
39. The
friction . coefficient · I·.
between plank and floor is µ. ·
The man applies, the·
m
maximum possible force on
the. string and the system
remains at rest. Then :
~ ------- (a) . frictional force between plank and surface is
2µmg
1+µ
(b) frictional force on man is zero
(c) tension in the string is Zµ mg'
.
1+µ
..•
",;..,.~ I,
the - shown diagram
~ 1 =.m 2 = 4_kg and m3 = 2
kg. 'Coefficieni: of friction
between m1 and m2 is 0.5.
The mass
is given a
velocity v and it just stops ' - - - - - - - - - ' = ' - " '
at the other end of the mass m2 in 1 sec. Let a1 , a 2 and
a3 be the acceleration m,, m2 and m3 respeGtively,
then:
36. In
mt
(a) fort < 1 sec, a1 = 5 m/ s2 , a 2 = a 3
.
.
(b) fort< 1 sec, a1 = Sm/s 2 ,a 2
= I. m/ s2 3
.
= a3 = O
(c) the value of vis 5 m/s
(d) fort> l sec, a1 = a 2 = a 3 = 2m/s 2
3 7 • . 0 is a point .at the bottom of a
rough plane inclined at an angle a
to horizontal. Coefficient of
. between AB 1s
. tana
- and
fri ctton
I
2
·
·
:
B·o . . 3taiia B . h
IS - - .
1s t e
b etween
. 2
r;-2···_·\
I
'A
.
r_i\ . •
.. - - '
B
'
j
,
,,
! ---·· -. . . ., ... _
,.
middle point of AO. A block is released from rest at A,
then which of the following graphs are· correct :
(a)LJ~]- il
r·-.
(b)
---·
-!-- _ -~t
(.,_ . ,.__ - - -
'_'!
38. In above question 37 :
(a) velocity of block at O will be maximum
(b) velocity of block at O will be zero
(c) velocity of block at B will'.be inaximum·
(d) -average velocity of the block is zero·
(d) net force on man is zero
·40. In the shown figure, friction -- · -- -- -- • ·-·· ,
exists betw-een wedge and block
and also between wedge and
m
I,
M
.
.
. .
:
fl_oar. The system 1s m
equilibrium in the ·shown ~ ------ - ··· ··· ··' __ _;
position:
(a) frictional force between wedge and surface is
µ(M+m)g
.
(b) frictional force between wedge and surface is mg
(c) frictional force between wedge and block isµ mg
(d) minimum coefficient of friction required to hold
w
__·- -\
· the system in equilibrium is ~
M+m
41. A block is projected with velocity v 0 up the inclined
plane from its bottom at t = 0. The plane makes an
angle 8 with the horizontal. If the coefficient of friction
between the block and the incline is µ: ~ tan a (a > 8)
then frictional force applied by the plane on the block
fort>
Vo
will be:
g [sine+ tan a case]
·
(a}" tanamg case
(b) zero
(c) mg sine
(d) tanamg sine
--. 42. In the shown diagram friction •· --- -- --·
exists at each contact · surface / ~
·_ m
_'
;
with coefficientµ and the blocks I _ .'
M ·
l
are at rest. Then :
'.
, ' e.
·'
:
. (a) frictional fo~ce between L_.. - ·-··:- · -· · - ·
wedge and surface is mg sin 8 case
(b) h~rmal force by the surface is (M + m)g
(c) friction force on m kg is mg sin8
(d) net force of m is zero
43. A sphere of weight W ~ 100 N is kept
stationary on a rough inclined plane
by a ho,izontal string AB as shown in
'figure. Then_: .
(a) tension in the string is 100 N
(b) normal reaction on" the sphere by the plane is
100N
www.puucho.com
1
Anurag Mishra Mechanics 1 with www.puucho.com
[}2s_;~_-.____ ....;__-'-'----"-""
·='.·,_·,_''af:,.,';,~··C..---~-----~---ME_c_HA.;_N_l(S_,1_,q
(c) tension in the string is lO~ N
(a) The car cannot make a tum without skidding.
2 + .;, 3
(b) If the car turns at a speed less than 40 km/hr, it
'
. on.the sp.here 1s
·
(d) ,orce
offri cnon
lOO
~
slips down.
·
(c) If the car turns at the correct speed of 40 km/hr,
the force by the road on the caris equal to mv 2 /r:
N
2+-v3 .
44. The position vector of a particle in a circular niotioh
about the origin sweeps out equal area in equal time :
(a) Its velocity remains constant
(b) Its-speed remains constant
(c) Its acceleration remains constant
(d) Its tangential acceleration remains c<_>nstant
. 45. ABCDE is a smooth iron track in the
t'M-~j-
vertical plane. The section ABC and
1 .
CDE are quarter circles. Points B and ~· -/ .
D are very close to C. M is _a small · L.. El•..:._••_,.·_'_.:,
magnet of mass m. The force of
. D·f :, '
attraction between Mand the track is
--·....; __ EL
F, which is constant and always normal to the track. M
starts from rest at A, then :
(a) If M is not to leave the track at C then F ;., 2mg
(b) At B, the normal reaction of the track is F - 2mg
(c) At D, the normal reaction of the track is F + 2mg
(d) The .normal reaction of the track is equal to F at
some point between A and C
46. A particle· i~ .moving alohg a circular path: The angular
velocity, linear velocity, angular acceleration and
centripetal acceleration of the particle at any instant
v, a, a
are ro,
0 respectively. Which of the following
relations are correct ?
·cal
(c)
roj_ v
ro J_ "it:
(bl
'
..
'
Cd)
roJ_a
vJ_a;,
(d)" If the car turns at the correct speed of 40 km/hr,
the force by the road on tlie car.is greater than ing
as well'as greater than mv 2/r.
49. A body moves on a horizontal circular road of radius r
with a ta!).gential acceleration
The coefficient of
friction: between the body and the rpad surface is µ. It
begins to slip when its speed is v, then :
a,.
·
(a) v 2 '=µrg
(b) µg
(c) µ2g2
v4
=-+a;
r2
(d) The force of friction makes an angle tan-1 (v 2 /a,r)
with the direction o( motion at the point of
slipping
50. A particle P of mass m attached to a
vertical axis by two strings AP and BP of '"""---,length L each. The separation AB = L, P
rotates around the axis witli an angular
.P
velocity 'ro'. The tensions in the strings
AP&BP areT1 and T2 respectively, then:
(a) T1
(b) T1
= T2
+ T2 = mro 2L
(c} T1 -T2 =2mg
[r".t.
47. Suppose a machine consists of a·
~ag~ at the end of one arm._Th~ arm
--:-·,,1·
IS hinged at O as shown m figure . 0,........- ...,,..;F. ·
s,uch that the cage revolves along a
/
·v_vertical circle of radius rat constant cf----:·-·-G.;;;:::··------)G
linear speed v = .,fir. The cage is so j \__
"[ll]~
attached !hat the man of weight W, ! B'".,_,_,_.... ·.
standing on a weighing machine L_,_}._ ·
inside the cage, remains always vertical. Then :
(a) The reading of his weight on the machine is equal
to W at all positions .
(b) The weight reading at A is greater than the weight
reading at E by 2W.
(c) The weight reading at G is same as that at C. .''
(d) The ratio of weight reading atE to that afA = 0.
48. A smooth circular road of radius r is banked for a
speed v = 401anjhi-. A car of mass ni attempts to go on
the circular road. The friction coefficient between the
tyre and the road is negligible. The correct statements
are:
(d) BP will reJ?ain taut only if ro 2'~2g/L
51. As shown below AB represents an infinite r---:---:::7
8
wall tangential to a horizontal
semi-circular track. 0 is a point source .of
light on the ground at the center of the
circle. A block moves along the circular
A
track with a speed V starting from the
point where the wall touches the circle. If ~ - - ~
the velocity and acceleration of shadow along the
length of the wall is respectively V and a, then :
o¥E--~·:
(a) V
= v cos (;)
vsec (vt)
R
(b) · V =
- .
(c) a= (
2
~ }ec
2
2
(d)
www.puucho.com
.'
,V2
=-+a,
r
a= ( ~ }ec
( ; }an(;)
2
( ; }an(;}
't'
Anurag Mishra Mechanics 1 with www.puucho.com
=t7""~
"'
",-
-
-·-
-~-·-
#--~
-·q"f- - - ·-:--
~f;(}!C~(Aj!A~~J;:~::·,.·
.
·:.'.:i:::••.i
.~' -. ~- .. ' ·-1-·· -~·"""'~--··-" ·--- -,·=---_-~
•-'--.-"--~
I -
,·
.; ' ,'\
.• ..-w.:•~-\--·
~
52. A curved section of a road is banked for a, speed. v. If
there is no friction b~tween road and type. Then :
(a) a car moving with speed v will not slip cin road
(b) a car is more likely to slip on the road at speed
higher than v, than at speeds lower than v
-(c) a car is move likely to slip on the road at speed
lower than v, than at speeds higher than v
(d) a car cannot remain stationary on road ahd willstart ~lipping
·
53. A tube of length 'L' is filled completely with an in
compressible liquid of mass 'M'' and closed at both
ends. The tube. is then rotated in a horizontal plane
about one of it's ends with a uniform angular velocity
'ro'. Then which of following statements are true :
(a) The force exerted by liquid at the other end is
. 1Mro 2L
,
2
(b) Ratio of force at middle and point of the tube will
be 4;1
(c) The force between liquid layers linearly with the
distance along the length of tube_
(d) Force is constant
54. Aparticle of mass m describe circular path of radius 'r'
and its radial or nmmal or centripetal acceleration
depends on time_ 't' as aR = Kt 2. K is +ve constant.
Then:
(a) at ~ time 't' fore~ .acting on particle is
'
,
m-,/kr + k 2t 4
(b) Power developed at any time t is mkrt
(c) Power developed a~ any time t is mk~'2 /r3/ 2t
(d) Tangential-acceleration is also val)~ng. ·
55. Aparticle of mass' m' describes circular path of)adius
'r' ·such that its kin~tic energy is given by [( = as_ 2 • 's' i_s
the distance travelled, 'a' is constant : , , · · ,
(a) Power ·developed at distance; is' proportional io s2
(b) Tangentiai'accelerationis proportional to 's.,
(c) Radial acceleration is proportiqnal to s3 - ·
(d) None·of these
.
Three
particles
·describes
circular
path
of
'radii
r1 ,' 12
56.
and r3 with constant speed such that all the particles
take same time to complete the revolution. If
rot,ro 2 ,ro 3 be the angular velocity, v 1 , v 2 , v 3 be linear
velocities and_ a1 , a 2 ,a 3 be linea'r acceleratior; tha~ :
(a) ro 1 :ro 2 :ro 3 = 1:1:1
(b) vi :v 2 :v 3 ,=·r1 :r2:r3
(c) a 1 :a 2 :~ 3 aal,1:1 . _· ·,
(d) a 1 :a 2 :a3 =r1 ·:r2 :r3
57. A particle of mass m describes a circular path of radius
'r' such that speed v·= a-Js ( S is distance traveled).
Then power is proportional to :
(b)
(a) S
',Js
(cl s312
(d) None of tliese
58. A ring of radius' r' and mass per unit length' m' rotates
with an angular_ velocity 'ro' in free space then :
(a) Tension in ring is zero
(b) Tension will vary at all points
(c) Tension is constant throughout ring
(d) Tension in stri~g is mro 2 r 2
A body moves on a horizontal ~ircular road of radius r,
with a tangential acceleration Uy, Coefficient of·_
friction between the body and road surface is µ. It ·
begin to slip when it's speed·is v, then :
(a) · v 2 = ,trg
v2
(b) µg =-+ar
r
v4
(c) µ 2g 2 =_._+a;'
r2
(d) The
force
of
friction
makes
an
angle
tan -J ( ~ ) with directiqn of motio~ at point of·
a 1. X r
slipping.
60. A simple pendulum has a bob of mass m and swings
with an angular amplitude qi . The tension in thread is
T. At a certain time the string makes an angle 0 with .
the vertical (0 S: <I>) : · '
(a) T = mg case for all values ofe
(b) T = mg c,:,s0.for only 0 = qi
1
(c) T=mg,fo~0=cos- [½c2cosqi+l)]
(d) Twill be larger for smaller values of 8
61. A particle of mass m moves along a circle of radius 'R'.
The modulus of the average vector of force acting on
the particle over the distance equal to a quarter of the
·
'
circle is :
(a) zero if the particle moves with uniform speed v
0.
(b)· '
2
'
'
mu if the particle moves with uniform speed v
itR
2
(c)
z,./imv if the particle moves with unifonil speed v
11R
:,
(d) ma ·if particle moves with constant tangential
acceleration 'a', the initial velocity being equal to
. zero
particle'~' moves afbqg a circle of radius R = 50 cm,
62.
so that its radius vectllr 'r' relative to the point 0
rotates with the c'ilns&Jmt a~gular velocity ro = 0.4
J
·
.
rad/s. Then :
(a) lirn;ar velocity of particle is 0.2 m/s
(b) . linear velocity of particle is 0.4 m/s
A
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
'\
(~~ magnitude of net acceleration is 9.08 m/ s 2
(d) acceleration of particle is :tero
63. Two bodies are moving with constant
speed v clo~ise andi,:!ll"e initially
diagonally OPP.CJSite. The },~rticle B now
achieves a tangential acceleration of a
m/s 2 • Then: : '
·
·.
,:©
.,
(a) they c~llid~ after tin}e ~
(b)
rrii
A.-,v.
'~
R
67. Two blocks· of masses mi = 2 kg [
and
= 4 kg hang , ,over a
. massless pulley as shown in the
figur~. A force F0 7 lOON acting at
~
the axis of the pulley accelerates ,;kg
the system upwards. Then : ·
I
4kg
(a) acc.eleration of 2 kg mass· is <-.....;..cc.-_..;
15~/s 2 ·:
•
.
+. v· B·
(b) acceleration o_f 4kg mass is 2.Sm/s 2
~~y collide afte; time ~ 21;;
(c)' · relative velocity just before collision -is .JrcaR
(d) -~~lative velCJcity just before collision is .J2rcaR
64. A P1\r1;ii:le P is attached by means of two equal strings
to 'two points· A and B in same vertical line and
desct:1be~ horizontal ~ircle_ with uniform angular speed .
~ {2i"where AB= h.
·_
. ·
~,;:''
(a) T1 > T2
(c) T1 :.T2 ~
68.
69.
,·,
•
;/5_ : -J3 '.
(b) T1 : T2 = 5: 3
(d) T1 = T2
65. A particle is &cted upon by constant magnitude force
P~il>endiculaf to it which is alw;iys perpendicular to
velocity ofj>4rticle. The motion is taking place in a
plane it follows that :
(a) vela~!~ i~ constant
(b j accel~riition is constant
(c) KinetiJ'~~etgy is constant
(d) ii lllOVes in circuiar path
66._ A parti~le 9f mass m moves in a.conservative force field
along' aifis where the potential energy U varies with
position coordinate x as U = U0 (1- cos ax),U0 and a ·
being positive constants. Which of the following
statement is true regarding its motion. Its total energy
is U O and ~tarts from X = 0.
(a) !i]e 'cceleration is constant
(b) It's speed is maximum at the initial position.
70.
x
(c) It's maximum x coordinate is~
2a
rd) It's maximum kinetic energy is U0
.
71.
(c) '.'cceleration of both the masses is same
(d) ·"\!cceleration of both the masses is upward
' ·,
Which of the following is / are incorrect:
(a) If net normal force on a surface is zero, friction.
will be z¢ro.
'(b) Value ofstatic fii~tion is given byµ ,N.
(c) Static friction oppo~es relative motion between
two surfaces is contact.
(d) Kinetic friction reduces velocity of an object.
A spring block system is
.
placed
on
a
rough
?orizontdaltfloor.dTh: bhlock ( ..
1s pu 11 e owar s ng t to ~---·--~-give spring some _elongation and released. Then: .
(a) the bloc1' may s~op before the spring attains its
natural length
(b) the block m\lst stop with spring having some
compression
(c) the block may stop with spring having some
compre_ssion
(d) it is not possible that the block stops at mean
·position
In the above situation the block will have maximum
velocity when:
(a) the spring force becomes zero
(b) the frictional force becomes zero
(c) the net force becomes zero
(d) the acceleration of block becomes zero
A book leans against a crate on a
table. Neither is moving. Which
'. :1
of Lhe following statements !
concerning this situation is/are
incorrect ?
(a) The force of the book on the crate is less than that
of crate on the book
(b) Although there is no friction acting on the crate,
there must be friction acting on the book or else it
will fall
www.puucho.com
.·
i m.:,;:,::;
· ...-ii].
.
·
,J::1:;,,,i,;. :
r.so-cratel
.,_~
.j
Anurag Mishra Mechanics 1 with www.puucho.com
(c) The net force acting on the book is zero
(d) The direction of the frictional force acting on the
book is in the same direction as the frictional
acting on the crate
72. An iron sphere weighing 10 N rests in a V shaped
smooth trough whose sides
an angle of 60° as
shown in the figure. Jhen the reaction forces are:
73, In the sy~tem shown in the figure m1 > m2'. System is
held at rest by thread BC. Just after the ·thread' BC is
, burnt:
r, . . ' .
form
·
14
..
li;
~·
G ___ ____1~---'~____
(a) initial acceleration of m2 will be upwards
(b) magnitucje of initial acceleration of both blocks
(a) RA= ION andR 8 = 0 in case (i)
(b) RA = l0N andR 3 = ION in case (ii)
20
dR
10 N , . .
("')
(c) RA=. .f:3Nan
8 = ../3 m case m
will be equal to ( mi -. m2 ) g
·
m1 + m2
(c) 'initial acceleration of m1 will be equal to zero
(d) magnitude of initial acceleration of two blocks
will be non-zero and unequal.
60°
60~
.c_(ii-'-i)_-.__...,
(d) RA = l0N andR8 = 10.N in
all the three cases .
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
: 230
Compreh~nsion Based Problems
I
-::
s.S'A;l(E
--,
-"·
3. The tension on side ofheavier\nass will be:
1 ...: ~,~---
pA
(a) m1g
(c) '2m 2g
(d) 2m 1g
3
3
4. The tension c;m side of lighter ll_lass will be:·
Effect of friction between pulley and thread :
In ideal cases i.e., when pulley and strings are massless and
,no friction exists at any contact surface, then tension in the
string is constant throughout its length. But consider a,
massless pulley and massless string but friction exists:
,between pulley and string With coefficientµ. Then tension
:at the two .ends of the pulley will be different. As .shown in.
figure, consider an element of string :
2
2
µdN i
8
8
Tcosd2 ~
' (T+dT)casd2 ,
:
,··· e d;e___ ~_T+dT
.
m,
--···:· de· .
de.
.
· 2
,_
t
2
2
2
·., ._
.?.Ji~~
twq,
2 .
2
JliTdT = l"oµ. de
T2
2
2·
=> In
·(T.~ J= µ1t
=}
T
~
= eµ:c
T,
!Suppose coefficient of friction between the string· and'
.
1 .
pulleyis µ. = - .
'-·-· · · - · - •. , lt.
1. What should be the ratio of heavier mass to lighter
mass for no motion ?
(b)
I
.
2
2
.____ __ ,------ - ,-- - - ~I
'
,._
de
de]·
dT=µ [ T·-+O+T=µTease
eL
,·r-rr'~fi;t
'• ,-
37'
de
[r sm-.
, de + dT ·Sm-+
· . de T sm~
. de]
dT -cos-=µ
(c)
.,
3
2
dTcos de= µ[er+ dt) sin do+ Tsin de]
(a) e
4m1g
, Cons!<le; tl)~-situ~ti~~.sho~ in figure in which a block 'A'
of mass 2 kg is plac~d over a.biock.\B' of inass 4 kg. The
combination of the blocks are 'placed on a inclined plane of
,inclination: 37° with horizontal. The:coefficient of fyiction
between block B and inclined p1ape is µ.; and in lletween
the
b\oci\5 is µI. 'f~e system "is r.eleased from rest:
. (Take_ g· "';'lcim/ sec 2 )
·
·
•
·
:
'
'
,,' ' '
- - --: ~ .... ,-
· (massless string)
de.=µ dN
dTcos2
'
(d)
·3
:
,/ .
,(T + dT)cos- -Tcos- -µ dN = dr/a= 0
,
· (b) m 2 g
(c) 217\2g
PASJJl\'.G'E
,+.dN
;
(a) m1g
,
dN = (T + dT) sin de + T sin de
(b) m 2 g
~
e
(d) e"
2. If m2 = 2em,, D.1.en- acceleration of each mass is :
(a) g
(b) g/3
(c) eg/3
(d) zero
1. Ifµ. 1 = 0.8,µ 2 = 0.8then:
(a) both blocks will move ,together
(b) only block A will move and blockB remains at rest
(c) only block B will 1n~ire and block A remains at
rest.
(d) none of the blocks will move
2. In the previous question the frictional force between
block B and plane is :
(a) 36 N
(b) 24 N
(c) 12 N
(d) 48 N
3. If.µ· 1· = 0.5,µ 2 = 0.5, then :
~a) Both block will move but with different
· accelerations
(b) Both block will move together
(c) Only block A will move ·
(d) Only block B will move
4. The frictional force acting between the two blocks in
the previous question is :
(a) 8 N
(b) 6 N
(c) 4 N
(d) 0
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
2311
[ FORCEA~ALVSI~- ~-~-- -__: .• -_- ___ -_ -~~5. The acceleration time graph for 2 kg block is:
5. Ifµ 1 = 0.4,µ 2 = 0-5 then:
(a) Both block will move but block A will slide over
the blockB
(b) Both block will move together
(c) None of them will move
(d) Only block A will move
6. The frictional force acting between the blocks in the
previous case will be:
(b) 6.4 N
(a) 8 N
(d) zero
(c) 4 N
7. Ifµ 1 =0.5,µ 2 =0.4,then:
(a) Both blocks will move but with different
acceleration
(b) Both blocks does not move
(c) Only block A will move
(d) Both blocks move together
8. The frictional force acting between the blocks in the
previous case :
(b) 6.4 N
(a) 8 N
(d) zero
(c) 6 N
1/31----/
~ t
(c)
6.
,)---!
,
; ~'
I
F =0.5t
:
µ2 =
/
-..-
(d) 3.2 N
1/3~
''
6
28/3
28/3
t
: -1,
41---~~.-j
3 -------- .. .Jh.;
(a)
I.hi
6
\
28/3
-- -
•f,
3
(b)
I__
6
•
28/3
.... ?I'"
·-1;
(c)
I ;3
....-··· I
I
,_
6
(d) None of these
8. The friction force between the blocks and time graph
is:
sec is :
(c) 3.6 N
(d)
7. .The frictional force acting between 3 kg block and
ground w.r.t. time will vary as:
3
3. The frictional force acting between the two blocks at
t = 8 sec.
(b) 3 N
(a) 4 N
(d) 3.2 N
(c) 3.6 N
4. The frictional force acting between the blocks at t = 10
3N
1/3~
t
6
J __
(d) 6, 6 sec
(c) 8, 6 sec
2. The relative slipping between the blocks occurs at t =
(a) 6 sec
(b) 8 sec
28
(c) - sec
(d) Never
(b)
(b)
a
(c) : 1/3~-----;
0
'
t
'I
1. The motion of blocks 2 kg and 3 kg will begin at time
t = -,- respectively :
(b) 6, 8 sec
(a) 8, 8 sec
(a) 4 N
28/3
8
a
'
~- ---
i i
6
8
(a) : 1 ••••
/3
,
~ :
,-···
t
·
6
,
µ1 =0.2
-~.,
a.as-;tlXJ""\
1/3-----~
(d)
''"
6
;In the given figure, the blocks of mass 2 kg and 3 kg are'
placed one over the other as shown. The surface are tough
with coefficient of friction µ 1 = 0.2,µ 2 = 0.06. A force
F = 0_5t (where 't' in sec) is applied on upper block.in thei
:direction shown. Based on above data answers the
'following questions. (g = 10 m/sec 2 )
a
The acceleration time graph for 4 kg block is:
3
: PASSAGE
I
a
(a)
·41----=-!3 ------- -- 0, i
0 !
!
6
www.puucho.com
28/3
---· --- ·-
.
t'
I
Anurag Mishra Mechanics 1 with www.puucho.com
(dl None of these
r··-- - .... --,---. ! ' . . .,
•.-,n
--.,.
, P}A,S_Sll,.G;_~
,•
~1
':--'!.?.
1
4
'--,.
.,,.7
•:
G:
\i . ·_·l~l-.
:I
x!
I
··----~-~--~
-~-===c.....---
1. The maximum velocity of block will be :
I
:;,'
J
(cl 3 m
(dl
2 .
,----,
.
2
--······
2
(bl 2mg sine
(dl . 2mg cose
oft O
is
µ~g2 + (ral2
(al
µ~g2+(ral2
ra 2
{mg
-v~
.
-'. • , :
ofa
[A very sma_II c_ube of mass 2 kg is pla_ce·d·. on the surface
furinel as sho\vn in figure: The funnel. is,totating,about'ifi;
-.;ertical axis,of syrnmetrywith"'iingulat velocity'ro': The~all
of funnel mai<es"an angle 37° with:horiiontal. The distance
of cube from the axis,_of rotation is' 20 cm and fric'°i;i,:m
coefficient·is µ. (Take g = lO_m/s~l .,
.. . .
0
. . ,---j).,
. I_..,..__
(bl 2mg sine
·, i
(dl 2mgcose
;, I
r:::20cm
·I
.. ';)
--··-. :. - _~...,,---
1
s1sti- olJ:
S
-- ~~
If \ 5
! ir:il
, ,~n.
then value
.'-er====
4, Frictional acting on the block just before it comes to ·
rest :
(al mg sine
(cl mg sine
=t 0
~;:~N;e;;~;s_~{- - · ~ -
~m
3. Frictional force acting on the block after it comes to
rest:
(al mg sine
. (cl mg sine
- 3. If the bead start sliding at t
given by:
(cl
(al
(bl
sine
cose ·
(di
tane
(cl
2. Maximum distance traveled by the block :
(bl 2_m
(al lm
Jg
2. Friction force acting on bead at timet (< t 0 l is given
by:
(al µmg
(bl mr(atl 2·
'
Jg
Jg
..Ji
j
._. _.M~CHANICS,1
(dl µm~g2 + r2(cxtl4
Irt the adjacent figure, x-axis has been taken down the
inclined plane. The coefficient oHrictioh varies with ·x as
µ = kx, where k = tane. A block is released at O.
I
•-
'.
(bl mr(cxtl 2
2 +_(_ra_l~
2
(dl m~~g~
(al mg ·
(cl m~rg~2 -+_r_2-(cxt_l_
4
L
.
.- ., -
j232
'·
,
:The figure shows a .r9d _wl\ich ~t;trts 'rotating with an~lar
Iacceleration a about verticru:axis·passing through one ,of its
;end (Al in horizontal plane. A bead_of.mass mjust fit's· the
'rod .and is· situated ·at a clistance '·r' from.end A: Friction
;exist between rod ahd the bead with coefficientµ. As 0e
1angular velocity of r~~ increas~s the b~ad starts sliding ove~\
[!P-5'_.r_Qc;l _(siiy~fter.!lJn
..~.lo),
~---·.
e
;',::.
::'l
'--~~~--~-;~.:__:l.__·_,,'-.--'.~'--·___?'..-''
'; '/"
'j
"•!
-.'_ ~:~_.; ;_)·/."
le·''·
,
1. The friction force acting between the block ancl surface
(if µ = 0.3l of funnel at ro = 5 rad/ s is _:
(al 6.6 N
(bl 4 N
(c) 2.2 N
(d) zero
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
I FORCE ANALYSIS
2331
2. For what value of OJ, there would be no frictional force
acting between the surfaces :
(a) .5 rad/sec
(b)
H
rad/sec
.
(c) ..J?, rad/sec
(d) -./40 rad/sec
3. The maximum value of angular velocity for which no
relative slippjng occurs and also direction of frictional
force is : (takeµ = 2/3)
(a)
(b)
(c)
(d)
J¥
J¥
N
N
rad/sec; down the surface of funnel
c;,
it,
jA body of mass m = 1.8 kg is placed on an inclined plalle,
Ithe angle of inc)ination is a = 3,7" 1 and is attached to., the
ltop end of the slope with a. thread which is parallel to the
slop. Then the slope is - moved With a horizontal
acceleration of a. Fraction is negligible.
---7
1. The acceleration, if the body pushes the slope with 'a
rad/sec; up the surface of funnel
3
'
force of - mg is:
4·
rad/ sec; down the surface of funnel
N
(b).
H;
ii1i'si:(i '-. 1,
i.m
L------·'-----'-i__,_ :
rad/sec; up the surface of funnel
(a)
relative slipping occurs and also the direction of
frictional force acting µ = 2/3 :
(c)
{iu
-4i
4, The minimum value of angular velocity for which
(a)
fA
" ..s··sii,G'iE'·
·- .__ --
(d) .
J¥
ffs
l
~~ ,: . ,_j)
J,.__,.._.,..,,L-;.;J.: __ _
A car is moving with speed v and is taking a tum on ~
circular road of radius 10 m. The angle of banking is 37°.
T_he driver -wants that car does not sli,P on the road. The.
[coefficient of friction is 0.4. (g = l0m/ sec 2) _·_ __
1. The speed of car for which no frictional force is
produced is :
(a) 5 m/sec
(b) s./3 m/sec
(c) 3-/s m/sec
(d) 1_0 m/ sec
2. The friction force acting when v = 10
and mass of
car is 50, kg is :
(a) 400 N
(b) 100 N
(c) · 300 N
(d) 200 N
3. If the car were moving on a flat road and distance
between the front tyres is 2 m and the height of the
centre of the mass of the car is lm from the ground,
then, the minimum velocity for which car topples is :
(a) 5 m/sec
(b) s./3 m/sec
(c) 3-/s m/sec ·
· (d) 10 m/sec
nvsec
~ m/s 2
(b) 0.5 m/s 2
3
(c) 0.75 m/s 2
(d) ~m/s 2
6
2. The tension in thread is:
(a) 12 N
(b) 10 N
(c) 8 N
(d) 4N
3. At what acceleration will the body lose contact with
plane:
40
(a) 3 m/s. 2
(b) 7.5 m/ s2
(c) 10 m/ s2
(d) 5 m/ s2
fifsi~G}j g@
~
A lift can move upward or do~ward. A light inextensible
string fixed from ceiling of lift with a frictionless pulley and
tensions in string T1 • 1\vo 'masses of m1 and m2 ~re
connected with Inextensible light string and tension in this
string T2 as shown in figure. Read the questions carefully!
and answer.
·
·
= m and lift is moving with constant
velocity then value of T1 :
1. If m1 + m2
(aJ·;,,mg
(c) :,; mg
www.puucho.com
(b) =mg
(d) > mg
Anurag Mishra Mechanics 1 with www.puucho.com
--~--~··;o
,....2_3_4_~~---~··£'-----..:..:....'-'-"----...:......::--:.:::::···===~:::::·.·!:'=;r~."'--~~--MEC~~N·cs:!J
2. If m1 is very small as compared to m2 and lift is moving
with constant velocity then value of T2 is nearly:
(a) m 2g
(b) 2m1g
~;;5,5f~;~~
T~
,c 1~·~·-·~
I~ th:·~e~sho~:~~assqfthe trolley is 100ifand it,
can,move'.without friction on the.horizontal floor. Itslengthl
. (c) Cm1 + m2)g
is 12in. The mass of the gidis:sokg; friction exists between
.Cd) zero
the· shoes of the girl and. the: trolleys upper surface, with
3. If m1 ; m2 and m1 is moving at a certain instant with µ ;1/3. '.J:'lfe girl can run·witl). a D1aximum speed ;,9m/s on
velocity v upward with respect to lift and the lift is the surface•ofthe trolley, with respecttothe surface..Att ~ o
moving in upward direction with constant acceleration the girl ~t!l.rts nmJling from left:· to thti _right. ~-e trgMy: was
initially, sfationary. (g ; lQm/l) .
, I
(a < g) then speed of m1 with respect to lift:
(a) increases
(b) decreases
(c) remains constant
(d) depend upon acceleration of lift
•..
.
.,... .e1
!"'· re,
~A~.tll:~:!
i
ml Ii!
1. The minimum time in which the girl can acquire her
'
\!JJ ~-
.,·.,'
. .
I
A ,,;hot putter with a mass Rf 801<Kpushes the iron bill of;
mass. of 6. kg from a stancliitg position acc!'!lerating it;
uniformly.form rest at an angltof 4s with the'horizonta!I
1
during a, time. interyal of
'isec:onds. The bail foar,es his1
0
2.
in
~~!~~e?;:;~;~,:~~ ap~ve_;~e~evel ground :md hi!s thej
1. The accleration of the balll in shot putter's hand:
2
(a) 11.Jz m/s
3.
(b) 10W2m/s 2
(c) 9W2m/s 2
(d) 9.Jz m/s 2
2. The horizontal distance between the point of release
and the point where the ball hits the ground:
(a) 16 m
Cb) 18 m
(c) 20 m
(d) 22 m
3. The minimum value of the static coefficient of friction if
the shot putter do.es not slip during the shot is closest
4.
5.
to:
(a) 0.28
(b) 0.38
(c) OAS
(d) 0.58
·~,.~
6.
maximum speed, for no slipping, is:
(a) 1.5s
(b) 1.8s
(c) .2s
. (d) None of these
The total kinetic energy of system (troliey + girl) at the
instant the girl acquires her maximum relative. speed
with respect to troliey, is:
(a) 1350J
Cb) 1250J
(c) 2475J
(d) None of these
The displacement of the trolley by the time the girl
reacltes, the right end of the trolley, i_s:
·
(a) 6m
Cb) 12m
(c) 3ni
(d) 4m
The minimum time in which the girl can stop from 9m/s
relative speed, to zero relative speed, without causing
her shoes to slip is:
(a) 5/3 s
Cb) 4/3 s
(c) 9/Ss
(d) None of these
At a certain moment when the ·girl was acceierating,
the earth frame acceleration of the trolley is found to be
1 m/ s2. At this moment, the friction force between the
girl's shoes. and the trolley's surface is: ,
(a) 200N
Cb) 150N
(c) l00N
(d) None of these
Suppose the girl accelerates slowly, at a constant rate,
and acquires the relative speed of 9m/s only when it
reacltes the right end of the trolley, then, what must be ·
the earth frame acceleration of the girl ?
(a) 2.5m/s 2
(b) 2.25m/s 2
(c) 1.125m/s 2
(d) 3.375m/s 2
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
i FORCE ANALYSIS
--- --·- ---- -
'--,-·--~---#~----·-· ~--4 - - · ~ ------------ __ . ____ ._, ---- ----- - -
1. · A motorcycle moves around a vertical circle with a
constants speed unaer the influence of the force of
~
~
gravity w, friction between wheel and track f and
.
.
m
(A) Minimum value of--'so that m (P) --~
M
3
,slides down
(B)
,~ •'r, :.. ~::
•
, (P)
:. t
,
(B) iD.ir<(qted
I
•
'
(D) 'Ratio of vertical component of (S) 5
acceleration of m and acceleration
,of M:,
I
:cen\i'Ei when value ·in·
i~~n:~~ro < .
j
(C) · fl;'otal reaction force, by' (R)
!track •
(D)
:
.
·
',When. ;notion is.
;f + ~
4. A river is flowing with speed 3 km/hr west to east. A
man swims with speed 5 km/hr in still water. Man is at
south bank of the river. Match the column-1' with
direction of velocities of man w.r. t. ground in
column-2.
! -'
.
along, (S) · ;~
'
.
.
;veiticaltheval'Ue is zero;·
m
:slides up
(C) :Value of!!!. so that friction force on (R) 3
5
M
1m fa zero
'N .
towards;(Q) IN +f
( ••, •
. I,
M
value of~ so that m (Q) 1
~
normal reaction between wheel and track N :
(A) :Conitanrmagnitude·
1Minimum
~
·1.-,+w+
,
c.,f
1
2. A block is projected with an initial velocity v Block on a
long ~Qnveyor belt moving With velocity V Block (at that
. : 'instant) h~ving constant 'acceleration aB,It. Mark the
correct option regarding friction after long time
(friction
coefficient
betweeri
block and belt =µ). If:
•' ,,
,,
.
'
(A) ,Man swims at an angle· (P)
1127' from river flow
i
I
(B) ,Man swims right angle (Q)
(A) 'v 81;ik = 2v'a,r, and da,r,
'
.
'
=0
·
(C) ;varock. = 2va,r, and a 6,r, = µg (R)
·
·
I
'
(B) iv block = 2va,It and aae1, > µg 1 (Q)
'
'to river flow
, (P) zero
'
(D).'. JvBI;,ci:' = 21/~elt an_d aaelt < µi; , (S)
J, static friction
;co< f, < Al
Ji
(C) ·Man swims at an angle• (R) '
143° from river flow
limiting
(S)
friction
jK kinetic friction
3. The inclined s·urface is rough withµ = .!. For different
.
2
values of m and M, th~ system slides down or up the
plane or remains stationary. Match the appropriate
entries of column-1 with those of column-2.
5. A particle is moving on a straight line. It is initially at
rest.
v
= instantaneous velocity
P = instantaneous power
F = force
S = displacement
t = time
Mathe the possible expression of the quantities in
column-1 with the situation in column-2
' 3
=constant
=S
(P) P
•v 2 =t
'(Q) -p
oc
(C)'v 2 =S
1(R) F
=constant
(A) ,v
(B)
(D) v
=t
V
1
(S) F=V
,(T) p =t
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
MECHANfcs.n
[236-- .. --~-'- - - - - - · - - - - - - - - - - - - - ' ' - - - - - - 6. Match the column:
'' -
..
,
if sine
-----·-.,-~
···- - . -- -- -,
~
µ=O 11
!L(. ~··
F case
F
y
Mg
F = lOON, m = 7.5kg
.,Ic~t~fi!J.,\:
(A) .0 = 37°
, (P)
if is upwards
I
i
1
(B) 0 = 45°
(C)
.:eI = 53
(Q)
If is downwards
•
6
: (R)
If is static
, (S)
If is kinetic
..
I
J
'
(P) aA
(B) Just after spring X breaks
(Q) ;aB
I
(D) Just alter spring Z breaks'
··-y·~-- -- --·-----;
T
=0
=0
:ac = 0
!CR)
( C) Just after string Y breaks
.
Lift can move in y-axis as well as along x-axis. A ball of
mass m is attached to ceiling of lift with inetensible
light rope and box of mass mis placed against a wall as
shown in figure. Neglect friction everywhere.
f
•
(A) !Just after string W breaks
Ii (S) :aB
= Uc
9. In the situation shown, all surfaces
are frictionless and triangular
wedge is ' free to move. In
x
column-2, the direction of certain
vectors are. shown. Match the
a
direction
of
quantities
in '.mliirtm=nmmilim
Column-1 with possible vector in column-2.
m
!acceleration of
;block X relative to:
1
1ground
(A) ,In figu_r_e lift is moving along x-axisl (P) ;zero
!then.value ofT may be
I
I
.
I
(B) :Lift moving toward right along, (Q) > mg
!x-axis with decreasing speed, then:
;value of N may be
1
(C) :Lift is moving in upward direction! (R) < mg
:(y-axis) then value ofT may be
l
(D) ,Lift is ,moving in down;ward: (S)
= mg
:direction with constantvelocitythenl
value of T may be
8. In the diagram strings, springs and the pulley are light
and ideal. The system is in equilibrium with the strings
taut (T > 0), match the column. Masses are equal.
(B) iaccel.eration ofblockX 1' (Q)
relative to wedge
i
.
/
(C) :normal force by block! (R) : ·
;on wedge
e
· r \ :
I
I
I
'
l
·. :
:
'
'
:
-
:
'
I
(D) Inet force. on the wedge ' (S)
I
•
10. See the diagrams carefully in Column-1 and match
each with the obeying relation (S) in column-2. The
string . is massless, inextensible and pulley is
frictionless in each case. a=g/3, m= mass of block T =
tension in a given string, apulley = acceleration of
movable pulley in each case, acceleration due to
gravity is g.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
1 FORCE ANALYSIS
237'
·:. :.- =- :. . :- =- - :.:.:··=·-=·==-=--=-=-=-=:::::::::::::::::;;;;-------·
-- -.. -------·- -- --· -- - -- -----·-'
<1
.=-·::::-·:..:-·;...::-=·-:..:.·.:..:---=---.:...:·
Column-1
1----~-
Column,i!
,j_'
12. Column-! shows certain siruations and column-2
shows information about forces.
(A)
(A) '
(B)
( Q)
a
Situation
(P)
,Fi+ F + F
(Q)
Fi
(R)
F1
(S)
Fi_+ F2
is
centripetal force.
2
3
apulley ,; a
Front view of a car roundi\lg a
curve with constant speed
T
(B)
m
(C)
(R)
T>mg
is
friction
sta\ic
Passengers in a rotor not
,sliding relative to rotor wall
,cylindrical rotor is rotating
with constant angular velocity
about its symmetry axis.
T
m
a
(D)
(S)
Force
on
fixed
support
T1 > (3/ 2)mg
(C)
_,
F,
can be in
direction
opposite to that
shown in figure.
a
11. A block is placed on a
-c·
rough horizontal surface. A
F~~0<9~!
constant force F is acting
-rm-=-2-,/i-lkg_____ µ=1
!
on the block as shown in ,mn1111min~
'
the figure.
Column-! gives the magnirude of force F and
column-2 gives information about friction acting on
the block. Match the entries in column-! to all possible
entries in column-2.
. ..
, , Column-1
Cohi'mnc2.
." ' _.,.
(A)
lSN
(B) 20N
(C)
25N
(D) 30N
·Particle kept on rough surface
of a bowl, no relative motion
of particle in bowl, bowl has
constant angular velocity.
(D)
.
(P) Static friction
9
.
. -
Car moving on a banked road
with constant speed, no
·sideways skidding
1
(Q) Kinetic friction
(R) Zero friction
(T)
(S) Limiting friction
(T) Magnirude of friction is equal to
:magnirude of normal
www.puucho.com
('
~
0
Anurag Mishra Mechanics 1 with www.puucho.com
I 238
l)IECHI\NIC~:U
AN9WER9
·--·
_.,.,,,.-~,
. ~ev~:1: Qnly ()ne Al~~~n_ative is C~rrfi!ct~
1.
Cd)
9.
(c)
17.
Ca)
2.
25. , (b)
'Cc)
4.
lcb)
10. : Ca)
11.
,Ca)
I'
12. !Cb)
Ca)
19.
:(b)
20.
27.
Ca)
I
'
1
3.
Ca)
18.
26.
. Ca)
Cc).
5.
13.
'Ca)
21.
i
!Ca)
Cb)
6.
I
,Ca)
' (d)
7.
•:c~)
(b)
:(c)
15·_ ,(a)
16.
22. :Cb)
23 . 'Cb)
24 . ttd).
14.
•
28.
8.
I
;(]:,)
29. :Cd)
30.
Cb)
31. ,(b)
32. !(b)
.(b)
'
•(b)
38.
(d)
39. Il(d)
40.
i(c) \
33.
Cc)
34.
:(b)
35.
41.
(a)
42.
(b)
43.
·Cd)
44.
lee)
45. : (d)
46.
:ca)
47. i(c)
48.
:(d)
,,
49.
(c)
50.
Cb)
51.
Ca)
52.
!cb)
53.
' Ca)
54.
(d)
55. :(d)
56.
I'
1(a)
57.
(b)
58.
(d)
59.
Cc)
60.
!Ca)
61.
(c)
62.
64.
;Cb) I
I ::::' , j
(c)
36.
'
I
i
65.
Ca)
66.
Ca)
67.
Ca)
68.
i(d)
73.
(c)
74.
(a)
75.
Cc)
76.
(b)
37.
!
.(a)
63.
69. ,(b)
70. ,Cb)
71.
77.
1Cd)
78. '(d)
79.
86.
i
1(d)
80.
87. '.Cd)
: 88.
:ca)
96.
(a)
82.
(d)
83.
:Cc)
84.
;Cc)
85.
89.
(a)
' 90.
(a)
91.
(d)
92.
,Ca)
93,
:ca)
94. '(d)
95.
97.
·(a)
98.
CaJ
99.
'Cd)
100.
Cc)
101. 'Cb)
102. Ca)
103. :CcJ
1.
(b, d)
2.
(a, c)
3.
(c)
4.
7.
(a, c)
8.
(a, c)
9.
(a, b, c)
10. :(a, b)
13.
(a, d}
19 . . Cb,c)
14. (a, b)
15. '(a)
20. (a, b, c)
21.
,ca, b)
:ca, b).
. Ca, b)
'
5.
;Cb, d)
28. 'Cb dJ
'
29. :Cb)
I
j'."36. : (~; .c,d)t
33. (a, c, d)
34. iCa, c, d)
35. ·C~)
Cb, c)
' 38. (a, c)
39. :(a, b, c, d)
40. !Cc, dJ
41.
:cc)
(b, d)
' 45. :(a, b, c, d)
! 46. •(a, c, d)
47.
' Cb, c, d)
51. '.Cb, d)
' '
'
I
, 52.
;ca, d)
(a, b)
61. .(c)
67.
(a, b, d)
56. (a, b; d)
' 62. (a, c)
68. , (a, b, c, d)
57.
'.Cb)
, 63. ·Cb, d)
69. i(a, c)
58.
,Cc, dJ
:c
64. l _a, b)'
70.
lee, d)
73. :(a, c)
www.puucho.com
t,
'
.
· ._ - ...
I
r
1·1,;.
'
,
i(a, c, cl) ,
·i
: 54. 1Ca, b} : . .. ,
· I 60. iCb; c, dJ. )
;
; '
'
"
-.-
1'
; 66. !C~;c,.~l_
.
I
72 I(~, b; c);J
71. 'Ca, b, d)'
65.
1'
:
j 48. ;~, d):- ~,.
'
'
!Cb, c);, :< . '
I_
l 42.
i~,·cj
,cc, dJ
·'
' ~,
30. 1Ca, c)
l
1'
53. Ca, b)
59.
18. l (c)
j 24.
I
55.
'i
:Ca)
32. '(a, c)
50. , Cb, c, d)
I
23.
I
31. ,(a,d)
49 . . Cc, d)
.
'
CaJ.· I
12. Cb)
I
I
(c) " i
22. .(b, c)
27.
44.
ichJ
I. . '.
!Cc, d).
26. (c, d)
43. , (a, c, d)
!(b)
17.
Ca)
16.
t
Cd)
6.
11. ;Ca,c,d),
25. :(a, b, c)
37.
. I 72.
.(~) i'
81.
'
L.. .
• I
I
''Cc)
(a)
I
CcJ
F"
. 'I . .
;j
,I
,J
I
J
Anurag Mishra Mechanics 1 with www.puucho.com
ifORCfANAL!SIS _
'"; ,---~~,-,
_,
23~j
'";h-1$&',~,--,,, ------·-~-~.t)i4•o;.¼i0.\i :.;;; _____ ~
Level-3: Comprehension Based Problems ,:.~
---
-·-·
'
.
-·~-
- -~---
Passage-1:
1. (a)
2. (b)
3. (c)
4. (d)
2. (a)
3. (b)
4, (a)
5. (a)
6. (b)
7, (d)
8. (d)
2. (c)
3. (c)
4. (a)
5. (c)
6. (c)
7, (c)
8. (a)
2. (b)
3. Ca)
4, (b)
2. (b)
3. (a)
2. (b)
3. (a)
2. (b)
3. (d)
2. (a)
3. (a)
2. (b)
3. (c)
2. (b)
3. (b)
2. (a)
3. (d)
5. (c)
6. (b)
Passage-2:
1. (d)
Passage-3:
1. (d)
Passage-4:
1. (b)
Passage-5:
1. (d)
Passage-6:
1. (b)
l
4, (a)
Passage-7:
1. (b)
Passage-8:
1. (d)
Passage-9:
1. (c)
Passage-10:
1. (c)
Passage-11:
1. (b)
4. (c)
==;.,~~~l~110!!P,!! ;,;~-.;!~~~~.
1. A-S; B-P, S; C-Q; D-R
2. A-P; B-S; C-R; D-Q
4. A-P; B-R; C-Q
5. A-P, S; B-P, S; C-R,T; D-Q, R,T
6. A-P, S; B-P, R; C-Q, R
7. A-Q, S; B-Q, R, S; (C) P, Q, R, S; D-S
8. A-Q, R, S; B-S; C-P, S; D-P, S
9, A-Q; .B-P; C-R; D-S
10. A-Q, R, S; B-P, Q, R; C-P, Q, R, S; D-P, Q
11. A-P; B-P, S, T; C-P, Q,
12. A-P, Q; B-P, Q, S; C-P, Q, R; D-P, Q,R
www.puucho.com
3. A-S; B-Q; C-P; D-R
S; T; D-Q, R, T
Anurag Mishra Mechanics 1 with www.puucho.com
. -.. ......
1·
24_0
_____
'_.,.·..-.:
~·· ·_ ·-·, ~--=-' . .
',., """~,.. -, ........
-< _· .:::,:·:t:'.".':rt:t~..:;'~;:;~~~
• v,,,,,, ~,.,~ - · ~ ~ - - - " ' " -
c··· "---._·~"'". · ·...,.?·-.,,, :·:.:,::;i
· ..,-. ·ii=
N=T'+T
·
ir·+·1< . :·.·:.~!f·:·\::·}<.: ._:, •·c·
;;'; '·,.,:·N',;,if; _.-:.,'·:.s·:_:,, •. ~-'
L
11·0 IO Al . , . ·
e:ve O : . n y · ne temiilt1ve rs orrect
1. [d]
1
Cycle and cyclist moves with uniform velocity this
means that net force on this system is 0.
:. Inclined plane applied force' mg' vertically upwards
so that net force become 0.
2. [a]
Earth is applying a force of magnitude Sg downwards
, while falling.
_' :. From Newton's mrd Jaw block will apply a force
Sg N upwards,
3 .. [c]
Let
AB = diameter = D
(LACE= 90°) ·.
AC =Dcosa
Time to reach C = t
1
·2
-xg cosa xt
2
~
= Dcosa
t=f!
... (i)
.!xgxt' 2 =D
=>
t·=f!
i.(' a,J,g
gcosa
2
·4_ [b]
For equilibrium
~---~---~
T' =2T; >,.
Jl:4
"." .
·~
... (ii)
From eqn. (i) and (ii)
t = t'
t~
B
_j
[. ~: gs:J
Time to reach B = t'
=>
C
tTI
,£, •t,.
N = 600 - T
=>
,
... (i)
... (ii)
T = 600 = 150 N
4
5. [c]
~~ : 2 tN
Both blocks will moves together
21 = 3x a
a= 7m/s 2
=>
T=lx7=7N
Net external force on block A = 7 N
6. [a].
1
Tension at all points will be F
=> rope
is
not
moving,
acceleration will be 0
F-T=0
F=T
8.· [c]
T~20w
Reading in spring balance
=T/g
T -4
tension in thread
connected to spring
is
in
Here
system
equilibrium and T = 20 x g
.
20g 20
read mg=-=
g
[~L
www.puucho.com
V
•
Anurag Mishra Mechanics 1 with www.puucho.com
\;FO! RCE ANALYSIS ,~'J.,;
"'"'-'-··~-'----~=··'-----'-~'-"'..,,,~_....,,--9. [c]
Tension T = !Oxg
Also
T'=T=_lOxg
reading in both the spring
T' T
= - = - = 10kg
g
.
g
10 .. [a]
Magnitude of F1 and F2 may be equal or may not be
but their direction cannot be same because F1 is
accelerating and F2 is decelerating..
11. [a]
Deceleration of body A
~
dA=(MAg+f)
MA
Similarly dB =·MBg + f
MB
2
Now,
v = 0=u 2 +2ah
u = same for both bodies
u2
u2
~
hA=-=
2dA 2(g +_L__)
.
MA
u2
u2
hB=-=
2dB
15. [a]
T=mg
... (i)
[.
.
.
Mg= zr case
... (ii) '.·T~e.
\a
, , T...
e.
T .~i" . . . .
From eqn. (i} and (ii)
1 · . Mg ·: . •, mg
Mg= 2mgcose
I~--~-· . ,•-- ,, , ~---,
M = 2mcose
(as case< 1)
M<2m
16. [b]
kx=ma
k
a=-x
m
It is a straight line. Here
X is the compression in
block.
In our question X = X O - x
Since X is decreasing with ·x i.e., spring is coming to
natural form and X0 is initial compression.
17. [a]
Let
F= kv
mg-kv=ma
kv
a=g--
m
z(g+ ~B)
MA >MB
hA > hB
12. [b]
When cable is ·cut down then chamber will fall freely
under gravity, wedge and block both will also fall
freely under gravity.
:. acceleration of both will be g ,J.
:. block will remain at top of wedge
13. [a]
with time velocity will increase since initial
velocity was 0.
'a' is decreasing also after certain time
a=O
mg
v=k
when velocity = mg then a= 0 and. ball moves with
k
constant velocity.
18. [a]
While going upward a = F - mg·
m
h =.!..at
2
2
~
t1 =
T'=m 1g
T'
.
m2g--=m 2 a
/2h
v~
T'
2
m
t2
~
= ~
f-;,-
... (i)
2
While moving downward
, F+mg
a=---
.
-m 3 g
= m 3a
... (ii)
Form eqn. (i) and (ii)
as
a'> a
a=
t1 > t2
(m 2
-
m 3 )g
m,2 +m3
'
14. [c]
In both cases initial relative velocity of elevator = 0
and g,J.
· :. time will be same
Putting value of ci
( m m·)
T'=2m 2 l+ 3 - 2 g
. m3 +mz
www.puucho.com
(•
Anurag Mishra Mechanics 1 with www.puucho.com
2m 2 x2m 3
m1g =.
m2 +m3
4 . 1
1
23. [b]
g
Mg-T=Ma
T-mg-=ma
(M-m)g=(M+m)a
(M-m)g
a=
-=-+~1
m2
m3
19. [b]
mg-T=ma
3
given
Tmax
=4mg
a.
=f4
mm
(!"f + m)
r'
for minimum value of acceleration 'T' should
be max.
.
:r
l'
m(M+m)g
T =mg+----~
M+m
T= 2mMg
M+m
..'
.
~ ..
.n,g
T-2mMg =2mg
M
Total downward force on pulley =2 T
20. [a]
=4 mg.
24. [d]
Block B will come to rest when V!!locity of block A
velocity of block B
cc}
J~12t dt
. '..(i)
6t 2
Now let m kg sand is put
(M+m)g-B=(M+m)xf
.
6
cc}
t
... (ii)
25. [b]
From ~qn. (i) and (ii), m =~ M
5
21. [b]
·
~
cc}
...
...Fnet = 0
~
.
T2=W2+N2
=0, 0.5
=0.5sec
(,~:~;IlJ·
T
And get
N cose =W sine
N=Wtane
22. [b]
.
... (i)
a+ a'= Sg
4
T-100g=100a'
Solve eqn. (i), (ii) and (iii)
T+W+N=0
also, W and N are at right
angles
also
= 3t
T--600= 60a
~
...
=J~3dt .
t .:___ _ ~]
System is in equilibrium
:::::}
t
cc}
=
... (ii)
... (iii)
= l 9 soo ~ 1218 N
16
.
26. [a]
Tcose = mg
...
For. moving with constant velocity F = 0
...F+mg=0
_,
...
A is in x-direction
.
...
:. For net.force to be 0, F should be in +ve y-direction
......
Now
v xA_= mg
.
... (i)
j
rT-!8 -. -r:,~
. -· . -. - -.·.l
- •• .. r
net
, ·- :•
. A
[____~ ·.
cc}
''
,
B
..
. .
/.
m~2
T'-mg cose = --·= 0
r
T'= mgcose
vAsine = mg
. .
v=·~
A sine
;
For min. v, sine should be.maximum
cc}
v= mg
A
www.puucho.com
2
•
= cos 0 = ( ~ )
T'
3
T
4
2
... (ii)
Anurag Mishra Mechanics 1 with www.puucho.com
1
IS-'-'-~"------~-"""'-'-----~~''""·::1""·t1"':;c..?...
I.• ....
· _ _.....:..,_•..•i~-'~}~'-'--dv
y--_.,l--_.
a=-=0
I o ..
i...:..FO...:R...:C_E_AN...:A._LYS;.,.___
27. [a]
As shown in figure
vsine = Vo case
V = Vo Cote
28. [b]
'j
0;58
.-
I~_
ij.
, ,.
sEf
dt
F=0
=}
33. [c]
-+
-+
Finclin~d + Fgravity
-+
~ F inclined
=0
--+
-+
= - Fgravity = - Mg
34. [b]
Since the block is held held against a wall, the
coefficient of friction will be equal to the weight of the
block. Hence
µ=mg = (0.1 kg) (9.8 !IlS-2)
= 0.98N
I
'-'·--2Tcose = F
For any mass
Tsine=ma
T sine
.a=--=
35. [c]
N = Mg & Fp,non
F sine
m
2cosem
F
Fx
a= -tan8 = ----,===
2m
2m.J a 2 - x2
29. [d]
flmax =µMg
Clearly the magnitude of net force acting on the block .
from the horizontal surface is
F = ~f2 +N2 = ~f2 +M2g2
---
But=}
=}
=}
o,;;J,;;µMg
o,;;J2 ,;;µ2M2g2
M2g2,;; f2 +M2g'2,;; M2g2 +µ 2M2g2
Mg,;; ~f2 +M2g2 ,;; Mg~l+µ 2
36. [b]
The force constant is inversely proportional to length.
If the length 1 of the spring is cut into x and 1- x such
that
x=2(l-x)
then
x = 21/_3
From the inverse relation, we can write:
I
30. [b]
k1
l
l
3
-=-=-=-
dv
.mv dx = (ma - Toe)
J;mvdv= J;cma-kx)dx
k
~a,
I
0
kx2
2
gm/ and
2ma
k
21/3
37. [b]
Since the blocks cannot accelerate in horizontal
direction therefore the nom1al interaction force
ber,,reen the blocks as well as between 5 kg block and
the wall is F = 1000 N. Again both the blocks
accelerate downward with acceleration
s2
0=max-2
X=--
X
therefore the relative acceleration between the blocks
is zero. Hence the friction force between the blocks is
zero.
.
31. [b]
f, = µkN
38. [d]
=µdF+Mg)cos8
32. [b]
Slope of displacement-time graph gives v~locity which
is constant here
v = constant
If a block is released on ·an inclined plane of inclination
8 and having friction coefficientµ with the block then
the acceleration' a' of the block is (assuming tan 0 > µ)
a = ..!. (mg sin 8 - µmg cos8) = g (sin8 - µ cos8)
m
www.puucho.com
.
--
Anurag Mishra Mechanics 1 with www.puucho.com
.- d
Hence greater the value of µ lesser is the value of
acceleration irrespective of mass of the block.
39. [d]
A block begins to slide on an inclined plane ifµ = tan 9
irrespective of mass of the block, where µ = coefficient
of friction and' 9 = angle of inclined plane with
horizo1_1tal.
40. [c]
fl~ax =µN =
(¾)(lO~)
N~Fsin30
30
Fcos 30
3g
25,.J3
=--Newton
2
Since 1_let force (excluding friction) acting on the block
· is 20 N upwards therefore f = 20 N downwards.
41.
[af
·
·
For tile insect to be at equilibrium
Ffr = mg sina
or
µN = mg sin a
or
I+ (mg coscx) = mg sin a.
Hence, · · cota = 1/µ = 3.
42. [b]
.flmax = µN = (0.5)
(45) = 22.5 newton.
Since magnitude of net external force except friction is
25 N, therefore, .
f = 22.5 N
. lal=.25-22.5 = 1.25 m/s2.
and
2
43. [d]
Tension in the ·string, T = Mg
.
Ther~
two forces acting on the pulley. The force T
acting horizontally and the force (M + m) g acting
vertically· downward. The resultant of these force is
( ~CM +m) 2 +m 2 )(gl.
'are
44. [c]
If T is the tension in the string, then
T = mg
(for outer masses)
2f cos9 = :/2 mg
(for inner masses)
2(mg)cos0=:/2~ ·
or
cos9 = 'lj:/2.
=>
9=45°
45. [d]
Let m be the mas of the body.
F1 = mg sin9 + µmg cos9
... (i)
F2 +µmg cos9 = mg sin9
. . :(µ)
=> mg sin9 +µmg cos9 = 2(mg sin9-µmg cos9)
=>
3µ cos9 = sin 9
=>
9 = .tan-1 3µ
46. [a]
With respect to platform the initial velocity of the body
of mass mis 4 m/ s2 towards left and it starts retarding
at the rate of
a= 2m/s 2
Using
v 2 =u 2 +2as
we get:
0 2 = 4 2 + 2(-2)(s)
=>
s = 4meter.
47. [c]
If F1 & F2 are not zero then friction force on m1 acts
west wards & on m2 acts east wards.
For m; to be in equilibrium F1 - ·f = 0
For m2 to be in equilibrium
F2-f=O => F1=F2=f
But
f S 10 N. Hence F1 = F2 & F2 S 10 N
48. [d]
Consider A and Bas a system. There is no vertical force
in upward direction to support their weight.
Therefore, the system cannot' remain in· equilibrium.
49. [c]
Limiting force of friction between A and B is
F1 =µ1mAg=90N
Limiting force of friction between B and C is
F2 =µ 2(mA +mB)g = BON
Limiting force of friction between C and grou~ci is
F3= µ 3 (mA + mc)g = 60 N
As F is gradually increased the force of friction
between A and B will increase. When F = 60 N block A
will exert a horizontal force of 60 N on C. Hence C will
be on the point of motion. Hence the least value of Fis
60 N.
50. [b]
The acceleration of blocks down the
incline will be g sin 9. Horizontal
component of this acceleration is
.·,,,,:
Nl
aH = aco.s9 and vertical component
a, ';'asin9
f.',"'' \ •...
aH = acos9 = asin9cos9
\mg
and
av =asin 2 9
For body A:
Mg-N=ma,
.
2
or,
N = mg - mg sin 9 = mg cos 2 9
and,
µN 2' maH
µmg cos 2 9 2' mg sin9cos9
www.puucho.com
',
Anurag Mishra Mechanics 1 with www.puucho.com
µ;:: tan0
0=tan-1 (µ) _
or,
51. [a]
Horizontal acceleration of the system is
F
F
a=-----=-
2m+m+2m Sm
L_et N be the normal reaction of the system is
Tcos45°= ma
or,
2F
N=2ma=5
Now B will slide downwards of
T =.fi. ma
mg-Tcos45°= ma
mg-ma=ma
a= g/2
T.=mg_
µN;:: mBg
so,
µ(~);::mg
p;:: 5mg
2µ
52. [b]
Friction force between A and B(=µmg) will accelerate
B and retard A till slipping is stopped between the two
and since mass of both are equal acceleration of B =
retardation of A =µg
. .fj_
55. [d]
Extension in the spring= AB -R = 2R cos 30°-R
=(.J3-l)R
V1=Vo-µgt
v 2 =µgt
and
Hence the correct graph is B. When the slipping is
ceased the· common velocity of both blocks becomes
• v 0 /2.
53. [a]
Free body diagram (RB.D.) of the block (shown by a
dot) is. shown in figure.
For vertical equilibrium of the block,
.
F
N = mg +Fsin60°= ,J3g +-./32
So, spring force = kx
c-./3 + l)mg c-./3 ..:1)R = 2mg
R
Free body diagram of bead is :
N =(F + mg)cos30°
= (2mg + mg)
2
Tangential force
f+- N. -:-:~--·
·1
.
'·
.
•
I;f"
1·,
\.fert1ca1
··-· .: .
.
r: :
. F cos60'_ •
; : ,' : Horizontal
.!1'19 + Fsin 60° ,.. : •. "
2
· 56. [a]
For no motion, force ·of friction
•
.J3 = 3-./3 mg
... (i)
I
:
=F sin 30° - mg sin 30°
=(2mg + mg)si~30'= m{
I>
i
••.,. ~.,.
Tangential acceleration = g /2
57. [b]
When the inclination of the slant side ·reaches the
angle of friction, sand will betin to slid do~. So, for
maximum heightµ= tan0 = · .
I
J;::Fcos60°
µN;:: Fcos60°
or
F
g;::or
2
or
F ~ 2g
or
20 N
Therefore, maximum value of F is 20 N
R
or,
58. [d]
T=Mg
54. [d]
Just after the release B moves downwards and A
moves horizontally leftwards with same acceleration
say a.As shown in the free body diagram of both A and
.
N =60g-Tsin60°
Also,
T cos 60° =µN
Solving these three equations
M = 32.15 kg
B.
www.puucho.com
.
Anurag Mishra Mechanics 1 with www.puucho.com
· MECH~~
59. [c]
For m2 mass
= m2 a
T1 =T2 + f.
m2 g-T1
Also,
... (i)
. .. (ii)
... (iii)
T2 - m1g = m1 a
. From above three equations
T should be maximum
1000-SOxlO = SOa
a= 10m/sec2
1 . . 2
Now
10=-xlOxt
=>
64. [b]
(m 2 -m 1 )g·-J
a=~~-m1 +m2
f~
2
t = .Jz sec.
fmax = 0.2xN
1000- 0.2x SOg = SOa
.
400
a=-=18
60. [a] .
so
' d=-xl8x(v2)
1
r;:; 2
Now
=18
2
:. distance between man and ~lock= 20-18 = 2 m
65. [a]
fmax = lOx 0.2 = 2N
Initial force = SN > 2N
block will move with
acceleration
For equilibrium of
3mg sin37°= f + 2mg cos37°
=>
f=2m
For man, mg sin37°+f = ma
6m+2m=ma
a=8m/s 2
61. [c]
a=S-21:-fmax S-2t-2
1
dv
-=3-2t
dt
v=3t-t.2
S =~[2n:...l]
n
2
a
Sn+1 =-[2(n+l)-ll
2
so,
~
l
N'.
~2t)
'
10
(•: at t = 0, V = 0)
v=O
t_= 0,3sec
:. at t = 2 sec block is moving
:. Jmax will_ act i.e., frictional force acting = 2 N
[2n - l]
[2n + 1] '
Sn+l
,--B~-
=>
66. [a]
Small block m will fall vertically as no external force is
acting on it.
67. [a]
=>
N'=mgcose
N'= 2mg + mg cos 2 8
= 2mg+ mg= Smg
2
fmax
=µN:=
2
S~ xµ
= mg case X sine= mg
N=.J3mg
2
1
µ =-= 0.20
fmax =.J3x ~mg= ¾mg
s
63. [c]
fmax = SOg X 0.2 = 10g
T-SOg = SOa
For minimum time acceleration of
man should be maximum
·1·~·,,,N',,1
..
.· ·.· •.·, .i""'. T'-'
f
I.'.__.
50g.
:. block will not slide
3
Since
f= mg _ mg =mg <f,
2
2
max
68. [d]
As shown in the figure the forces F and Mg passes
through the center of mass and so they have zero
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
[ FORCE ANALYSIS
torque. But friction will produce clockwise torque. So
for rotational equilibrium the normal should produce
an anticlockwise torque.
69. [b]
3mg,
B
[2m
70. [b]
Zf sine= F
mv 2
R
N =--+mg sine
.
1:9
I
74. [a]
By conservation of energy
mgR sine= ~mv 2
2
2
.
mv
--=2mgsm0
Tcose = mA
1-----f
R
F
,:
N
So,
75. [c]
Tcos8
Tease
T
T.
=3mg sine
Ratio =3:2
From Q. No. 74,
2Tjsin9
8
'
mv 2
.
- - = 2mg sme
R
.---------------7
i
···-······t-,::ra--··
F
I
2tane=-·
mA
A-F( .Ja2 - x2
X
- 2m
)
.-. I
I
i '~ 'ff'I
·1
mgRsine = ~mv 2 = K
71. [d]
2
mv 2
N»=mg--R»
mv 2
ND =mg---
2
. e =KSo, mgsm
R
mv
Ne=mg+-Rv
Re
2
mv
NE=mg+-RE
(where N x stands for normal reaction .at point x of
path and Rx for radius of curvature at point x.)
=}
NE >Ne =}·NE is maximum.
·: RE <Re
T1
T2 -T1
=. m1ro 2 r
=m2ro
2
(~)
T2
from eqn. (i) and (ii) we get : T,
3K ·
.
mv 2
Nsme=-r
Ncose =mg
76. [b]
.
Ncose~
v·2
tan0 = -·-
' ~8- Ns,]a
rg
v2
mg
4
l0x 10
2
3 300
V = 100 X - = = 75· v =
4
4
77. [d]
When car just topples,
contact at B will be no
more i.e., N 8 =0
Moment about A is just
zero
2 mv 2
=} mgX-=--Xl
(for m1)
(form;,)
.
N =3mgsme=R
3
-=
72. [b]
·: Centrifugal force = (mass of body) x
(Angular velocity of frame from
which
body
is
observed)
x (distance of particle from axis of
rotation) = mro ~a
73. [c]
and
·
2
... (i)
r
=}
g x r =v 2
=}
10 x 10 =v 2
... (ii)
2m, +m2
= -'----"2m,
www.puucho.com
=}
'
s./3 m/ sec
.,
i;>
,,.,,,,.
-·
..
~~ Ne
•
"----2m____.
NA
,
~
1m
B
-
v =lOm/sec
•
mg
A
mv2
r
,, .
.
Anurag Mishra Mechanics 1 with www.puucho.com
[248~~-·
,C,/
·' ~. ,·
·,: ••
Friction force = µN = µ m~
=}
Retardation=~ = (
µ and r constant.
7?.
.
7}
=}
2
>
~lm
retardation oo, 2
.
mv 2
Nsm8=--
r
Hence
,·,:~·;s,r.,E~~~~·~s-q
•'
.
.
____.;.:_~.~----~
IT)Q
2
T sine= 41t P 2 mr
... (ii)
But
sin8 =· r/1
and cos8 = h/l
. ·: Particle is to be in contact with the table only
N ;:,, 0
... (iii)
when
From eqn. (i), (ii) and (iii) and using values of sin8
and cos8
[a]
and
,,
Ncos8=mg
v'
tan8=rg
83. [e]
:. For correct value of u car does not slip even if there
is no friction. But for any speed, other than v above
condition is not satisfied and the car slips. This is also
true for a stationary car.
'-·~· ',
80. [b]
·: When total acceleration vector makes 45° with
. · radial acceleration, then
a, = a, =_2t
... (i)
dv
·
a, =-=2t =} v=t 2
dt
v2 t4
and
a =-=... (ii)
'
R
R
. from eqn. (i) and (ii),
.
t4
2t = -
=} .
R.
=}
I
!·
v'
r=-gcos8
t 3 = 2R = 8
'
•
..~
•
: .. (i)
v'
v3
r = lOMO metre.'
=}
84. [e]
- t = 2sec.
l
~~-2 ··.
mgcos8-N = - -
V
R
2
N = 3mgcos8-2mg
... (i)
if N > 0, then ball will be in contact to lower surface
and if N < Oit will be in contact with upper surface.
= vr = ( a;zJ
Angle between a 0 et and v is same as
angle between a.et & a,· :
a
tan ex== ___f_
a,
a= tan-
1
N >0
,.,
.·_·,- 8~--_
·a·_·(2"'
e ,' . ,, .
r•
=}
cos 8 > ~
3
[from eqn. (i)],
net,
- ' ·
.
conservation
mgR(l - cos8) = -1 mv.2
dv
a dl
av
a2
a--------' dt 2./f. dt a.ff.
2·
, a,
Using
energy
between A & B.
= a./f_
2
=}
'
mv 2
mgcos8=-r
Let, when particle is at angular position 8, then
distance travelled = 1.
and·
v =30m/s' •
Jy/'
81. [a]
But
"
,. . '
" ·--·-•-....:...
.. '' .....--t,_..:,.;...._
L
(X' •
•
, .• ,;:v.. a,
85. [e]
(~)
Given
82. [d]
N+Tcos8= mg
... (i)
=}
www.puucho.com
dv v2
=
dt
R
V dv
t l
J-dt
uoV
oR
-
f2=
... (i)
Anurag Mishra Mechanics 1 with www.puucho.com
[ FORq ANALYSIS
=?
R (__!__· Uo
.!) =
t
... (ii)
Again from eqn. (i)
dv ds v 2
-·-=ds dt
R
2
V dV
'"' ds
J
=?
====>
""---;;- o·R
t
v
90. [a]
e-2•
~
I
r@II
2(!~) (!:)
1:2
=
2.roA=COc
N sine= mro 2 r
... (i)
--=-
R-h
=?
.
mv 2
T-mgcose = - r
h
g
h=R-_L
=?
mv 2
- - > µmg
r
as well as mv both increases. Hence graph will be
r
(d), (c) is not acceptable because at t = O, T ;t 0.
(b) is not acceptable because the variation is not
linear.
dm=(7)dx
Hence (a) and (b) are both true.
Again
If there is tangential acceleration then for slipping :
µmg=m
V
r
when mass is released from displaced position, 0 starts
decreasing and v starts increasing. As a result mg cos0
> .JµriTrue
V
=?
mv 2
=mgcos8+--
92. [a]
Considering an element oflength
dx at distance x from axis of
friction.
(02
88. [c]
Car slips if
T
2
g
=?
=?
At any angular position 8
=?
and
2
ro r
tan8=-
=20
91. [d]
=
N cos8 = mg
2
R=20cm
=?
co= l rad/s
At new position
R = 10 cm
So,
v =Rro =lOcm/s
And
acceleration = R 2ro = 10 cm/ s2
II . A•'
. 87. [d]
=?
-=20
Rro
)
I
26 = <I>
COA
COc
v2
... (iii)
Uo
=?
=(gcotet)
R
86. [a]
=?
x
=21t11X)
4it2112
= Uoe21t
=~ (1 -
(·:v
v2
=?
dv v 2
=:> v - = ds
R
- J
from (ii) and (iii),
=?
tanet= gx
=?
V
K
at x = L, T = 0 (T for tension) (T + dT)
2
2 )1/4
=.Jµri l - _a_
( µ 2g2
·: (d} is also true.
89. [a]
m
T
X
0
LL
2
= JdT= J-xro dx
= -nt2 ( x22
I
=?
T
=?
T = -mro (x2 -L2) = mro (L2 -x2)
2L
2L
2
2
=?
.
N
93. [a]
Let 'F' be force of friction in each case for stopping car
by applying brakes
sina
.!
N cos
mg
i
2
mv
2
N cosu = - X
and N sinu = mg
mv 2
~ F. r
(i.e., work done by friction should be
.
greater than kinetic energy)
2
p;,, mv
=?
2r
www.puucho.com
... (i)
Anurag Mishra Mechanics 1 with www.puucho.com
MECHANI~
2
For turning the car
F :c, mv
N = mg - mv
2
... (ii)
r
The required force is less in case of applying brakes.
94. [d] .
Direction of speed is changing so velocity is changing
=> acceleration and force are also changing.
95. [a].
Length of thread = I
T
Mass= m
0
N ~ 0 in limiting case
Tcose = mg
Tsin0 = mrOJ 2
rng •
rOJ 2
=>
tan0=--
g
99. [d]
At position B acceleration is
only vertical.
For particle 1.
Let velocity at B·= v
from energy conservation
1
2
-M1v =MgL1
2
M v2
Also atB
T1 -M 1g = -1-
=>
.J3:,; !xsin60°xOJ
=>
10
From eqn. (i) and (ii),
T1
OJ :c, 10
Conserving energy at points
AandP
Net acceleration at
2
mv
=-
cos0 =
R
~
=>
B = ~a
2
+ a 2 it 2
= a~l + 1t 2
102. [a]
At the highest point, we will have
Mg +N = mv 2 /r
= 2mgcos0
T = 3mg cos0
when particle is only horizontally accelerated at this
moment
=>
T cose = mg => 3 mg cos0 x cos0 = mg
=>
2
R
= 2gcos0
·
=> T-mgcos0
= 3Mg
ic2
Normal acceleration at B = .....!!.. = a 1t
2
2
R
= 3M1g
vf = a!tR
97. [a]
-
= M 1g + 2M1g
100. [c]
As car is moving in anticlockwise
direction and have
,
tangerrtial acceleration .(swell as radial acceleration
:. Friction component should be along tangential and
radial direction
101. [ b]
2
2
1tR
VB "' V0 + 2a X
When particle is at point A
acceleration g ,J,
Point , B acceleration is
towards ot
:. acceleration varies as
i.e.,
clockwise
v2
T1
T
M
m
1
-1 = -1= - = Tz M 2 2m 2
96. [a]
mgxRcos0=~mv
... (ii)
Similarly for particle 2 :
T2
=> t = 20sec
OJ= ext
... (i)
L,
For block to leave contact e :c, 60°
2
will be different
R
1 )
e =· cos-1( .J3
Hence, minimum the curvature r, the maximum is the
normal reaction.
103. [c]
a
Net acceleration
of the bob in
position B has two components.
->
98. [d]
Since earth is also rotating
Therefore, both will have different velocity w.r.t.
centre of earth as they are moving in different
directions
~-.,',
(i) an = radial acceleratioh''(towards
'
'~
.'
BA)
~--~,.~(
(ii) a, = tangential acceleration (perpendicular to BA)
Therefore, direction of
(c).
.
www.puucho.com
ais correctly shown in option
Anurag Mishra Mechanics 1 with www.puucho.com
FORCE ANALYSIS
·251
2. [a, c]
Particle is not accelerated as seen from both the
frames.
:cc; frames are not accelerated w.r.t. each other
:cc; either both are inertial or both are non-inertial but
moving with same acceleration.
3. [c]
.
·- -· ·
, F ~ F,
1;- -
j'
=;
- ··----
for
t <0
For t > 0 system accelerates
:cc;
F-F2 =ma>0
F2 <F
F1 -F > ~
F1 >F
4. [a, b]
N-mg=ma
N=mg+ma
N>mg
=;
.
F2~FI
F = F1 = F2
:cc;
N
4
T = S0xlO
4
T = 125 N
=;If
For t < 0 system is in equilibrium
I --
= 70g
N
N=30g
30g+T-30g=30a
3T- 50g = 20a
4T = 10a
50g
a=-70
T = 150g
7
If boy
applies no force on rope T
:cc; free fall will be there
"i
ir'
a
·•.·.
!L_ ·___•· .mg
if a is +ve
i.e.,
elevator speeds up while
going up or speeds down while going down.
8. [a, c]
Tease . ~
.
~mgl
-x2+y2=h2
-2xvx + 2yvy
=;
T = w.r.t. elevator
=0
V
=-y-
V
x
cosB
=;
Vy= Vx COS0
=;
ay = ax cosB
a=O
It can move only when with uniform
speed
= aring x cosB
T = 2m x ablock
S2 is accelerated w.r.t. S1
=; relative acceleration of the twci frames is not zero
:cc; minimum one of the frame is non-inertial
at least one of F1 and F2 * 0
7. [a, c]
For equilibrium
N+T=30g
3T-N=20g
4T = 50g.
N = 30g- SOg
4
... (i)
... (ii)
... (iii)
ablock
6. [d]
F1 = F2 = 0
I
··--··- - .. . - __ J
r
5. [b, d]
is not possible.
=0
2mg T cos0
= m X aring
From eqn. (i), (ii) and (iii)
2g cosB
and T =
Zmg
aring =
1+2cos2 8
1·+ 2cos 2 8
9. [a, b, c]
When block does not slip
mg =N coscx
N = mg seccx
Since block m does not slip on block 2m
:. both can be taken as on~ system
N'=3mg
Normal reaction on 2m by ground
Also from figure 1
www.puucho.com
= 3mg
Anurag Mishra Mechanics 1 with www.puucho.com
M~CHANl~~-1 .]
.
14, [a, b]
, ~-N:cosa
Nsina
'
For equilibrium on man, net force
on him should be zero.
Also as shown in figure
m=f, =µN
,
.',, .mg·
.·
N sin a= ma,
N = mgseca
~
a=gtana
And from figure 2
,,
F=3ma
'.F
~
F=3mxgtana
F= 3mgtana
10. [a, b]
H2+x2=y2
Differentiating
2xxvx = 2yxvy
•
N'
'
'
/'/'.,
.
3mg,
15. [a]
Since small block m is not moving
w.r.t. wedge
:. Both can be considered as a single
system
which
is
accelerated
horizontally
N=(M+m)g
16. [a].
... (i)
N
.
(M·~·;;g,
N= mg
cos0
Vear= y xvblock
17. [c, d]
X
~x2 +H2
1/x
= Vear = ----Vbtock
X
X
--a====
--block
V '
2
2
~
V x1
vx +H
Differentiating eqn. (i) again,
2
2
xax+vx=Yay+Vy
given
~
a;=acar=O
v2X -v2y
--~=ablock
y
v2H2
~
-(H_2_+_x_2_)~31~2 = ablock
In equilibrium acceleration of each block is zero.
~
kx 2 =(m1 +m 2 -m 3 )g
Just after .the string is burnt only
T = 0 and no other force is changed
~ acceleration of
m1 = m2 = m 3 is zero
kx 2 -m4 g ·
and acceleration of
m4 =
,m4
12. [b]
Let acceleration of pulley·is a
T
~":_i_
- :T }9s·
--- ,
. . . . . 50
.
T·
~
= [ (m1 + m 2 ) - (m 3 _+ m4 )Jg
m4
OON
'
,· .- .t_.1_t
l-,,a ,-
100 • ,:
•
T-50 = S(a+ a')
T-100 ='l0(d-a)
zr = soo
From eqn. (i), (ii) and (iii)
R
.
T
.
T
... (i)
... (ii)
... (iii)
d=ss
2
13. [a, d]
Clearly if'B is stationary and pulley moves then block ·
will rise.
VB =u+vA
aB = 0+aA
18. [c]
At first B will move downward and C towards tight
with a constant acceleration and·v, =at·
The· moment when B touches ground A will lift up.
Now as C is moving toward tight A will rise and string
between BC will become loose. Therefore block C
decelerates with a constant deceleration due to the
tension generated in string between A and C. At a
certain moment v c = 0 (after this A moves
downward). C again accelerates in the opposite
direction upto the moment A reaches the ground.
19. [b, c]
Just after BP is cut
.
For block A-no force has changed
:. acceleration of m1 = 0
for m 2 downward force is being reduced
:. m 2 will move upwards
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
I FORCE ANALYSIS .
'253]
20. [a, b, c]
at
.
Acce1eranon = m
i.e., a straight line passing
through origin
dv at
-=dt m
at 2
1. c=J:--+•tl
In
2mg-mg = ma1
a, =g
2mg-T = 2ma 2
znd case :
T-mg=ma 2
a2
v=2m
at
m
23. [a]
In 1st case:
Parabola
=!
3
In 3 rd case : mg +.mg -T = mag
t
V=-X-
T-mg = mag
2
t
.
v = acce 1eranon x2
21. [b, c]
+--'t;,,-,,~-+ T2 sln 8
AO--o
T2 •
mg
T2 cos a
-,=~I
24. [b, c]
2h
tane=d
d .
cose=zF'
d2 +h2
mg
T2 sine= mg
T2 case= mg
T1 sincx = T2 sine
T1 coscx = T2 case+ mg
From eqn. (i) and· (ii)
tane = 1
e = 45°
=}
T2 = ..f2.mg
From eqn. (iii), (iv) and (v) ·
... (i)
... (ii)
... (iii)
... (iv)
R
4
sine=
.
... (v)
T
In 2nd
ma=2mg-mg
a=g
case :
T-ing=ma'
2mg-T = 2ma'
a'=!
'
3
a-·a'= 2g
3
~ s i n e+Tsln8
T
slowly
T
Teas 8
-
,
. mg,
TC0s,8
.
.
= ...!1!!L_
mgRd2
T=-h +2xh
4
= mg.J d 2 + 4h 2 ·
2tancx = 1 = tane
22. [b, c]
In 1st case:
.
2sine
as man moves upward e becomes small
sine decreases
=} T increase
2
=}
- - __'!2_ga__ _...1
4
.as man moves
2Tsine = mg
tancx = mg
2mg
1
tan ex= Ti= ~ = mg-./s
smcx
T, ..f2. = T2 X -.J5
T
4h
... (vi)
25. [a, b, c]
[]
2t-2T=0xa
=}
T
lift
=t
{1-)2t . ~.T =_t
TMT
For
m1
to.
off.
10
T=mg=lO
So
t=lOsec
Similarly for 2 kg block aN = 20 sec
26. [c, d]
The acceleration of mass' m' and' M' along the inclined
plane is g sine so the contact force between them is
zero.
So mass 'm' will fall freely with acceleration g and
acceleration of wedge will be g sin_9.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
I 254· ·'.
MECHANICS-I
'.·
27. [a; b]
As discussed in question No. 26 contact force between
'm' and 'M' will be zero. So contact force between
wedge and inclined wedge will be Mg cos0.
28. [b, d]
· (i).Let the force F be applied on m1 and both the blocks ·
accelerate without any relative acceleration.
fm,';. =0.lxSxl0= SN
F-J=Sa
adding:
F = 15a
a=F/15
J=10(;s)
Hence
3f
F-=2
3
15
Fmax = - fmax =~newton
2
2
Hence (b) is correct and (c) is wrong.
(ii) Let the force F be applied on m2 and both the
blocks accelerate without any relative acceleration.
f =Sa
]
F-f=l0a
adding:
F = 15 a
=>
If~~
F
a=-
~
J=s(:S)
f = mg sin0 - mg
2
1
30° s; 0 + 45°; sin-
=µmgcos0,
.::
4
1
( --)
2../2
.
. -l (2../2
~) <
0 < 90°
+sm
__
Hence (b) is correct curve between 0 and friction force.
30_. [a, c]
N =Fcos0+Mg
... (i)
... (ii)
fmax =µN = µ(Fcos8+Mg)
To just push the block
Fsin0 = fmax
=> Fsin0=µ(Fcos0+Mg)
~ -j-
=>
=>
=>
=>
=>
F=
~
_ _ M9,c._
µMg
sin0-µ cos0
sin0-µ cos0 >. 0
tan0>µ
tan0 > tan(tan-1 µ)
0>tan-1 µ
Hence the block can be pushed forward only if
0 > tan- 1-µ.·
15
=>
Hence
J
=>
F=3f
=>
Fmax = 15 newton
Hence .Ca) is wrong and (d) is correct.
29. [b]
At 0 = 30°, mg sin0 = mg/2 which is equal and
opposite to external force. Hence at this moment
friction force is zero. As 0 starts increasing from 30°,
the mg·sin0 component starts increasing. Here
Again as 0 decreases sin0 .decreases while. cos 0
increases, therefore, sin 0 - µ cos0 decreases.
Hence
µMg
increases.
sin0-µcos0
31. [a, d]
The free body diagram of blocks A and B is as sho'IVll
below.
( mg sin_0 - m;). will be compensated by opposite
!! f2 q_N_"
c,
. T,
"
I
mg
'
•
j
I
,•
friction force until
· mg sin0 ~ mg < µmg cos0
2
sin0-.! < µ cos0
2
N1 =mg
N2=2mg+N 1 =3mg
sin0 -µ cos0 < .!
.
1
2
f2=µN,=µmg
J/=µN2 =3µmg
Tr= f2 =µmg .
F =fr+ f2 +T1 = Sµmg
sin0- cos0 < -
2
../2 (cos.'.:
sin0 . 4
sin.'.: cosa) < .!
4
2
1
sin(0+%) <
2
•lt
• -1
0 <4+sm
(
1 )
2../2
www.puucho.com
... (i)
... (ii)
... (iii)
... (iv)
...(v)
... (vi)
Anurag Mishra Mechanics 1 with www.puucho.com
ANALYSIS·
·· •• "·
{•2p5j'
I•-FORCE
- ~ - - - - - - - - - ' ~ - ' - - - - - - - - ' ; ; , , , , · , C C . . ' " " ' , e . '- - - - ~ --------'------~=~
L..
32. [a, e]
N = mgcos0+Fsin0
Also if mg sin0 = F cos0 then the
friction force acting on the block
is zero.
f,rati,lmaximum = µN =µ(mg cos0+F sin0)
(The maximum static friction that can act on a body
under a given solution is known as limiting friction
under the given conditions.)
33. [a, e, d]
When v ¢ 0 the acceleration is
ains, = _!_m (mg sin 0 - µmg cos0)
=g ( sin0v dv
dx
=>
X
2
cos0
)
f = (mn)a
Let
Hence for
0 :-;; t < (mA
kmB
J
g(2- X)
../2
2
: µmAg ----1~8
I
I
8~
fvdv = 2,i2
gm f<Z- x)dx
~[ 2x2
kt
(mA +mn)
(mA + mn)mAµg
And ,,or t > -~-~~=
kmn
kt-µmAg
a,
mA
µmAg
and
a2=-mn
36. [a, e, d]
a1
0
x:J:
v =Jg../2 m/s
Also it is clear that for x < 2 the body accelerates, at ·
x = 2 the acceleration is zero and for x > 2 the body
retards till it comes to rest.
34. [a, e, d]
Hence F = JJ 2 ·+FJ
82
(mA + ms )mAJJ9
km 6
J
2../2
v: =
+ mn)mAµg
,------!
a,
= __f__(2- x)
0
= a2 = - - - -
Frictionrorce on m, = µmg = ~ ?~x
x 10 = 20
Hence f>FN &F>f.
L
o•
Also
F·= JFJ + f 2 + 2.FNf-2.FNf
For m 2
+/ 2 -2.FNf + 2.FNf
I
rn, D .
- - - _ _ _20
_ J'
:
m1 =
,
20
= Sm/s 2
4
v=u-at
O=u-Sxl
~
When m1 stops slipping over m2 ,
= J(FN .:_ fl 2 +2.FNf
Also using
= (FN - f) +o
> (FN - f)
FN - f < f < FN + f
Hence
35. [e] ·
For a certain maximum value of
P both the blocks move without
any relative acceleration. In this
range
For
As obvious from diagram that the masses m2 and m3
will not move, and de-acceleration of
= (FN + f)-o
F = JFJ
=--,:... T
·---.
= J(FN + fl 2 -2.FNf
Again
p
Adding: P = (mA + mn)a ~ a
a
r--;:::==:=:;:;;,
I
1
:
,.~J
~;;/~m""'""'"'~
J
www.puucho.com
P'
)\
\ . I'1:·
u = Sm/s
m3g
= 2xl0 = 2 m/s2
.m1 +m 2 +m 3 4+4+2
Anurag Mishra Mechanics 1 with www.puucho.com
1256.
ME~HII~
37. [b, c]
I· . •-.~I
i
t
'"-~
~ i•.
;..~s~ .....~..J
For motion between AB
g sin0
(downward)
a=-2
Since tan a=µ > tan0 so block will came to state of
permanent rest and then required frictional force will
be mg sin 0.
42. [a, c, d]
since mas 'm' is at rest so riet force on it will be zero.
Also friction force will balance mg sin 0, so its value
will_ be equal to sin 0.
For M + m as a system net normal force will be
(M
For motion between BO
gsin0
(upwards)
a=-2
Also the velocity is increase from zero to maximum
value at B and then starts decreasing with same rate
and finally become zero at 0.
38. [a, c]
As discussed the above question velocity is maximum
at B and zero at 0.
39. [a, b, c, d]
For man and plank as a system
T+N=2mg
N=2m-T
T = µN = µ(2mg -T)
T=~g
+ m)g.
43. [a, c, d]
For equilibrium
T,,/3 ~ f =50
2
Taking torque about centre
T=f
2
r(,,/3 + )=50; T = ~ = f
2
2+,,/3
-~----~ ·-;· --~:1
,.•-·····
~~,11.••' ,.•
l+µ
Friction force =µN =T
No horizontal acts on man, so no friction force will act
on man. Also he is in equilibrium so net force acting on·
man is zero.
40. [c, d]
I
I
1
___________ _J
50
2+.J3
10W3 + 150 + 50
=
2+.J3
r;:; ·
n
N=5w~+--
= 10W3 + 200 = l00 N
2+,,/3
44. [b, d]
Speed is constant and tangential acceleration is zero.
45. [a, b, c, d]
For ABC part :
mv 2
N+F+mgcos0=--q
... (i)
;~1
r
m
µ=-M+m
s~,/3
_._ 100
'·!
Tension in the tread =mg
i~N
Assuming (M + m) as a single
M m) .· T·=. ipgj
mass unit, the only external 1fr
l
+ m)g ·. · I
force acting is rightwards so !~.- - -(M
- · __;,;_.;,_- '-"'j
frictional force T = mg
Normal force between wedge and block is zero, so no
frictional force acts between wedge and block.
At limiting condition for (M + ml system.
T=µ(M+m)g
=>
mg=µ(M+m)g
I
\
,...
··---'•
N
41. [c]
If
block
moves
up,
downward
acceleration
i
=g sin0 +µcos 0.
So using
v = u + at
0 = v 0 -(g sin0+µg cos0)
t=
Vo
g (sin0 + tanacos0)
I
.
j--~J.:
.. I
,:
' " - - - - - - - - - •E
and from energy conservation :
v q =.J2gr (1- cos0)
www.puucho.com
".(ii)
Anurag Mishra Mechanics 1 with www.puucho.com
--
FORCE ANALYSIS
~-----------------
-
·--·.
From eqn. (i) and (ii)
atG
ate
48. [b, d]
N =F+ 3mg case- 2mg
(a) at C
=>
=>
=>
--- --- ----- -------- ------------ - - - - - ---
·: e = 90°
=>
N =F-2mg
(c) For CDE part
mv 2
... (iii)
r
From energy conservation
v = ~2gr(l + case)
From eqn. (iii) and (iv)
N -F = 2mg +3mgcos8
=>
N_= F+ 2mg + 3mg case
8=90°
ForD
=>
N=2mg+F
... (iv)
3
46. [a, c, d]
Consider point P on circle of
motion
-->
-->
-->
. .
mv 2
s1n8 = - r
Ncos8= mg
v2
Hence
cane= rt
(a) carwillnotskidifv = 40km/hr.Hence (a) is na_se
(b) if V < 40km/hr
v2
" --+
=>
r1 < r
=> (b) is true
(c) If
v = 40km/hr
mv 2
l
N=--··
r sine
mv 2
=>
N>-r
Also,
Ncos8 = mg
=>
N>mg
= V i, ac j, (1) = rok_
a=ak
-->
m: ---~-- . ·-ii~J".'9 __ ,
~t:';->_:::,
g sine
C.
--+ --+
~-~,
r1=--
True, when cosB.= ~ which is possible between A and
V
-->
R=mg
1V
N -F - mg case= - -
--+
(·:8 = 90°)
(·:8 = 270°)
R=mg
~~~·· , ...
N=F-2mg
F=N+2mg
F :i! 2mg
(·: N :i! 0 for M is not to leave the track at C )
(b)
at B
e = 90°
(a) ro .L v => true
-->
-->
-->
-->
(d) is true, (c) is false.
(b) w .La => false
(c) w .La, => true
49. [c, d]
(d) v .L a, => false
47. [b, c, d]
t· - .. - - - ~ - - -
v2
=> (d) is correct
a, =-=g.
r
at A
atE
-->
N-mg = ma, cos8
N = mg+ ma,
(·: 8 = 0)
=mg+mg=2mg=2W
N = mg - ma,
(·: 8 =it)= 0
3
for G and CB = ~ and " respectively
2
2
257
·: Friction force = m anet
:. N = mg
=> (c) is correct.
www.puucho.com
Car will slip down
(:.sin8<1)
=> N
=
mg
case
Anurag Mishra Mechanics 1 with www.puucho.com
[.2ss:
MECHANI~-·
50. [b, c, d]
r = ~1
_-(½r 1;
=
= L,
v2
... (i)
aR = -
r
v
r; = mro L
2
=
... (ii)
ro 2L
T2 > 0, only when - - ;;, g
~
co e:
Jf
aR
m(ro~L -g)
~
~
~ ~
2
~
'-¼
= Ftangential v
= mkrt
i.e., (d) is true.
55. [a, b]
dk
ds
P=-=2asdt
dt
2
p
aT
52. [a, d]
Since the road is banked for
speed therefore,
· mv 2
mg sine= - - cos8
~
= 2aS X {2a S = (2a)'\'2 S2
v-;;;
here a car moving with
speed v will not slip even in
absence of friction.
It speed is less than or greater than v the above
condition is not satisfied and car will slip, this is even
true for stationary car (v=0).
53. [a, b]
M
dM=-dx
L
dF = dMro 2 x
ml/2
dv 2a
=-=-s
dt
m
21t
21t
21t
-=-=-
co,
002
Ol3
= 1:1:1
= r1ro1, Vz = r2C02, V3 = r30>3
~
ro 1 :ro 2 :ro 3
~
v1:vz:v3 =r1:r2:r3
a1 :a 2 :a 3 = r1cof;r2ro~:r3ro~-= r1 :r2 :r3
V1
f"
v=~s
56. [a, b, d]
T is same for all three particles
r
L
2
From eqn. (i) and (ii)
2 2
2
e,hado(~{!:)= ~ sec (~)_tan(~)
= M ro 2 xdx
... (i)
½mv =as ~
_Vshadow = V SeC2 ( ~ )
lM
r
Power = F- v
X=Rtane .
dx
2
de
V.badow =-=Rsec e-dt
dt
e=vt de=~
R 'dt R
dF
=-=--=Kt
r
acceleration net = ~ Kr + K.2t 4
51. [b, di
~
= rkt 2
-dv = "Kr = ar = constant
dt
2
v
Krt. 2
2
(b) is true
From eqn. (i) and (ii), T2 =
2
· v=-JK.rt
1
~
= 1/8Mro 2L
54. [a, b]
Tcos30°+T2 cos30°= mro 2( 1; L)
T1 +
2
f
(c) is true, (a) is false
~
F =!Mro 2L
at x =L/2,
l
T1 sin 30° = T2 sin 30°+mg
T1 =T2 +2mg
~
~
2
x
at
:::::::)
57. [b]
V=a..fs
dV
a
dS
= 2../s
~
VdV a 2
a=--=aS
2
a2
P=F·V=-~..fs
2
58. [c, d]
Consider a small section of ring
X
·Tcos8
. .~
. .-Tcos8i
2 2
F=--Ol X
2L
F=Oatx=O
T,-! 'W!"T _
+C
~
T sin8
C=O
r·sinB
Zf sine= dmrro 2
www.puucho.com
...(ii)
I
Anurag Mishra Mechanics 1 with www.puucho.com
2s91
FORCE ANALYSIS
Average force = m x average acceleration
2v 2 -./z
=mX--ltR
62. [a, c]
r = 0.5
ro = 0.4 rad/ sec
v = rro = 0.4x 0.5 = 0.2m/sec
a= rro 2 = 0.5 x (0.4) 2 = 0.5x 0.16
= o.sm/sec 2
sine-ease is small
zre = (2r8m)rro 2
T = mr 2ro 2 = constant
~
59. [b, c]
at time ·of slipping f = µmg
f cose = mar
.
mv 2
fsme=--
f
r
!2
= (mar)2
+(
r
r
m~2
(µmg)2 = (mar)2 + ( m~2
63. [b, d]
For collision
Position of A = Position of B
ltR + distance travelled by a = distance travelled by B
v4
itR + vt
µ2g2=a:+r2
v2
tan8=arr
Also,
60. [b, c, d]
1-·-
.-. they collide after time t =
--- ---- -----
a
'
'
VB
For angle = <jl
T-mgcose
-~-
l
v = 0 (extreme position)
mv 2
=- -
1
From energy conservation
.!mv 2 = mgl(cose- cos<jl)
(T1 -T2 ) sine= mg
~
T=mgcose+2mg(cose-cos<jl)
T = 3mg case- 2mg cos<jl
T = mg cose ~ e = <jl
T=mg
mg = 3mg case - 2mg cos <jl
+ 2cos<jl)]
e = cos
~
~
~
-i[(l
~
~
=
average
[
_,
. _,
Vfina1-V·
··a1
1mn
]
time elapsed
vi-vJ
ltR
2v
2v 2-./z
ltR
=--=---
T1
-
T2
x 2g
h
2L
= mg x -
... (i)
h
2g L SmgL
T1 +T2 =mxdx4x-x-=-h d
h
... (ii)
5
From eqn. (i) and (ii) T1 = mgL
h
3
T2 = 2mgL
if e is small case will be large
T = 3mg cose - 2mg cos <jl will be large
61. [c]
4
(T1 +T2)cose = m x dx
2
a
== V
64. [a, b]
! ______ mg____
~
~ 2:R
va =v=at=v+-J21taR
·
I
~
2·
t=~2:
1~ ,Potential energy= O
I
= vt +.!at 2
h
T1 :T2
~
= 5:3
Clearly
66. [b, c, d]
Al_~--v
i\J21
(_____ J
F
= -U O a sin ax ~ acceleration is not constant for
K-U 0 cosax=0
K=U 0 cosax=0
xmax~V=O
K=v 0 cosax=0
www.puucho.com
·
U+K=2U 0
Anurag Mishra Mechanics 1 with www.puucho.com
~
....
.
ax= 2:, at this point Fis -ve. So particle comes back.
2
Kmax = V0
cosax = v 0
67. [a, b, d]
"
68. [a, b, c, d]
(a)' A cork is fixed in a take, net normal force on
curved surface is zero but friction is not zero.
(b) µ,N is value of maximum static friction.
(c) Static friction opposes tendency 9f _relative
motion.
•
acceleration of 4kg block(a 4 )
r.
Ib -,
~
+.
;.--. '#
~-
,. . --,·
~S. -V/
.e '.
n
.
f'
•
'
-
5
'
'I
o- 20 = 15m/s 2 i
5o- 40 = 2.5m/s 2
i
4
,
..
'
velocity of upper block will be increased by kinetic
, friction.
:3--. --.~~-T~-~~he~~i~n ·B;;e~:Pr~bl~m~ .
- -
- · - - - - - - · -
--·-..··-~
- - - - - - · -
• • •,
••,
•••
,
•••
'
_ _
h
_____
-
-- . -- ·--·-~
.1
.
. "
' ·J'
, --- '<.. ~ _:',,,
--
i.e.,
µ .< tan0
both block will move
m 2g-T2 = m 2 a
T-m 1g =m1a
T2 = eT1
m2g - eT1
.
... (i)
.
m,f
3
.·
N · N'
f
1
!
,
:.
'
2
..
--~ 4oxo.a·
-
L.4~·~--6-'.,' ..... ····---····-
j,m,g ...... , ,
!
·,
f ,~;,=0.5•60•D.8.:. I
.,·, =,24 · .
'
I
'
I
;
,I
Let both blocks move together with acceleration 'a'
12-f, =2a
24+ Ji ~24= 4a
12= 6a
a= 2 m/sec 2.
=}
f 1 =12-2x2=4N
i.e.,
f, required = f1 max.
:. both will move together
T =·4m1g
3
1. [d]
tan37°= 0.75
Here
µ 1 =µ 2 = 0.80> tan37° i.e.,µ> tan0
=} sufficient friction is there at each surface
:. No block can move.
2, [a] ·
•I
I
· . ' ~ \ i·
= m2a
=}
I
2QxQ.8°.
20.~ o,s
... (iii)
a=--=3em1
·3
·2m 2g
T2 = m 2g - m 2 g I 3 = - •
3
T-mg =
'
:
'
... (ii)
eT1 - em 1g = emaa
(m 2 - em1 )g = (m 2 'f- em 1)a
em1g g
3. [c]
: _" ;z:·N:' f:f,~~=~:5·:;6~~8'1
'---7
2. [b]
4. [a]
See previous question solution
· 5. [a]
µ < tane for both block =} both will move
F1max = 6x 0.4= 6.4.N
t,m•• = 60x 0.Sx 0.5 = 25
Friction force = (2 + 4) g x· sin 0
= 6x lOx 0.6= 36N
=24
3. [b]
Here
Moving
_.:__..J
1. [a]
'
Passage-2
2
2
Passage~~
4. [d]
I
I
• (d)
a~celeration of 2kg block (a2 )
•
. A f l n i t i a l at rest rough!
1
:
Rough
!
.
1
. •
µ 1 =µ 2 = 0.5 < tan3r·~ 0.75
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
i,t___FORCE
ANALYSif·.- ;,
,,_.--~:_:_~,,L';_ _,;_ _ _ _ _ __
for motion to begin F = f,
2f1 <= f2max
2f, <= 2f2max
=>
F <= f2max =. 3
t = 6sec
=>
2. [c]
12- f, = 2a
24+!1 -24=4a
12= 6a
a= 2m/sec 2
=>
f 1 = 12-2x2 = SN
But
f 1mll = 6.4N
:. both block will not move together and frictional
force acting = 6.4 N
6. [b]
See previous solution
7. [d]
I
Here µ 1,µ 2 < tan0 i.e., 0.4 and
0.5 < tan37°= 0.75
.-. both block will move
Let they move together with
acceleration ' a'
J,max =I.6xO.5=8N
.)t:.
N
1, ••.I
'1'~
I
'
12
N
I
I
16
I
N
I
I
iI
f1
24
_ -
,f
Jtax = 6Ox 0.8 x 0.4 = 19.2
12-fi =2a
24-19.2+ J, = 4a
12+ 4.8 = 6a
a= 2x 0.8 = 2.8
Putting value of a = 2.8
F1 = 12-2x 2.8;,, 12- 5.6 = 6.4 < SN
:. both block will move together
8. [d]
Let both blocks move together with acceleration 'a'
a.st - J, = 2a
...
f1 -3=3a
... (ii)
when f 1 equals to 4 N then relative slipping just likely
to occur. Putting f, = 4 in eqn. (ii)
4-3 = 3a
a= 1/3
=>
Putting
a= 1/3 in eqn. (i)_
f, =4,
· 28
We get
t = - sec
co
3
3. [c]
'
.
Since the two blocks move together for t = 28/3'
at t =8 sec no relative slipping occurs
both can be treated as single body
F-3= Sa
SxO.5-3= Sa
a= 1/5
=>
=>
1
f 1 -3=3x5
3
f 1 = 3+- = 3.6N
5
4. [a]
See previous solution
28
t = 10sec > -sec
At
3
Passage-3
·I · _
relative slipping between the blocks occur
friction force = 4 N
1~·-,
1. [d]
for both blocks
·2
'
I
, t,
;
,
·
i.1: l
..
·I
20
5. [c]
'
Upto 6 sec there is no motion 6 ·:, t :, 28/3 blocks move
together with
, .:
L ..... ---- - -··· ..
N 1 = 20, fimax = 4N
f .......N •.
r-- ·- - - · ' l·
l_~--'_N,··'_f_),
)~
.
.
l
L __ ... _. -~Q
.c.
-
-
~---
5
. I
__
f'Jmax =: 4N
O.St - 3
a = ---
, I
;
;:::::J
I ·~ , a . s t 1
-~ ·
1
N2 = 50, f2max = 4OX 0.06 = 3N
at first will be no relative slipping between blocks
since f1max > f2max.
www.puucho.com
J•·· -•i•/.
'· ;J '.
~,' '\',
___,
-·
...j-.f~---
(from qu_estion 2)
28 ,;;. t reIanve
. s1·1ppmg
.. occurs
-
3
•O.St ~ 4
a=--2
Anurag Mishra Mechanics 1 with www.puucho.com
f 262
l\\ECHANICS,1 · j
6. [c]
4. [b]
Just before coming to rest, maximum frictional force
will be acting fmu = 2mg sine
Passage-5
1. [d]
Upto6sec
a=O
(No motion is there)
0.St -3
6:,; t < 28/3
5
28/3 :,; t ·relative slipping between blocks
4-3 = 3a
a = 1/3 = constant
f2
= 0.St
3. [a)
f2 =3N
f mu = mrro 2
When
8. [a].
upto 6sec
6:,; t:,; 28/3
f 1 = 0.St (Since a = 0)
... (i)
both blocks move together
0.St - f 1 = 2a
J1 -3=3a
~
ft= 1.St + 6
28
For
3
... (ii)
:,; t f 1 = 5 N (maximum)
Then at this moment sliding just occurs
mr(at ) 2 = µJ(mg) 2 + (mra) 2
0
Passage-6
1. [b]
2
... (iii)
ro, •0.6'
•
roo2 •
Passage-4
1. [b]
,
At any position x
\/
.7
r<ii2 X 0,8
-~
N.. ' · i
•• 8
8. 8 :
·"·12
° ,
N sine= 2 x o.2ro 2
a=gsine(l-x)
Ncose= 20
N -16 = 2 x rro 2 x 0.6
. =2x0.2x25x0.6
N-16= 6
fmax = 22x 0.3 = 6.6N
12- f = 2>< rro 2 x 0.8
12- f = 2x 0.2x 25x 0.8
f=14N
dx
V
X
0
0
f vdv = f g sine(l- x)dx
~
Also
v2 =gsine(x-x2)
2
2 .
vwillmaxwhen a=O
~
Vmu =
sine
x=l
.Jg
f required
For
u
= 0,
X=
= 4N
2. [b]
2. [b]
OJ
X=2m
3. [a]
At
x=2,
µ = 2k =: 2tane
fmax
,'
N
/
=µN = 2tanemgcose
'
mg cos 8
[9' .s-4),
• ,·
()
.
= 2mgsine
Ncose= 20
N sine= 2x 0.2xro 2
f=mgsine
www.puucho.com
.16
,,_c'-;,e20e,__..J
a= g sine-µg cose = g (sine-kxcose)
uau = g sine(l- x)
~
.
(3-d,~g\Jre)
I
Only frictional force
gives the required centripetal force as it is the only
force acting along the surface of rod
f = mrro 2 = mr(at) 2
(t :,; 6sec)
6:,,t
goo
2. [b]
7. [c]
for
N,
N= .J~N_f_+_N_i
a = - - - (blocks move together)
I
N,
N1 =mg
N 2 =mra
·
.
Anurag Mishra Mechanics 1 with www.puucho.com
. 0.4ro2
tan0=-20
2
CO =~ x tan 370 = 200 x~ = 600
0.4 .
4
4
16
(0
= 1oJ6 = ~ .[{, =
4
2
v
=>
2
3 300
=lO0x-=-=75
4
v=
2. [b]
[75
f2
1· ~
.Ncos.8~f_.·
- ...·•
I · . .a,· r-l.
3. [a]
For maximnm co frictional force f acts downwards.
2
f=µxN=-XN
:•
1·
. _e
mg .•.: · •:
3
2
2
... (i)
N -16 = 2 x rco x 0.6
f + 12 = 2 x rco 2 x 0.8
(
:~-~·:·/
12
1_
__
r
..
... (ii)
=3
(25
4. [a]
For minimum value of angular velocity' f should ·act
upwards (i.e., up the surface)
~,.,~-
T~).rro2. : 0;6
~f
12A1s
2
..... (i)
2
... (ii)
N -16= 2x 0.2xco x 0.6
... (iii)
f=~N
(0
2
.==> NS
25
=-
CO=
9
-
9
Passage-7
1.. [b]
N cos8
.
v
.... '.
= l0xlO
trolley velocity= -v/2
vre1 = 3v/2
Fmax =µgm => amax = µg
and Vmax(rel) = 3v/2·=·9m/s
Vmax = 6m/s
·
6x 3
t = v/a.= - - = 1.8s
10
= 10/3
= 6m/s Vy = 3m/s KE= KE 8 + KE 7
1 ·
2
KE; = - X 50 X (6)
v
•
~sin8
tane = rg
3
1. [b]
If velocity of girl w.r.t. ground = v,
•
v2
4
= lOm/sec
2. [a]
mv 2
Nsine=-r
Ncose = mg
2
v
Passage-11
...,,.,, 0,8
12- f = 2x 0.2xco x 0.8
3
A
,mg
When car just topples, contact at B will be no more i.e.,
NB =0
M9ment about A is just zero
2 mv 2
·=> mgx-=--xl
2·
r
=> gxr=v 2 => 10xl0=v 2
=>
..
•
>rriv2
r
1m
B
co= VJ rad/sec
=>
·NA
Ne
from eqn. (i) and (ii)
2
25
(0
10
3. [d]
2·
-N + 12 = 2x rco 2 x 0.8
3
=>
. + f cose = mv- = ::..:..:.c..::.::...:.
50x100) X cose
N sme
(N cose - f sine= mg = 50 x 10) sine
=>
f = 500cose- 500sine
= 500x 0.8-500x 0.6
f = 500x 0.2 = 100N
,.,,_ ~ .·.
_,
•.
4
5--J3 m/ sec
.
.·
mg
:•.
..'
8
2
and
..
www.puucho.com
KEr
1
=-
2
.
2
x lO0x (3)
KE= 1350J
Anurag Mishra Mechanics 1 with www.puucho.com
ni
-=5
M
3. [d]
2
3
Dr
·=>
Dr+Dg
Dr . 1
-=12 3
=>
1
3
--
4, (A)
=> Dr =4m
4. [c]
When vrel = 9m/s =;> vg = 6m/s
Max. retardation = µg = (10/3) m/ s 2
Minimum time= v/a = 6/(10/3) =18/10 = 9/Ss
(B)
5. [c]
Force.on trolley 7 lm/s 2 x.IOOkg = IOON
This also the force on girl by newton's 3 rd law
6. [b]
D•
2
-=Dr
l
=:,
D•
=>
Dr+Dg
Dg .2
2
=> - = ~
, 12 3
3
(C)
Dg = Bm (in earth frame)
= 9m/s
=>
vg = 6m/s
v 2 -u 2 =2as
2
=:, 6 - O= 2 x ax 8
v,el
=:-
5.
=:,
a= 36/16
9/4=2.25m/s 2
~ ~. c~~ ~ -
.
~ ~Ma~~h!!l!/;I!'!!t~le~.!~!!1~\
P = k = constant
F = k = constant
Fv= k
ma=k
mav=k
m(v:}
a_> µg
~
F.
mm.
A
a<µg~f,
a=ug~f,
a=0~f;,,o
3.
-mg sin37°-µmg cos37°= Mg
dv
mv-=k
ds
11. Minimum value of F regarding to move the block is
2. Maximum acceleration due to friction is µg
So
=k
= µmg =lx2Jzx10= 2 0N
~l + µ2
J'j,
For less than· 20 N friction will be static always. For
20N friction can be static as well as limiting since
F, max.= µ N and µ = 1 so F = N maximum value of F
regarding to move the block is
µmg= 2J'i.g = ;!SN
So for F = 25N friction can be static, as well as kinetic
and limiting.
·
F = 30N friction can be . zero when force is acting
nearly at 90°. ( s_ince F > mg).
12. No slipping any where.
Net force is centripetal as v = CO!Jstant.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
:; ,K;: ~~~, , r
·,, •
,c,.b·.L;:i:i~t,
\
WORK AND ENERGY/
WORK DONE
(i) Work Done by a Constant Force
The work done on a body by a constant force is the
product of the force in the direction of motion and the
magnitude of displacement.
__, __,
W =Fscos0 = F· s
Examples:
1. Consider a block sliding over a fixed horizontal
surface. The work done by the force of gravity and the
reaction of the surface will he zero, because force of
gravity and the reaction act perpendicular to the
displacement.
N
~
F sine
-2-+
direction
rmmrmmrrrlmr of motion
F cos B
mg
Fig. 3.2
s
Block displaced by an
external force
~ r=go•
5
W=O
F
0
0
s
F
F
Sign of work depends an angle
between force and displacement
Fig. 3.1
2. Consider a body moving in a
circle with constant speed. At
-->
. every point of the circular path,
goo s
the centripetal force and the 1
'
displacement
are
mutually
perpendicular (Fig. 3.3). So, the
work done by the centripetal
Fig.3.3
force is zero.
3. The tension in the string of a simple
pendulum is always perpendicular to
displacement. Which place along arc
(Fig. 3.4). So, work done by the ,
tension is zero.
t
. . :r
Case I :
When 0 = 90°, then W = Fscos90°= 0
So, work done by a force is zero if the body is
displaced in a direction perpendicular to the
direction of the force.
www.puucho.com
Fig. 3.4
Anurag Mishra Mechanics 1 with www.puucho.com
!266
______
MECHA~t~s-17
Work done by a force is zero if the body suffers
no displacement on the application of a force.
A person carrying a load on his head and standing at a
· given place does no work.
Work done by a force is said to be positive if the
applied force has a component in the direction of the
displacement.
Examples of Positive Work:
1. When a horse pulls a cart, .the force applied by horse
and the displacement of cart are in the same
direction.
2. When a body is lifted vertically, the lifting force and
the displacement act in the same direction during
lifting.
->
5
1
i__ _
Positive work
__ Fig. 3.5 · ____
i
_j
3. When a spring is stretched, by an external force both
the stretching the external force and the displacement
act in the same direction.
Work done by a force is said to be negative if the
applied force has component in a direction opposite
to that of the displacement.
Examples for Negative Work :
1. When brakes are applied to a moving vehicle, the
braking force and the displacement act in opposite
directions.
2. When a body is dragged along a rough surface, the
frictional force acts in a direction opposite to that of
the displacement.
.
3. When a body is lifted, gravitational force acts
vertically downwards while the displacement is in the
vertically upwards direction.
r-· ·- ------ ----- --·--- --- -- ---1
(a)
(b)
(c)
(d)
Fig. 3.7
In (a), 0 = 0°, cos0 = 1 (maximum value). So, work
done is maximum.
In (b), 0 < 90°, cos0 is positive. Therefore, W is
positive.
In (c), 0 = 90°, cos0 is zero.Wis zero.
In (d), 0 > 90°, cos8 is negative. W is negative.
1. Work is defined for an interval or displacement.
2. Work done by a force during a displacement is
independent of type of motion i.e., whether it moves
with constant velocity, constant acceleration or
retardation etc.
3. Work by a force is independent of time during a given
displacement. Work will be same for same
displacement whether the time taken is small or large.
4. When several forces act on a body, work done by a
force for a particular displacement is independent of
· other forces.
5. A real force is independent of reference frame.
Whereas displacement.depends on reference frame so
work done by a force is reference frame dependent.
Unit of Work
In SI i.e., International System of units, the unit of work
is joule (abbreviated as J).
One joule of work is said to be done when a force of one
newton displaces a body through one metre in its own
direction.
ljoule = 1 newton x 1 metre= 1 kg x 1 rn/s 2
= 1 kg ms-2
t
Work done by a force when an object is displaced
along a general path ..
.
->
5
system is scalar product of F and differential change in the
Negative work
b •..
·-·-
·-·· _
->
The differential work done dW by any force F on a
->
->
.
position vector dr of point of application of the force
Fig. 3.6 _____ _
Fig. 3.7 shows four situations in which a force acts on a
box while the box slides rightward a distance d across a
frictionless floor. The magnitudes of the forces are identical,
their orientations are as shown.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
IWORK AND ENERG'i'.
+
The work done on the system by the force
component Fx as the system moves from X; to x f is
the area under curve between X; and x f.
W =
Fxdx+ fyYJ Fydy + f:l Fzdz
J:I
'
'
'Fx
'
X
Flg,3.8
_ _ _ _ _ _ _ _ _ _J
........
= F-dr
dW
--+
A
A
Each term is the area under the curve of the graph of
that force versus the corresponding coordinate.
,.,_
dr =dxi+dyj+dzk
f
= fi
W
•
•
ke~«~il_C~
,.
(Fxi +Fyj +FzK)
-(dxi+dy j+dzk)
!An objecUsdisplaced;:;~::~:~ vector;t1 =·~2 i+3);
=ff Fx dx + ff Fydy +ff Fzdz
I
,
l
l
If force F is constant,
W = Fx
dx+Fy f; dy +F,f; dz
to ;t2 =(4):f 6fc)m under aforce•
the. work do[lej)y_this iorce.
J;
Solution : W
=Fx(Xf -X;)+Fy(yf -y;)+F,(zf -Z;)
....
........
= F-b.r
---+
or
W,0 , .1 =
-+
f.1-iF,-dr + f.
--+
l
---+
'1
f
-+
--+
F2 · dr + J.1 F, 3 -dr + ...
An object'is displaced from point A(2m, 3m, 4m) to il point
under
'ci
_constant
.force
IF~
i.~
IB(lm, 2tii1 3m)
3) +4 k)N. Find.the work done by this force in
this process.,
. .
, .
(2
Solution :
-+
--+
r,
(2 i+ 3j +4k)-(dxi+ dy j + dzk)
= [2x + 3y + 4z]Clm2m3mJ
(2m3m4ml
;(·.
······•·.·..
;
W=f} F-dr
3
·= J(lm2n\ ml
(2n\3m4mJ
perform any work., - - - - - - - - - , , - - ,
=-9J
...
~.
··- ......
__ Fig. 3.9 (a)
:2 (3x2dx + 2y dy) = [x3 + y 2J(f ~?
= 83J
ds along curved path; therefore centripetal force does not
..
....
= f
....
F,
2
'1
---+
f--+
1
f_; (3x i + 2.Y.i) - (dx i + dy j + dz k)
.
= F, + F2 + F3
f....
....
= Ji Ftotal • dr
=W1 +W2 +W3 + ...
Total work done on the system is work done by the total
force or algebraic, scalar sum of the work done by individual
forces.
+ When a particle moves along a curved path, the
work is done by tangential forces only.
W=fF,ds
Centripetal force is perpendicular to small displacement
I
---+
....
---+
Wtotal
= f! F- dr
r,
=
where b. r is displacement of system.
+ When more than one force acts on system,
F,ota1
1--+
F= (3x 2i + 2Y.i)N. Find
.(
Illustration for Work Done ·
(i) The Fig. 3.10 ~hows a smooth circular path ofradius
R in the vertical plane which is quarter of a circle. A block of
mass m is taken from position A to B under the action of a
constant force F that is always directed horizontally.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
'•'"
~-,·::,-,·,. tr·/:
i:.·
:,i
,r;-----··"·7--:·7
?-------~'--,- 13 ·
::·
I
- - - - ~ - - ~ - - - - --·-..
l
--~;.;/4...l
Fl
.
"
MECHANf($-1
'·1
.{:.~h~-~---~-~·~·~=-~·- ~ "-:,~.;-:; •, - - -
: ''
tRl,
I [
~'-'::+F
A
Fig,.3.10 (a)i ,,
...... = fFdscos0
WR= JF-ds
or
· W = J:Fdx
(dscos0 = dx)
or
W =FR
As the block moves from A to B, the displacement of the
block in the direction of force is equal to radius R. -
--:a·-~
,R
!: , •: ;l: ..
... ...
dW = F-ds = Fdscose
Thus
~
: :.:. r
dsj(! d~
=F(Rda)cos (~-%)
dX-,.,, ,
~--Fl~g.. 3.10
.
. or
(E]...:_ .,
dW = FR (~os.C: + sin.<:) da
-./2
Therefore, the work done by the constant force F is
/
. W'=FR
(ii) If the block is pulled by a
force F which i~ always tangential to
the surface. In this case force and
displacement are always parallel to
each other. The displacement of the
block in the direction of force is ~ R.
W = ~[
2
2
J;12 cos%da+ J:12 sin%da]
W = FR-.J2
Conservative and Non-conservative Forces
or
A conservative force is one whose work done is
independent of the path taken by the system, or whose work
done along a closed path is zero.
We can write the above definition in the mathematical
form
2
Thus, the work done by the force is_
w. ·= f.c1 ose.d path i. ds
W=F(:)=~FR
(ili) Block is pulled with a co,nstant force F which is
always directed towards the point B. In this case angle
wp~th1
where
= wp,th2
= ofar a c~nservative force·
for a conservative force
:
between force vector and displacement vector is varying
'~~-,
R
• <?'"fP- - -
,;
!
I
-'
·'
...
. In Fig. 3.12 (b) the a,ngle between F and d sis~- Block is
...
lnitlal,position and-final.Positidn, ·
'
Fig. 3.13
.·-· ---·- -·
-
·-"
If the work done by a force around a closed path is not
zero or if the work done by a force as a system moves
between two points depends on the path taken between two
points, then the force is called non-conservative. Work
www.puucho.com
at angle a from vertical. The magnitude of ds is R da. The
relation between 0 and a is
""'"
-
-----
-
<
Anurag Mishra Mechanics 1 with www.puucho.com
----·--------------- --"-----------------·~·-~
done by frictional force as a block is dragged along the
ground depends on the path, length, therefore it is a
non-conservative force.
r:liE -- -~ -.- . r-;i ~
[,_; C?~~~l;?z.;}~j 3 ~
r-·--·· ·. ·
-- ---- -·-···
--------------:--7
;A block is being pulled slowly along a frictionless incline1
;[Fig. 3E..3(a}J.
.
.. .• ..... __ . _ __ __ .
j
'
!
·, 2691
p
I WORK AND ENERGY
'
1-----~
:
F
A
C
S2
,
,
'
A
S3
W,0 ta1 is not equal to zero; therefore frictional force is
not conservative.
CONCEPT OF POTENTIAL ENERGY
When a conservative force acts on a system it changes
energy of system. Energy associated with conservative forces
is called potential energy. Only conservative forces have
potential energy functions associated with them. Since
conservative forces are function of position only, therefore
potential energy functions are functions of position of the
system.
Formally we can say that the work done , by a
conservative force on the particle is the negative of the
change in -potential energy of the particle.
J:~ dU = - J:: F(r) · dr for a conservative force
-c---- b
····>
c- ------
b. ------•
Urf -Uri =
Fig. 3E,3 (a)
( a) Show that the gravitationalforce is conservative . .
(b) Now.consider the incline tb be rough to show thatt~ej
ftjc_tion,alforce is non-conservative.:: .... _ _ . _· --~--_;-.J
+
Solution : (a) In Physics the phrase "slowly" implies
-+
that the body moves in equilibrium, i.e., L F,oral = 0. We
arbitrarily choose a triangular path ABC as shown in Fig.
3E.3 (a). Work done by the gravitational force can be
calculated separately along each of the paths AB, BC and CA.
W AB = mg xLcos<1>,
where <I>= 90°+a
W AB = -mgLsina = -mgh
W8 c = mg x bcos90°= O
Wrn = mg x hcos0°= mgh
= WAB + Wsc + WCA
= -mgh+ O+ mgh = 0
which proves that'the gravitational force is conservative.
Another important point to notice is that
w,otal
WAB +Wsc =WAc
i.e., if the block is taken to C along ...... -;
path A ~ B ~ C or along path A ~ C,
work done is same. Work done by
gravitational force does not depend on
i
path taken, it depends only on initial
!
and final positions.
(b) We
consider the closed
path A~ B~ A.
WAB = (µkmgcosa)Lcos180°
= -µkmgLcosa
W 8A = (µkmg cosa)L cos180°
= -µkmgLcosa
w,otal = wAB + WBA = -2µ kmgL cosa
Frictional force is always opposite to displacement,
therefore it is negative.
,
'
-f't F(r)·dr
r,:
Work done by a
conservative
force
does
not
express
absolute
value
of i
potential energy at a
point, . it
express (
change in potential
energy. We choose a
convenient reference
point and assign it zero ,
potential energy, then
we obtain
1
~U=
J:
Initial position
Fi~~ 3.11 _
dU
= U(r)-U(r0 ) = U(r)- 0
whereU(r0 ) is reference point energy.
In example 3, negative of work done by conservative
force mg is change in potential energy of block as it is
dragged from initial point to point,
u 1 ...,u, = -(-mgh) = +mgh
u 1 =mgh+u,
If we choose a reference level at the base of incline and
assign it zero value, we obtain
U =mgh+O
' --1-- -- ·~·--·--- ----- --,
.(b)j
www.puucho.com
•
1
•Path
. followed
by particle
.~
I
• Fin~I. ;
_/ pos1t10n,~
,'
z
F,lg. 3.15
---
Anurag Mishra Mechanics 1 with www.puucho.com
1210
MECHANICS-I
We can assign any value to potential energy ofreference
level, e.g., if we choose U; = 100 J, then
u1 = (mgh + 100)
Note that
Note that potential energy is either equal to negative of
work done by conservative force or it is equal to work done by
external agent.
CLASSICAL WORK-ENERGY THEOREM
l!.U=Ut -U;
= (mgh + 100) - 100
=mgh
Le., l!.U remains unchanged whatever be our reference
level.
+ A particle is moved from initial position to final
position under the influence of gravitational force.
I!.-;= (x1 -x;) i+(yj-y;)J+(z1 -z;)
.
....
I
t<
Consider a particle moving along a general curved path
under the influence of an external force F.
From Newton's second law,
dv
dv
F,=m-=mv... (1)
dt
d,
mv 2
F
=
... (2)
n
R
From eqn. (1),
F,ds
= mv dv
i------
F=-mg j
.... ....
Wene,gy = F · I!. r
1
,
;,• I-axis
I•
-~··'/ Finaf
'position ·
= -mg(y f
.k-~-F
-y;)
Gravitational Potential Energy (GPE),
Ug =-Wgravity=mg(yf -y;)
GPE increases if elevation of body increases, i.e.,
Yt >y;,
GPE decreases if elevation of body decreases, i.e.,
Yt <Y;
+ Elastic potential energy of spring: Consider
the block in the figure being pulled by an external
agent. The block is being pulled slowly, i.e., the
block is in equilibrium. ·
n-aXis
Fig.3,17
On integrating the above eqn. from initial pos1t1on
where velocity is v; to final position where velocity is v 1 .
.!.J 51 F,ds =
2 Si
2 -.!:mv 2 ..
f-" 1 mvdv = .!.mv
2 1 2
Vj
l
Note that only tangential forces perform work, so that
fs,S1 F,ds = w,otal = I!. KE
....
+
.
t::~:::::::;J,~.[1'::::F
Fig. 3.16
...,------------------~
....
external
['Fspring [=[ Fexternal [
-t
where
W
F,pring
·
~-
W external
,.-t
=-kxi,
"
Fexternal
= kxi
+
=Jx1 (-kxi),dxi=.!.kx 2 -.!.kxJ2
~
.
= ff (kx i) · dx i
2
'
2
+
'
= .!.kx,2 _.!_kx2
2
2 '
If we assume initial stretch in spring zero, Le.,
X;
The above equation is the classical work-energy
theorem, which states that work done by all the
forces acting on a particle is equal to change in
kinetic energy of the body. If work done on the
system is non-zero, energy is transferred ,to the
system. IfW10ta1 is positive, the kinetic energy of the
· system is increased', If w,otal is negative, the kinetic
energy of the system is decreased.
Note that CWE theorem is independent of nature of
forces acting on the system and the path followed
by the system.
The work done by all the forces can be classified
into two categories, namely, work done by the
conservative forces and work done by other forces.
W total = W,ons. + Wother = I!. KE
=0
From definition of potential energy,
w,pnns = -½kx2
= -I!. PE
W other = I!. KE - W cons. = I!. KE + tJ. PE
W cnns.
wextema1
= -lkx2
2 .
Spring force is conservative force, therefore negative of
work done by spring is change in potential energy of system
as the ~pring is stretched or compressed.
Elastic Potential Energy (EPE) ,U, = .!. kx 2
This is the general form of the work-energy theorem,
which states that work done by the "other forces" on the
system is equal to the sum of change in kinetic energy and
change in po[ential energy of the system.
2
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
rwoRK AND ENERGY
.271 I
Conservation of Mechanical Energy
In an isolated system of objects that interact only
through conservative forces, the total mechanical energy of
a system remains constant. The total mechanical energy of
the system is defined as the sum of kinetic energy and
potential energy.
Principle of conservation of energy states that
Concept: 1. Net work done by static friction is always
zero.
2. If block is placed on a conveyor belt that. is a'
accelerating, there is no displacement of block relative to belt.
Work done by static friction on block is positive whereas on '
belt it is negative.
E, =Ei
or
_,
K;+U,=K1 +u1
i\.KE+i\.PE= 0
D
or
Le., change in total mechanical energy of the system is
zero.
Concepts: 1. The total work done by all forces is
wtotal
= wnoncon. + wcons.
.
This total work equals the change in the system's kinetic!
• _energy
I'
Wnoncon. + Wcons. ~ liK:,
I
.
The net work done by conservative forces changes the 1'
system's potential energy
'
'Wnoncon. = Af( - Wcons. = M - (-i\. U) = L\. (K + U)
2. The net work done by the non-conservative forces in an,
isolated system equals the change in the system's total·
mechanical energy.
·
-s
~ ~ - • • fstatic
Fig. 3.19
Case II : When a force F
~
which is sufficiently large to
r-;:::-i
F
overcome friction, i.e., F > fmax· 1...
, ,___ L:_J __....,
Here, the work done by the
F > fmax: fmax = µsN
friction force is negative because
fk = µ,N
force
of
friction
and
Fig. 3.20
displacement are in opposite
direction. If F > µ ,N; friction is kinetic.
Case III : In Fig. 3.21 shown when the block A is
pulled with a force F. The friction force and displacement
are oppositely directed in case of block A while in case of B
they are in the same direction. The friction force does
negative work on block A and positive work on block B.
Wnoncon. = i\. (K + U).
3. The net work done by non-conservative forces during
any process equals the decrease in the system's internal,
_1,
~-s.
en erg)( U int.
Wno neon. =- li Uint·
4. The total energy of an isolated system-equal to the
kinetic energy 'of its particles, the potential energy'
associated with conservative forces acting within the,
system, and internal energy-is conserved.
.
Ll(K + U
+ U 10,) = 0.
Some Conceptual Points for Numerical Solving
(A) Work Done by Friction
Case I : Consider a block
=
5 0
placed on a fixed surface. When
F,
a block is pulled by a force F 1
- - • 1'
0
1
which
is
insufficient
to .,•,____ 1
overcome the friction, i.e.,
F < f max· Here, the work done
by the friction force is zero
Fig._3.18
because displacement of body is
zero.
Fig. 3.21
(B) Work Done by a Spring Force
(i) The work done by the spring force for a
displacement from X; to x I is given by
or
W
s
= -~k(x12 2
x2)
'
(ii) A spring stretched from its equilibrium position
by an external force.
Fspring and x are antiparallel, wsping < 0
Fexr and-x are parallel, Wexr > 0
l':::!!'.:'•;;;;~,
o
x<O
m
x>O
F5:pring
10 = natural length
L-[J-F,,,
During extension
sFsprin~F.,,..1
. Fig. 3.22
www.puucho.com
-s
Anurag Mishra Mechanics 1 with www.puucho.com
[272 -··-----·--·- -·
(iii) A spring is compressed from its equilibrium
position by an external force.
Fspring and s are antiparallel, Wspring < 0
-
F,xt and s are parallel, W,x, > 0
-·-------· ___ . --· . ,.,_
- .
!
Concept: The work done by spring force depends on the:
'initial and.final state of spring only. The net work done by the'
:spring force is zero for any path that returns to the initial
I
••
pos~uon. _ _
.
_ _
.
(C) Work Depends on the Frame of Reference
Displacement in a given time interval depends on the
velocity of the frame of reference used to measure the
displacement, hence the work also depends on the frame of
reference.
The Fig. 3.23 shows a cart moving with a constant
velocity v O along the positive x-axis. A block is pushed with a
constant force Fon cart. The reference frame x y' is attached
with the trolley and the frame zy is attached with the
·Y
y't
!
And, the work done with respect to ground, i.e., in the
zy frame is
W =F(X1 -X,)
Coordinates in the two reference frames are' related as
X1 =x 1 +x' 1 andX1 =x1 +x' 1
therefore,
W =F[(x1 -x1)+Cxi-x 1 )]
or
W =W'+F(X1 -X1)
where
X 1 - X 1 • is the displacement of the
trolley with respect to ground.
(D) Work due to Internal Forces (Friction)
Although resultant of internal forces for a system is
always zero but network due to internal forces for whole of
the system may or may not be zero.
,---,-•Fextemal -
m,
fstat!c --~~--'-,--,~fstatic [on block 2]
'ran black 1]
No friction
There is no slipping of block m1 on block m2
Fig. 3.24
x'
_DF
"'-=-~~-:;::=;~;,Ji!®!)
: (!@
V
In Fig. 3.24 Fext,rna! acts an block m 1 such that m 1 does
not slide on m 2 but has tendency to slide.
Displacement vector of m1 and m2 will be same and let it
bes.
Work due to friction on m1
0
i-x•.-:
:
I
:
'
I
a~----------.x
l
'
Xj
X1
W1
Initial position of block and cart
with respect to ground.
W2
ground. The Fig. 3.23 (a) shows the initial position of the
block in a coordinate system attached to cart represented by
x, y' coordinates and a coordinate system attached to
ground x,y.
And, the Fig. 3.23(b) shows the final positions of the
block.
y
y'
tf®!J
X'
~1-"o
?- x'f- x'1 ---!
'
force;
W = W 1 + W2 =
W=O
0
X1
-CJ, )s + f, (s)
Concept: If relative displacement of one body of system;'
w.r.t. other body along the direction of internal forces is zero,'
then total work due to internal forces is also zero.
'
I
Total work· performed by static force of friction for a
'system is always zero.
'
Let us see a case where work done by internal force is
:
I .~----,~·----cc·-+.x
I,
= (J,)s
If m1 and m2 are part of the same system then f, is
internal force and total work performed by this internal
not zero.
'
f
= -(J,)s
Work due to J, on m2
Fig. 3.23 (a)
I
m,
~
x,
r---r-_,...,. Fe,temal
F
m1
~
kineUc+--,-~-~-=,,=.. Fkinetic [on block 2]
Ian black 1!
m2
~
Final position of block and
cart with respect to ground.
Fig. 3.23 (b)
The work done by the constant force with respect to
cart, i.e., in the X y' frame is
W'=FC:x:1-X;)
... ...
Fig, 3.25
www.puucho.com
No friction
block m1 slips on m2, s 1 and 52 displacements
of block m1 and m2 respectively
Anurag Mishra Mechanics 1 with www.puucho.com
If m1 slides on m2 and s1 > s2 then s1 - s 2 is the
displacement of m1 w.r.t. to m2
W1 = work due to Fk on m1
W1 = -(Fk)s1
W 2 = work due to Fk on m2
W2 = -(Fk)s2
Network performed by Fk (internal forces) for system is:
W = W1 + W2 = -Fk(s 1 -s 2 )
W = -Fksrel not zero because there is some relative
displacement s,.1 a/?ng the direction Cl_f_ i-~~e~al ~°..r':'::. --·-
The
onj
Concept:
network performed by kinetic friction
the system is..· always negativ.e. and {t depends on relai:iv.e
displacement betwee1:_the contact sicrf(lces: _________ ·
(E) A block of mass m is
projected with an initial velocity v ~
towards a fixed spring of stiffness k
attached to the wall as shown in the
Fig. 3.26.
The work done by the -.-.---,-------~·cc·-:-spring force is negative beca~se
r::::l_ ·;"::oooo,' ·.
the force exerted by the spnng [ .
mmlmclm mmt1m~
is opposite to the displacement j .
of the block.
[' .
Fig. 3.27 . .
Let x 0 be the maximum ·-·-·-·-··-·-···--·
compression in the spring, then work done by the spring
!F~~,,~~ ·
·~1!
regain its natural length _'
W
= -~k(x12 2
x2)
'
.1
2
From work energy theorem, we get
W=MC=Kf -K,
mg
(b)
(a)
Fig, 3,29
.
---··-··" J
The work done by gravity is Wg = -mgl(l- cos8)
The work done by pseudo force is Fps = mal sin 8
The work done by tension is Wr = 0, because tension is
perpendicular re-displacement.
At the extreme position the velocity of the bob is zero.
Applying work-energy Theorem, we get
Wg + Wps + Wr = MC, at the extreme
position block is at rest
-mgl(l- cos8) + ma1sin8+ 0 = 0
2
or
g[2sin ~]~a[2sin~cosU.
8
or
a
tan--= '2 g
or
___
1
8=2tan-
(:)
--
·- ',
'
!
J'quilibrium. --··-·· •
-~kx~ = O-~mv5
2
2
R
When the spring gets completely : .-:-- : - :-Motio-n -~:.
compressed, then it begins to original / . •dx ~ : ·
length during this phase the spring :
F~
force F and the displacement dx of the l\\imm,miu\~
block are in the same direction. The I •
Fig. 3;2s ·· ·
work done by the Spring force is L. - - ·
.
.
.
. positive. Form work energy theorem we get
.• ,
1
2
W =-mv
2
.
lL---~ ~,--- _
..
2
x 0 .=v 0
·~a
Concept: In this case pendulum will execute oscillations!
of angular amplitude
.
,.
8.= tan·' a/g
As; · you have learned· earlier equilibrium;'. is at''
the·
1
I,a= tan- a/g that· this artgl~ is double io thaC'at
,,,
W=--kx 0
or
-m,·T.:
-·- · - - . ·----·- -.
l ·mq~i. ,./:
pendulum bob in equilibrium position was discussed. Now
we wish to find the maximum deflection 8 of the pendulum
from the vertical.
0
(F) Work Energy Theorem in a Non-inertial
Reference frame
A pendulum of mass m and length 1 is suspended from
the ceiling of a cart which has a constant acceleration a in
the horizontal direction· as shown in the Fig. 3.29. We have
previously solved a problem in which the deflection of
·
_ --·--'---···--·-·----···-·-
.l
(G) How to Apply Conservation of Energy Equation?
A block of mass m falls r·- .. [ffi]-- - · · - ~ ·· · --- -··
from a height hon a massless 1
spring of stiffness constant k. 1
Let
the
maximum 1
compression in the spring be · ·
, •
l
_hI .
_:
@;,i:l:··,:,~lt::.fi~.firence;
Lev.el ;
x. Weforassign
the energy
reference
level
potential
at ..c
k''
the position of maximum · . .'.. .
·
compression, reference level
. (a)
· ··. (b)
..,.'
can be assigned arb itrarily,
according to convenience.
i:..-~-----··'r_.:g._~- 3~. .
From work energy theorem, we get
W,pring + w,ravity = ·O
1
.
--kx 2 + mg(x+ h) = 0
2
or alternatively
www.puucho.com
I
·
_______ J
Anurag Mishra Mechanics 1 with www.puucho.com
, r,
Applying energy conservation theorem, we get
K,+U,=K 1 +'U1
O+·mg(h +x) = o+~kx 2
= 371 cos0 + 3mg sine- 2mg
= mgl 3,J2sin(e +
2.
x
2
T is maximum at e = 1t/ 4
·
Tmax = mg[3J2-2J
-2(7 )x-2(:g )h = 0
illustration 1.
After solving qua:;ti['c e q ~ ] e get
x=k
l+v1+mg
¾)- 2]
.
r-~
A plank of mass M and
["':0'."""" ,'"
~~~~~~!t~~:::c:~ ~
-
.
M
.
I
~:~thL
!~~oo!h!
.small block of mass m is _,__, F,g. 3 -31 {~)
...
projected wii:h a velocityv 0 as shown in the Fig. 3.31 (a); The
I
'
coefficient of friction between the block and the plank is µ,
plank is very long so that block eventually comes to rest on
: If block is released slowly it will stay at Xjj. If th~ ·block is
it.
1droppedfrom h ;= .Q then the defo'rmation in the ;pring is just
(i) Find the work done by the ftjctioh force on the block
;_d.Q'!b,le_th~,stah,;' deforrrmtiQf!._ . , ...ci.i "", ' ·' - · ' ·
during the period it slides on the plank. Is the Work
krix· a- · ...--.. 'l·e_:r-:;-7,"';:-,,.
positive or negative ?'
_
~~--- .l'DJ:>0.v-~~~
(ii) Calculate the work done on the plank during the same
period. Is the work positive or negative?
-~ pend11l~~ b~b of_;,_;;s ; ~ ;~j,~~1ed at rest. A constanj
(iii)
Also,
determine the net work done by friction. Is it
horizontal force F mg starts acting on it. Find :
'' .
positive or negative ?
!
(~) the:m~~um angular deflection of the string.· ·
Solution : PrQblem solving strategy:
:
(b) the. max/1:'um
J~11,jo: in the string: . Step 1. Apply Newton's law, determine acceleration of
;
- - ' ,l
\ .
..
blocks.
',
.
Step 2. Determine instantaneous-velocity of blocks·.
Step 3. When slipping slops blocks have common
· · velocity.
• i
I
The free body diagrams of the block and the plank are
'
shown
in the Fig. 3.3l(b).
,I Fig, 3E.4 {a)
r________ _ ,
f- .
Block :
a1 = - =µg
Solution, (a) Let at angular deflection e and let velocity
m
be v, from work energy theorem change in kinetic energy=
Instantaneous velocity,
v 1 = v O - µgt
work done by all forces
Plank:
. a,-= l_ = µmg
M M
~mv 2 = -mgl(l- case) +Fl sine
2
N, .
a1
= mgl [-1 + cose + sine]
+7 Motion
Maximum angular deflection v = 0 =}' e = 90°
f=.µmg
.
i
.... ·
....· -,
.
.~
Concept:Eqtiilibriumposition of block is at x0 , ; mog_
.
.
.
k.
=
l
p~.1~
/'
:
·,
.
--~----~---·---------~-~--'---'
1
-- ••
. !.
;
mg,
'.
,,' ' '
Fig. 3.31 {b)
/sin 8
.....-·
,)case+ · .... /
''
_µmgt
.-.
M
Finally, both the block and the plank start moving
together, i.e.,
v1 =v2
, F=.mg
Instantaneous velocity,
'
:· - mg
Fig, 3EA {b)
(b) Tension at angular deflection e
·
mv 2
T-mg (cose+sine) = - 1
=}
T = mg cose+ mg sine+ 2mg(-1 +case+ sine)
then
or
www.puucho.com
'I
. Vo
V2---
-µgt= µmgt
'M
t=
Mvo
(M+m)µg
Anurag Mishra Mechanics 1 with www.puucho.com
= 5; 4 t(t 2 dt) = 4 J;t 3 dt
mv 0
. .·.
and, the fi na1 common ve1oc1ty 1s v = - .
.
M+m
(i) The work- done by friction on the block is equal
to its change in kinetic energy; i.e.,
W1 =Kt -K,
1
2
1
= 41
w, = .!_
2
=
m(
mvo
m+M
dx
Ve1ocity v = -
)2 - .!_Mv~
2
·
:t·mM(M + 2m)v~
2
=
4
0
4
(24 '_ 04) = 16 J
_
Method II. From work-energy theorem, W = c. KE
3
X = t /3
2
=-mv --mv 0
2
2
or
~12 ~
(M +m) 2
The work done by friction on the block is negative.
(ii) The work done by frictio_n on the plank is given
=t
;
dt
2
•t = 0, Vi = 0 = 0
At
At
,t=2,vf =2 =4m/s
Work done
W =
2
1
2
2
2
m(vf -v;)
=!x2(4 2 -0)=16J
2
by
r
--··,..-~l:-,.,,.__
b.fi:;~a~lti~~¾ 6 ~
.. -
:A force of (3
.
··--- . . --•-
- · - - - - -··
• - - - - - 'I
i-1.sj)N acts on 5 kg'boczy. .The body is at al
:positionof(2 i-3])m~ndis travellingat4 ms·'.. .Th.eforce:
i
.
,
,.
,....
I
Jacts on the body until it is at .the position ( i + 5 j)
;Assuming no other force does work on' ilie bod); -the finali
!,,peed of the body.__________ . ··-·" · ·-· ____ ·---·
m:
Solution: Given, mass of the body= 5 kg
__,
F=3i-1.Sj
Force
The work done by friction on the plank is positive.
(iii) The net work done by friction is
W=W1 +W2
1 mM 2
=----Vo
--,
--+
s
·- -·-· ··--..
.
3
lposition x as a function of ti,;,e t is given by x
;
3•
.
-
'
A
,._
,.
2
A
= (-i + Bj) m
2
2
'i
I
A
(21- 3j)} m
W=F·S=-m(v -u)
= !..., ·xis in\
3
A
= {(i + Sj) -
From work energy principle
--+--+
1
-------···--:-:"--=-·. --·--- ----- '1
Under the action of force, 2. kg· body moyes such that it.ii
I'
3•
Now displacement
r~~Exam;:;;11~-Q~
tfr---- ~- .,"'7'7-1!L.::J~
--
-
a=si-lOj
2M+m
The net work done by friction is negative.
r- --
-
=>
v=Mm/s
!metre and tin•second: Calculate the \\'Ork.done by theforce in'
•the first 2second. _______' ... · --~ ____ ·__ _- ____ . _· I
k-==xcai~M~~1
·· .·
r-,'"'?"""'-=c-"\£1'6~~.);.'""Sc"c/~
Solution : Method-I: Based on basic expression for
work done
W = Fdx
¼spryng block system is placed on a rough horizontaz" surface;
iliaving" coefficient of friction· µ. Spring is given initial'
:elong.ation 3µm.·g. /k (where m = mass of block and k =spring!'
;constant)- and the block is. released from rest. For· the
[subsequent motion find:
'
f
t3 .
as
x=3
on differentiation, we get .
~
v=-=t dt
I
,
\
= t 2 dt
dv ·
a=-= 2t
dt
F ;... ma= 2(2 t) = 4 t
Resultant Force
Work done by force W =
fi dx
-----
-- ,.-
I
I
'
1
j
.
'
m. i
:~1//ll/HIIJJJl,l
dx
dx
-
91k
1 3µ
m
!; ~
i,o---oi
L - --- ".'9.·.~~-7 .
·- -
(a) initial acceleration of block
(b) maximwn compression in spring
,___ (c). maxirl_l_'!m spee~ gf th~. ~fock_<· •.
www.puucho.com
I
Anurag Mishra Mechanics 1 with www.puucho.com
MECHANl~S-1 \
276
Solution : (a) From Newton's second law,
. we get-· '
ma= 3µmg-µmg
=> a= 2µg ·.
(b) From work-energy principle_Wspring + Wfri,tion = 0
.
·1
2
2
'
.
.
'
-k(x0 -x1 )-µmg(x 1 + x 0 ) = 0
2 ,.
. :1.
•
.
or
x 0 - x 1 = 2µmg/k
=>
x, =µmg/k
· (c) Speed will be maximum where net force is zero
µmg= rox
=>
x ~ µmg/k (extension)
Now from work-energy principle, we have
'
2,
lk(µmg)
2
.
(3 2
'k
.,.
)-2+,tmg (µmg)=1mv 2
2
k ·
· (rn
.~ 2µ,g ~T
_Solvj~~ we ·get,
...-ir',
-1 2
Solving
~2gh
'
v=--=2m/s
.
7
.,'
'Ilvo blq_ck$ having md.sses 8 kg arid 16 kg are connecteq tp the
two erids of a light spring. The system is placed on a smooth
horizoht~l floor. An inextensible string aJ,o connects B with
ceiling as J/iown in figure at. the initial moment. Ihitially the
spring hq§,tts natural length. A constant horizontal force Fis
appliedtq,,(he·heaviei- block as_shown. What is the maximum
possible. vti!ue of F so .the lighter ,block doesn't loose. contact
~---iA
.,
withgrolf~1-
.v
1•
[where h is the distance fallen by block of mass m ]
4m
,
I
_A_
F
)In the figuf;~JiolJ'n, tM mass of th~ hanging block is m, while
that of th~ ffi(iitk resting on the floor is 3 m The floor is
horizontal aJ~ffi¢fioriliis and all pulleys ideal, The system is
initially held,sfafiqnary·, with tl/.e i1tclined thread making qn
angle a= ~Q'f,·wJQN~e. h,orizont,~L The, blocks, f~e now
released from rest ·a1frJ, allowed to ,move. The hanging block
lfalls through a h¢igljt_'(49/5) m' befor;e hitting thef/.oor. It is
!found that'·the va[ue'..of a becomes'-60°, when the hanging
block hits th'e flciot. :F(nd the speed with which the hanging
block hits the·jlocir. ";
...
Fig. 3E.9 (a)
Solution : Draw FBD of B to get extension in spring.
Instant when block B just looses contact with ground net '
force on it is zero. ,
·
kx-Tcosa = O·
·Tsin9+N-mg = 0
to riseN = 0
0
T
' ·'j--~·~--7
N
8
..
'
.
•
mg
J-kxl
'fig.3E,9 ( ~
',
;.
kx . ·a =mg
--sm
cos9
mg
80
X=--=--ktan9 kx(4/3)
~
,_,
Flg.3E.8
'----~~·-~:------.,,.-------'---------'
Solution: First apply string constraint velocities are
related as v 3mcos9 = 2vm
At the moment of strike
Vam = 2vm sec60°= 4Vm
Let·
Vm =v
=>
v 3m = 4v
From work energy theorem Wg,avity = t.KE, we get
.
=>
1
2
1
mgh=-mv +-(3m)(4v)
2
· 2
.
k
If spring has to just extend till this value at their
extension it should be at rest.
Now we apply work energy theorem to get
2
.
www.puucho.com
7
60
&=1kx2
2
F= 30N
Anurag Mishra Mechanics 1 with www.puucho.com
.. 277J
-----··· -- ---------·-(a) A 2 kg block situated on a smooth fixed incline is
·connected to a spring of negligible mass, with spring constant'
k = 100 Nm-1, via a frictionless pulley. The block is released.
from rest when _the spring is unstretched. How far does the·
block move down the incline before coming (momentarily) toj
rest? What is its acceleration at its lowest point?
;
(b) The experiment is repeated on a rough incline. If the block!
is observed to move 0.20 m down along the incline before. iti
comes to instantaneous rest, calculate the coefficient of kinetic'.
friction.
A ring of mass m = l kg can slide over a smooth vertical rod, A'
light string attached to the ring passing over a smooth fixed
pulley at a distance ofL = 0.7 mfrom the rod as shown in Fig.'
3E. ll (a). At the other end of the string mass M = 5 kg is
:attached, lying over a smooth fixed inclined plane of
inclination angle 37°. The ring is held in level with the pulley
and released. Det_ermine the velocity of ring (in m/s) when,the
·string makes an angle (a= 37°) with the horizontal.:
[sin 37° = 0.6]
k = 100 Nm·1
·· ...
37° ..
37°
Fig. 3E.11 (a)
Solution : Let xis the vertical distance covered by the
Fig. 3E.10
Solution : (a) At the extreme position blocks stops.
Applying work-energy theorem, we get
·
ring. Then x = L tan 37° = 0.7 x ~
4
L
-=L'.l
4
M = distance moved by block M
L'.l = Lsec37°-L = L(sec37°-l)
mgsin37°=_!ks 2
2
2 X 10 XS X ~
5
L
= _! X 100 X s2
2
on solving s = 0.24m
Acceleration at its lowest point
ks- mg sin 37°
a=--~--
v,
Fig. 3E.11 (b)
m
100x0.24-2xlOx~
=--------"'-5
2
a= 6m/s
(b)
or
=}
6m/s
2
v r = velocity of ring,
v M = velocity of the block at this instant
= 6.KE
mgssin37° + µmg cos37°xs = .!ks 2
Wg + W friction + Wspring
From work energy theorem, we get
W g,-a>ity = L'.KE
2
_!ks= µmg cos37°
2
3 1
4
2x lOx--- x l00xs = µ x 2x lOx5 2
5
gives s = 0.20 m
12- 50s
µ = 16
µ=s1
... (1)
5
2
mg sin37° _
Now, from constraint relation
4
VM = vr cos37°= -vr
-mgx+Mgt'.lsin37°+.!mv; +.!Mvt = 0
2
2
On solving eqns. (1) and (2), we get
v, = 0m/s.
... (2)
'
'From
what _minimum height h must the system be released
'when spring is. unstretched so that after perfectlv frtelastic:
,collision (e = 0) with ground, B may be lifted off ,he ground
1_(Spring _constant= _k) .
·
Solution : Just after collision with ground
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
- -
,- 278
•-
•w•-···--•
·--··
,
··- ~- •--•• ••
MECHANICS-I
~-------···-·------------------1
m
Et =-mv
2
A
1
I
2
From conservation of energy,
Initial state
E, =Et
1
2
= -mv
2
~---v = ~ 2gL(l - cos8)
mgh
or
(a)
Cb) Now we apply Newton's second law at the
lowermost point.
mv 2
LFy =T-mg=-L
(b)
At the moment
of lift off
Agaln when
spring Is relaxed
m fv
v>0
or
X
=mg+
= 2mg/k
~~~~g[~J
2m
(c)
Fig. 3E.12
(d)
;
-
--
14
)J>
•A boy throws a ball with initial velocity u at an angle of
projectiori e from a tawer of height H_ Neglecting air
·resistance, find ·
(a) hoiv high .above the building the ball rises, and
l(b) its speed just before _it hit§_ the_ground.
_!-_mv 2 + mgx+_!-_kx 2 = _!-_m(2gh) + 0+ 0
2
2
2
1
2
-mv 2 > 0
=> h > 4mg/k
.--f
i _-E:>f~-1'!'.Pj~=-~
A pendulum bob of mass m and length L is released from angle:
8 with
the vertical_ Find
(a) the speed of the bob at the bottom of the swing, and
(b) tension in th_e suing i;,t_that tj17J_e. ___ _
'
Solution: (a) We can apply conservation of energy to
bob-Earth system because gravitational force mg is
conservative and tension is always perpendicular to velocity,
it does not perform work We choose reference level at the
loWermost point, i.e., Ugi
-- --. -
0.
::;
-
--
Solution (a) Only gravitational force acts on the
ball, which is conservative; therefore we can apply
conservation of energy. We assign reference level at the top
of the building, i_e,, Ugi = 0. At the topmost point, the ball is
moving horizontally with velocity u cos8.
Initial total mechanical energy
E-
I
..
_
X
2
V
~
From conservation of energy, we have
·····-. -"· ---·R·· •f=ence
'k'.T'
level
2
= O+_!-_mu
2
Total mechanical energy at the topmost point
1
2
2
Et = mu cos 8 + mgH
L
·- ...
-····-
Fig, 3E,14
T
·--·
(a)
L
1
Applying COE,
•••••••••• -.. ... ......._.........
m, 2gL(l- cos8)
= mg(3_:_ 2cos8)
~
••. m
L
--- tx [extension]
· c:ptwi
=>
mv 2
T=mg+--
E, =Et
mg
_!-_mu 2
2
(b)
Fig, 31:.13
Initial total mechanical energy,
E, = mgh = mg(L-Lcos8)
Final total mechanical energy,
or
= .!_mu 2 cos2 8 + mgH
2
u 2 -u 2 cos 2 8
H=-----
2g
(b) If vis the speed of the ball at the ground,
1
www.puucho.com
2
Et =-mv -mgH
2
Anurag Mishra Mechanics 1 with www.puucho.com
279
: WORK AND_ENERGY
From conservation of energy, we have
E, =Ef
1
2
1
2
-mu =-mv -mgH
2
2
v=~ru~2-+-~-H-
m
L_:g_~f!.~J:?J~ fwl>
Consider an Atwood machine with both the masses at the
,same level as shown in Fig. 3E.15. Use the principle of
lconservation of energy to find
·
(a) speed of either of the masses as a function of its position
and
(b) the acceleration of either of the masses.
:A block of mass m hangs on a vertical spring. Initially the
spring is unstretched, it is now allowed tc fall from rest. Find
( a) the distance the block falls if the block is released slowly;
(b) the maximum distance the block falls before it begins to
move up,_
Solution : (a) When the block falls slowly, it comes to
rest at a distance y O, which is referred to as the equilibrium
position. From, condition of equilibrium,
LFY = ky O - mg = 0
•
Yo
Reference
l~Yf'
___________ Reference
level
•
y
Solution : (a) We choose reference level at the initial
position of masses,
E,=U,+KE,=0+0=0
1
2
E1 = m1gy + m 2g(-y) +-(m1 + m 2 )v
2
Mass m1 moves above the reference level, so its
potential energy is positive. Mass m2 moves below the
reference level, so its potential energy is negative.
From conservation of energy,
E, =Ef
= m1gy + m2g(-y) + ~(m1 + m 2)v 2
or
( :: : ::
mg
Yo=,:
= O+ O+ 0
Final total mechanical energy,
2
E1 = E f
1 2
0 = -mgy m +-ky
J2gy
= 2ay
2ay =
Lower extreme
From conservation of energy,
2
(b) Since acceleration of Atwood machine
constant, we can use the kinematic equation
v2 =
'
--- :------.
'
'
1 2
E1=-mgym+-ky +0
On solving for v, we obtain
or
...--------.
..--- ·---·.
mg
Equilibrium
position
(b) When the block is released suddenly, it
oscillates about the equilibrium position. Initially the speed
of the block increases then reaches maximum value and
then decreases to zero at the lowest position. In this
situation the block oscillates about the equilibrium position.
The block is released from rest, therefore its total
mechanical energy initially,
E,=Ug +U, +KE
2
v2
Upper extreme
(amplitude)
A
____!__ _____ _:_ . Equilibrium
:
:
position
'. ••••••• : A
Fig. 3E.16
Fig. 3E.15
V -
Xo
.L
.-----. ··..
--- .:' -------,·..
'
'------·.
,· -------.
is
or
2mg
Ym =-k-
At a general pointy, the total mechanical energy is
1
2
1 2
=-mv -mgy+-ky
(m2 - m1 J2gy
2
m1 + mz
www.puucho.com
2
Anurag Mishra Mechanics 1 with www.puucho.com
.~
~~atneJJ?.J 1a ~
··,
----
--~---- --------------
· · . M~CHAm.~i!_l
'.In Fig. 3E.17, the· mass m 2 te.sts on a.rough table. The mass
:ml is pushed. against the spring to 'which it is not attached.
!Force constant of the spring is k, coefficient of friction is J!k·
1
(a) Find the speed of the blocks after the spring is released
and m 2 ·has fallen a distance of,h.
'(b) If the spririg is attached to the block and it falls a distance
j h before coming to re.st, calculate• the coefficient offriction
µk.
--(
· Solution: (I) For individual bodies :
!-·-"-- · ~~.
,
,
L ---·-
'i
;5 -
.•
·,'
l
•
Initial mechanic_al energy
l
2
Energy dissipated by friction=µ km 1gh.
From work-energy theorem,
l>W = M
=
1 - [l2
2
kx2
(m1
v=
kx
-
.
+ m 2 )v 2 -
m1 ~ m2
(b) When the blocks comes to rest, the final kinetic
energy of the system is zero.
Initial total mechanical energy,
E;=O
1kh2 -m gh
E1 =2
-t
2
2
Sv
2
-t
-t
-t
-t
-t
-t
.
~~a:.me.!~~:~
I
iPind vefoci~·;f A and B when A---.-~--a-_b-~-u-~_to_t_o_u-ch-th--~-gr-ou_n_·d~,
IA/so verify that work done by tension 011 the whole system and ·
between A q1]d_B /§_~r_o~_..
·
.
. ·
.
i
.
I
•
/':I--;,-,
,!'
·•
:
.! ~1-------~
:
·;
2
A
i
g=10ms-2
37°
J
. Fig. 3E.19 (a)
L ___11J~_;"j_kg,_rijB.5._l_O/cg_,_.- - - - - ~ " -
1· ,
,..,
Solution:
...,
/v /=/u I
Net speed of block
2
2
2
vB = ~u +u -2u cos37°
= o-[½kh 2 -m 2gh]
µk=
1
+ x
V=2
= ~2u
1
m
2
fA·dsA+fB·dsB =ObecausedsA =dsB butfA =-fB
t, W = M
From work-energy theorem,
µkm 1gh
-t
m 2gh ]
2m 2 gh - 2µ km 1gh
.
Final total mechanical energy,
z1 X 10v
(II) We know work done by static friction will be zero'
because action-reaction will be in opposite direction but
displacement of contact point will be same. Thus
!N
Now on solving the above equation, we obtain
2
= 1Sx2+(-S)x2
Thus, 15 x 2 = (1/2) x 10v 2 + (1/2) x sv 2
When each block has moved through a distance h, the
final mechanical energy
1
2
Ef =.kE1 +Ug1 = (m1 +m 2 )v -m 2 gh
, µ km1gh
WB
v=2
!2 kx 2
E. =
!
1:W = 30 = Af<Esys =
-..------c--------'
Solution : (a) From work-energy theorem, the
energy dissipated by friction equals the change in
mechanical energy. We consider the table, blocks and spring
as a system; then w ext = 0. We assume initial potential
energy of the system to be zero.
WA= 5x2
'.
~1s
-·. ----·-- .,1
Fig. 3E.17
--- ___ _.h,,==------=--~ .....
·
i
l[§J
!
m 2g - -kh
2
m1g
www.puucho.com
2
-
iu
2
~ = v,Jf
•
Anurag Mishra Mechanics 1 with www.puucho.com
woiiiANii ENERGY- ·
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 281
_ '_'.j1
1
I
---------
-
-------
V
Fig. 3.32 (c)
Fig. JE.19 (b)
By energy conservation,
Decreasing in P.E. of block= Increasing in K.E. of wedge
+ Block
1
2
1
2
mgh=-mv +-mvn
2
2
1
2
1
2 2
mgh=-mv +-mv 2
2
5
·= -Nsin8x + Nxsin0cos8 = -Nsin0x
Nxsin0cos8
·Net work done by normal reaction
·Nsin0x - Nsinf!x = 0
__ _.. __ _
5 x lOx 2 = .!.10v 2 +.!. X 5 x ~v 2
2
2
5
12 ,
J
µ= 0.1
A
SxlOX2=-v2
•u=O
B
µ=O
v=/¥
Velocity of wedge= 5~ mis
Veloci~ of
block
Smooth
,Tsin a
I
I
X
2m
Fig. JE.19 (c)
'
. .I
II
=vl =v¥xi=Fl =2/¾m/s
Fig. JE.20 (a)
Concept: Work done by tension:
If.ind velocity of A, B_and C wl,en_ C has_d~q;_nded 2 f7!.
Tsin 8
(1) On wedge
W
Solution: Here work is done by kinetic friction
between A and B so it will not cancel out. But by tension on
A and C will cancel out.
= (I'-TcosS)x
'(2) On the block
'rcosS(x- xcosS)- TsinS xsin0
:
=TXcosS-Tx
-~T:
X
x sine
Net
W = Tx- 1xcos0 + Txcose -Tx = 0
By normal reaction between
'1
3.32 (a)
Fig. J_!:_.20J~)
AandB
N
'(1) On the wedge
,
NsinS-x
,(2) On the block
We =l00x2-Tx2
Total work = 100 x 2 - 1 x 2
N
X
99x 2 = .!. x 10v 2 +.!.x 1 xv 2
2
2
N cos8
Fig. 3.32 (b)
100
, Fig. JE.20 _(c):
- -·· !
:-N~in0(x_~_co~8)_+ (-Ncos_SxsinSJ.
WA =Tx2-lx2
.. I
v2
99 2 2
x x Jv=6m/sJ AandC
11
Concept: Thus except ten.sion, normal and static:
friction even if we write work because of action and reaction\
1_,m_ a _syste"! it is_ not necessary that total work will be_zero. '
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
I 282 ____ _
___ra_EC_HA_NICS·!J
Finding displacement of B
a8 = 0.5 ms-2 , u = 0, t
From A and C, t = 2/3
1 1 4 1
S=-X-X-=-ffi
2 2 9 9
_,_,
1
1
F-s=1X-=-x2xv
9
or
v8
1
=-
3
ms
Solution: f
1N
B
= 3Mg
µMg
4
3Mg
4
4
=> µ = 3
--=--
Fig. 3E.20 (d)
2
2
-1
Work done by friction force when chain completely slip
off the table.
df =µdMg
114 M
dW = dfx= µ-dxg
0
l
f
You can see that work done by kinetic friction on A and
B is not cancelling out completely.
WI =3Mg(x2)114 = 3Mg!
l
2 0
32
Concept: Work energy theorem is valid only from'
inertial frame of reference and we must try to stick to inertial;
frame while using it.
Now decreasing in PE
= increasing in KE
~ta'
:But if we observe from non-inertial frame the
write work done by pseudo forcefor dist. s
(a)
m
,
We should
·
,
'From ground frame
T-mg=ma
~'
T=m(g+a)
(b)af
W = [m(g + a)-mg] = mas
'
1
2
T;
mas=·- x mv
PE,-PEJ =
(-9Mg1)-(32
(c)
j
mg
ma
I
Fig; 3.33
J
2
+Wf
= I_Mv 2 + 3Mgl
2
32
7Mgl = I_Mv 2 + 3Mgl
32
2
32
I_Mv2 = 4Mgl
2
32
mgl
2
From frame oflift
·
T = m(g +a)
Total work= 0
'Total change irt KE = 0
Mgl)
2
1
2 mv
V
= I.-Jg[
2
As we have learnt from previous problem if some forces
are acting on a body
W1 +W2 + .... +Wn =KE1 -KE,
If some of them are conservative and others are
If chain starts slipping find
its KE when chain
becomes completely straight.
.
·
';A
1
, Fig. 3E.21 (a)
Solution:
w. = (KE 1 -
KE,)
But
w.=-(U1 -U;)
-Uf +U; = KE 1 -KE,
KE 1 +u1 = KE,+U;
Find U by using calculus emphasise that
1,
if we have tried to find work due to gravity Fig. 3E.21 (bl_
directly, then it would have been very
difficult as compared to the solution we are giving.
non-conservative, then for conservative forces we can write
P.E.
LW, + LWn, = KE f - KE;
L{-(Uf -U;)}+LWn, =KE1 -KE;
LWn, = KE! -KE, +L(Uf -U;)
Term on RHS is often called mechanical energy.
[gi~m;p}grn1~
""
--,:::;~.c.~
--·~'·--,
~
'.Find how much m will rise if 4 m falls awaY:
Blocks are at rest and in equilibrium.
L~-a.9.mel~ ~;>
Chain is on the verge of slipping, find the
'velocity of the chain, when it has slipped.
Solution: Applying WET on block of mass m
1
~g +W,p =Ki -K,
Let finally displacement of block from equilibrium is x.
Fig. 3E.22 (a)
-mg(Smg +x)+I.k(25m2g2)_I.kx2 =0
k
2
k2
2
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
I-WORK AND ENERGY
2831
2
2
lk X 2 +mgx----=
15m g
-
2
2k
in the direction of normal. For O < 0 < ~ N will never be zero
0
2
3mg
x=-k
. 1acement from IIllll
. . :al 1s
. --+-5mg 3mg = 8mg
D!Sp
k
k
k
~~q~el¢-:~
r----
2
as both mg cos0 and mv
are positive. Hence it will be
R
contact and will have circular motion.
Using work energy
mv 2 mu 2
-mg[R(l-cos0)] = - - -
2
- -.---- ~-- -
:Find velocit;y of ri11g .wh~I)_ §PJ:i_ng becomes..J!orizonta/.
2
v
v2
v2
2
2
= ~( m~
-
mgR(l- cos0)
J
= u 2 - 2gR(l- cos0)
= u 2 -2gR+2gRcos0
2
N
-
----
l0 =4m
Fig. 3E.24 (a)
mu2
-- -----
= mg cos0+---2mg + 2mg cos0
R
Solution: m = 10 kg,
k= 400N/m
Natural length of spring= 4 m
Decreasing in PE = Increasing
N
1
-kxl+mgh =~mv
2
2
2
Normal will not become zero.
If we want to find minimum
value to reach B there is no need to
see the equation of normal all that
matters is speed.
2
.!x400xl 2 +10xl0x3=.!x10v 2
2
2
·
200 + 300 = 5v 2
5v 2 = 500
V
2
= m[u
-2gR +3gRcos0]
R
.
0<0<~
in KE
1
m(u - 2gR + 2gR cos0)
= mg cos 0 + ~
--~R-~--
At0=~
2
2
0 = u -2gR+ 2gR(O)
= ,/100 = 10 m/S
C
y;::;Q: .
B
i
!ii) ;
I
A
Fig. 3.34 (~-- _!
= mg (R)
u = .J2gR
Case I: u = .J2gR it will just reach B.
Motion: A~ B~ A~D~ A ~B
At B, N = 0 but it will not loose. constant.
Case II: u < .J2gR
The body will not reach B but its velocity will become
zero before B.
e.g., Letu = .,/iii
0 = gR-2gR+ 2gRcos0
1
cos0 = -
u = .J4gR. This is wrong. Why?
At any 0 with yertical.
:. At 60° the body will stop. The body will not remain
stationary as its tangential acceleration will not be zero.
VERTICAL CIRCULAR MOTION
Consider a block projected on
inside of a vertical circular track.
What is the minimum speed to reach
BandC.
·
1 ' 2
(B)-mu = mg(R)
2
.
A
u = .J2gR
Solve for (C) like this
1
(C)-mu 2
2
C
L__ Fig. a.aa <at_
2
· mv 2
what if0 > ~
N = mg cos0 = - -
2
R
mu 2 ·
N = mg cos0 = - R
This equation is valid through out for
0 >~as cos0will go.negative and component ofmg will act
2
.
Here the normal will become zero before velocity. TWs is
why .J4gR was wrong as we were considering speed and not
normal where as to reach C it is necessary that 'N' does not
become zero.
Find minimum speed to reach C.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
MECHANICS-I
284
Q=u 2 -SgR
,.case III:
[0
= It]
a·
u = ,JsgR
u = ,JsgR ·
2
v ·= SgR-2gR-2gR=gR
Minimum possible val;e of 'N' and 'v' is at 'C'.
V
2
fj :'~""'
Q v=O;T=-ve
A
= ..Jiii
Fig. 3.37 (b)' ,
,
mv . alid .
As mg=--.ISV
A~ P~Q~P~ A~ P'~ Q'
Case W: u = ,J4gR
v = 0, T ~ -ve
.
R
So the body will continue moving m
circular motion.
u = ,JSgR implies the body has just
completed circular motion.
£!~· 3;34'(bl
Note: We check for' 1t' as cone has maximum negative
,..------~----'===:::;
·
value. If N is not O at this point thenfor all 0 < 1t the normal
will never be zero.
,___
I
----------------'
~ase W: u > ,JsgR
The body will freely move in a circle and 'N' will never
be zero.
C~se V: ,JsgR u > ,J2gR.
·Th~ norm at will become zero some
__ .
,_,,
where between B and C. At this point [ ~ c
•.'. ·,;'.,i
v ¢ 0. It will leave circular motion and
--······:·-.-~- <'. ,_'
will become projectile because
..N=O • ,,' .
symmetry will no more be there as in
B'
the next instant velocity will decrease
Fig, 3.35
further for which N should be negative
which is not possible and so it will leave
circular motion and will have projectile
o.
·6
motion.
For a mass tied by a, string about 0.
Here instead· of normal 'Tension' is the
,r·
..•
.
[
worrying factor.
.
T = 0 ~ String is slack and ' m' will
_Fig. 3.36
leave circular motion.
All previous cases are valid similarly.
B
Consider a pendulum bob
connected with a rod.
Rod (Rigid)
Case I: u < ,J2gR - Pendulum
Case II: u = ,J2gR will reach B and
m
come back.
Fig. 3.37 (a) ,
>
'
.
-~
Case m: ,J4gR > u > ,J2gR. The body will continue
moving in circular motion as tension of a rod can go
negative which is allowed as then the rod instead of pulling
the body will push it.
·
The body will stop· at the top.
Case V: u > ,J4gR ·Forever will do circular motion.
Concept~:.·.
Case J: !]vb < ,J2gr, th~n the velocity vanishes before
tension T, then the particle will oscillate belo,w the horizontal
diameter without 'leaving the circular path, but the particle
will not rise upto the horizontal level of 0.
'
Case II : If vb = ,J2gr , t~e velocity v and tension ·T
vanish together,'then the particle will rise upto the horizontal
level of fixed .point O and_ will os~illate along semicircle..
Case III ;,.[(vb > ,Jsgr I then tension as well as velocity
does not vanish ;,ven at the hig/ie$t point and" the pa,;ticle
completes ci~cl~ successfully.
·
'
Case W.: tfvb = ,Jsgr, then velocity will not vanish at
the highestpoini where as the tension will just become zero in
this case the string will not slack due to velocity particle moves
orward and due to string constraint circle · just gets
completed.
·
Case V : ,J 2gr < vb < ,J Sgr; then tension vanishes at
so7:1e point; i.. e.,. the strilJ/r b,ecom~s slackened but thf ve!oc.ity
being not zero,,,
.
.,
.· .
., . ,_
l'--:'.-::---------;::::=::;:- .
Where
v, = ,Jrg cos~
--
Velocity of projection at the bottom, is given by·
vb= ,Jgr(2+ 3cos~)
Motion on the Outer Surface of a Fixed Smooth
Sphere
A small particle is released
an outer surface of sphere
outer surface of a smooth
sphere, starting from rest at
the highest point.
Which force makes particle
move along circle ? What is
role of normal equation ?
\____
·
_!:!g. 3.3B
Resultant force towards ·
centre is
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
------ ------------2
mgcosP-N
mv
=-
Case III: If $g < vb <
r
,contact with the outer side of tube at P but, it is constrained to lmove within the tube, hence will change side of the tube ani
'will start mo~ing on the inner side within the tube as shown,
:in the above figure. For constrained motion inside tube thei
'minimum value of v_b for_ complete circular motion is fiir- i
A:
2
N
-Jsri, the particle wil_l leavej
mv
= mg cosP---
r
To avoid loss of contact; N 2' 0
mv 2
mgcosP--- 2' 0
r
v 5a .,f'rg_c_o_s~p
v critical = v max :::; ,/rrg-c-os-p~
If the velocity of the body becomes critical at an angle p,
then
from work energy theorem, we get
or
= dKE
1
2
1
2
mgh = -mv --mu
Wgravity
2
or
where
v
u = 0,
2
2
2
= u•
+ 2gh
[!g~~~01?~~:[251>
v =v, = .,frgcosP
h = r(l - cos Pl
= rg cosp = 2gr (1- cosp)
2
cosp = -
:A block of mass _m starts from rest-with spring urtstretched on,
:a rough incline_ Force constant of spring, k = 8 N /m,'
v;
!coefficient of kinetic friction,µ k -=
3
R
t-'
Fig. 3.40
I
I -
½- What is the speed pf the,
'[blo,/< when i~h_a§_s_lid a tijstsmce x_ = 0.5 m down the incline? '
= cos -1 -2
3
Vertical distance of this point where the particle leaves
contact with the circle;
h = r(l - cos Pl
(cosp
=
i)
Solution : From CWE theorem, dW
Work done by gravity,
Wg
Work done by frictional force,
w1 = -µkmg cosex
= !ill
= mgxsine
Concepts:
Motion of a particle inside a circular Tube: 1
In this case body will s_tart moving from the lowest point'
A on the outer side within the tube with velocity vb- '
[
i
B
Fig. 3E.25
c-
D
Work done by spring force,
W
s
dW
2
= -~kx
2
= mgxsin9- µkmg cos9x-~kx 2
2
1
/ill= -mv 2 -0
Fig. 3_39
J2ri,
Case I : If vb <
the particle will oscillate about-A)
within the tube on the outer side.
[
Case II : If vb = ,/2ii, the particle will oscillate in the,
2
or
1
2
.
'
1 2
-mv =mgxsme-µkmgcosex--kx
2
2
-_,(l)
Note that we have not counted elastic potential energy
of spring in /ill. Instead, we have counted work done by
spring in dW. We can write the above equation in another
way.
www.puucho.com
,semicircle O\Ll on the outer_s(de_11lithin t/te_.tul,_e,_ _
:
i
Anurag Mishra Mechanics 1 with www.puucho.com
MECHANIC£!]
= liUg + iiU, + t.KE
IiW = W friction
AE
or
µmg cosex = (-mgxsin0- 0)
+(½kx O) +(½ mv O)
2
2
-
-
Total frictional work done as the chain completely slips
off the table
m
Jl-nl
=-µTg
... (2)
We have assigned initial position of block as reference
level.
Mathematically eqns. (1) and (2) are same.
On inserting numerical values in eqn. (1) or (2), we
obtain v = 2m/s.
l,:axam,,..-f~
.........,~~=--~S~~~
O
X
dx
1·
=--(l-n)nmgl
2
Note that different elemei,ts on chain move different
distances on travel, that is why we have calculated work
done on a small element and then integrated it for the entire
chain.
-The] IA unifo~ chai~-~flengthl and;;~-;,, ;kep-;o~-~-s;,,ooth;
!A~hain~j;;;;;;··;,,-;;-~;il;ngt_h_l_l_ies-on~·; rough table.
;chain just starts to slip when the overhanging part equals n th
:fraciioi, of the chain length. If the chain is slightly distributed
iso, that it completely' slips off the table, what is t_he work
'.pe,formed by the friction forces. · '
:table. It is released from rest when the_overhanging part was/
jn th fraction of total length. Find the kinetic energy of the 1
chain as it complete /y_~li12.s..off_tliLtabk ....
17'"'.'.~-1
I , ---· ---- -·------------- -·-;:-··1
!
· ,
\
Reference '.
!'
1,
(1-nl)
'
X
lI
dx
L _________
m
T(l-nl)g
.' -
Fig. JE.26
.
·-----,,-.--·--~-------------~-
Solution: We will calculate coefficient of friction first.
Initially the chain is in impending state of motion. From
conditions of equilibrium:
Equation for part on table :
:r.Fx =T-µN = 0
m
:r.Fy = N (1- nl)g
1
or
T
m
= µ 1 (l -
... (1)
nl)g
I
!
!I
--- ------- -I
1
_:=:: ~;:'. J_E:2~ (~}_,_.:::~
Solution: We assign reference level in the table; thus
the potential energy of part of the chain on the table is zero.
As the chain slips more and more, the length of the chain
goes below reference level, thereby decreasing potential
energy of the chain. This loss in potential energy is
converted to gain of kinetic energy of chain. .
.
Method 1: Consider a small differential element dx at
a distance x from the table.
Potential energy of this differential element
m
=-Tdxxg
Total potential energy of the hanging part of the chain
nlm
J
-'z
.,
.
= -~mgn 2 l
... (2) .
·
m
m
From eqns. (1) and (2), µ - (I - nl)g = - nlg
l
i
=- o Tgxdx
Equation for hanging part :
m
:r.Fy = T
nlg = 0
l
n
or
µ=1-n
Now we consider a differential element dx at a distance
x from 0. Frictional force on this .differential element
= µ dxg. Work done by frictional force as it slips distance
7
'
I'
a
2
When the chain has completely slipped off the table, its
potential energy ·
I m
=- ozgxdx
J
=-~mg!
2
.
·Loss in the potential energy= -~mgn 2 1-(-~mg1)
2
2
-
X
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
rwoaiiAND ENERGY
287
= .!.mgl[l- n 2 ]
2
= gain in kinetic energy
Method 2: Consider a small differential element on
the edge of the table. When it falls through a distance x,
work done by gravity while the chain slips completely
- rl-nl m
dx
- Jo Tgx
=_!_mg (l- nl)2
2 l
A
Fig. 3E.2B (a)
Solution: Since friction is absent, we can apply the
law of conservation of energy. Centre of gravity of a
semicircular arc is at a distance (21t/r) from centre.
=.!.mgl(l-n 2 )
2
Initial potential energy = (11.irr) g ( ~)
According to CWE,
AW= AKE=_!_
2
mgl(l- n 2 )
Final potential energy = (11.1tr) g (-;r)
Method 3: Potential energy of a body of finite size is
calculated from the height of centre of gravity of the body.
When the chain is completely slipped off the tube, all
tlie links of the chain have the same velocity v.
r
I
I
I
I
rrr/2
I
I
I
I
i
Fig. 3E.27 (b)
For the sake of convenience, we assign reference level
on table, therefore potential energy of this part is zero.
~nlg
Fig. 3E.28 {b)
Centre of gravity of hanging part is at a distance nl from the
Kinetic energy of chain = -1 (11.irr) v 2
2
table. Centre of gravity of uniform body is at its centre.
When the chain completely slips off the table the centre of
gravity is at 1/2.
7 )g (-;l)
Initial potential energy= (
nl
._____,__,
'---,,-'
m
h
.G.
2
From COE,
11.irrg
(~it) = (11.irr) g (-;r) + ½(11.irr) v
2
From which we find
Final potential energy= mg(-½)
Loss in potential energy= (
7 )g
nl
(-;I )-(-mg½)
= .!. mgl(l- n
2
. - ..
-·. . r;;i
L..:~-'59,'}}P_~(?u 2s
2
)
= gain in kinetic energy
;>
A heavy, flexible, unifonn chain of length irr and mass 11.1tr lies
in a smooth semicircular tube AB of radius r. Assuming a
slight disturbance to start the chain in motion, find the
velocity v with which it will emerge from the end B of the tube.
L-J;:°~A~PJ~ :
_2~L-->
A chain of length l < rtR/2 is placed on a smooth
hemispherical surface of radius R with one of its ends fixed at
the top of the sphere.
( a) Find the gravitational potential energy of the chain.
Consider reference level at the base of hemisphere .
(b) If the chain slides down the sphere,find the kinetic energy
of the chain when it has slipped through an angle e.
(c) What will be the tangential acceleration of the chain
when it starts sliding down.
Solution: (a) We consider a differential element dl of
a chain at an angle 0 with the vertical, that subtends
differential angle d0 at the centre and its mass is
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
dm = m(Rd0)
l
Potential energy of differential element is
PE= ( 7 R de)gR cos0
Potent.ial energy of chain = J 1/R ( m R
'
'
0
l
da) gR cos0
(Note that the chain subtends an angle Z/R at the centre.
·
of the chain.)
r.,
·!. ·. ·. "·.
'·--·;'
\,,i pendul~iii bob is. suspend~d~n;~ flat. car that ~~~ei: i!;ith
/velocity v0.'111e'jlat car-is stopp?d.by a bump~r:. ·
·~ ·: •
;ca) What is the angle through,\Yfi~h the pendulum swings.
(b) If the. sw/ng angle is 0 =·60° /llld l = 5 m, what \Yas the
\ initial speed of the flat'car?",.'' : .
' · 's,f ' · .
. , ~ _ ; _ , ·.
a
!~41
<
'
,::·r:·.·.::
d
Vo .
t~
.
f ti "' 1;~
+-~
t, ~
Ji,& ·A>%!-,,.·,',.. -.·-·---_··
L_-~:__,_,
(b) Final potential energy when the chain has
slipped throujlh angle 0 is
-~i;c::::;:~7
.....
2•
.i
I
._
•• •••• .-··••• ·
R~f~;~~~~~~~~
1
A ------
level
Fig. ~ ~ ~ - - - ( - b ) - - - ~
Solution : When the flat car collides with the bumper,
due to inertia of motion. the bob swings forward. No work is
. done by tension of string on the bob, therefore energy is
conserved.
·
KEA+ PEA= KEB + PEB
r1
.!.mv~ + 0 = O+ mg (l-lcos0)
I
2
l
v~
or
= 2gl(l -
cos0)
2
= 4gl sin 0/2
\1..---~~-""'-~--l
Fig: 3E:2f(b)
I
J
Uf = s+1/R
0
or
(m)
1 Rd0gRcos0
2
= m~R [sin(a+¾)-sine]
From conversation of energy,
Ui = U f + KE
0 = 2sin-'(
z#)
... (1)
... (2)
On substituting numerical values
e = 30°' ! = 10 m, g = 10 m/s 2,
we obtain
KE=U,-Uf
'
V
2
=
z.Jg[ sin!!.
2
= ~R [sinCD+sin0-sin(0+¾)]
·, (c) Tangential force on differential element dm,
dF, = dmg sin 0
Resultant tangential force on chain
= JdF, =Jt~(7Rd0)gsin0
= -mgR [cos0] gR
l
= 'f(:-cos (¾)]
IA 'p~ndul~inb~b-can swing alJnga circular'p~th i~1ismopth
inclined plane, as shown . in Fig. · ;3E.3il, 'iwhere
m = t.2/<g; l = 0.75m, 0 = 37°: At the lowest poifwo,t. the
circle the,t'e'nsion in the sting• is.T .= 11 ON.· Determine: ·
'(a) the speed af:the bob dt the)i:Mestpoint,
1
Vi) thesp~ed of the. bob at theflighes,t point on the.'cifcle'. and
1cc) Jhe
in th?: strimcqt1thd{ighest positiim,.:.'~2 .J
>· . · ·
tension
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
2897
(!ORK AND ENERGY
1- -
1:
Ii
- - - --
·-
-
-
-
-- ------------
------·,
a,
(. :bi-:\
,'
ii
'I
---- -- -
I
·· .. .,; +o
lI
8
l:'---=1
- - - -- --- I
--- ---- ---------_--- -·- --- -- - -------/
·A small toy car of mass m slides with negligible friction· on a1
;"loop" the loop track as shown in Fig. 3E.32. The toy car starts 1
Ip-om restat.apoint H:above the level of the lowest point of the,
It.rack :
(a) If H = 2R, what normal force is exerted by the track on:
i the toy car at point q?. What are the speed and normal;
force at point r ?
'(b) At what hright will the ball leave the track and to what·
I maximum height will it rise afterwards?
(c) If H = 4R, what is the speed and normal reaction at point;
i s? ,---
'I
Ii
·-
----
I
(a)
!
(bl
''
1·-
,
i[ i
Flg.3E.31. __ --- -- - - - -- ---------- ---·-- - -- -------·
Solution : (a) From Newton's second law, at the
lowest point, we obtain
2
TA - mg sine= m~o
or
2
v0
TAI
... (1)
.
=--glsme
m
= (ll0)(0. 75 ) -(9.8)(0.75)(sin37°)
(1.2)
= 64.34
or
v 0 = 8.02 m/s
(b) At the highest point,
Iii --------- --- -- - --- -- ---- p
I'
j
'
j
i
I
_......,,
N
'
H
:ii j
, ,
'
' "" :,
.L ........... ;~;·
mv 2
... (2)
1
From energy conservation between position A and
position B,
.
TB +mgsme=--
u__________
I
mv 2
N-mg=-
1
or
2
or
v=..}4gR
... (2)
On substituting expression for velocity v in eqn. (1), we
obtain
N = mg+4mg = 5mg
Similarly; we can obtain velocity and normal reaction at
point r.
From Newton's second law at point r,
0
----
= KEB +UB
1
2
•
-mv 0 + 0 = -mv + 2mglsme
2
= I.mv 2 + 0
'H
. "!9.'
_
Fig. 3E'.3_1 (c)
--.-- -- -- -
KEA +UA
... (1)
Note that velocity at point q is not known, therefore we
apply conservation of energy.
KE,+U,=KE 1 +Ut
O+ mg(2R)
.
: '.
~__!i~~!:~~ - . ______;....,...,...,_...__,,,.___.}.__:
R
C--•
'
mg
Solution : (a) From Newton's second law, at point q,
---·--- ,7
i'1
Reference level(b)
2
2
2
v = v~ - 4gl sine
= (64.34)- 4x (9.8)(0.75)(sin37°)
=46.7
or
v = 6.83 m/s
(c) From eqn. (2),
mv 2
TB= ---mg sine
l
(1.2)(6.83) 2
(l.2)(9.8)(sin 37°)
= 0.75
= 67.56N
N
= mu
2
... (3)
·R
From conservation of energy;
KE,+U,=KEt +Ut
0 + mg(2R)
= I_ mv 2 + Pg(R)
2
= ..j2gR
From eqns. (3) and (4), we obtain
N
= mv2
R
www.puucho.com
... ( 4)
V
m(2gR)
R
= 2mg.
Anurag Mishra Mechanics 1 with www.puucho.com
'(,
..•,-,..
(b) The toy car will lose coiltact with the track at
the point where normal reaction vanishes. Let contqct
breaks at an angle.a. with the vertical.
·r--
. ----- ----··-
-_.. .. ·..f;,·· ..
... ,: ....... t, ... Jt ._
\·:r_: __ .. ,._, _:_:-... _
.
:e /
r;.·
Reference:
!
· ieve1
-
-
•
-M
es
From Newton's second law,
1- • •
mv 2
mgcos0-N = - R
When contact breaks, N 0
Thus,
v 2 = gR cose,
On substituting expression for
v
= 2gR(l - cos0).
On equ\lting expressions for v
.
-.
2
,
... (6)
we obtain
2
2
cos0=v =-gR
3'
3
After breaking contact with track the toy car moves ou a
parabolic trajectory as·a projectile. ·
·Now we apply conservation ·of energy between highest
point of trajectory and point where contact breaks.
· KE; +U; = KE 1 +U1
1 2 =-mv
1 '2 +mg
'
h~mv
2
2 1
=v 2 cos 2 0
· where
2
-
'
'
.. ,(8)
, .. (9)
vJ in eqn: (8), we o~tain
fsmooth ·e,rcu/ar track as slzo_Wl!•clt): Fig, 3E.33.:Jf·;i,;;;!;}!<,I
determ_ine.: tlfe. r~qufed speed -~. ~o'J:lzat_ t~e ball i-et:itnJ Iii
the poznt-bfp_r0Ject1.0n. ,W7\at ~;t1!e 1f/lnl(11U':1 Vallfe,oli.:rto:1
~h,ch the·?fl/Jcan reach t~J_~t7!t,~pro1ect1on ?, :. ; '':; I
A,f
0
•
J
I
'j
C
---'-----:-:"Iii..:
·
:C-( :~
"I
,·
j
,, :."
·~j'l,:_,,.'
'
A\111=""'-'-'='-="~~
Reference
level
,I
I
l
Fig. 3E.33
.·
( • "'-' ';; 1
,..,__,_ _
~_,,,,~,----------~,AA'>M-,~
2
=R+~R+2R= SOR
3
27
27
(c) From Newton'~ second law,
mv 12
N+mg = - R
.!_,.NJ
iA srrtaU~°.~! 1is·J-~lle<i'wit~; s~ee~'!:u'fro,m; point,\J:,~1~~-~)
vJ
2
::l
R
.... (7)
= -gR
and cose = 3
3
On substitµtirig these values in eqn. (7), we obtain
·.
h=2R
,
. ,
27
· Therefore maximum height from base of track'
=R+RcosfJ+h
',Vi
··:-~
1 -mg= 3mg.
N =--
2
2
M
-.. ·;, , ·
mv 2
... (5)
From conservation of energy,
'KE; +U; = KEr+Ut
1
O+ 2mgR = -mv 2 + mg[R+Rcos0]
or.
·r· '•
vJ = 4gR
or
=
20
''.,'.:\~,!
~lg. 3E,32 1(J)',:,.
L,-----L--- ····.1.~,·-•._:
,'
,. _ _ _
'Re"ferenceO level
From conservation of energy,
KE;+U;=KE1 +Ut
1
2
0+4mgR-= mv 1 +2mgR
2
Fig. 3E.~2 (c)
-
O
1,,.
,.
i_ ___ .
J
I
Solution: After reaching point C the. ball becomes a
projectile with vertical displacement 2R and horizontal
displacement 3R. Let the velocity at'c be Ve.
Motion from C to A :
From conservation of energy between points A and C,
KEA+UA=KEc+Uc
1
or,
www.puucho.com
:i
· .1
2
.
-mu + 0 = -mvc + 2mgR
2
2
·
2
u =-V~ + 4gR
--
Anurag Mishra Mechanics 1 with www.puucho.com
~-
LWORK AND ENERGY . . - -- - --
- - ---- --·7
291
-
- - • -
--------- J
9
=-gR+4gR
4
is:
U=~Jii
2
or
(b) Minimum velocity with which the ball can reach
point C is Jgi{, for which u must be ~SgR.
Motion from C to A for
Ve
= fiii
:
Solution : (a) Minimum velocity at lowest point for
completing circle is u min =
As
u=S-J2m/s, l=lm
2R =_l_gt2
X=Vc Xt
Jsii
2
=ffexrf
or
or
t
Therefore Xmin
=2R
A particle attached to a vertical strin;r of length l m
projected horizontally with a velocity 5,,/ 2 m/ s.
1
( a) What is maximum height reached by the particle from the:
lower most point of its trajectory.
(b) If the string breaks when it makes an angle of 60° with
downward vertical, find maximum height reached by the·
particle from the lower most point of its trajectory._ _
=ff
U
= Umin = -!sii
To complete the whole circle is satisfied
= 2R
Hmax
-·
= 21 = 2m
I
!A block of mass m is pressed again.st a spring offorce con.stant;
,k. The block after leaving contact with the spring moves along;
'a_ "loop" the loop track. The sliding surface is smooth except,
lfor rough portion of length s equal to R as shown in Fig.;
,3E.34, where the coefficient of friction is µk. Detennine the'
minimum spring compression xfor which the particle will not
lose contact with the track?
'
Fig. 3E.35
•
-~--
- ___ j
-w
(b) By work energy theorem from A to B, we get
1
2
1
2
1
0
-mu
=-mv
1 +mg (1-cos60)
2
2
v 1 = .,J4Dm/s
Height from the lowest point
H
-
2 . 2 600
= 1(1- cos60°) + v, sm
2g
Fig. 3E.34
,_
Solution : We know that minimum velocity required
at B so that the block can complete the loop is v B = ~ SgR.
Work done by friction when the block moves along the
rough portion = -µmgs.
From work-energy theorem,
LlWnon-conservative = AKE+ ll.U g + 6.Us
- -µmgs=(½mv~-o)+(o-½kx
or
1 2 1
2
-kx =-mvB+µkmgs
2
2
'A particle is .suspended by a light vertical inelastic string of,
length l from a fixed support. At its ';9.0-librium position it is_
projected horizontally with a speed -.J 6gl. Find the ratio of the.
ten.sion in the string in its ho1izontal position to that in the
•string when the particle is vertically above the point of
support.
Solution
2
)
By work-energy theorem,
.!.m[vf -u 2 ]
2
= -mg(l)
+-i:rl---
v, =-J4gi.
or
Thus, when at horizontal position,
tension is T1
or
r,
or
T2
mg
mv 2
=--
1
T1 = 4mg
At the topmost point, velocity is v 2
www.puucho.com
. v2
___ Fl~- 3E.3_6
Anurag Mishra Mechanics 1 with www.puucho.com
1
--2
2
Solution:
At extreme v
-m[v 1 -u ] =-mg (21)
2
V2
=-fiii.2
mv 2
T2 +mg = - =2mg
l
T2 =mg
Thus,
T1
-
- - ------ --------- -----
~ mv~ = mgl(1-cos0)
and T2-mg=mv~//
Vo
mg
mg
[£?iti~i~l'?-_ ,Gl>
MECHA.Nics;~
----------~
=0 At vertical position
·~
=4:1
T2
-
(a)
(b)
. Fig. 3_1,_.3~
A small ball is hung as shown on a string of length L
(a) If v O > .j2gL, find the angle 0 ( < 90° ) [ in terms of,.
v 0, g, L] ;With the upward vertical at which the string'
1
becomes slack.
(b) Find the value of v O [in terms of g, L] if the particle passes'
through point of suspension.
Given
T1 = mgcos0
T2 = mg+ 2mg(l- cos0) ·
T2 = 2T,
mg(3 - 2 cos0) = 2mg cos0
3-2cos0 = 2cos0 =:> cose = 3/4
'
'
A heavy particle hanging from a string of length l is projected!
horizontally with speed
Find the speed of the particle at:
,the point where the tension in the string equals weight of the•
lpa,pcle. _
____ . ____ ·-·- ______________ . _
_,
..Jii.
Fig. 3E-37 (a)
Solution: (a) At the angle 0, when the
string becomes slack
mv 2
- =mg cos0
... (1)
L
.!cmv~ =.!:_mv 2 +mgL(l+cos0)
2
2
..Jii -fiii.
Solution : Speed at bottom =
<
1
1
2
mgl(l-cos0)=-mgl--mv
2
2
mv 2
Also,
T-mgcose = - 1
... (2)
Solving eqns. (1) and (2) gives
v 0 = .jgL(2+ 3cos0)
v 2 2gL
=:>
cos0 = 0 3gL
-Lcos0=vsin0t-.!:_gt 2 Cy-direction)
2
~h=/(1-cos0)
9
T
:
Fl~_- 3E._37 (b):
•
1---+
I'
/v
Af,
·vo=-,/(gl)
I
i
mg
F!l!- 3E.3~_
But
J
T= mg
mv 2
- - = mg - mg cose
1
... (4)
.!:_ mv 2 = mgl (l - cos0)
i.e.,
Solving eqns. (3) and (4) gives, tan 0 =./2.
v 0 = .jgL(2+ 3cos0)
=:>
!
.
V
(b) After the string become slack, the ball follows
the path of projectile.
For it to pass through point of suspension
L sin0 =v cos0t
(x-direction)
... (3)
... (1)
2
2
eqn. (1),
v 0 = ~gL(2+-,J3)
=:>
1
1
mgl(l - cos0) = - mgl - - mgl(l - cos0)
2
.
1- cose =.!:_
3
:A simple pendulum swings with angular amplitude 0. The:
Itension in the string when it is vertical is twice the tension in·
_it,; extre1J!<lpositi1m. Then find the value of cos0 :
www.puucho.com
V
=.,fgl/3
2
=:.
2
cose =-
3
Anurag Mishra Mechanics 1 with www.puucho.com
··- . . ... --- . - · 2~f3l
IWORK AND ENERGY
- - - - · - ---~- - ---"---"--··--~-~---·-----··-------- ·,,,,.,_, - · - - - - __ ,; ·..J
POWER
Work done per unit time is called power.
Instantaneous Power
is defined as,
P = lim t-.W
.M....+O Llt
or
P=dW
dt
The work done by a constant force F is W
..., ...,
Thus
..., ...,
= F· s
P=d(F-s)
dt
~ i~
=F·-
dt
..., ...,
or
P = F-v = Fvcos0
The SI unit of power is J/s. 1 J/s = 1 W
Power Delivered by Pump
Consider a pump that lifts water from h meter deep well
and deliver at the rate of (dm/dt) with a velocity of v.
Suppose dm amount of water is delivered in time dt.
The .work done
dW
= (dm)gh+.!(dm)v 2
2
Power delivered,
p = dW =
dt
(dm)[gh
+ ~]
dt
2
(G) Potential Energy Diagrams: Stable and Unstable
Equilibrium
For a conservative force in one dimension,
..., ...,
We can see that at the bottom of the curve the slope is
zero and so the force component is zero. When x > 0, the
slope is positive, so the force component Fx is negative
indicating that force is directed toward - i When x < 0 the
slope is negative and the force component becomes positive
or directed toward + i
Fig. 3.42 shows a
one-dimensional
~ ,.t. X
potential energy curve.
U(x) i
Position of :
Think of a potential
Equilibrium;
energy curve as a roller
coaster ride; you are the
object riding without
friction over the track
X,
ax 1
x2 b
: Region : Regioh
(you must remember that
iwhere :where(
the actual particle motion
' slope is 'slope '
is along a straight line).
negative is positive
In the region where the
Fig. 3.42
slope is positive, there is a
negative force. The force is directed toward left on the
particle. In regions where the slope is negative, the positive
conservative force accelerates the particle to the right. So
.the range of values of x for which the potential energy curve
appears "uphill" to the particle, it slows down and the region
where U(x) appears "downhill" the magnitude of the
particle's velocity increases.
The total energy is constant and can be represented as a
horizontal line on the graph. Because E = U(x) + K, U(x)
must be less than or equal to E for all situations: U(x) s; E.
Thus, the minimum value which the total energy can take
for the potential energy is E O (see figure). At this position x 0
·the mass can only be at rest, it has potential energy but no
kinetic energy.
,JFt.
•
:· Li<x>
dU=-F·ds =,-Fxdx
I
'
F =-dU
dx
X
EsH.------------
The force is negative derivative of the potential energy
function. Graphically the force is negative of the slope of the
line tangent to potential energy curve. For example, the
potential energy function of a spring-block system is
U = (1/2) 2 • By differentiating U, we get
kx
=-: =-!(½kx )=-kx
E1 H---'--,..._---br-1(
X/,i
2
Fx
U(x)
Total
energy E
01/
'-stopeof
tangent
positive
Slope
tangent
negative
I
II
_,
Fig. 3.41
-----M~o~""-,• - - - -
••
..>fa
Eo1..--,_ ___;___-:...:::,,_.-,:-1 :
X4
X3
,Xo
X1
____ f.lg. 3.4_3
As K = E - U(x), the kinetic energy at any value of xis
represented by the distance between the E line and the U(x)
curve at that value of x.
Consider an object with total energy E 1 • At position x 3
and x 2 the total energy will be the potential energy; the
velocity is zero. If x > x 2 or x < x 3 , the potential energy
K
would be greater than E, meaning
,
= .!2 mv 2
< 0 and v
would be imaginary which is /4ossible. The points x 2 and
x 3 are called turning pc:µnts of the motion. Similarly,
www.puucho.com
/
Anurag Mishra Mechanics 1 with www.puucho.com
j294
MECHANICS-I j
a and bare turning points in Fig. 3.43. A particle with energy
E
is confined to the region a ,;; x ,;; b.
A particle with energy E 2 has four turning points but the
particle can. move in only one of the two "potential energy
wells" depending on where it is -initially. For example,at a
position x 4 , U > E 2 which means v would be imaginary; a
particle cannot reach it.
For energy E 3 , there is only, one turning point since·
UcxJ < E 3 for all x > x 5 • A particle initially moving to· the left
will have variable speed as it passes through the potential
wells but eventually stops and turns around at x = x 5 • If
then proceeds to ·the right indefinitely without return.
At x = x 0 , "the slope of potential energy curve is zero; the
force Fx = :...dU/dx is zero and the particle is in equilibrium.
A particle is in equilibrium if the net force acting on it is, zero.
If the particl~ is displaced from x = x 0 , the force is directed
back toward ·x = x 0 • The equilibrium at x = x 0 is. stable
e~uilibrium. If a particle returns toward its equilibrium
position whe~ 'displaced slightly, is said to be in stable
equilibrium.
Solution : Setting U(x) = 0, we get
a
tot;;
-dU
For particle at x = x 4,Fx = ~ = 0. When x > x 4 the
X=-
z1/6
The force is negative derivative of potential energy
function.
The potential energy has its minimµm value when its
slope is zero.
On setting Fx = 0, we get x = a; The·minimum occurs at
x = a, which is the average spacing between atoms in such a
molecule. The minimum energy of a molecule is slightly
greater than the minimum -U0 , so the energy needed to
separate atoms is slightly less than U0 •
~~am.~·le~f"4il;;>
IA particle of mass 2 kg is moving under the influence d] aforce
llwhich. a. lw.·<zy.s acts towards 3·the. c.·en.·tr·e·· and whose po..t.ential
energy is given by U(r) = 2r joule. If the body is moving in a
,circula~ofqitof;radius Sm, then.find its energy.·· '. ~ .
slope is negative and the force Fx is positive and when
x < x 4 the slope is positive and the force Fx is negative. The
force is in the direction that will accelerate the particle
toward jower potential energy, but the force is away from
the equilibrium position. The maximum at x = x 4 is a point
of unstable equilibrium. The object will accelerate away
from tpe equilibrium position if displaced slightly.
· For a· particle at x = x 6 the force is zero for some
distance, the object is in equilibrium. A small displacemen,t
results in zero force and the particle remains in equilibrium,
called neutral equilibrium.
du
Solution:
F=- dx'
F
-d(2r 3 ).
dr
mv 2
F'=-6r 2 , F = - -
r
Required ce_ntripetal force,
'
2
mv
= 6r2
r
mv 2
= 6r 3
k~~~'~"J
40 ~
·--4ft:~:::;-
KE =.!mv 2 = 3r 3 PE= 2r 3
2
'
Total energy = PE + KE
Total energy= Sr 3
jThe forse b,el:}1/een two awms_,in a diatomic molecule cqn be
represented approximately by the potential energy function .
Total energy = 5 x (5) 3
TE= 625J
-r-::c""
.
r· -..
]':, .·,,., : · _·U. = U '[(a)12-_-2'(a_).6]·
~1
· · •,
.
-
.
ciri1
0
X
'·1 .. '
X
'
~her~ U0
a are constants. (d) .: At what, value of x; Is the
potential energy zero? (b) Fiitd the force Fx. (c) At what value
:of xis the potential energy a minimum?
i' '
L
l·._.
n=----!
-U,
x~~
Fig. 3E.40
x'
[E_~~tn-~~
IA single ~onversationforce Ei~)acts on a i.o k'gpµrticJe that
moves.along 'the x,axis: The potential energy U(x) is g/ven'by: .
.
•
· ·
U(x)=20+(x..:2) 2 .
where x. is. in meters. At x = 5.,0.m-the particle has a)iinetic
energyof20J.
·
:
.
,· ·..
,
(a)· What is the mechanical energy of the system?
'(b) Make a plot of U(x) as" a function of 'x for
...:1om.:;; x s 10m.
'(c) The'least value of x and
,
'(d) The greatest value of x between which the particle can
I
'
move
~e) The maximum kinetic energy ~fthe particle'and''
'w
The value ofiat which it occurs.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
295\
U(x)-- -
'(g) Determine the equation for P(x) as a function of x.
(h) _Fo,· wh,rt(finite) value _of x_ilpes f_CxL":' 01 -
Solution: (a) At x =
s, PE= 20 + (5 -
l U3
----------------------
U2 ------------,·----
2
2) = 29 J
, u, ------------
ME = KE + PE = 20 + 29 = 49 J
! Uol------1
E
U(J)
:---· --·---·· 1a4 -
ol--------'.-,"---c~-d~-x'.
(b)
U(x)
xi
10
-
Uo
I
3,38
E
_______ Fig, ,3E,42 _ __
X
(b)
a
U(x = 0) = 24J
-u,
Umin(X= 2) = 20J
(c)
U(x = 10) = 84J
U(x = -10) = 164J
(c) and (d) When
U(x)
PE=ME=:>KE=0,v=0
49 = 20+ (x-2) 2
-~9
(x- 2) 2 = 29
----
E
x-2 = ±5.38
=>
Xmin = -3,38m, Xmax = +7.38m
(e) KE is max when PE is min ( = 20)
=>
KEmax = 49 - 20 = 29 J
(f) KE max when PE min at x = 2 m
-b/2-a/
''
!
''
L ··-
,_. _____ !i
Solution
given by
dU
(g)
b
F=--=4-2x
dx
Total mechanical energy of the particle is
E=K+U
K=E-U
'
:diven below (figure) are examples of some potential energy,
:junctions in one dimension, The total energy ofthe,particle
'indicated by a cross on the ordinate axis. In each case, specify,
'the regions, if any, in which the particle cannot be found for'
;the given energy. Also, indicate the minimum total energy the;
particle must have in each case, Think of simple physica!I
'contexts for whic~_th~~ l?.'!!entia[~"::rgy_~hapes are relevant.
is:
U(x)
As kinetic energy K is always positive, particle can exist
only in that region where U < E,
(a) For x > a, U > E
K becomes negative. Thus, particle cannot exist in the
region x > a.
(b) For any value of x, U > E, therefore, the particle
cannot exist in any region
(c) In region x < a and x > b, the value U > E ,
K is negative, The particle cannot be exist in these
region~-- _ ____ _ __ _ ____ _ ___ ____
___
7
'
inegative.
I, Uo
-------·----------
i
, E
I
!
Concepts: (a) Kinetic energy of particle can never be:
'
-~---'-------x
0
a
Fig, 3E.43 (a)
(b) Total energy of particle can be negative _
(c) Potential energy can be negative
_(d)_IUl<IE!for K > O _____________ _
___ _J
www.puucho.com
i
I
Anurag Mishra Mechanics 1 with www.puucho.com
-- ---=-'-·~-,--··----------- -------~-- -..
(H) Internal Energy Sources and Work
Concept: We will consider self-propelled objects, e.g., a_
car~ frog, helicopter, people, etc., that have their own internal'
energy sources. Each can be accelerated by a net external force
(F = ma) arising from its interaction with surroundings. As a:
rule, such a force does no net positive work on the active':
non-rigid body, and W "' t;KE. No energy is transferred to the
body from the environment via the reactionforce even though'
I
that force accelerates the body.
'
The energy required to walk, climb, skate or jump comes/
from the internal energy stored in the person. When we jump,·
the upward reaction force that accelerates us acts at the
stationary foot-floor interface. If the floor is rigid, there is no
motion of the point of application of the force, and no work is:
:done by the floor on you, Wnoor = 0, even though t;KE > 0 In',
reality the floor sags slight(y while. it exerts a normal force on'
us and W floor is positive though very small, since displacement
'is very small.
. -_
. ·--.:--
_. MECHANICS-I I
w,
·
Concept: In an accelerating car the types on the drive
I
,wheels push back on the ground; the ground pushes forward
:on the tyres and the car accelerates forward. But the region
'.between the tyre and the road is motionless, and no work is
·done on the_ car by the ground. The car does not derive its,
energy froni' the ground, it just pushes off it; the energy·
!equivalent to L\.KE comes from the fuel via the engine.
·
-. - .
. -- - ·-
·11vo particles of mass m and 2m, connected by a massless rod,·
slide on the inside of a smooth circular ring of radius r, as•
shown in Fig. 3E.44(a). If the assemb(y is released from rest:
1
when 8 = 0, determine
(a) the velocity of the particles when the rod passes the'
:
horizon ta/ position,
_(b) the maximum velocity Vm,. of the particles.
2m
Fig. 3E.44 (a)
Solution: (a) In the absence of friction the energy of
the system is conserved.
KE,+U,=KE1+U1
Fig. 3.43
O+ 2mgr = .!mv 2 +.!(2m)v 2 + mgr(l- cos45°)
2
2
+2mgr(l - cos45°) ... (1)
A swimmer's hand pushes back on the water and the:
water pushes forward on the hand, accelerating the person.:
The hand does positive work on the water; the force it exerts is!
in the same direction as the displacement. On the contrary,.
the water pushes in thefonvard direction on the hand. It does I
negative work on the swimmer. W w, < 0 even though
t;KE > O The water gets energy from the swimmer, it gets'
.
.
I
·warmer.
i
---:..--~----:~
-~ ~ -~::-~-,;:-===-~=-=:=t==~==-:::- -~-~~ \
'
~%:i,~o;;~~~~.
or
~v 2
or
v
2
= 3grcos45°-gr
= 0.865.,/ir
(b) At any general position 8 of the rod, the
conservation of energy between initial position and final
position gives
2mgr = mgr(l - cos8) + 2mgr(l - sin8)
+.!mv 2 +.!(2m)v 2
2
2
or
~mv 2 = mgrcos8+ 2mgrsin8-mgr
2
v 2 =~gr[cos8+2sin8-l]
3
or
Fig. 3.44
The swimmer is a self-propelled source and uses water
to generate a reaction force so that she can swim. If you
suspend a motor boat in air with its engine running at full
speed, will it move? The fuel provides the required energy
but without water to push on, the boat cannot accelerate.
... (2)
... (3)
... (4)
,,_(5)
For v to be maximum, the expression in bracket must be
maximum,
i.e.,
www.puucho.com
~ (cos8 + 2sin8 -1) = 0
d8
Anurag Mishra Mechanics 1 with www.puucho.com
!' WORK AND ENERGY_________ _______________ _
297'
- I
V
x=-1
5
(b) From conservation of energy between position B
co'
•O
.!.
..jzii. = ~Sg(I- x)
3
or
-~""'
=
and C,
I
~
KEB +UB
l __ ·________,_., _-'-R-c:eference level·
= KEc +Uc
0+ mg(l -1 cose) = .!:_ mv5 + mg(l- x)(l- cos<j,)
2
Fig. 3E_,~4j_b)
or
or
- sine+ 2cose = 0
tane= 2
from which we obtain sine=
Js
and cose =
.Js
(c) From conservation of energy between position B
andE,
KEB +UB = KEE +UE
0+ mgl(l- cos60°) = 2-mv~ + mg(l- x)
Substituting these values in eqn. (5), we obtain
v
or
2
vmax
=
v 0 = [2gl (1- cose) - 2g(I- x) (1- cos <j,)]1/ 2
or
2
1gr[.Js + Js-1]
or
Or
= 0.90BJji
v~
= 2gx- 2glcos60°
-VE=[
2g(x-½)f
2
·e.-xam~·!e :I~·----"f::> ii.---
i
L~':- ·-· - c..::·:-
0
:,
Tize figure shows a pendulum of length l suspended at a'
,distance x vertically above a peg.
,
( a) The. pendulum bob is deflected through an angle e and!
then released. Find the speed of the bob at the instant'
shown in Fig. 3E.45.
.
~
A
O···-····F,
.·:-<...;
t
,Tlvo blocks are connected by a massless string that passes over
One end of the string
.is attached to a. mass m1 = 3 kg, i.~., a distance R = 1.20 m·
'from the peg. The other end of the string is connected to a.
block of mass m2 = 6 kg resting on a table. From what angle
'.e, measured from the vertical, must the 3 kg block be released
iin order to just lift: the 6 kg block off the table?
:a frictionless peg as shown in Fig. _3E.46
/:•• 0: X
.. ::::··
B
(j°
iI
\E
·:;_·········;·····
Smooth peg
~·.. /
i ·......
. .... ... :::,-.._;__....-:: ... ~.- ...
..
.....
C Reference level
Fig. 3E.45
I
e = 90°. For what x
(position of peg) will the pendulum complete the circle?
(c) The pendulum is released when e = 60°. What is the'
velocity of the bolJ as it p_as~".5 pp.,_iti,on E, ___ _
(b) The pendulum is released when
Solution: (a) As we have learned earlier, the
minimum velocity required at the lowermost point so as to
complete the circle is given by the expression
v=.Jsii
From conservation of energy between position of
release and position C,
KEA+UA=KEc+Uc
0+ mgl = 2-mv 2 + 0
2
or
v = ..Jzif.
For just completing the circle,
Fig. 3E.46
Solution: This problem involves several concepts.
First we will apply conservation of energy to find the speed
of the block m 1 at the bottom of the circular path as a
function of e and the radius of the path, R. From Newton's
second law we will determine the tension at the bottom of
its path as function of given parameters. Finally, the block
m 2 will lift off the ground when the upward force(tension)
exerted by the cord just exceeds the weight of the block. We
take bottom of ·me circle as reference level. From
conservation of energy, we have
KE-+U- = KE! +Ut
'
'
1
0+ m1g(R -R cose) = - m1 v 2 + 0
2
or
www.puucho.com
v 2 = 2gR(l - cose)
... (1)
Anurag Mishra Mechanics 1 with www.puucho.com
;;:, ~1_.,;~-' -..
-
:.~-
Applying Newton's second law on block of mass m 2 , we
have
..
.
.
2
For angle' e to be real,
,(4m) 2 -4x6mxM>'0
3M
or
m>-
V
LF. = T - m1g = m 1 R
m 1v
2
2
T=m 1g+--
or
Fromm= 2M, eqn. (4) reduces to
12cos2 e '-Sease+ 1 = 0
... (2)
R
As the string is massless, tension T is constant
throughout. When m 2 just lifts off, the normal reaction
becomes zero. For block ·m 2 , we have
... (3)
T= m2g ·
From eqns. (1), (2) ·and (3), we get
ZgR(l - case)
m 2 g = m 1g + m 1 ~ - - - .
R
case·=
3m
1 - m 2· 3 x 3- 6 = .!.
or
2m1
2x 3
2
or
e = 60°
t:i§·~~J 47-~
i.Fig. 3E.47 (a) shows a circula" ring of mass M that hangs in a
lvertica'1\7,1~ne.. .' Two beads pf mass m are·., released .
ll'#multa;eouslyfrom th(top ofthe ring in opposit~ ~irectiohs:
There il.no frictional f<!rce between the bead and the ring.
3
M. If m = 2M, ~i
Show ..that. the ring
will start i~· rise,' if m
.
. > 2
cose
~
I
l
'.
.T
._ ~ fl
s.
.· I
'I
N
l""@xcimj1.lg,£j 48 I~
:A force acting on a certain particle\-7
r
·
•
•.
', 1'""
r · ·l
•
:.•
m
in the xy=plane. This force F is .
given'by the_idcpression ·· · ·•
••
M
l_·______<•_>_ _ _. ·_F,_lg_._3~-~!_;_ _ _ _
~
x
where
. y
and
~
.
are· expressed, in
''
'
Fig. 3E.4B
.. .,·,
'
metre,.If'.F is a conservative force?
Exp]ai~'jour' answer. ·
'
'
•
'
e
''
,,
•
''
,~::--·1
~,,_
.: __J
Solution: For each of the paths from
done is given by
·
o
to
c,
work
,.
"
--+
"·
.....
where F = xy i + xy j and ds = dx i + dy j, so the·dot
--+
--+
.
-
product F· ds = xydx + xy dy. The path OAC consists of OA
and AC. Along OA,y = 0 and dy = 0, and along AC, x = 1 m
and dx = 0.
So,
WoAc=W0;._+wAc=O+f~ydy
__
i_.'.:,
=ly2
' f'
.
R
=lx2I'
=.!.J
2 ' 2
0
·Along the straight line OC, y
--+
2
--+
.
F-ds·= xy.dx+ xy dy
.!. mv = mgR(l- cos0)
v 2 = ZgR (1- case)
... (2)
From eqns. (1) and (2), we.get
·
.iv= (2-cos0)mg
From the· force diagram of the ring, we see that, at the
instant the ring begins to rise, tension in the string reduces
to zero.
2Ncose=Mg
... (3)
From eqns. (2) and (3), we have
2(2- 3cose)cos0mg = Mg
... (4)
or
6mcos 2 e-4mcos0+M=0
=½J
[
The path OBC consists of OB and BC. Along OB, x = 0
and dx = 0. Along BC,y'= 1 m and dy = 0.
So
WoBc=WoB+WBc=O+ 0 xdx
From energy conservation, equation for bead,
2
0(0,0)
2
Solution: Figs. 3E.47 (a) and (b) show force diagrams
of ring and bead .respectively. Let v be the velocity of the
bead at'this position.
mv 2
... (1)
LF = N + mg case = - n
-t<-.-.-~
..... x(m)
. A(1,0)
F=,(xyf +xy j)(lN/m~)
w;,,; Ji-d-;
l
mg
~/ m
y(m)
B(0, 1)>----~ 'C(1, 1)
on the particle's' position
Idepends
i
~~/f\
2 6
~-=-=:=.~ .;;.!!~.:;p-..=;r,~~.
--+
!),
1
As cos e = ½ occurs first, so the required angle is 0 = 60° .
ivhat,(lriilttqftom the verticCl/J/ifsJ,_gppens?__._.:_
·1~
1
= -,
Hence,
= 2x
2
~
x, so dy = dx and
dx.
f
. W~c = F · ds =
•
f~ 2x dx
2
1
2
=l tl
0
=¾J
1
Although W oAc = WoBc = - J, the work done Woe along
2
OC is not equal to WoAc
.
p
or WoBc· The force
is
non-conservative, because work done between two points
depends upon the particular path.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
•. ' IWORK AND ENERGY
\
>
•
'
r . ---~ . --- ---------. --- - --
.1- ·
'
IL
~--
------------
·only One Alternativ~ is Correct
--~~--------- -
-
2. Consider two observers moving with respect to each
other at a speed v along·a straight line. They observe a
block of mass m moving a distance 1 on a rough
surface. The following quantities will be same as
observed by the observers:
(a) . Kinetic energy of the block at time t
(b) Work done by friction
(c) Total work done on the block
(d) Acceleration of the block
3. The force acting on a body
moving along.x-axis varies with
the position of the particle as
shown in figure. The body is in
stable equilibrium at:
·
X
=X 1
= X2
(c) Both x =x1 and x =x 2
(d) Neither at x = x1 nor at x = x 2
4. A uniform chain has mass Mand length L. It is lying on
a smooth horizontal table with half of its length
hanging vertically downward. The work done in
pulling the chain up the table is:
(a) MgL/2
(c) MgL/8
i~ .· .
(b) '"'
(b) MgL/4
(d) MgL/16
5. A blodc is resting over a smooth horizontal plane. A
constant horizontal force starts acting on it at t = 0.
Which of the following graph is correct:
' ''
--- . .
:
r-L·-----·--7i
I
l,"
.
0 IJ1sp(ace~ent
i
j
!
'
______ t-+,
,~·~I
IKE '
. . .i
jKE
Cc)
~ ~
_,, ..
I
. ._
(d) I
.. '
'
!. .
.
J.I
O... ~sp!~men~l
'•
·I
6. If the block in the shown
arrangement is acted upon by
a· constant force F for t ~ 0, its
.
maximum speed will be:
(a) Fl Jmk
(c) Fl .J2mk
X
------------~- .. -
-···---
l
--
r·~·-----,
,li,
: -·· • ' . )
1. A small block of mass m is kept on a rough inclined
.surface of inclination 8 fixed in a lift. The lift moves up
with a uniform velocity v ,ind the block does not slide
on the incline. The work done by the force of frictio!l ,
on the block in time t will be:
(b) mgvt cos 2 8
(a) Zero
(c) mgvt sin 2 8
(d) mgvt sin:?B
(a)
(b)
------- -- ---
(b) 2FI ..J,;ii
(d) ..fiiiI ..Jmk
7. A block hangs freely from the end of a spring. A boy
then slowly pushes the block upwards so that the
string becomes strain free. The gain in gravitational
potential energy of the block during the process is
equal to:
(a) The work done by · the boy , against the
gravitational force acting on the block.
(b) The loss of energy stored in the spring minus the
work done by the tension in the spring.
(c) The work done on tlie block by the boy plus the
loss of energy stored in the spring.
(d) The work done on the block by the boy minus the
work done by the tension in the spring plus the
loss of energy stored in the spring.
(e) The work done on the block by the boy minus the
work done by the tension in th!! spring.
8. A particle of mass m is moving in 'a circular path of
constant radius r such that its centripetal acceleration
is varying with time t as a, = k2rt 2 , where k is the
constant. The power delivered to the particle by the
forces acting on it is:
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
(b) mk 2 r 2t
(d) zero
9. A self-propelled vehicle of mass m, whose engine
delivers a constant power P, has an acceleration a =
(P/mv). (Assume that there is no friction). In order to
increase its velocity from v 1 to v 2 , the distan~e it has
to travel will be:
' m 3 3
3
(a) P (v~ -vf)
(b) -(v
-v )
m
3P
'
(c) ~(v~ -v{)
3P
1
2
m
.3
(d) -(V2 -Vi)
3P
10. A stone tied to string oflength I is whirled in a vertical
circle with the other end of the string at the centre. At
a certain instant of time the stone is at its lowest
position and 'has a speed u. The magnitude of the
change in velocity as it reaches a position, where the
string is horizontal is:
fiii
(a) ~u 2 -2gl
Cb)
(c) ~u 2 -gl
(d) ~~2(-u2---g-p
11. A ball of mass 5.0 gm and relative density 0.5 strikes
the surface of the water with a velocity of 20 m/sec. It
comes to rest at a depth of 2m. Find the work done by
the resisting force in water: (take g = 10 m / s 2)
(a) - 6 J
(b) + 7.5 J
· (c) - 9 J
(d) - 10 J
"12. · A particle of mass 1 gm executes an oscillatory motion
·
on the concave surface of a spherical dish of radius 2m
placed on a' horizontal plane. If the motion of the
. particle starts from a point on .the dish at a height of 1
cm from the horizontal plane and the coefficient of
friction is 0.01, how much total distance will be moved
by the particle before it comes t,o rest:
(take g =" 10 m / s2)
(a) 100 m
(b) 1 m
(c) 10 m
(d) 0.1 m
13. A spring is compressed between two toy-carts of
masses m1 and m 2 (spring is not atta.ched to the toy
carts). When the toy-carts are released, the spring
exerts on each equal and opposite average forces for
the same time t(t ~ o). If the coefficient of friction isµ
between the ground and the carts are equal, then the
displacements' of the two toy-carts are in the ratio:
(a) ~ = - m2
(b) ~ = - ~
s2
(c)
m1
s;
=
Sz
.
-(m2)2
m1
s2
(d)
m2
(½)ky
2
. (c) (½)k(x+y)
(a)
2
(½)k(x2+y2)
(d) (½)ky(2x+y)
(b)
17. A ball P is projected vertically up. Another similar ball
Q is projected at ari angle 45°. Both reach the same
height during their motion. Then, at the starting point,
ratio of kinetic energy of P and Q is?
(a) 0.50
(b) 0.25
(c) 2
(d) 4
18. A particle of mass m is moving in a horizontal circle of
radius r unde: a centripetal force equal to (- r~ }
where k is a positive constant. Then if kinetic energy,
potential energy and mechanical energy of the particle
are
and
respectively. Which one is correct?
KE, PE ME
(a) KE=(~), PE= -(!5.), ME=-(~)
2r
r
2r ..
(b) KE=(~) PE=-(~) ME= zero
2r'·
2r'
-·
(c) KE =zero , PE =zero, ME =zero
(d) KE =(~), PE=-(!), ~E =(;r)
·
19. A 10 kg block is pulled along a
frictionless surface in the form of ,an
6.O~··:O. F
arc of a circle of radius 10 m. The ~·
..··
applied force F is 200 N as shown. If
the block started from rest at point P,
then its velocity at Q will be? (take g = 10 m/ s2 )
(a) 14.7 m/s
(b) 15.7 m/s
(c) 16.7 m/s
(d) 17.3 m/s
20. 1 kg block collides with a
horizontal massless spring of
force constant 2 N/m. The
block compresses the spring by 4m. If the coefficient of
kinetic friction between the block and the· surface is
0.25, what was the speed of the block at the instant of
collision? (take g = 10 m/ s2 )
.
!r~--.
.·
,
(-=-·.~:-,._;0~
.J
~ = ~(m1 )2
s2
body is permitted to fall instead, through what
distance does it stretch the string?
(a) d
(b) 1.5 d
(c) 2 d
(d) 3 d
15. A running man has half the .kinetic energy of '! boy of
half his mass. The man speeds up by 1 m/sec and then
has the same kinetic energy as the boy. The odginal
speed of the boy was:
(a) 2 m/S
(b) 9.6 m/S
(d) 7.2 m/S
(c) 4.8 m/s
16. An elastic string of unstretched length L and force
constant k is stretched by a small length x. It is further
stretch by small length y. The work done in the second
stretching is:
m2
14. A light spring is hung vertically from a fixed support
and a heavy mass is attached to its. lower end. The
mass is then slowly lowered to its equilibrium position:
This stretches the spring by an amount d. If the same
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
IWORK AND ENERGY
· (b) ifz m/s
(a} 7.2 m/s
(d) 10 m/s
(c) 4.5 m/s
21. A stone with weight W is thrown vertically upward
into th_eair with initial velocityv 0 • If a constant forcef
due to air drag acts on the stone throughout the flight
& if the maximum height attain by stone is h and
velocity when it strikes to the ground is v. Which one is
correct?
(a) h =vg(i+ !)/2g,v=v 0
(b)
·
mv
=--.2r,
W' =zero
r
,
mv 2
(c) W = - - , W = --.2,rr
·
r
r
(d) W = zero, W' = zero,
mv 2
26.
~::::e !:sv; ;gu:t:~
!)
-f} v=v 0~+J
W
W+f
(d) h=v 0
2 /2g ( l + .
W
W-f
1
.
-----7
(a) i
I
(c)
l,
LT
i
·O~e.
pig.
?a]
r
directed towards the center. If work done by this force
in moving the body over half the circumference and
complete circumference is W and W', then:
mv 2
,
(a) W =--. itr, W =zero
_!
r-~F.
. . -·:
7C
_
I
',2mg---.·
(d) '
.
:
1
8
1t
1t
l:_ _____2
'
•..•...
!
r-·~F-.. --- ·:
'21.
0
e
(b) ·;
1
,
,
!
iy-
! _0
:E.
. 2
L--------- -
"
a
0 . 'E.
2
·a
"
- - - - _, ___
--«
1. 'F~---··
.... ·1
I
•
3mg
' 2
I mg
(c)
j
!~--.
I
"
:F~- '
~--
2
8:
:
27. In the Q. No. 26, if M = 2 m and friction exists
between the circular track and the horizontal surface
then, which of the following lot best represents the
variation of frictional force versus the angle 8:
:·F~--····
.I
I
a constant speed v. The force on the body is mv and is
·i
I
. l -- ½-
I
IL_,. ___2. _______
value of h will be:
(b) 2R
(a) R
(d) 3 R
(c) 2.5 R
25. A body of mass m is moving in a circle of radius 'r' with
r
(b)
al
"
j3mg____
(a) Vv > VE > Vp
(b) Vp > VE > Vo
(c)vv=VE=Vp
(d)vv=VE=Vp=O
24. The mass m slides down the ; 1
track and completes the , !
"------~
... -
,c
..J
__D_______'. __ I= ______x_ ..~I
'. h
:
O
L__. ____ 2 _ _
(a) 1
:~~~ s~~f~~e. rii~~i~7~~:
,3mg ...
l_ .._. _L _ _...1
~ ( B ) y ~ (C)y~!
1
~
---·-1 !~F
··.1:
i
.F.
!smg~--- :
22. A stone of mass m, tied to the end of a string, is .
whirled around in a horizontal circle (neglect gravity).
The length of the string is reduced gradually such that
mvr = constant. Then, the tension in the string is given ·
by T =Ar", where A is a constant and r is the
instantaneous radius of the circle. Then, n is equal to:
(a) + 2
(b) - 2
(c) + 3.
(d) - 3
23. A block of mass m released from rest from point O as
shown below. The velocity of the block at the lowest
points are v O , v E, v F respectively. Assume coefficient
of kinetic friction between surface and the block is
same in all cases. Then,
0
~
.
- -
sliding down a smooth
and stationary circular
ii/Ht'
track. Which of the · -------- -·· . - - - following graph best represents the variation of
magnitude of the force applied by the track on the
mass and the angle 8?
(c) h =v 0
2 /2g ( l + - ,v=v 0~-J
--
-·--- - - -
--~m~it··~·.::.:-:-M. - :.'
fl, . .
,mm
2g(1 +;} v = zero
h = v~ I
2
(b) W
(d) \
!
I
·
0
1t
·-------
- _ _ _j
-
.2
: '
:
:
1t
'
'
8:
:
i
-·---- - ·- --'
28. A particle of mass m is whirled in a vertical circle with
the help of a thread. If"the maximum tension in the
thread is double its minimum value then the value of
minimum tension in the thread will be:
(a) 6 mg
(b) zero
(c) 3 mg
(d) can't be found
A
particle
of
mass
mis
located
in a one dimensional
29.
potential field where potential energy of the particle
has the form U(x) =~ -~ where a and b are positive
X2
X
constants. The positi_on of equilibrium is:
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
-, __
(a)
_!!c
.-2d~
,
____
(b) 2b
.
a,
(d), 2a
,:·
(b) pA( h 1
~h
(c), ~
' '
": b
,
,' b
'
Two,
cy)im/_ri'cal
vess~ls
of
equal
cn;,ss-sectioµal
area, A
30.
contairi water upto height ,h1 and h 2 • The vessels are
interconnected so that the levels in ,them become
equal. The work ,done by the force of gravity during
the process is:
(a) zero
I; 2
2
)\
2
(c),
pA ( h
h ) g
(d) pAh;h 2 g
31. A block of mass 100 g moved with,a speed of 5 m/s at
the highest point in a closed circular tube of radius 10
,cm kept in a vertical plane. The cross-section of the
tube is such that the block just fits in it., The block
makes several oscillations inside the tube and finally
steps at the lowest point. The work done by t\Ie tube
on the block during the process is:
(a) 1.45 J
(b) - 1.45 J
(c) 0.2 J
(d) zero
32. A heavy stone is thrown from a cliff of height h with a
speed .v. .The_ stone will hit ground with maximum
speed if it is thrown:
(a) vertically downward
(b) verticaliy.upward
(c) horizontally
(d) the speed does not depend on the initial direction
33. Two springs A and B(kA = 2kB) are stretched by
applying forces of equal magnitudes at the four ends.
If the energy stored in A is E, that in B is:
(a) E,
(b) 2E
2
(c) E
', ,
'
,.
.
(d) 1!_
.
4
;
,
34 .. , Two equal masses are attached to the two ends of a
spring of spring constant k. The masses are pulled out
;YI)l~etri~ally to ~tretch the spring by a length· over
its natural length. The work done by the spring on
each mass is:
(b) -~kx2
(a) ~kx 2
x
2
(c) ~kx 2
2
(d) -~kx 2
4
(a) total energy
(b) kinetic energy
(c) potential energy
(d) none of these
,
37. The work done by ,in the forces (external and internal)
on: a,·s)istem equals the change in:
(a) total energy
(bl kinetic energy
(c) potential energy
(d) none of _these
38. .................... of a two particle system depends only 'on
the separation between the two' particles. The most
appropriate choice for the blank space in the· above
sent~nce. is:
(a) kinetic energy
,(b) total mechanical energy
(c) potential energy
(d) total energy
39. A block of mass m slides down a smooth vertical
circular track. During the motion, the block is in:
(a) vertical equilibrium
(b) horizontal equilibrium
· (c) radial equilibrium
(d) none of the above
40. A particle is rotated in a vertical circle by connecting it
to a string of length l and keeping the other end of the
string fixed. The mininlum speed of the particle when
the string is horizontal for which the particle will
complete the circle is:
(a)
(b)
(t) .J3gl
(d) .JSgl
41. In the shown diagram mass of ~ - - - - k
A is m and that of B is 2 m. All
the surfaces are smooth.
System is released from rest
with
spring
unstretched.
, Then, the maximum extension
(x~) in spring will be:
(a) :mg
(b) 2mg
.
k
k
:.JiL
.Jiil
(c) -3mg
(d) 4mg
k
k
42. In above question, speed of block A, when the
.,
(a)
(c)
4
35. The negative of the work done 'by the conservative
internal forces on a system equals the change in:
(a) total energy
(b) kinetic energy
(c) potential energy
(d) none of these ...
36:'· The work done ·by the external forces on a system
equals the change in:
, ,' :,.,,
,
.. ,,
extension in spring is
2gt
2g'\/3lZ
{2m
.
X .·
, '
___!!!_,
2
is:
(b)
(d)
2gP2
~4m
g 3k
43. A chain of length L and mass Mis arranged as shown
in following four cases. The correct decreasing order
of potential energy (assumed zero at horizontal
surface) is:
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
\\VOIIKAND ENERGY ' :: ., :~
l
.
I,
':
• • -
r_' _. ' '. ~:i:,~-~-I:i.~: :,·.•,~
~-:. :_ c.,·_:·=j ·,
~ (i)
· ·
(ii).·'
' . ' ~ ' · ·,
(iv)
.
'~i--/:~,·_:'.
~la~~~U~:~~~;
~
'':J)"./
_.
·
-'----'
(a) i > ii > iii > iv > v
· (b) i .= ii·;. iii > iv>. v·
(c) i = ii > iv > iii > v
(d) i = ii > iv·> V > iii
A
block cif mass m is pulled· by a· constant p9w~r P
44.
placed on a rough horizontal plane. The _friction
coefficient· between the block and surface is µ:· The
' '.' ·
maximum velocity of the block is: ·
p .
'
µP
(a)-.
mg
(c)
..
...!_
µmg·
(b) -
mg
',·
(d) _P_
µ2mg
45. Forces acting on a- particle moving in a straight line
~aries with the velocity of the particle ;s·F =~;where
V,
.
u is constant. The work done by ,this force 'in time
interval M is:
· ··
(a) ooit
(b) ]:_Mt
2
. (d) u 2t.t
.
.
A
pendulum
of
mass
1
kg
and
length
1
=l
·mis
released
46.
from rest at angle 60°. The power delivered by all the
forces acting on the bob at an angle 0 = 30° is:
(g=10m/s 2 )
·
(b) 13.4 OJ
I
I
I
I
~--·=
k
(c) 3mg
2k
(b) 3mg ,k ...
(d) 6mg ·,
k'
48. In the shown figure, the mass in
sticks to the string just after it strikes
it. Then the-minimum value of h, so
that the lower mass bounce off the
ground during 'its rebound is: · · ··
(a) 2mg
, · , (b) ,3mg
·k
. ·,"k
(c) 3mg.,' .
2k
. (d) 67/l&·
' '
k'
1_1'
(a) 2
(c) 3 ·
50. A spring mass system 'is held at rest
with the spring relaxed at height h
above the ground. The· ;minimum
value bf h for which the system has a
tendency to rebound ·after hitting the
·ground is (assume zero coefficient of
restitution for lower block and
ground):
(b) 3mg
(a) 2mg
, ' k
' k
a
'
(c) 3mg
, 11
l;;L.~.'.. . .m]
(d) 6mg
'
. k'
''
51. In shown figure, the trolley ~ - - - - - ,..-,--i
I
"·t-·_
starts
accelerating with .
~Lm
al
acceleration
a.
The ·· , .
,
·
.J
.maximum angle deflected by ,,...,,,Q,;,,4J,p
, thread from vertical will be:
-7
(d) 5 OJ
47. A system consists of two identical· \
'
cubes, each of mass m, linked.
m . - ·•
together
by
a
compressed
weightless spring of force constant
k. The cubes are also connected by .! ,
i
a· thread which is burnt at a certain I
m
moment. The minimum value of
hlitial compression x 0 ; of the spring for which lower
cube bounce up after the thread is burnt-is:>
(a) 2mg
49,
0~ 'am:i~z!t!~
_1,:,·.((Dm
: rTl •• , c·_
·, --. surface as shown iri:figure. Two
_,smallsphereseachofmassm;just ;·M, ..
fit ih the tube one released from 1
•
1
the. top. If tlie tube' looses contact : "'"'""'"'"'"'" ,
' · with the ground ate·= 60° .then
the value of m/M:
,' . ,
2k•
(c) 2a.M
(a) l.'34 OJ
(c) 0.670J
~~!J
____,.~--~-----'-',_;___._._.--~--'----='.
-
(b), tan-I(;)
(d) tan-I(~) .
....
.
.
52. A force F = -K(y i + x j) (where K is a positive
•. constant) acts on a particle 'moving in the X -'Y plane.
· · Starting from the origin, the particle is taken along the
' ' positive x-~s to the point (a, 0) and then parallel to
the y-axis to the point (ri, ci). ·The 'total_ work.done by
the force F on the particle is:
(a) -2Ka 2
(c) -Ka 2
Cb)
,C\li
2Ka
2
Ka
2
53. A particle free to move _along x-axis has potential
energy given by U(x) = K[l- exp(-x) 2 ] for
~= $ x $ ..,.,, where· K is . a positive c.o_µstant of ,
appropriate dimensions. Then:
(a) At point away from the origin, the particle is in
unstable equilibrium,
·
(b) For any finite non-zero ,value of x, there is force
directed away from the origin
·
(c) If its total mechanical energy if K/2, it. has its
·
·
www.puucho.com minimum ri at the origin
Anurag Mishra Mechanics 1 with www.puucho.com
____-----
r-308
,
.
..
------ --- -- . ------ ··-- ------ -----------·------~--· ------·-- --
(a) Work done by
F is 12olz J
_,
1
'
(b) Work done by F2 is 180 J
_,
(c) Work done by F3 is 45,c J
_,
(d), F1 is conservative in nature
19. The potential energy U in joule of a particle of mass 1
kg moving in x-y plane obeys the law
U = 3x + 4y,
where (x, y) are the co-ordinates of the particle in
meter. If the particle is at rest at (6, 4) at time t = 0
then:
(a) the particle has constant acceleration
(b) the particle has zero acceleration
(c) the speed of the particle when it crosses y-axis is
10 m/s
(d) co:ordinate of particle at t ~ l sec is (4.5, 2)
,. ______ .. ,
simple
pe'ndulum ,., .
20. A
,.
consisting ..of a mass M
l ' :a.
attached in a string of 1.
I £ :
:
length L is released from ·
rest a~ an angle a. A pin is
located at a distance '1' i .
e
below. the pivot point. - .. ______ -- ..... ..:.•. When the pendulum swings down, the string hits the
pin as shown in the figure. The maximum angle 0
which string makes with the ·vertical after hitting the
pin is:
(bl cos-' cosa +
(a.) cos--i(Lcosa+l)
L+l
L-l
(c) cos-' cos a
(dl cos-' cos a
.
L-·l
,
L+l
•
!.
(L l)
(L -1)
(L -1)
.
21. An object is displaced from a pointA (0, 0, 0) toB (lm,
_,
.
lm, 1ml under a force F ~ (y i + x jlN. The work
done by this force and the nattire of the force is:
(al 1 J, non-conservative
(bl 1 J, conservative
(cl zero, conservative
(d) zero, non-conservative
22. A particle iµass is tied to an ideal string and whirled in
a vertical circle of radius L, where L is off-course the
length of the string. If the ratio of the maximum to
minimum tension in the string throughout the motion
is 2 : 1, then the maximum possible speed of the
particl~ will be:
1
(al .,jl lgL
(bl ..Jsif,
(cl .,jlOgL
(dl .j3if.
23. The following plot shows the variation of potential
energy (U) of a system versus position (xl. From the
graph we_ can interpret th.at:
·-- - - -MECHANlcs:_!J
--------·--·-----
------- ----
(al Point D is position of
neutral equilibrium
(bl Point B is position of
unstable equilibrium
(cl Point C is position of
stable equilibrium
(dl Point A is position of
neutral equilibrium
24. A smooth narrow tube is in
form of an arcAB of a circle of
center at O and radius R. is
fixed so that A is vertically
above O and OB is horizontal:
Particles P and Q of mass m
and 2m respectively' with an
-·----- --··
-1
u
I
I
:A
i
C
I
D,
B
l . - - - - _,_ - -
X'
'
-·--MO-·----'
'/~-
R:
'
'
.
c............
O
.
!
.
Q
B
ideal string of length":,, co,nnecting them is pla_ced as
shown in the figure. ·The speed of the particles as P
reaches B will be:
(al ~2gR
.
3
2(1 + 1t)gR ,
(bl
~2gR,
3,c
,,
(dl ~21t;R ,.
3
~
25. In a children's park, there is a slide which has a total
length of 10 m and a height of 8 m. A vertical ladder is
provided to reach the top. A boy weighing 200 N
climbs up the ladder to the top of the slide and slides
down to the ground. The average friction offere.l by
the slide is three tenth of his weight. Then:
(al The work done by ladder on the boy as he goes up
is zero
(b) The work done by ladder on boy as he goes up is 1600 J
(cl The work done by slide on boy as he comes down
is.:. 600 J'
·
f
(dl The work done by slide on boy as he comes down
is 1600 J
26. A particle of' mass m is kept at the top' of a smooth
fixed sphere. It is given a horizontal velocity v then:
(al it will start moving along a circular path if v < .jiR
(b) it will start moving along a circular path ifv > .jiR
(cl it will start moving along a parabolic path if
V < .jiR
(d) it will start moving along a parabolic path if
V > .jiR
27. The total work done on a particle is equal to the
change in. its kinetic' energy:
'
(al. always
.
(b) only if the forces acting on it are conservative
(cl only if gravitational force along acts on it '
(d) only if elastic force along acts on it
(c)
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
. : "309]
IWORKANDENERGY ________________ - - - - - - - - 28. A particle is acted upon by a force of constant
magnitude which is always perpendicular to the
velocity of the particle. The motion of the particle
takes place in a plane. If follows that:
(a) its velocity is constant
(b) its. acceleration is constant
(c) its kinetic energy is constant
(d) it moves in a circular path
29. You lift a suitcase from the floor and keep it on a table.
The work done by you on the suitcase does not depend
on:
(a) the path taken by the suitcase
(b) the time taken by you in doing so
(c) the weight of the suitcase
(d) your weight
30. A particle of mass m is attached to a light string of
length 1, the other end of which is fixed. Initially the
string is kept horizontal and the particle is given an
upward velocity v. The particle is just able to complete
a circle.
(a) The -string becomes slack when the particle
reached its highest point
(b) The velocity of the particle becomes zero at the
highest point
(c) The kinetic energy of the ball in initial position
1
2
was -mv
2
= mgI
coordinates of the point being x and y, measured in
metres. If the particle is initially at rest at (6, 4), then:
(a) its acceleration is of magnitude 5 m/ s2
(b) its speed when it crosses they-axis is 10 m/s
(c) it crosses the y-axis (x = 0) at y = - 4
(d) it moves in a straight line passing through the
origin (0, O)
34. A ball is projected vertically upwards. Air resistance
and variation in g may be neglected. The ball rises to
its maximum height H in a time T, the height being h
after a time t :
(1) The graph of kinetic energy Ek of the ball against
height h is shown in figure 1 •
(2) The graph of height h against time t is shown in
figure 2
(3) The graph of gravitational energy Ek of the ball
against height h is shown in figure 3
h
E0
0
T
(2(
(1)
(d) The particle again passes through the initial
position
31. The string of a simple pendulum can with stand a
maximum tension equal to 4 times the weight of bob
suspended to it. The string is made horizontal and bob
is released from rest then:
(a) String will break somewhere during the motion
and will then follow straight line path
(b) String will break somewhere during the motion
and then follow parabolic path
(c) It will complete the vertical circle
(d) lv!otion will be oscillatory and string will not break
32. A particle of mass m is at rest in a train moving with
constant velocity with respect to ground. Now the
parti~le is accelerated by a constant force F0 acting
along the direction of motion of train for time t O• A girl
in the train and a boy on the ground measure tl).e work
done by this force. Which of the following are
incorrect?
(a)' Both will measure the same work
(b) Boy will measure higher value than the girl
(c) Girl will measure higher value than the boy
(d) Data are insufficient for the measurement of work
' done by the force F0
33. The potential energy in joules of a particle of mass 1 kg
moving in a plane is given by U = 3x + 4y, the position
l,. __
- - - - - - - _ _ _. , _
-- -
I
h_
IL..---'-------+1
H
0
(3)
-------- - - -
(i) - Which the figure shows the correct answers ?
(a) 3 only
(b) 1, 2
(c) 2, 3
(d) 1 only
(ii) In the above situation the block wili have maximum
velocity when:
(a) the spring force becomes zero
(b) the frictional force becomes zero
(c) the net force becomes zero
(d) the acceleration of block becomes zero
(iii) Two particles move on a circular path (one just inside
and the other just outside) with angular velocities ro
and 5ro starting from the same point. Then :
(a) they cross each other at regular intervals of time
·· are opposite
·1y
-27th
w en th"
eir angul ar ve1ocines
4ro
directed
(b) they cross each other at points on the path
subtending an angle of 60° at the centre if their
angular velocities are oppositely directed
(c) they cross at intervals of time...::.. if their angular
3ro
velocities are oppositely directed
_
(d) they cross each other .at points on the path
subtending 90° at the centre if their angular
velocities are in the same sense
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
__. ,,
.,
-~..-....'....,.;..'.;, . '
.
. .
·;: ,ME~HAN@:(1
~
--~----__.,,;_:,.,L.L.
Comprehensio~ ~~e-d Problems-.--~
" - ---,<~·-··--·. -----· -----·-·-----·----·-----· ---···=-··--·-··-·-·--....
,,
.'
'
.
--- ---------·---
3. The maximum speed of the particle is:
(a) 1 mis
· · (b) 29 mis
·- "' ·r,, ·-P ,;1,
n ;,;
~l'' j'•
~,
p;as-S.A161E
..; ...,,.. ~,,, _
i'..nJ l.4 Lb:
~
4.
spring the stiffness of 100 N/m;and m = 1 kg friction
exists between mass 2 m and surface with coefficient
W =s O,s;' Tlie sy~tem is released with spring from its
reiax~d. position. Based on aJ;,ove data, answer ):he
following question: (take g .= 10 m/ s2 )
.:;]
5.
--~,
6.
1-"0m""-*;
7.
1/
I·
.·
:,~I=-_:____ ;-__:~_____
L_:__,_.
j
8.
1. Maximum extension in spring is:
(a) 8 cm
(b) 4 cm
(c) 36 cm (d) 20 cm
2. Magnitude'. of work done by gravity during the motion
of system is:
(a) 0.8 J
(b) 1.6 J
(c) 0.4 J (d) 4 J
3. Magnitude of net work done by spring after the system
!
fil s's:h'GTE
·-
is released for motion is:
o:s J
(d) -Jss mis
The minimum speed of the particle is: ·
(a) 1 mis
(b) ./40 mis (c) -Jss mis
(d) zero
The maximum value of potential energy is:
(a) zero
(bl 20 J
(cl 29 J
(d) 49 J
The least value of x (position of particle is) will be:
(b) - 2
(a) zero
(cl -../29+2
(d) ../29+2
The largest value of x will be:
(a) zero
(b) - 2
(cl -../29+2
(d) ../29+2
The position of equilibrium and its nature is:
(a) x ·= 2, unstable
(b) x = 2, stable
(c) x = 2, neutral
(d) no equilibrium position exists
(c) ../29 mis
fu the iho~figure, the sprfug and string is id;al. Th~
(a)
~_J
- (b) 1.6 J
4. Frictional force acting on the mass 2 m when it finally
comes to rest is:
(b) 8 N
(c) 12 N
(d) zero
5. After what displacement of mass 2 m, its velocity
becomes maximum?
(a) 4 cm
(c) 2 cm
-~-·
-
i, -,_ i»Tii~-- ~
,i~~;;' ~
l..{.....
'
'
:I
' -
.
.
· A block of mass m moving with a velocity v 0 ·on a/
smooth horizontal surface strikes and compresses a
spring ofstiffuess k till mass coines to rest as shown in
the .. figure. This phenomenon is observed by •two
observers :
A : .~tanding on the horizontal surface , .
B : standing on the bloc!,
(c) 0.32 J (d) 2.40 J
(a) 16 N
,..
""v,--~~
I
oJ;j;JyJJ~J"'
1. To an observer A, the work done by spring force is :
fr.xJ. acts ~n a. m = 1 kg particle
A s4!gle;. conservative
.
. moving ~long-the x-axis. The potential energy,UcxJ i~
given
bf - ·
·
·
·
Ucxl = 20.+(x- 2) 2
.
where xis in·mettes. Atx = 5 m, a particle has kineti
~~energy of.;!O J._____
---- . ---.---~
1. The total mechanical energy of the system 'is:
(a) zero
(b) 20 J
(c) 29 J
(d) 49 J
2. The minimum.potential energy of the particle is:
(a) zero
(b) 20 J
C~) ;29 J
Cd) 49 J
'~ '
(a) negative but nothing can be said about its
magnitude
1
2
(b) --mv 0
2
(c) positive but nothing can be said about its
magnitude
1
2
(d) +-mv
0
2
'
2. To an observer A, the work done by the normal
reaction N between the block and the spring on the
block is:
1
2
(a) zero
(b) --mv
0
(c)
www.puucho.com
1
2
+-mv
0
2
2
(d) none of these
Anurag Mishra Mechanics 1 with www.puucho.com
,W~RK AND ENERGY.
3. To an observer A, the net work done on the block is:
(a) -mv~
(b) +mv~
1
2
(c) --mv 0
(d) zero
2
4. According to the observer A :
(a) the kinetic energy of the block is converted into
the potential energy of the spring
(b) the mechanical energy of the spring-mass system
is conserved
(c) the block loses its kinetic energy because of the
negative work done by the conservative force of
spring
(d) all of the above
5. To an observer B, when the block is compressing the
spring :
(a) velocity of the block is decreasing
(b) retardation of the block is increasing
(c) kinetic energy of the block is zero
(d) all of the above
6. According to observer B, the potential energy of the
spring increases :
(a) due to the positive work done by pseudo-force
(b): due to the positive work done by normal reaction
between spring and wall
(c) due to the decrease in the kinetic energy of the
block
(d) all of the above
1
~-':?T' --·-:
0
:
PASSAGE ···
'
•
-
2
f
K
i
1. Which of the following Jaws/principles of physics can
be applied on the spring block system ?
(a) Conservation of mechanical energy
(b) Conservation of momentum
(c) Work energy principle
(d) None
2. The correct statement is :
(a) The block will cross the mean position
(b) The block will come to rest when the forces acting
· on it are exactly balanced
(c) The block will come to rest when the work done
by friction becomes equal to the change in energy
stored in spring
(d) None
PAS S,A Gl
This diagram depicts a block sliding along a,
frictionless ramp in vertical plane. The eightl
numbered arrows in the diagram represent directions,
to be referred_ to \Vhen answerin~ the questions.
:
B
' *12
1. The observer B finds that the work done by gravity on
1
-~~Jj
.I
and released.
u]I·
(c) -mgat 0
2
'
K
'
(a) .!:.mg t5
2
• -
an elongation less than Zµmg but more than µmg
A block of mass mis kept in an elevation which starts 1
moving downward with an acceleration aas shown in·
figure. The block is observed by two observers A and'
B for a time interval_ t O• •
•
2
-·
A sprii;ig block system js placed ·on a rough horizontal
floor. The block is pulled towards right to give spring!
PIISSAG~
the block is :
2
4. According to the observer A :
(a) the work done by gravity is zero
(b) the work done by normal reaction is zero
(c) the work done by pseudo-force is zero
(d) all of the above
..• i
I
2
1
(d) --mgat
0
2
(c) - mgat 0
2
··: ~"' 7
'. I 6
1 2 2
(b) --mg t 0
•.·..
2
(d) _.!:.mgat5
2
3
4
Ill
I5I . /··-
··---··
______ "_J
!
2. The observer B finds that the work done by
pseudo-force on the block is :
(a) zero
(b) -ma 2 t 0 (c) +ma 2 t 0 (d) -mgat 0
3. According to observer B, the net work done on the
block is:
1
2 2
(a) - - ma t 0
2
----··-··-
1. The direction of the acceleration of the block, when in
position I, is best represented by which of the arrows
in the diagram ?
(a) 2
(b) 4
(c) 5
(d) None of the arrows, the acceleration is zero
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
r312-
j
MECHANICS-I
2. The direction of the acceleration of the block when in
position II is best represented by which of the arrows
in the diagram ?
(a) 1
(b) 3
(c) 5
(d) 8
3. The direction of the acceleration of the block (after
leaving the ramp) at position III is best represented by
which of the arrows in the diagram ?
(a) 2
(b) 5
(c) 6
(d) None of the arrows, the acceleration is zero
PASSAGE
The kinetic energy of any body depends on the frame
of reference of the observer. The kinetic energy is
given by l/2mv~1• Similarly the displacement of the
object from different frames of reference will be:
different. But the forces acting on the body remain
unchanged. So work done by the forces as seen from:
different frames will be different. But work energy!
theorem will still be hold in every inertial reference'
frame.
·
For example, if a block of mass 2kg is moving with;
velocity of 1 m/s towards east on a rough surface its
1
' I
KE=-x2xl 2 =1J
2
If it comes to rest, its KE = 0.
Work done by friction= K 1 -K, = -lJ
If we observe it from a frame 2 moving with 1,m/s
toward east, its initial velocity will appear to be,
1-1=0.
3. Choose correct statement :
(a) In ground frame, work done by friction on ground
is positive
(b) In ground frame, work done by friction on ground
is negative
(c) II: frame 2, work done by friction on ground is
negative
(d) In frame 2, work done by friction on ground is
positive
.--
2
Final velocity = O- 1 = -1
'PASSAGE
i
The potential energy at a point, relative to the
reference point is always defined as the negative of
work done by the force as the object moves from the
reference point to the point considered. The value of
potential energy at the reference point itself can be
set equal to zero because we are always concerned
only with differences of potential energy between two
points and the associated change of kinetic energy. A
particles A is fixed at origin of a fixed coordinate
system. A particle B which is free to move experiences
an force F= (- 2a +
r3
1-) t due to particle
r2
A where
t
is the position vector of particle B relative to A.
It is given that the force is conservative in nature and
_potential energy at infinity is zero. If B has to be
removed from the influence of A, energy has to be
supplied for such a process. The ionization energy E 0
is work that has to be done by an external agent to
move the particle from a distance r0 to infinity slowly.
H_ere r0 is the equilibrium position of the particle.
(a)
~-f
2
(b)
-~-f
2
(d)
r
2
Final KE=_! x 2x (-1) = lJ
2
.? vl/or_k don~ by friction= 1- 0 =_lJ .. _ .. _
1. According to passage:
(a) In 2 nd frame, force of friction was opposite to
displacement
(b) In 2 nd frame, force of friction was in same
direction as displacement
(c) In ground frame, force of friction is in same
direction as the displacement
(d) None of the above
2. What should be the velocity of an observer so that he
will report the work done by friction on the block to be
0:
1
(a) I.m/s W
(b) -m/s E
(c) lm/s W
......
1. What is potential energy function of particle as
function of r:
Initial KE = _! x 2 x 0 2 = 0
2
---
(c)
-~+f
2
r
r
r
~+f
r
r2
r
2. Find the ionization energy E O of the particle B :
r
p2
(a) -
20:
(b)
2p2
~
3, If particle B is
o:
p2
(c) -
40:
p2
(d) -
o:
transferred slowly from point
P1 (-J2 r0 , -J2 r0 ) to point
P;(!Q..., !2...) in the .zy-plane
-J2 -J2
by an external agent, calculate work required to be
done by it in the process:
ca)
9p2
640:
(c)
2
(d) lm/s E
www.puucho.com
L
640:
Cb)
L
160:
(d) None of these
Anurag Mishra Mechanics 1 with www.puucho.com
313
WORK AND ENERGY
MATCHING TYPE PROBLEMS
1. Match the following:
____
Column-1
Column-2
.__ ___ - ----·- - ------ __
---- ---- - -··--·
(A) Work done by all the forces (P) Change in potential
energy
,
(B)
(C)
B
: QR
F~U~
A;C
Work done by conservative (Q) Change in kinetic
forces
energy
Work done by external (R) Change
in
forces
mechanical energy
(S)
None
___Column-1
_______
.,:..:.,..,,_.
p:
X
(P) p
B
(Q) Q
(C) C
(R) R
(B)
Column-2
---- - --
-------
(A) A
X
···-
(S) None
2. A particle is suspended from a
string of length R. It is given a
5. A body is moved along a straight line by a machine
delivering a power proportional to time (P = t ). Then
velocity u = 3.jgii at the bottom.
Match the following:
'.
Column-1
.
(P) 7mg
(B)
(Q) ~5gR
(C) Tension in string at B
(R) ~7gR
(D) Tension in string at C
(S) 5 mg
Displacement
proportional to
is (Q)
t2
(C) Work
done
proportional to
is (R)
t3
(B)
6. A pendulum is released from
point A as shown in figure. At
(T) None
some instant net force on the
bob is making an angle e with
the string. Then match the
following:
3. A force F = kx (where k is a positive constant) is acting
_on a particle. Work done:
.
C
0_·:/ :
..
··- .. __:___ ... ·
A
B
Column-2
Column-1
···--·.
(A) In displacing the body (P) Negative
= 2 tox =
t
to
Column-2
(A) Velocity at B
fromx
Column-2
(A) Velocity is proportional (P)
----- ---- ----
Velocity at C
match the following:
.· ..
Column-1
Column-1
(B)
In displacing the body (Q) Positive
fromx=--4tox=-2
(C)
In displacing the body (R) Zero
fromx = -2 tax= +2
Column-2
------------------------
4
(A) Fore
4. F-x and corresponding U-x graph are as shown in
figure. Three points A, B and C in F-x graphs may be
corresponding to P, Q and R in the U-x graph. Match
the following:
= 30°
-------·-(P) Particle may be moving
betweenB &A
(B)
Fore = 120°
(Q) Particle may be moving
between C & B Particle
is at A
(C)
Fore= 90°
(R) Particle is at A
(D)
Fore= 0°
(S)
www.puucho.com
Particle is at B
(T) None
Anurag Mishra Mechanics 1 with www.puucho.com
. MEotANiCS-1 :
10. Initially spring are in natural length. An application of
external varying force F causes the block to move
-7. Match the following:
' c"otiir'i\~~1 .. · ____
1--,---'---···~---~~.:::_·
Column:2
slowly distance x towards wall on smooth floor :
1
(A) Electrostatic potential' (P) Positive
'energy
'
'
(B)
Gravitational potential (Q) 1Negative
energy
(C)
Elastic potential energy (R) Zero
.
.
\ Column-1
;:;,. . , . .
1----·~· -~-'-----·.
(D) Magnetic potential (S) Not defined
energy
8. A particle of mass m kg is ,displaced from one given
(A)
Work done by S2 on block (P) ' Zero
(B)
Work done by.S 2 on S1
(Q)
:
point to another given point under the action of
several conservative and non-conservative forces
(Neglect relativistic considerations). Now match the
following.
:·~::>~-~~nz~:~9~l~flin~f·· ,
Coluni~--2 __ _
Colutjlri~,~--
(C) ,Work done by Fon block ,(R)
_.!_(
k1k2 )x2·
' 2 k1 +k2
•.!_(
k1k2 .Jx2
2 k1 + k2
(D) Work done by S1 on wall '(S) ·1
k I k 22 x 2
(A) '.Displacement of particle (P) Path dependent
(B) ,work
done
:conservative force
by, (Q) :Path independent
11. Column-1 represents potential energy graph for
certain system. Column-2 gives statements related to
graphs.
(C) ·Work
done
by (R) 'Frame dependent
]non-conservative force
Column-1
.
,.. ·
Column-;z,.,
~,,iJi,.._-~~-
"i, '-~ ~--"'---- _ . _ . . _ ; . , _ . _ _ _
(D) Angular displacement
(S) Frame independent
, (T) Dependent on location
.......................... 2mgl=E2
9. In the figure shown, upper block is given a velocity
6m/s and very long plank, velocity 3m/s. The
following quantities are to be matched when both
attain same velocity.
·
- rciUgh-
·-
•
(B)
e
vs 0 graph for a bob hanging
vertically from a string with its
lowest position as reference level
and 0 is angle of string from'
vertical line
~kg-6m/s
smooth
--~-~
mis,
(B)
(A)
(Q) For a small
U(x)
1
work done by friction on 1 kg (P) Positive
'block in joule
'Work done by friction on 2 kg
..............................•... E,
.• .•••..•..•.. •.. ..•..••..•..•
(Q) Negative
:Plank in joule
E2
------ -------- --------- E1
(C) !Magnitude of change in' (R) 3
momentum in N-s of 2 kg plank
(D) ,Change in KE of system (S) 7
consisting of block and plank in
, i)
joule
1
is E 3 , it.is not
possible for
;the body to
have any
turning. point
in its motion
U
- .
;____ ~:,~~mm,Ji;;;,3
(P) If total energy
........................... 2.5 mg!=E3
of observer in a given
frame
j.
U(B)
(A)
A pmticle moving along x-axis
with potential energy function as
2
U(x) = [1- e-x ]
(T) 2
www.puucho.com
:displacement
about point 0
potential
energy
function is
'quadratic in
'variable
plotted on
x-axis
Anurag Mishra Mechanics 1 with www.puucho.com
r-----.
, WORK AND ENERGY
r.-----'---._-_··.c_·-c..·-'----'--"-'-~-=;_:__:__;_:_:_=__;_--::;-·c.___:::;--
U(x) .
(C)
. -
(R) For a small
displacement
about position
0 motion is
simple
harmonic
. . . . . . . . . . . . .. . . . .............. E,
, ••••• Ez
--"::_-':-1-=_"a-+--'+"'a+-'--=•x
............... ······· ....... E,
-- ·- -
-- -
- --· --- -
---- -
--
- - -
-- -
-·- - - --
13. In column·l, a situation is depicted each of which is in
vertical plane. The surfaces are frictionless. Match
with appropriate entries in column·2.
0;•C<iluinn~1
Column~2' ,
·--'--·
-·-·-------~·--'- (A) .Bead is threaded oh a (P) ,Normal force is
circular fixed wire and is
zero at topmost
,projected from the lowest
point
of
its
trajectory
point
-------
-~- -~--
;!,•
'.'
,Potential energy function of a
panicle in an arbitrary force field
(D)
(S) If total energy
is Etoral < E2
U(r)
panicle
executes
periodic and
oscillatory
motion for all
energy values
greater than
energy atO
. ·····························E,
a
-----·
Ro b
······E2,
'
'
..............
.' ·········E1
Graph represents potential
energy for a particle
5m/s
(B) _Block loosely fits inside the (Q) Velocity of the body
fixed small tube and is_·
projected from lowest point'
.-··
(T) Point Q is
position of
stable
equilibrium
12. A bob tied to an ideal string oflength 1is released from
the horizontal position shown. A peg P whose height is
adjustable, can arrest the free swing of the pendulum,
as shown in figure.
!
y:f:
Peg°"'\
'
···'-O-······
(C)
Block
is
projected (R)
horizontally from lowest
point of a smooth fixed'
cylinder
~=1m
.•
....,'
6m/s
'.
(D) Block is projected on a fixed: (S) Normal force is
,hemisphere from angular
position 0
(A)
21
21
For what range ofywill the (P) <y<string wind up on the peg,:
15
3
remaining taut throughout'
the swing
(B)
Forwhatrangeofywillthei(Q) •0<y <~
'pendulum
become:
:
5
'projectile
(C)
For what value of y will (R) ~ < y < l
mechanical energy always
5
remain conserved
l
Acceleration of the
body is zero at the
topmost point of its
trajectory
''
~
I,
.. •
-JZomls
•
I 11 I \Ill\\" I I\ II I\\\
is zero at topmost
'point
of
its
trajectory
'(S)
l
~
_ _cos 8 = 2/3
21
-<y<!3
3
~'
www.puucho.com
radially outward at
-topmost point of
·trajectory
Anurag Mishra Mechanics 1 with www.puucho.com
[316 _____ -------- __ ------ __ -- --~------- ~--"'
. .
'
Column-1
14. A block of mass m is tied with an inextensible light
string of length!. One end of the string is fixed at point
0. Block is released (from rest) at A. Find acceleration
of particle during its motion in vertical plane at
positions specified in column-1 and match them with
column-2. Given that A and O are at same horizontal
level.
A---i._-_----,0'
<·
.
''--"
'
.·,
Column-2 ·
---I
· (P) Acceleration is horizontal
<-----·-------(A)
Highest point
(B)
At lowest point
(C)
At
0; tan- 1 (-.J2) (R) 'Acceleration is vertically
,with vertical
,downwards
; (Q) :Acceleration is vertically
'upwards
I
, (S) ·Acceleration has both
:
horizontal and vertical
components
!
-
-AN8WER8
- - -- . -·
-
,.
. -
-
=-Lf:vel~~:- O~iy ~~e ~A!te~na.tii~-is :c~~r~~ ~"3.
(a)
7.
(c)
8.
(b)
(c)
14.
(c)
15.
(c)
16.
(d)
21.
(c)
22.
(d)
23.
(c)
24
(c)
(a)
29.
(d)
30.
(c)
31.
(b)
32.
(d)
(a)
37.
(b)
38.
(c)
39.
(d)
40.
(c)
44.
(c)
45.
(a)
46.
(d)
47.
(b)
48.
(c)
(c)
52.
(c)
53.
(d)
54.
(d)
55.
(b)
56.
(b)
59.
(b)
60.
(c)
61.
(d)
62.
(b)
63.
(c)
64.
(a)
(b)
67.
(d)
68.
(a)
69.
(bl
70.
(a)
71.
(b)
72.
(c)
{dl
75.
{cl
76.
(d)
3.
(b)
4.
(c)
5.
(c)
10.
{d)
11.
(c)
12.
(b)
13.
(a)
18.
(a)
19.
(d)
20.
(a)
25.
(d)
26.
(b)
27.
(b)
28.
33.
(b)
34.
(d)
35.
(c)
36.
41.
(d)
42.
(d)
43.
(c)
49.
(a)
50.
(c)
51.
57.
(a)
58.
(b)
65.
(c)
66.
73.
{al
74.
(c)
9.
(b)
17.
=
'
i
6.
2.
1.
'
'
{al
1:_~v~1-2: f.!or.e-i:11a~ o_n~ Alt~r-n_~tiv~ i~t~r~ c;;rr~~i' ·
1.
(a, b)
2.
7.
:(a, d)
8.
13.
,(b,c, d)
(a, b, d)
:(a, c)
3.
(a, b)
4.
,(a, c)
5.
(a)
9.
{b, c)
10.
(b, d)
11.
(b, d)
12.
6.
(b, C, d)
·(a, c)
14.
(a)
15.
(b, c)
16.
(a, d)
17.
(a, d)
18.
(a, b, c)
:<c)
19.
(a, C, d)
20.
'(c)
21.
(b)
22.
i(a)
23.
(b, C, d)
24.
25.
(a, c)
26.
(a, d)
27.
·(a)
28.
(c, d)
29.
·(a, b, d)
30.
(a, d)
31.
;(d)
32.
'(a, c)
33.
,(a, b, c)
(a)
(ii)
(c, d)
(iii)
,(b, c,d)
I
34. (i)
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
=r-
-~-- ----.
; WORK AND ENERGY
,.__ - -
-·-
-- -·- ..--
-----· --- ----
------·--·-
-
-- -
L~vel-3:C~~~r!he~~io~ B~s~d ~~o~l~~s ,_:>
Passage-1:
1, (a)
2. (b)
3. (cl
4. (c)
5. (a)
2. (b)
3. (d)
4, (d)
5. (d)
6. (c, d)
2. (b)
3. (c)
4. (d)
5. (c)
6. (b)
2. (a)
3. (b)
4. (d)
Passage-2:
1, (d)
7, (d)
Passage-3:
1, (b)
Passage-4:
1. (c)
Passage-5:
1. (c)
2. (c)
Passage-6:
1. (b)
2. (a)
3. (b)
2. (b)
3. (c)
2. (c)
3. (b)
Passage-7:
1. (b)
Passage-a:
1. (b)
=a:Match_i·!'~-~pe_ P_~bi_e_rns~
1. A - Q, B - S, C - R
2. A - R, B - Q, C - P, D - T
3. A - Q, B - P, C - R
4. A - R, B - S, C - P
5. A - P, B - Q, C - Q
6. A - Q, P; B - T; C - R; D - S
7. A - P, Q, R; B - Q, R; C - P, R; D - P, Q, R
8. A - Q, R; B - Q, R; C - P, R; D - P, R, T
10, A - Q; B - S; C - R; D - P
9, A - Q; B - P, S; C - P, T; D - R, Q
11. A-P, Q, R; B-P, Q, S; C-P, Q, S; D-Q, S
13.
A - Q, S; B - P, Q; C - P; D - Q, R, S
-317 -
------- ----------------------------- _________ J
---- ·---- ---- - - - - -
12. A - Q; B - P, R; C - Q
14. A - R; B - Q; C - S
www.puucho.com
8. (b)
Anurag Mishra Mechanics 1 with www.puucho.com
r3Ts
·. · ·
L_:__.._:_..__:___ _;_:;="'-'-·,_·-·----~·-:_,.~,'":.:.:...~---~-'·-,!
=!!~:-1i9~~
0
0n~.
Alter!!~~fs.c:;~
7. (c)
ti.KE = 0 = Work done by boy + Work done by gravity
+ Work done by spring
Work done by boy = - (Work done by gravity
+ Work done by spring)
1. (c)
As shown in figure F = mg sin 0,
vertical displacement in time
t = vt
Work done = Fvt sin 0
= mgvt sin 2 0
8. (b)
a
2. (d)
Kinetic ~nergy of a body depends upon the reference
frame and so does the work done. Since two observers
are not accelerated w.r. t. each other so they will
observe same force acting on mass and so they will
observe same acceleration of block.
3. (b)
At x 2 , if we displace the body in +vex force acts in -ve
direction and if we displace the body in - ve x, force
acts in +ve direction. So it is a position of stable
equilibrium.
4. (c)
W =U1 -U1
f~ v dv = f: Pdx
2
m
KE= .!.m(2as) = mas
2
6. (a)
Speed will be maximum where a = 0
kx=F~x=F/k
1 2 1
kx - - k:x: = - mv 2 (by work energy theorem)
2
2
k
3
3P
From energy conservation
Now since the two velocity vectors
shown in figure are mutually
perpendicular,
hence
the
magnitude of change of velocity
will be given by
It,. ;I = ,l-u-2_+_v_2
Substituting value of v 2 from
equal (i)
2
F2 1
2 1 F2
F
-=-mv +-k-~v=-2
k
2
v 2 =u 2 -2gl
KE= .!.mv2 = .!.m(!:.c)2 - _F_2t_2
2 .
2
m
2m
F
F2
2
·V=krt
Therefore, tangential acceleration, a, = dv = kr
dt
or
Tangential force, F, = ma, = mkr
Only tangential force does work.
Power= F,v = (mkr)(krt)
or
Power = mk 2 ; 2t
vdv
P
9. (b)
a=-=dx
mv
10. (d)
m
= k 2 rt 2.
m( 3
3
m
3
s
Px = - V2 -Vi) ~X = -(V2 -Vi)
(b)' P=Fv=F-t=-t·
(c)
or
=0-(-~g %)=Mt
5. (c)
(a)
'
2
~
r
or
= k 2 rt 2
.Jmk
www.puucho.com
2
iti. ;I= ~u +u
2
-
2gl
.... (i)
I
'
r
:
'.
,-'.
.:
,
1·
-·- .,, .. -~.... ••. !
Anurag Mishra Mechanics 1 with www.puucho.com
11. (c)
By work energy theorem,
AKE = Work done by (gravity
resistance force)
+ buoyant force +
Because to reach same maximum height their vertical
velocity should be same,
1
2
- mv 1
-2- - =
_! = 0.5
1
2
-mv
2
2
o-.!mv = mg x 2- 2mg x2+W
2
2
2
18. (a)
1
W = --mv 2 +mg x2
dU
dr
2
-k
r2
k
r
--=-=}U=--
1
= --x 0.05x 20 2 + 0.05x lOx 2 = -9J
2
12. (b)
µmgs= mgh
Since radius of sphete much larger than the
displacement of particle' so it can be assumed to
perform linear motion.
1
0.0lxmg xs = mg x (1cm) =}S= --cm= lm
0.01
•
mv 2
k
1
2
k
--=-=>-mv
-2
r
r
2
2r
k k -k
Mechanical energy = KE + PE = - - - = -·-.
2r r
2r
19. (d)
By work energy theorem,
1
2
2 mv
-
= w1 + Wgravity
0
r
13. (c)
= 200xPQ-mg2
r
= 200x r- lOx lOx2
½10v 2 ;=150Xl0
=}
=}
14. (c)
In first case, mg = kd
d= mg
=}
k
v = .J300 = 17.3m/s
20. (a)
1
2
1 kx2
-mv
-0=-
2
2
+µmgx
.
2
.!x 1 xv = Lx 2x 4 + 0.25xlx l0x 4
2
2
u = ..J52 = 7.2m/s
2
In second case,
·
1
2
2mg
k
mgx = - kx =} x = - - = 2d
2
.
·
15. (c)
21. (c)
If original speed of boy is v O then
For vertical motion
2
..... (i)
.!Mv = .![.! M v~]
2
2 2 2
1
2
1M 2
Also
-M(v+l) =--v 0
2
2 2
From eqns. (i) and (ii),
2
v+1)
"l
( - V - =2=}V= ..fz-1
u=..f2+1=2.41m/s
From equation (i),
v 0 = 2u = 4.Sm/s
17. (a)
. 450 V2
v 1 =V2SIIl
= ..f2
=
Wv~
2g(W + f)
v~
2g(l+ f/W)
Also for whole motion; work done by gravity is zero.
1
2
1
2
2fh = -mv 0 --mv
2
2
2
2
·4fh 4ghf
2vU
Vo-V = - - = - - = - m
W
W+f
v = v~(1-.....3L) = v~(W- f)
2
16. (d) ·
1
2
1 2
W =-k(x+y) --kx
2 ·
2
1
=-ry(2x-l:y)
2
.
h
=}
..... (ii)
·
W+f
W+f
v=vo~W-f
W+f
22. (d)
2
2
We know
· mv
m ( - c ) (asmvr=c)
T=-.
-=-
=}
T
www.puucho.com
oc
r
r mr
3
r- => n = -3 ·
Anurag Mishra Mechanics 1 with www.puucho.com
~[3_2_0_ _ _ _~--~~-----~~~-~---:~·''"'--'--'--~""---~M_·~_HA_N_l~~S_,l,j
23. (c)
For any incline plane . if the block
slides down the incline plane, work
d ~ friction ~ µmg cos0
..Jx· + y• = µmgs
which
is
independent of y and 0. So in all the
- case, since µ and 's' are same so loss in PE will be same
· so final KE and so the speed will be same.
24: (c)
mgh
31. (b)
Work done
= mg(2R)+.!mv 2
= Loss in energy
1
2
=mg2R+-mv
2
_ 2mgR mgR _ Smgk
mgh +----2
2
. SR
h=-=2.SR
2
= 0.1 X 10 X 2 X 0.1 + _! X 0.1 X 5 2
2
= 0.2 x 0.1 x 125 = 1.45 (negative)
2
25. (d)
32. (d)
Since force is always perpendicular to velocity, it will
always do zero work.
26. (b)
By energy conservation, we have
1 2 =mg
. R . 0
-mv
sm
~t'!Jv·
....
N·
/fL_
2
mv
=>
2
-R
·
,
I
.
= 2mg sm0
.
Because kinetic energy does not depends direction of
projection.
33. (b)
For spring A:
1- 2
F = kAx =>E = -kAx
2
.
For spring B:
F,=kBx'=>EB =½kBx'
mg.
1
p2 l p2
EB =-kB-=-2
ki 2 kB
mv 2
N=--+mgsin0
R
N = 3mg sin0
lk 2 x 2
1
= __A_= -(2kAX 2 ) = 2£
2 k~
2
27. (b)
Hz force on track due to mass m
34. (d)
= N cos0mg sin0cos0 = ~mg sin20
Work done by spring
2
= frictional force
Tmax_-Tmin = 6mg_
35. (c) Potential energy
= Zfmin
36. (a) Total energy
3 7. (b) Kinetic energy
38. (c) Potential energy
Tmax
Tmin =6mg
29. (d)
x3
For equilibrium,
F=O
X
30. (c)
When level become equal then h =
h 1 +h 2
2
I
Work done by gravity change in potential energy of the
40. (c)
By conservation of energy,
1
2
1
3
-mv = mgl+-mgl = -mgl
2
2
2
=>
V = .j3ii.
water ·column
= 2(Ahp)g!:-[Ah,pg!!J..+Ah2Pgh2]
2
,
.! kx 2
4
Since the biock has acceleration both · in vertical,
horizontal and radial direction. So it is not in
equilibrium.
x2
__!_(b2a) = 0=>x = 2a
x
b
2
2
39. (d)
F=_dU =-2a+_I!_
dx
= __! kx 2
Work done on each mass = -
28. (a)
We know,
Since
=>
2
_2
2
.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
321'
,.
41. (d)
46. (d)
By conservation of energy,
J
42. (d)
By conservation of energy,
2
.12m2
1
2 - 2 mg-=
Xm
+v +-mv
lk(Xm)
2
2
2
2
2
_!.kx! +~mv 2 -mgxm
8
'i
2
: ,
.
--· ---··:.- ·L____ ·--· _.-·· -- '
)
O
'
';
I
By conservation of energy:
2
O= ½mv -mg(Rcos9- ~)
=0
KE= mgR( cosa-½)
p
~mv 2
= dKE = mgR(-sin9 dB)
dt
dt
= mgRsin9co
2
.
1
=lxlOxlx-xco=Sco
43. (c)
2
L
(i)
= Mg - = 0.SMgL
2
(ii)
=Mg!:_ = O.SMgL
47. (b)
Extension in spring to lift the lower block x
.
2
r-·-- - - - -~-... -- " "'1
= Mg 2R = 2Mg !:_ = 2MgL
1t
1t
(iv)
\
,
~
0
••• i
' '
I
_!.kl6m2g2 +~mv2-mg.4mg =0
8
k2
2
k
(iii)
'
•
; PE~ O····· ...,. .
k
2
•
I
-2mgx+.!.kx 2 = 0 => X = 4mg
j
= 0.2MgL
k
+mg!
7t2
1t
= mg
' k:
= _!. MgL = 0.4MgL
lt2
(v)
=Mg(
=
I
R-2:)
! -·-- -·- -·---- _J
(1t-Z)MgR = 1t-Z MgL
1 14
7t
2
2
lt2
7t
= ·
By conservation of energy,
1 2 = mg (mg)
1 (mg)
-mgxo +2kxo
k +2k
k
MgL = 0.114MgL
1 2
2kxo -mgxo
(v) < (iii) < (iv) < (i)= (ii)
44. (c)
kx5 - 2mgx0
Fv=P
p
v=F
Xo
for maximum v, F should be minimum which is
equal µmg just to drag the block.
p
vmax = µmg
F=~
45. (a)
V
=>
dv
m-
dt
Xo =
)
k
3m2g2
2k
=O
2mg ±~4m2g2 + 12m2g2
2k
2mg ± 4mg 3mg .
=-2k
k
48. (c)
Errata: The mass of the lower block should be m.
By conservation of energy:
-
=-o: => Jmvdv =I o:dt
,
D
-
- --~-
!ht __ ····-··· ..
V
2
( m~
=
3m2g2
- --"--
=
= o:t
Af<E = Mt = Work done
www.puucho.com
--
tmgi
·_
,,,, '
--~--''
Anurag Mishra Mechanics 1 with www.puucho.com
0
a
tan-=2 g
.
e =2tan-1
(i)
ma
.
'---==·· -~
.52. (cl
49. (a)
Conserving-energy for any sphere_
'
1
.
0=-mg(R-Rcos60°)+-mv 2
.
.
. '2 •
2
.
·
•
'
,,
.I(".,
,
N=Mg
·
.
·
... (i)
mv 2
. For any sphere N + mg cos 60° = - .
R
N +mg·,; mg~ N
2
w·=
53. (d)
,·'r.::: ·: ~. mt .·_ :~• •
A
A
,._•
A
F = K(yi + rj)
r" dW, = -Kr"
Jqo
Jqo d(:xy) = -K[:xy]~g
0
0
· W=~Ka 2
...
• J
A
dW = K(ydx+ xdy) = -Kd(:xy)
'
~ - µ ~,-~·
-
....
....
and
'
. '!!_=gR==>v=..{gR
-. 2
2
As shown in figure, ·for tube. 2N cos 60° = Mg (when'
it just lifts off) ·
=F. ds where ds =dxi + dyj + dzK
....
dW
= mg
2
U(x) = k(l- e-x )
It is an exponentially increasing graph of potential
energy (U) with x 2 • Therefore, U versus x graph will be
as shown.
From the graph it is clear that at
origin.
Potential ·energy U is minimum
(therefore, kinetic energy will
be maximum) and force acting
on the particle is also zero because F = - : = (slope
... (ii)
2
From eqris. (i) and (ii),
. m
mg.
-·=Mg==>-=2
2
M
of U - x graph) = 0
Therefore, origin is the stable equilibrium position.
Hence, particle will oscillate simple harmonically
about x = 0 for small displacements.· Therefore,
correct option is (d).
(a), (b) and (c) options are wrong due to following
reasons:
..
F =-dU
()
a At equilib'
num pos!Uon
dx =o·1.e., slope
U - x graph should be zero and from the graph we can
51. Cc)
By
work · energy
theorem
from
trolley
maLsin0 = mgL(l- cos0) asin0 = g2sin 2 0/2
see that slope is zero at x = 0 and x = ± =.
Now among these equilibriums stable equilibrium
_position is that where U is minimum (Here x =0).
Unstable equilibrium position is that where U is
maximum (Here none).
Neutral equilibrium position is that where U is
constant (Here x =± =).
Therefore, option (a) _is wrong
(b)For any infinite non - zero value of x, force is
directed towards the origin because origin is in stable
equilibrium position. Therefore, option (b) ·is
incorrect.
(c) At origin, potential energy is minimum, hence .
kinetic energy will be maximum. Therefore, option
(c). is also wrong.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
~KAND ENERGY
ds = vdt
s ~ rv 2 dt
F=-dU
dx
dU=-F.dx
54. (d)
or
3231
-~--------~----...~
-------~--~--
=
U(x)
f
-J; (-kx + ax 3)dx
3. (a, b)
Whenever the displacement is towards the relaxed
position, the spring does positive work.
kx2 ax4
U(x:r=--2
4
4. (a, c)
'fik
f-;i"
a= L;v = Lt
U(x) = 0 at x = 0 and x =
U(x)
= negative for x
m
55. (b)
Let x be the maximum extension of
the spring. From conservation of
mechanical energy:
Decrease in gravitational potential
energy = Increase in elastic potential
energy
P = fv = -
k
W 1 =W2 _=W3
1
2
KE=-mkRt
= -J; Fdx =J; (kx)dx
dKE
P=-=mkRt
dt
Paverage
~2
mkRt
= -If,O mkRtdt = t
2
8. (a, c)
=0
-+ -+
Therefore, the correct option is (a).
=7:;el_::;~,~!~~:n,,?~ A~ma!i".!'Fj;~~-;;~~
1. (a, b)
Final KE will be larger than initial KE. Larger the initial
KE larger will be the final KE.
2. (a, b, d)
mdv
Fv=P=;--v=P
dt
mvdv =Pdt
dw=f.'ds
Since body is hauled slowly, so
f = mg sine+µmg case
W = (mg sine+ µmg case) ds
f
= f mgds sine+ f µmgds case ds
= f mgdy +f µmgdx
= mgh+µmgL
= 2Pt =;v~J2Pt
m
µg( m, + J
2
= _du
kx2
v2
+ ~zg =
R
U(x)=-2
U(O)
]
ptton c
6. (b, c, d)
Work done by gravity and tension force is equal to
20J.
7. (a, d)
v2
2
- = kt 2 =; v 2 = kRt
dx
As
[O.
For mass m2 to· move, Kx = µm 2g
By work energy theorem on m1, Px - µm 1gx - 2Kx = 0
1
p = µm,g
56. (b)
Gravitational field is a conservative force field. In a
conservative force work done is path independent.
f~CxJ dU
[Option a]
5. (a)
2
F
t
m
__
lmv2 __ m __
f2t2
F2
2 m2 .
= ~kx 2
k
57. (a)
From
2
P =µm,g+-Kx= 0
2
2Mg
x=--
or
m
J
>~
From the given function we can see that F = 0 at x = 0
i.e., slope ofU - x graph is zero at x = 0. Therefore, the
most appropriate option is (d).
Mg x
=- s ~ r312
m
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
9. (b, c)
For no sliding rimg =.µ(ri - l)mg
=>
,---'-·.,.----,
'. .........x_ .,.._ '!
µ =.c3_
ri-1
If x length is remaining on table,
work done by friction force for
displacement qf ,ch: will be:.
1~··,
; , •,' 01
j· . ,", 'qi
1.· · · · oJ
I.·
13. (b, c, d)
Resultant force on particle
= F - ,,:ng
= (2-az)mg-mg = (1-az)mg
.
- - = (1-az)mg
0-1
dz
=(z-~ J
-dW=µ-xgdx
.
l
W = rc~-l) µmg xdx = µmg (fl -1) 2 12
Jo
1
21
(a) For maximum height v
µmg(ri -1) 2 1
=
2
10. (b, d)
(b) If two blocks are sliding over eacl, other, kinetic
friction does +ve work on one end. negative are
otheL
·
(d) Work energy theorem is valid from non-inertial
frame and we have to consider the work
done
by the pseudofo!ce
. .
. also.
11. (b, 4),
For KE to increase power should be positive ,
=>
2
Z=H=a
Velocity at
-=-
H
2
'
;;;
h- •"'
2
v:
=>
m
.
l _._'_
' F
,·f~-·'
mvdv
':'
fug
=0.
1
a
(b)
(c)
\
_, _,
·f.v>O=>e < 90°
P = .J2Km => p ~ -.JK
F = -dU = (-'-2a
dr
For l=
r3
+~)
02
r2
dr
For
'
2
4
r
3
2·
dr
2
r
3
r
-o)
.
=2(~axb
_b)=2.!?.=~(+ve)
3
3
2
2a
r
r
2
'2(3a_i,)=o.
3
r
r
=>r=3a
. b
r.
3
3
·
2a
b
2ab
b
F=---+--=---+3
2
(3bar (3ba
27a . 9a
b3
= 9a 2
-
u2
'Jg sine.
17. (a, d)
r
So U is minimum.
So it is a positjon of stable (steady) equilibriu!".
· '.
dF -d 2u
For maxnnum - = - = 0
dr
dr 2
·
=>
2(2g sin0)S => S
v=..Jii
. ' ' 2a.
r=-,
. b
d U
·
r
=u 2 -
16. (a, d)
To. complete the circle tension at
top point should just become zero
mv 2
for this - - = mg
l
=>
b
2a
2a
0, · , ,. 2 =; :-:g =>.r = -b
. r
r
.
2
d U_ +6a _·2b _ 2 (3a
14. (a)
!fit slides down with constant velocity,fcirce of friction
= mg sine
For motion up the plane a= -2g sine
2b'
b3
2
27a = 27a 2
www.puucho.com
kA > kB
lk
2
•
1
.wA =- Ax ;WB =-kBx
2
·
2
=>
WA >WB
For same force;
.
·F
F=kAXA =>XA = kA
2
Anurag Mishra Mechanics 1 with www.puucho.com
IWORK AND ENERGY, _·
23. (b, c, d)
At stable equilibrium position U is minimum, at
18. (a, b, c)
Work done by F1 = 20 x 6-J2 = 12()-J2J
Work done by F2 = 30 x 6 = 180J
21tR
Work done by F3 = 15 x = 45itJ
4
19. (a, c, d)
-du
(a)
F =-=-3
X
cJx
unstable It is maximum and at neutral position it is
constant.
2
dy
(c) For particle to cross y-axis x = 0
1
2
X =vxt +-axt
'
-+
v = 0-(3 i+4j) x 2 =>\vi= lOm/s
6x = 0 -
1
f.y = -0- -
2
.!_ X 3 X 12 = -1.5
2
X 4 X1
2
= -2
Co-ordinate = (6 -LS, 4 - 2) = (4.5, 2)
20. (c)
Since speed of mass is zero at C
and B, so they must lie on same
horizontal surface.
AP AO-PO
cos0=-=--PB
PB
Lcosa-l
L-l
=-~--
i
O
~LJ-'
21. (b)
J
W = F.ds =
=
W
J(y l+ xj)-(dx l+dyj)
J(ydx+ xdy) = Jd(xy)
= J((\O)
f(l,I) d(xy) = (xy )at = 1 J
22. (a)
Tmin = 6mg
Tmax = 2Tmin (given)
Tmin = 6mg, Tmax = 12mg
Since tension is max at lowest points
We know
and
Tmax
-
mv
2
T=mg+-L
mv
---1
i- ~
L~--~--
~ x 200 = 60 N
10
Work done by ladder on boy is zero
because while ladder applies force
on boy, his point of application does
not move.
Work done by slide = Work done by friction
= -60 X 10 = - 600 J
For resultant velocity
(d)
3
Frictional force =
2
...
I
____-_2'!'J
2(1+it)gR
25. (a,c)
-,6=0-.!.x3t =>t=2sec
,..
ml
'
v=
2
-t
.,
1tR 1
2
mgR = 0-2mg-+-(3m)v.
2
2
·
3;,,v 2
- - = mgR(l + it)
-du
_,
•
•
Fy = = -4=> F = -(3 i+4j)N
2
:-----·-·1
.
24. (c)
By conservation of energy,
C
27. (a)
The statement of work energy theorem.
~8_. (c, d)
Kinetic energy will be constant because, the f~rce w/11
do zero work. This is the case of uniform circular
motion.
29. (a, b, d)
Since work is done against gravity which is a
conservative force, do work done is independent of
path followed.
30. (a, d)
For mass m to complete the vertical circle, the string
becomes slack at highest point.
31. (d)
r
:·rs::-/,° PE=O
T - mg sin0 = mv2
r
=>
I
T · .,.
1 •
.!.mv 2 =mgrsin0
2
-· - - - - -
-
l ''.
[8
.
~
V
· · .mgsin8 !
mg
I
-·~-__:
=>
T = 3mg sine
maximum value of T = 3mg and given that string can
with stand a maximum load of 4mg. ·
:. it will not break.
2
12mg =mg+L
=>
-.
= ~llgL
-
www.puucho.com
'
Anurag Mishra Mechanics 1 with www.puucho.com
=·
326
MECHANICS-I j
'--~;;:;;:;;:::::::;;::;:::::;::;;;:;:::...---~----~---,--,
~·------"-'-----'
•..
Leve~~2 sompreh~~n Based Prob~em~
mg(2x)- 2µmgx-.! kx 2
2
2
4, (d)
=O
zero
6. (c, d)
Particle will move between the points where KE
becomes zero or its PE is equal to total energy.
Thus,
49 = 20 + (x- 2) 2
or,
(x-2) 2 = 29
X=2±,,/29
20x-16x- 50x 2 = 0
x= 0.08 = 8cm
2. (b)
8. (b)
Gravity does work only on the hanging mass and it is
equal to = mg(2x) = 10 x 0.16 = 1.6 J
3. (c)
Work done by spring force = energy stored in it
1 2 1
= -kx
= - X lOOx (0.08) 2
2
2
= 0.0064 X 100 = 0_32 J
2
~~
As shown in figure f, + 8 = 20 => f,
Vn
= ~7gR
Further,
TB
mv 2
= __
B = 7mg
Again,
v2 = u1 - 2ghAc
R
(a)
At any instant where 2m has been displaced by x.
mg - T = 2ma
zr - kx - 2µmg = 2ma
From eqns. (i) and (ii), we have
mg
=!!a!ih}~~}vpe ~~~lem~
= 12 N
fmax = 0.8 X 20 = 16N
¢
·
change in potential energy.
2. v~ = u1-2ghAB = (9gR)-(2gR) = 7gR
I
T
au
F = - - = -2(2-x)
dx
For
F=0,x=2m
Since at x = 2m, PE is minimum so it is a position of
stable equilibrium.
1. Work done by conservative forces is negative of
-------~7
120~81
So
= 20J
.!mv 2 =29=>v=.J5Bm/s
By work energy theorem
s.
= 2m so, Umin
Kmax =E-U = 49-20= 29J
Passage-1
1. (a)
4. (c)
2. (b) , U will be minimum at x
3. (d)
= (9gR)- 2g(2R) = SgR
.... (i)
Ve =~5gR
.... (ii)
Further,
mv 2
Tc+mg =--c
R
T
2
Tc =4mg
• EJ.:: icx
.....,.
a
3. From x = 2 to x = 4, force is positive and displacement
2µmg
2mg - kx- 2µmg = 2ma
For max velocity
a=O
2mg(l-µ)
==>
X;:; _
_ __
20 X 0.2 = 0.0 4 = 4cm
k
100
Alternate method:
Velocity blocks will be maximum when their
acceleration become zero. From mass 2m
kx+l6= 20 => X = 4cm
Passage-2
1. (d)
at x = Sm= 20 + (5 - 2) = 29J
Total Energy = 20 + 29 = 49 J
2
is also positive. Hence, the work done is positive .
Similar logic can be applied to other parts also.
4. A is the point of stable equilibrium, so potential energy
is minimum. Similarly, point C is the unstable
equilibrium position, where potential energy should
be maximum.
5.
p ~t
W
=f Pdt =f atdt
or
W oct 2
Since, work done is equal to change is KE
Hence,
v 2 oc t 2 or v oc: t
Further,
or
or
www.puucho.com
v = -ds
dt
ds~tdt
soc t 2
ds
-~t
dt
(by integration)
Anurag Mishra Mechanics 1 with www.puucho.com
,_w_o_RK'-.
r A_N_li_EN_ER_G'""~--.C---''----------~------'",.-."-·'-"---·-·:_·',-'.\_,,~;;~.,...:;::_0.
6. Angle between net force and the string can never be
obtuse. It is 90° at A, 0° at B and acute in between.
13. (A)
u = Sm/s
N•
2
2
v =u -2gR(l- cos0)
(BJ
U
< Umm,
V
= 0 at 0 =
Ii
•
,._.T
:
···s;i / ... ~-·-~-;~i- ;.~.j ··--~\
...... ,,, ..
..---a
...•,.
~-------~r:~~A~~
..
,,.
mv 2
u =6m/s < umin
f7 1::-·--1
:
:
YB'·
IT!lil
At 'C' : T - mg cos8 =- l
L.---
Ir
_.-......,:
-· 0 •.
l.~-
At 'If: v = 0,. No centripetal ~cceleration
So acceleration is downward (Due to mg)
At 'B' : T and mg both are vertical so acceleration is
vertically upward (centripetal acceleration)
I
!(
-;-'~'..O ;
A
mg
1-·-
(C)
14.
COS-l(¼)
u=-J20m/s<umin,
r,~;'..3,~I;]
mg Zcos8 =.!mv 2
2
mg,;
s··,
1
. ms1
'-
After leaving the cylinder it will follow projectile path.
(D)
... (1)
••• (2)
From eqns. (1) and (2),
T - mg cos8 = 2mg cos8
T =3mg cos8
IfT cos8 =mg then vertical component of acceleration
· will become zero.
(3mg cos8) cos8 =mg
cos8=J_
./3
.!mu 2 =.!mv 2 + mg R (1- cos8)
2
2
v=O
~tan8=-J2
. .
So at 8 = tan-1 ( .,/2) acceleration has only horizontal
component.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
' ·I
\
-
I
I
IMPULSE AND MOMENTUM,'
.- ,-'
IMPULSE
When a force Facts on a body for a very short interval, it
is called an impulsive force. The effect of impulse is
characterized by a vector quantity represented by the
-->
symbol J.
-->
-->
(:E Fext
Impulse of force F exerted on an object is
-->
J
=f,1'
-->
Fdt
'
Principle of Impulse and Momentum
Newton's second law may be expressed in the form
-->
Fext
-->
or
i.e., Initial momentum of system = Final momentum of
system
which is the principle of conservation of
momentum. We may state it as:
If the net external force exerted on a system is zero
Fdt
d
= O), then the momentum of the system is conserved.
When no force acts on a system from outside the system, we
say that such a system is isolated. The internal forces
exerted by one part of the system on another part have no
effect on the linear momentum of the system.
----- --- --------- --7
fF
-->
=-(mv)
dt -->
=d(mv)
t2 --+
.:
f<1 Fext dt = mv2- mv1
or
On rearranging the above equation, we obtain
... (1)
... (2)
Conservation of Momentum
If F.xr =0 for a system, then we may write impulse
momentum equation as
-->
(a)
,
'
-->
(b)
,
'
"
Fig. 4:1
_____ . ----·----------·
The classification of _a force as internal or external
depends on the choice of system. For example, consider a
system of Fig. 4.1 (a), consisting of masses m1 and m 2 and
-->
the string that connects them. The force T1 exerted by the
-->
string on m1 is an internal force; so is the force T2 exerted by
-->
-->
i
"i
T,
L ______ --------- -
For a system of particles we may add vectorially the
momenta of all the particles and impulse of forces acting on
the particles. We may write impulse momentum equation as
-->
:
,
= final momentum
-->
I::.............
ll
-->
Initial momentum + impulse of force (Fexr)
:Em v 1+ :E Imp 1__, 2 = :Em v 2
:'
-: .·-- - _____ j
--+
--+
I.
•••••••••••••,,.-bsystedm
:
: oun ary,
-->
-->
the string on m2 . The force F, m1 g and m2 g are external
forces. These forces are exerted by agents that are outside
•
-->
the system; e.g., force m1 g is exerted on m1 • and thus on the
:Emv 1 =:Emv 2
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
3297
IMPULSE AND MOMENTUM
dP-
1
--
dt
-,
Fj
-,
= F j,int + F j,ext
The forces exerted by particles within the system may be
written as:
-,
N-,
Fj. int
= L, Fj,,
fal
i#j
-,
where
Fj, . stands
for the force exerted on particle
number _j by particle number i. The term with i = j is not
included in the sum, because a particle cannot exert a force
on itself.
The rate at which forces change the· system's total
momentum:
-,
dP
d
dt
dt
( )
N-,
--,= - I,·Pi
L F internal == 0
-,
·
= sum of forces acting on particles j
-,
-,
-,
(b)]. The forces F and m1 g are external forces as they are
shown in Fig. 4.1 (a), but Tis an external force as well
because it is exerted on the system by the string, which lies
outside the system boundary.
Conservation of Momentum for a two Particle
System
Consider two particles
isolated from surroundings,
that interact with each other.
From Newton"s third law, both
the particles must exert equal
and opposite forces on each
other, i.e.,
•
-,
system as a whole- by the earth, which lies outside the
system boundary.
Now consider the system consisting ofm1 only [Fig. 4.1
-,
j=l
F21 + F12 = 0
-,
-,
dP1 dP2
d -, -,
-+
- = - ( P1 +P2)=0
or
dt
dt
-,
-,
P1 + P2
or
dt
= constant.
-,
F21 is force exerted on particle 2 by particle 1, it will
-,
change momentum of particle 2. Similarly F,_ 2 is force
exerted on particle 1 by particle 2, it will change momentum
of particle 1. But momentum of system, that includes
particle 1 and particle 2 will be conserved if the basic
= sum of external forces acting on particles
+ sum of internal forces acting on particles
The first term is the total external force :
-,
-,
-,
Formal Proof of Momentum Conservation
We model a system as a
large number of particles [Fig.
4.2
(b)],
labeled with
numbers, 1,2,3, ... ,N, whereN
is the total number of particles
•3
'",
in the system. The particles
o
·j·.
and the forces they exert obey
Flg.4.2 (b) ' . ;
Newton's laws. The particle
[){~
l
l" '.
with number j has mass mj, velocity
-,
'vj,
and momentum
-,
= m j ,; j. The total force acting on particle j is Fj, which is
the sum of forces exerted by objects outside the system,
-,
= L,Fj,en
j=l
condition L F ext = 0 is met.
'f'j
N -,
Fen
.
Fj, ext , and forces exerted by particles within the system,
dP
-, -
- = Fext + sum of internal forces
dt
acting on particles
The sum of internal forces is the vector sum of every
force acting on every particle. In accordance with Newton's
third law these forces occur in pairs, and since we are
summing over all the particles in the system, we are adding
both forces in each pair. Thus the sum of internal forces,_
equals zero.
Then,
-,
dP
--+
--+
--+
-=F.;.,+Fint =Fext+O.
dt
We have shown that the rate of change of a system's
total momentum equals the total external force acting on the
system:
-,
Fj, int. Then Newton's second law, applied to an arbitrary
particle number j, has the form :
www.puucho.com
-,
dP
-,
-=Fext.
dt
Anurag Mishra Mechanics 1 with www.puucho.com
!330,
I·
Concept: For an isolated system, there is no extern~!, .
!force.and.the$y~tem's tota/.11wmenwm cannot change, When
jextemal forces do act on a system, ·any change of its
momentum resultsfrom_an. impulse.delivered by the extern~!)
!forces.
·
,
!
ruustration 1.
Consider two toy
caru equipped with
spring bumpers. ,The
caru are tied together
with a string, while tlie
springs are compressed,
When the string is cut,
the carts are pushed
apart so that they move
apart
-m
opposite
directions. Friction in the
wheels and between
wheel and ground is
negligible.
Each cart is acted on
by gravitational force
and normal reaction in
Flg.4.3
vertical direction; these - - - -~>
external forces add to
zero . horizontal direction the spring exerts force on each
cart. That is why we cannot choose cart A or cart B as our
$Y5tem · for application of conservation of momentum
principle. · If the $Y5tem includes both the caru, then
I
.ln
.
_,
I: Fext = 0, because the spring force is internal to the system.
PAf = 0, PBi = 0,
From principle of conservation of momentum,
_,
P1
or.
_,
·
..,Befpre ..
.. =·.
'' . • • ..I•
'"' ,.,. _After~· ,..
I '•, •. ,
' '" :-: '·,.I 1, •
'-------'----·
-+
-+
ma Vat+mp Vpi
•:
Flg1'4:4
>
= ma -+Va/+ mp -+Vpf
_,
_,
0+0= ma VaJ+mp Vpf
. _,
_,
or
mpVpf
Va/=
ma
The negative sign indicates that the astronaut moves in
the direction opposite. to the direction of motion of the
pencil.
I
.
Consider a rifle that fires a bullet with the speed of 500
m/ s when the rifle is fixed rigidly. If the rifle is free to recoil,
the bullet leaves the barrel of the rifle with a speed of
500 m/s relative to the backward recoiling rifl~- In this case
the velocifY. of the bullet relative to. ground is less than 500
m/ s. While applying conservation of momentum equation,
we must specify all velocities relative to a single,
inertial coordinate system.
When assign subscript R for rifle, B for 'bullet and E for
Earth, we have-
_,
_,
_,
·' ": .. (1)
VBR =VaE-VRE
= P;
•
Relative Velocity and the Conservation of Momentum
Rearranging eqn. (1), we get
-+
VBE
or
which shows that the final velocity of cart B is opposite
to the velocity of cart A.
·
ruustration 2. Consider ·an astronaut trapped in
space, isolated from surrounding. He can reach his
spacecraft if some velocity is gained by him somehow.
Suddenly he realises that he has got a small pencil in his
pocket. How can he acquire velocity (momentum) ?
We take the system to consist of the astronaut and the
pencil as shown in Fig. 4.4. We assign the positive direction
of the x-axis to be the direction of throw. The gravitational
force acts on the $Y5tem, which indeed is external force.
However, this force is directed along the y-axis, it will not
change momentum along x-axis. we· can thus apply the
conservation of momentum to this system. What happens to
the force exerted by the astronaut on the pencil while
throwing it?
'-+
-+
= VBR+VRE
... (2)
If the bullet is fired with velocity v relative ,to the rifle
and the rifle recoils with velocity V relative to Earth,_ then the
absolute velocity of the bullet (relative to Earth) is the vector
sum of the vectors
.
~BR
.
~u
and ,IRE as shown by eqn. (2).
,
M'' s.
.
alt
H
tm · .
I~___.
Bullet
Rifle
•· Fig:"4;5
Thus,
l,; 8 £1= v-V
From the Jaw of conservation of mo~entum, we have
0= m(v-V)-MV
or
www.puucho.com
V=~
M+m
Anurag Mishra Mechanics 1 with www.puucho.com
-------'"~1j
:::IM::_:PU~LS:.:E:.:::A:::.:ND:..:M:::.:O:::M:;:EN::.:T.:::UM:::......--.:.-'-.L........:.___ _ _ _ _'----_ _ _._L:__._ _ _-'-'_....:...._ _ _
t..1
illustration 3. A man
of mass m standing at the ..
end of trolley of mass M
jumps off with a velocity ure1
(a)
relative to final state· of
trolley. The recoil velocity of ..
M
the trolley may be obtained
by applying momentum
conservation.
Flg.4.6
If velocities of the man
and trolley are assumed with
respect to ground as v 1 and v 2 , respectively.
t:-"ral ·:
I~
P,
P1
= urel
XB
= Xo/6
XA
= Xo/3
(b)Conserving momentum -mv A + 2mv B = 0
Writing work energy eq. for the system
kA
(1/2) mv~ _ 2
kB
(1/2) 2mv~ .-
1
1 2 1
2 1
2
-kx0 =-mAvA +-mBvB
2
2
2
VA =2VB
1 2 1
2 1
2
-kx0 =-mA(2vB) +-mBvB
2
2
2
121
2l2m2
-kx
=-mx4vB +-x
xvB
2 0 2
2
work done on B by spring is charge in K.E. of B
(c)
Form concept of relative velocity, we get
V1 +V2
& uCM = 0
XB
_1-"1
=0
= mv 1 -Mv 2
Applying momentum conservati~n, we get
XA
'
I =· ,., •
•
P1 =P,
· mv 1 -Mv 2 = 0
m(u,e1 -v 2 )-Mv 2 = 0
or
m(-xA) + 2mxB
nL.i
a
=0
=0
=2
Thus SCM = 0, since Fext
kx5
1
2
-x2mvB = -
murel
2
V2=--
m+M
6
~$;(~~~~
b:Ecxp,in:el!,~.~
==
~·=L=)~
lJn the Fig. 4E.l (a) shown thesp_ri_n_g_is_c-om-pressed by' ; 0' a~d]
1
released. '.lwo blocks 'A' and 'Bl.. of masses 'm' and '2m'i
respectively are attached at the en~ of the spring. Blocks are
i:..·small cube ~fmas; m slides down ;~{rcular path of radius
cut into a large block of mass M. M rests on a table and both 1
blocks move without friction. The blocks initially are at rest
and m starts from the,top of the path [see Fig. 4E.2(a)], Find
the velocity v of the cu~g.s_it leQY.<!§._lhe block. ·
kept on a smooth
hot~iind
released.
c.~~:_4E.1~
;
( a) Find displacement of block A by the time compression of.
the spring /s reduced to x0 / 2.
(b) Find the ratio of KE 4 blocks A & B by the time
I1 compression of the spring reduced to x 0 /2.
(c) Find work done by force of spring on blockB when spring
~aches natural length, _ _~ - - - - - - - - - '
I
Solution: (a) Let xA and
block A and B respectively. .
XA
+xB
xB
denote displacement of
RI
l~I
L_______ll_F_ig_.4_E·~------- ___ j
Solution: Let velocity of
the block is u when the cube
reaches its bottom to leave with
velocity v. By conservation of
momentum
in
horizontal
directions.
0= mv+Mu
·mv
= Xo/2
U=--
M
By conservation of energy, mgR
mgR
on solving, we get
In the absence of Fextemal CM remains fixed.
www.puucho.com
1
2
= -1 mv 2 + -Mu
2
2
= .!. mv 2 + .!.M
v=
HJ
2
2
(~~v)
M
2
Anurag Mishra Mechanics 1 with www.puucho.com
MECHANl~S;I
1332
r-----·-·-
------- ~--··----·----:
i71vo blocks of masses m and 2m are connected by a relaxedi
/spring with·a;Jorce constant' k. The blocks rest on a smooth'
ihorizontaltable. Att = 0, the block on the left is given a sharp
!irhpulse "]"_towards the right, a~d_the blocks begin to slide
lcilong the. table_ (see Fig. 4E.3). ,Find the maximum
.
/compression in the spring.
i.
.
-'
r·~1 .
j
Fig. 4E.3
i ,.
•
---------_..,__..._..____...____ ==----,"""-",;;....,"'"''°"'-' - - - - - - - ~
=-J~m
-.
m
3k
~:~tsv~~~~rz:,~oj~::s:.~.~~n~j
1
~o:o~Lr~~in?.-~i;u~-height h that car, alsp move along the plane. The pui:k 'begins
to slide up the'"hill" [see Fig. 4E.4(a)]. lf the hill is initially at
rest, iyhat value of v provides for the maximum subsequent
;velocity u of the "hill"_? Assume that all swfaces' are
·r-~, -~----- -i
..;._)
[ffi1
h
~
I
11
i!
h
'
II'/
M
,
lL
i . -----:---~·"--;- .
l
=>
1
2
1
'
--·---,-_--'
_!Mv 2
2
X
+
kinetic energy of block
_.!,;,v 2
2
kinetic energy of wedge
.!mv 2 =.!(v 2 +v 2 )+.!Mv 2 +mgh
2
2 X
y
2
X
Where v Y' vertical component of velocity of smaller
block at height h.
We have
vx
=(~)
m+M
2
:
I
2gh]
:
V
U=-
4
and
2
2
' ----
121
221
mv
--mv =-m(v +v )+-M - - +mgh
(
2
2
x
Y
2
m+M )
.
2
2
) 2+Mm+m
Vx2 +vy2 = [(M
----~v
·
(M+m) 2
·
-mv =-4mu +mgh
2
2
2
.
_!(v 2 +v 2 )
2 X
y
-mgh=
~ - m 'i
i
II JJ,.·.,•·.·
mv= 4mu
Fig.4E.5
-
'.,·
Solution: For the maximum subsequentvelocityu, the
puck comes to state of relative rest at the top of hill.
L..~--- ____'.~''!: ~:-~~~
M
Solution: (a) In the absence of any external force
acting on the system in x-direction, its momentum remain
constant and velocity of C.M. remain constant. The velocity
of C.M. is
mv+M,xO
mv
VcM·=
=--=vx
m+M
m+M
At the instant block is on the vertical part of wedge, both
the wedge and block have common velocity in x:direction.
From work-energy theorem, we get Wgravity = AKE
- ·
:_'__.__ :._
~
.
.
----,
ii -., .... ·--·-' ____ ,__ .. '
' ' ' '.': ·j-
Fig. 4E.4 (a)
II fiii1
II
i'
>;' ·: :, "
fi.E2<@m~J.~If~;;>
_i__ ~
!i
(a) Find the speed of the Sl!,l~ll~r, mass when it brea]<s, off the
I : larger mass at height h. ".,_'
, .
· :.
I
'(b) Find'the maximum height'(from:the ground) that smalleq
. /__ .mass -ascends. ____" -____ ,___ y: __ · + · ' ,·. ,c __ ·.' ·~--. :$ ·-J
X=Vo~¾7
; I
~-·- !
- - - -~
l
I .
.!mv 02 =.!kx 2 +.!3mv 2
2
2
2
C
'.lftionless._ ·.
ilv.=t,
:l:vx
At maximum compression both the blocks will move
with same velocity v,
· mv 0 = (3m)v,
..
. =x
Maximum
compress10n
,- ---- . ------' ...... --··-- ----------,
;Fig 4E.5.shows a small block. of m.as.s m. placed o_ver. a.wedge of)
mass M. _The block is pushed ·on the wedge at, a -speed v.
1
'.Assume that all. the. swfaces are fri
. . 'ction. less, inin.·al ve/ocity ,of.I'.
jblo_ck is horizontal. Wedge has gradual curvature ,, . ,'
·
'
J
-=Vo
m
Solution:
j
mv
2mv ·
or v -= ~Bgh
--=--+mgh
2
16
3
[Revisit the problem after studying C.M. and solve the
problem in C.M. frame.]
I
v'=,;v 2 +v 2
x
Y
= [(M2·+Mm+m2)i,2
(M +m)2
1/2
2gh
]
(b) After breaking off we can apply work-energy
theorem.
Let hmax is the maximum height, v Y = 0 ,
1
2 1
2 1
2
--mv = --mv x +-Mv x + mgh max
www.puucho.com
2
2
2
Anurag Mishra Mechanics 1 with www.puucho.com
-- -
- --
- ---
i.IMPULSE AND MOMENTUM
where
_______ _
333:
mv
m+M
vx = - - -
A smooth wedge of mass M rests on a smooth horizontal:
surface. A block of mass m is projected from its lowermost
point with velocity v 0 [see Fig. 4E.6 (a)]. What is the'
maximum height reached by_the block?
\,li, -
-··
Solution: As long as the block moves from A to B, the
reaction on the wedge presses it to the wall. When the block
reaches the lowermost position, its velocity from energy
conservation is
V
.M
Fig. 4E,6 (a)
Solution: At
the
instant the block breaks
contact with the wedge, they
have common x-component
of velocity. In addition, the
block
has
a
vertical
component of velocity. Due
to this vertical component,
the block rises upwards till
the vertical component of
velocity vanishes.
From
momentum
conservation along x-axis,
or
i
h
Initial position
P, =Pt
mz~2gr
+
ffi2V2
E,
2
2
m2V2
m1V1
(m+M)
... (2)
... (1)
= E1
mzgr=--+-2
2
... (1)
... (2)
On solving eqns. (1) and (2) simultaneously, we obtain
two solutions
V1 = 0, Vz = ~2gr
and
_
V1 -
~2gr
2m 2
m1 +m2
2
or
= m1v1
From energy conservation,
From energy conservation between initial and final
positions of block,
1
2
1
2
-mv 0 =-(m+M)v +mgh
... (3)
2
2
or
= ~2gr
When the block moves along the right half of the wedge,
during its upward journey as well as downward journey the
reaction of the block on
the wedge is towards
right as shown in Fig.
4E.7(b).
Therefore
N'
N""
during the entire motion
of the block from B to C
B
and C to B, the wedge is
accelerated
towards
Fig. 4E.7 (b)
right. Thus to find the
maximum velocity attained by the wedge at the instant
when the block passes separated from the wall,
Fig. 4E,6 (b)
mv 0 =(m+M)v
mv 0
v=
~---·-·~---·~
Fig. 4E,7 (a)
_ m2 -m 1
2
1
1( m ) 2
-mv
0 =- - - - v 0 +mgh
2
2 m+M
Vz -
l2gi
"y4:,t
m1 + mz
The first Solution corresponds to the instant when the
block reaches for the first time at point B. At this instant the
block moves with velocity v 2 and the wedge is at rest. The
second Solution corresponds to the instant when the block
has the maximum velocity
h=~[m:M]
l_E~g;~:Bl~~J7;,>
A wedge of mass m 1 with its upper surface hemispherical in
shape, as shown in Fig. 4E. 7 (a), rests on a smooth horizontal.
surface near thewall. A small block of mass m2 slides without
friction on the hemispherical surface of the wedge. What is the
'maxi.mum velq_city gttg/_ned.by the w~dge?
www.puucho.com
(v1lmax
2m 2 ~2gr
m1
+
m2
Anurag Mishra Mechanics 1 with www.puucho.com
!334
'
,
·~
IA ball B is suspended from a string oflength l attached. to- a
cart A, which rn'ay roll on 'a frictionless surface. Initiall,y .tlie
cart is at restandthe ball is given a horizontal velocityv 0 [see
Fig. 4E.B(a)]. Determine:
I
Cart A mA
~
K
I
·concept: Solve problem in CM frame
B•
me
v,
2 mA +m8
Fig. 4E.8 (a)
'
'
~ mAg
' h
mAmB ) Vo,i -1 ( ---"-~-
(a) the velocity of B as it reaches the maxim=um~_h_e_ig_h_t,_·_·
(b)Jhe mcajmum hejghtreached bythLbJlli
_I
· Solution: We choose ball and cart as our system. No
external force acts on the system in x-direction; therefore
momentum along x-axis is conserved.
The· ball will continue to move upwards until its velocity
·
'
relative to the' cart is zero.
Wgravity
.Af(II
+
.'.,_
-pi Agh
=0- 1
mAms v~
2mA +ms '
Note that for mA >> ms,
(vB)f =(vA)f =0
2. '
h= Vo
and
2g
i.e.,
_,
Ball B oscillates as a pendulum with A fixed for
_,
or
VB= VA,
When the ball reaches maximum· height, the cart and
ball move horizontally with same velocity at the extreme
position.
~----•~-,,-----~r,----.-,-~·
(v,J, = ~
~--'-,''(v,Ji =O
A --- --···
A
mA << mn,
(vslt =(vA)f =v 0
and
h=0
A and B move with the s,ame constant velocity v 0.-
@iExam:~~~
1 - -..
., _,': --. _._.,
'
" ---·: ::,
'.lwo identical wedges of massM are smoothly conjugate<i. The
wedges are free to move· on a smooth ·horizontal surfaqe, A
block of mass ni is released from a' ·height h on one of the
~egg<!§_(see Fig; 4K~J~-----·-~----~~....,
·_
, , :t+ ( ),-( J............
Reference level
V
B - Ve - v~ t.·-
B
.
:. Initial position
.:J:ifl~I position
Fig. 4E.8 (b)
pf= mBvB
= mavo
P/= mB(vB)j +mA(vA)f
= (m 8
+mA)v
From conservation of momentum,
m8 v 0
or
Fig.4E.9
= (mA + m8 )v
mavo
v=-~~mA +mB
In order to find maximum height reached by the ball we
will apply law of conservation of energy.
1
2
E, = mAgl+-m8 v 0
'
(b)
(a)
P, =Pt
2
(a) Show that the height h to which ,the mass m ascends :the
I
.. .
. .
M 2, '
· '
l nghtwedge,,s,hmax. = · . 2 h.
·,
, .
I
,·
. (M + m) • , •
(!,) Wh_gt_17.!f!,iU.'f!.!io..:.(M/.!!!l result§._i!l)lmax.--=..h/3:,~
Solution: (al .When the block reaches the bottom of
the left wedge, we can 'apply energy and momentum
conservation.
P, = 0
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
- - - · - - - - - - _ _ _ _ _ 33[)
->
->
P1 =Mv+mu
=Mv-mu
P; =Pt
Mv=mu
(b)
Sy·s·~r~-- ·:a:
t<;;~;~ :::.~;,: '"~/
----
Y
:
:
v.\
,.•• .;•
:
:
-
P;=mu,P1 =(M+m)v
P; =Pt
mu= (M +m)v
... (3) .
1
2
E; =-mu
= ----.
h
2
h
=
(M +m)
max
... (1)
Similarly from energy conservatfon,
E; =Et
1
2·1
2
mgh =-Mv +-mu
... (2)
2
2
When the block reaches hmax on the right wedge, the
block and the wedge will move with common velocity. The
vertical component of velocity of block reduces to zero at
this moment.
M2
hmax
As
hmax
h
(1 + m/M)2
h
4
=-,
(1+:r=4
m=M
or
RECOIL, DISINTEGRATIONS, EXPLOSIONS
In these cases the internal forces are exened by or on
the particles that compose the system during very shon
intervals. External forces, such as gravitational force, are
negligible in comparison to the large internal forces.
Generally the problem stans with a system of two or more
particles with no relative motion. Then some stored energy
is released causing the parts of the system to fly apan. The
total momentum of the system at the instant before
disintegration or explosion is equal to the total momentum
of all the particles immediately after the event. ,
The total kinetic energy of the system is not conserved
(similar to inelastic collisions). The source of released
energy may be chemical, mechanical or nuclear sources.
li:E~q~e;le}wl~
==- - · - i ~
IA rocket-·; p~~je~ted- vertical~-;;;,~ards. It explodes at the/
2
1
2
E1 .=-(M+m)v
+mghmax,E;=Et
2
1
2
1
2
-mu
=-(M+m)v
+mgh max
2
2
... (4)
topmost point of its trajectory into three identical fragments .
One of the fragments comes straight down in time t 1 while the 1
other two land at a time t 2 after explosion. Find the height at
which the explosion o.,:curred in terms of t 1 and t 2 ?
)
V2
V3
From eqns. (1). and (2), we obtain v = mu
M
I
!'
2
2
1, (mu)
1
-M
+-mu
= mgh
2
M
2
From eqns. (3) and (4), we obtain
and
... (5)
~ = (Mn~tm)
Fig. 4E.10
c___
.!.mu 2 =.!,(M+m)(~) +mghmax
2
2
M+m
mgh
max
2
1
1
mu
=-mu
--(M
+m) ( -)
2
2
,
M+m
--
-------·=------~
momentum,
2
or
-
Solution : At the topmost point of the trajectory, the
momentum of the system is zero. From conservation of
2
and
------ --
.. ,(6)
Now we divide eqn. (6) by (5) to obtain
m
mu2
(mu)2
1
hmax _
· (M + m) _
(M+m)
-h-(mu) 2 m
1+mu 2 + - M
M
hmax
M2
h
(M+m) 2,
m1V1
as
+ m2V2 + m3V3 = 0
m1 = m 2
v 1 +v 2 +v 3
= m3
=0
... (1)
The second and third fragments reach the ground
simultaneously, therefore vertical components of v 2 and v 3
must be same; secondly, v 1 is downwards, the vertical
components of v 2 and v 3 are-~ (i.e., directed upwards).
.2
•
For first fragment,
www.puucho.com
2
h = v 1t 1 + ~ 1
... (2)
Anurag Mishra Mechanics 1 with www.puucho.com
2
h =- v,t2 + gt2
For second fragment,
2
From eqns. (2) and (3),
V1::::
g(t?-tf)
2t, + t2
h = gt,t2
and
... (3).
2
(.t, + 2t2)
2t, +t2
2
IMPULSIVE FORCE
When a force, of relative higher magnitude acts for
relatively shorter time, it is referred as an impulsive force.
An impulsive force can change the momentum of a body by
a finite magnitl.!de in a very short time interval. Impulsive
force is a relative term. There is no clear differentiation
between an impulsive and non-impulsive force.
1. Gravitational force and spring force are always
non-impulsive.
2. Normal, tension and friction are case dependent.
3. An impulsive force_ can only be balanced by another
impulsive force.
1. Impulsive Normal : In case of collision, normal
forces at the surface of collision are always impulsive:
N 1 =Impulsive;
Normal reaction due to ground is N 2 = Non-impulsive
•
----+
,,
Consider a large ball. colliding with small ball
N1 ,N 3 = Impulsive; N 2 = non-impulsive
,,;
JI
'
It
''
Fig. 4.8 (b)
~--- ----------.
2. Impulsive Friction-: If the normal between the
two objects is impulsive, then the -friction between the two
will also be impulsive.
Both normal force N 1 and N 2 are impulsive
~
,,, ,'.
I
N,.: '
(a)
Fig: 4:9 (a) ·
' ,'111l1m;±m,m±mol11i
,,,
Friction at both surfaces is impulsive if it exists.
'.,,
Collision Of blocks
,,,··(bl:',,
~~
,'h__-
.
;''
~N1_.:_
~2 ...
·LL{-}--N, •:·: . ~19
,.
_,..,:.,.:._~_
.
·1
"''',"
(c)
.
Fig. 4.7· · J',·
m2g.
}
,·
.
Consider a ball dropped on a large ball.
Both normal forces N 1 and N 2 are impulsive
Consider collision of large ball with small ball
Friction due to N 1 is non-impulsive and due to N 2 is
impulsive
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
IMPULSE AND MOMENTUM
~- -:: _:--~~---- 337]
For block
J 2 =mV
Solving, these three equation, we get
Concept:
Impulsive Tensions In a string: When a string is
jerked out equal and opposite tension act suddenly at each:
end and impulses act on the bodies attached with the string in
the direction of the string.
·
... (iii)
V=.1::'.
3
Illustration 5. Two identical block A and B, connected
by a massless string are placed on a frictionless horizontal
plane. A bullet having same mass, moving with speed u
strikes block B from behind as shown. If the bullet gets
embedded into the block B then find:
I
-
, ~
T is non-impulsive
-
c~-!~
lmhndn111i1111111111~111;
Fig. 4.12
All normal are impulsive but tension i
T is impulsive only for the ball A
Fig. 4.10
One end of the string is fixed: The impulsive which .
acts at the fixed end of the string cannot change th<
momentum of the fixed o/;ject attached at the other end. The
object attached to the free end however will undergo a change;
in momentum in the direction ofthe string. The momentum
remains unchanged in a direction perpendicular to the string.,
In this direction string cannot exert impulsive forces.
The velocity of A, B, C after collision.
Impulse on A due
tension in the string
Impulse on C due to normal force of collision.
Impulse on B due to normal force of collision.
Solution : (a) By conservation of linear momentum
u
v=(a)
(b)
(c)
(d)
3
(b)
Bath ends of the string attached to movable,
abjects: In this case equal and opposite impulses act on the,
two objects, producing equal and opposite changes in'
momentum. The total momentum of the system therefore
remains constant, although the momentum of each individual
object is changed in the direction of the string. Perpendicular,
to the string however, no impulse acts and the momentum of1
each particle in this direction is unchanged.
In case of rod: Tension is always impulsive.
In case of spring: Tension is always _no_n:impulsive. __ _
(c)
(d)
w
· mu
fTdt=-3
-JNdt=m (3u -u J=-2mu
3
JCN-T)dt = fNdt- fTdt = mu
3
=>
fNdt=2mu
3
!>Jg!e: - - - - - - - - - - - - - - - - - - - Impulsive forces are those forces which can have very large
value in very small time, e.g., Tension, Normal and friction.
When impulsive forces act then momentum along the
direction of force cannot be conserved.
illustration 4. A block of mass m
and a pan of equal mass are connected by
a string going over a smooth light pulley.
Initially the system is at rest when a
particle of mass m falls on the pan and
sticks to it. If the particle strikes the pan
m
with a speed v, find the speed with which
the system moves just after the collision.
· Solution : Let the required speed
m
,m
is V.
Fig. 4.11
Further, let
J 1 = impulse between particles and pan
J 2 = impulse imparted to the block and the pan
by the string
Using impulse = change in momentum
For particle
J 1 = mv - mV
... (i)
For pan
J 1 - J 2 = mV
... (ii)
illustration 6.
J
(a)
Fig. 4.13
(b)
In. (a) momentum cannot be conserved in vertical
direction just after collision while in (b) it can be conserved
just after collision.
In (a) Tension will reach a large value in small time and
M
fTdt
* 0.
0
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
,; . "'
'i ,--,
'
" ' "':"..t __ ,.~.,.
·J~
:Li3=.·:::3=.8_~---:-':.a·:_..··---------~--:----··-·-'~'-'_·.~-' ..:'...:..L...2~, :··,·.:'<~·
. k~~r;.m:~,;IJ?J 1i" ~
V1
.. j'Iwo, ,blo~'A and B qre jofned'by means of a ·sl'.1~ked';~
·, ·,.··1
.
, .
·2kg
,'
A
•;,,.:-;~,- .
Mu-mv 0
=
M
Sm/s
90x 5-l0x 30 450~ 300
=-~~-9Q
'90
U=
lpassmg, over .c;z massless pulfoy as shown m dtagrprrz .. :The ·
'isyste111 is feleruedfrom rest and_itHecomes tautwherfB falls a .
Idistance' 0.5; m; ·then
";.'
· ;•
•, ,,
:,
" "
. V1
,,
.
Energy of explosion;, K. E. 1 ~K. E.;
·(1
1
=
1
·
2)
'
)u2
-Mv 12 +-mv
_ -~(m+M
. 2 .. 2 2 . 2
2
1 kg .
Flg.4E.,11,
J1;·a.:? Fi.nd
t/t~ ~-()mmon. velocit;x. of
become taut.
··
:
.
\•.\
- ' -
o blocks just aft~r. } ~
·,
rt,) ·Find the n"tagnitude of impulse on the pu_lley by the c!ari'tp
Lluri/Jg'J!J.~.. small interval w.l!il~
strj!)gJ?.ecom~ t@t.,, · ,J
''
. Solution: Veloc(ty of a just-before the string is tl!ut.
1 (m+ M)[Mu
+ niv~
=-~--"2
.
·M
·
· _2 ] .'
U·
. ·
·
10
™'_:_
= .!_(m+.M)[mv~:·] =· .!_(100)[ X 900] = 5000 J
·2
·
M
2
90
CENTRIPETAL ACCELERATION REVISiTED
.
'
.
Consider a particle
.
VB ';' .J 2g,1h /= .fw 1}1/s .
of mass .m moving at
(aScominon velocity= V ·_
speed v on a frictionless
'
;,,.. ·
Ko
surface inside a fixed
v = v.8 /3 =-·-m/s
horizontal
circular loop
. .
.3
.
of radius ras shown in
: · (b) Magnitude·of impulse ori A·
Fig. 4.14. . Assuming
=·Magnitude of.impulse on B
collision -of particle
with , the . hoop as
:=
=
N-m/~
elastic, we can see that
. · .. 3
3
.
.
the · magnituae
of
pulley~ iz-1mpulse
on A= ~.Js
N-m/s
momentum· of · the
. :; .Impulse'on
.
:
3
·particle is constant but · : ·;.;
·'
its direction is changed
r
with each collision.
, -_,:
Flg:4:14 ,, ·:.
i
r::. . . - .
. ,Thus momentum of the '---'-----=-,-~---~
P,ig. 4E.12 shows a; ·
' r . . . .
i. ·. · particle ·1s not conserved. When viewed from centre the
IMan Rock~t:Lau'ncher of totl:!l mass =¥ =90 kg
'
,.·
',
successive collisions are separated by angle 0. From Fig. 4.14
-l~·=Sm(s;'.massofrocket"=.TTf,=lOkg' . ·.
. · , _, ."~... -~·, . · we
see that
·
·
·
·
!Muzzle velocity
'
. ,, of Rocket =Vq_'.= 30 rn/'s
l
1(.Jro- ..Jw). ~Fla
:ijexdt1m:m,~ · ··
l I
·--------c----· j
+
'
.I
•
·.. '
"' "" ~,", ,
•
Fig. 4E.12
,
. _...
•
. ·•
':~\•>j. ,f·
··.
I
(a) What wi.ll'be m~n_'sand rocket's v,e_locity'after firin,g_.· . .
'{/,J__fill~_of._f2CP.losio"' .
·
·:i.,-1..J ..
Solution: Man fire rock~t with muzzle velocity= v 0 ••
·
· and.man is v momentum in
.. Let·velocity o(rocket
is~~
1
.horizontal direction remains. ccihstant.
· Initial mo111eritum = final lllomentum
(m+M)u =Mv1 +fmCvo +u)
i
"" -;.;~r-·~r~lj .
,
a
PJ'
=mvcos
~ i_-mv.sin _2 j
'
The change in momentum .1 P of the partjcle is .
·
-'>
_,
->
·
(9),
.:lP = ~1 -P,_-· .= -2mvsin - j
2
r
·4
· (0)
IAPI=
, 2mvsin
. . . -2
The hoop exerts an im~_ulse on.the p~c)e that changes
its momentum. .
(m.+M)u=Mv 1 ~mv 2
v 2. = v 0 +· u~.
l
~
F _·Llt=Af'=2mvsin(!!.)
avg._
,
2
· Thedistan~ betw~~ns~cc!!~siveco~sio.nsis2rsin,(!} .-
www.puucho.com
7
Anurag Mishra Mechanics 1 with www.puucho.com
_l. IMPULSE AND MOA1ENTUM
3391
Hence the time interval between successive collisions is
2r ,
ut =-;;-sm
A ',
(8)2
.
As the panicle continuously remains in contact with the
hoop, the time interval M' approaches M, the time
associated with the contact of panicle with the hoop.
Substituting M' for M, we get
2r sm
.
Favg.· -;;-
(8)2 =2 . (8)2
mv sm
.F
or
avg.
= mv
From conservation of momentum, the new speed is
given by
Ic2mi~2 _Ic2mlv'2=Z.mg2 3
2
2
r
From Newton's second law the force of the hoop on the
panicle is mass times acceleration. Thus we can say that
v 2/r is the magnitude of the centripetal direction.
·
~~~i>~k~:e~
[nv~-bodies.of.mctsses. m and 2m are conne~t~d by·~ light
\inextensible string passing over a smooth pulley and released.
,Afterfour second a mass m is suddenlyjoined to the ascending
iboey. Determine
(a) the resulting speed, and
(b) how much kinetic energy is lost by the descending body .
I when the boey of mass m is added. ·
2
9
~,!;f~am'RJ~! 14 ~
r . . . -- ---"·-
.,
1After falling from rest through a height h a body of mass m'
!begins to raise a body of mass M (M > m) connected to it;
Jthrough a pulley.
.
j
( a) Determine the time it will take for the boey of mass M to j
return to its original positio~.
·
, .:·
(b) Find-the fraction of kinetic energy lost when the body of
I mass'M is jerked into motion (see Fig.4E.l4,).
:
i
1-·- ··-----
J
.
0
I
3
3mv+0=4mv' or v'=-v=gm/s
4
Note that the firial momentum of the system is 4mv'
because the only effect of the pulley is· to change the
direction of tension in the cord, the sense of motion of both
the bodies is same.
(b) The loss in kinetic energy of A is
~----,1
i
;i:
i'
.
'I
'
~·~,: +v'
2m
·
2mg
!<4s
i i
L' i
a= O
string tautens
. i
·t>4s
(a)
(b)
(c)
-I
I
(b)
I Before the
2m
i·
I
mg
mg
(a)
After the i
string tautens'
:
i
. I
Fig. 4E.14
,
-
- - - ~ - - - - - ~ - - - - - - ~ - ~ - -- -~-~---- .. J
--""'=-··---
FlgAE.13
. • .....k.a'
Solution: (a) For. the sake of convenience the
proble!JlS involving pulleys can be solved by including'blocks
and pulley in the system. This single body has mass
mA + m 8 and is_.acted upon by a single force (mAg - m 8 g).
Fcir t < 4 s, the system is accelerated
I
As mA = 2mn" = 2m, so._the·equationof motion is
Solution : (a) The speed of the body B just before .the
string becomes taut is v= .J2gh.When the string is jerked,
large impulsive reactions are generated in the string. At this
moment effect of gravity is negligible. So momentum of the
system is conserved ·at this instant. Let v' be the common
speed of the two bodies· after they are jerked into motion.
From conservation of momentum, we have
.
mv = (M + m)v'.
in- v
or v' = M+m·
. The acceleration ,of the system is
Alternatively,
·The speed at t
. . M-m
or a=--.--g
M+m
The acceler~tion is negative, opposite to
Let the system return to original position at time t;
,
1 2
O=v't+-at ·
'J:F =mg-Mg= (M + m)a
= 4 s is
v•:
· v=D+at=4g m/s·
.
3
The addition ofmassatt = 4s_is equivalent to a collision
between the system and a body of mass m which is at rest.
.
or
www.puucho.com
2 ! •
2v'· 2m
@I
t=--;;-= M-mfg
Anurag Mishra Mechanics 1 with www.puucho.com
i..:'f2~··_,_'tc_·....,''-··='.;::;'·:.:J!c~"-~-~;·_-'-'-_·~-""-'-'---'---'-----'--"-'-----~-·--'----M-"E~..ci<-ij~_N~
·Li3:....4:..:0_c.::·~·,
(b) The fractional loss of kinetic energy is
. .!.mv 2 _
2
_!CM
2
1
2
-mv
+m)v' 2
·
I,m;X;i+ I,m;YJ+ I,m,z,k
=
i
i
i
M
M
M+m
2
. CENTRE OF MASS
Consider. twci particles unequal mass connected by a
massless rod ·(Fig. 4.15). If a force is applied between the
lighter particle and the centre of mass, the system rotates
clockwise [Fig. 4.15 (a)]. When a force is applied between
heavier particle and centre of mass, the system rotates
anticlockwise [Fig. 4.15 (b)J. When a force is applied at the
centre of !llass, tlie system moves in the direction of force
· without rotation [Fig. 4.15 (c)].
The overall motion of
Lighter Rarticle .... ~.- - a system can be described
in terms of lr point called
1· _ ··:
centre of mass. The : (a) .
centre of mass of a ~
system
represents
translational . motion of .
the system. It moves as if
all the mass of the system
is concen~ated at this
point. If tli.e resultant
external force on- the (
,",
.
f·- · .
:
r
•
''
t,J-J•
•
,_
where t; is the position vector of the ith paritcle.
r:=x;i+y 1J+z 1k
Centre of Mass of a System of 'n' Discrete Particles
Consider a system of n ------··--·----1 v
point ~asse~ m1 , m2 , m3 , ... m~ i
whose position vectors from I
origin O are given by I
--+ ,
• I
r-+1 ,r-+2 ;r-+3 , ... rn
respecnve
y.
Then the position vector of the
centre of mass C of the system
is given by
[,.;ff~b-
.
. -,
t· .i. ··b···-~·
[.icJ'
· · _ ·_ .~
• ·.
/: . :. : .:Fig, 4,1s
··--· ___
n
-,
>al
-+
rCM
-,
Psystem = Mv CM
For a syst\'m of many· particles in three dimensions
the position of centre of mass w.r.t. any fixed axis is
determined from
.
Lmixi
XCM =.m1X1 + m2_X2-+m3X3 +...+mnxn = _,_;-·mi +m2 +m3 +...+mn
Lmi
Similarly for y- and z-coordiilates of the centre of mass,
we may use the equations
I,m;Y;
Y. . _,_i_ _
and
-·
CM
.M
M
+ In vector notation, the position vector of centre of
'mass. is
_,
· ·
· :
.
I,m,
I
The system behaves as if the resultant external force is
·applied to a single particle of mass JI/located at the centre of
mass. The total momentum of the system is the same as the
product of the mass of the system and the velocity of.the
centre of mass point. .
.
Fig. 4.16
'
system is E Fext and the
total mass of the system
is M, then
_,
. '-> • LF
-· en
.'. a CM-,-~
·+
'--
r'CM';-=XCMi+YcMj+ZcMk
where M(=
= -1~
£.
'M >al
i 111;)
-+
mi ri
.
is, the. total m~ss of_ the system.
Position of COM of-Two Particles
Centre of · mass of two
r
particles of mass m1 and m2
COM
separated by a distance r lies
~
·
·1 Th
,m,
,~2
between the two.parnc
es.
e
I• - •I•
•I
1
2
distance of mass from any of the
'
'
particle
(r)
is
inversely
Fig. 4 -1:L, _
proportional to the mass of the particle_ (ml
,
1
i.e.,
r~m
or
or
and
Here,
www.puucho.com
.l
Anurag Mishra Mechanics 1 with www.puucho.com
r - ---
.
L IMPULSE AND MOMENTUM
- - - · - --
and
r2 =distance of COM from m2
From the above discussion, we see that
MOTION OF THE CENTRE OF MASS
2
- -,CM = dtdrcM
- -, = dtd[Lm
f ·r'. ]
the two particles of equal masses.
Similarly, r1 > r2 ifm1 < m 2 and r1 < r2 ifm 2 < m1 , i.e.,
COM is nearer to the particle having larger mass.
+ To find the centre of mass of an object with
continuous mass distribution, we replace the
summation with integral
V
= Em/d~/dt) = "f.mjvi
M
f1 dm
Since mi
--->
M
or
XcM
--->
fydm
=~,
YCM
VcM
fzdm
....
must be true of v CM.
Concept: If the net externalforce·acting on a system is
zero, its center of mass moves at constant velocity.
The net force acting on a system equals the rate of
change of its total momentum. Thus if the system's mass is
con~tant, we have:
....
. ....
-->
dP
d
--->
dvCM
Fnet,ext = = -(MvCM) = M - dt
dt
dt
i
Im/1,
--->
Fnet,ext
=~'~M
U=MghcM
CENTRE OF GRAVITY
Any object can be assumed to be composed of a very
large number of point masses. Gravitational forces on all the
particles can be considered to be parallel, all of which
combine to produce a resultant force, the weight of the body.
Centre of gravity of an object is that point where the total
weight can be imagined to act, i.e., a single force (weight)
acting at the centre of gravity produces exactly the same
result as· having gravity act on all the point masses
constituting the body.
Position of centre of gravity can be calculated from the
expression
L,migxi
= ~'=---
Ii m,g
,
Concept: If gravitational field is uniform over the body,,
·g cancels out in the above expression, and the expression·
·reduced to that for centre of mass of the body.
:
If g is constant, the centre of gravity coincides with the;
centre of mass.
M
Since P changes only if external forces act, the same
From the definition of centre of mass,
Xca
M
... (1)
Concept: For any symmetric object, the centre of mass
lies on an axis of symmetry and on any plane of symmetry,
this is valid of mass distribution is uniform.
The gravitational potential energy of a system of particles
in a uniform gravitational is the same as if the entire mass is
assumed to be concentrated at the centre of mass.
U= Im,gh, =gim,h,
or
p
=--=-
....
r=xi+.Y.i+zk
hcM
--->
(LPj)
= ~ , ZCM = M
where dm is a differential element on the body whose
position vector is
i
M
vi is the momentum Pi of the jth particle :
rCM = - - -
f xdm
341 i
- - - - · - -.. ---<
When the centre of mass of a system of particles is
moving its velocity can be obtained as
r1 = r2 = .! if m1 = m 2 , i.e., COM lies midway between
-t
_,_ , __ - - -
'
=M
....
. .. (2)
acM.
Newton's first and second laws apply to a system as if it
were a single particle located at the CM. So far we have
treated complex objects-rockets, automobiles-as particles
but have not given logic behind it. These theorems show
why the particle model is correct.
For example, when the external forces acting on a
system are entirely due to gravity, the total external force is:
....
--->
•
F,xt = "E Fg,j = "f. (mi g)
....
--->
=(Lmi)g=Mg.
Comparing this with eqn. (2), the acceleration of the
system is a'cM =
"g.
In the absence of air resistance, a
system's CM falls just like a particle, regardless of what the
system's individual pieces are doing.
The · center of mass reference frame as zero
momentum reference frame
Concept: A reference frame with its origin at the center
·of mass of a system is called the center of mass reference
frame. If the motion ofa system is describedfromthe center
of mass reference frame, we find that the total momentum of
the system is zero.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
!342
The position vector of the center of mass is
·1
--+
,
--+
.
=--Lmiri
rCM
mtotal
i
· where m,0 ,.1 is the mass of the entire system of particles.
Ifwe choose the origin for the coordinate system to be at the
center of mass itself, then the position vector of the center of
mass point is. zero; that is,
_,
• 1
rCM
= Om,
'
or
We. already µsed · the correspon.ding relation .among
position vectors (Fig. 4.17) 'in th~ derivation of eqn. 1.
. We may use eqn. (1) to calculate the linear momentum
of a system in its CM frame. The linear..momentum of
particle j in this frame is:
·
--+
--+~
--+
p.CM
=m·V·CM
=m•(V·-Vci,i)
J,
] }o
I
··
mtotal
i
where the 1•, locate each particle with respect to the
center of mass as the origin. Since m,0 ta1 0 kg, , the
summation must be zero:
0 = L,mi~'t
i·
'
•
_,
=Pi-mivCM.
The total linear momentum of the system in the CM
frame is:
·
·
*
Differentiating this equation with. respect to time, we
find
.
.
)
_,
0=-~m-?,4,,1 l l
..
}
--+
--+
--+
=:!:(P.J -
PCM =:!:P-J, CM
But
--+.
--+
'
--+
m-Vci.i)
J
--+
=Mv °'1, so PcM = 0.
P1ab
Be~ause the total momentum vanishes in the CM frame,
it is .sometimes called the center of momentum reference
frame.
'
v!,
where
is the veiocity of the ith particle relative to:the. Kinetic l;:nergy of a System of _Particles
The· center of mass also provides a useful simplification
center· of mass frame. Each term in the sum is the
.
_,
' :when finding· the total kinetic energy of a: system of many
moinentu~ P; of the respectjve particle in the center of
moving particles. We clioose a coordinate.system.with origin ·
mass frame, and so the summation. is the total momentum
0
at some· convenient point in space and locate a typical
_,
.
1,
.
P total wn CM of the system when measured with respect to the · particle in the system by means of a position veciqt
.
ceriter of m~ss; that is,
originating at O and another position veci:or1 i originating ~t
--+
--+ .
0 = L,Pi = PtotalwnCM
the center of mass; as shown in Fig 4.17. The two vectors· are
related by
·
Concept: Thus the total. 17!e>mentum of the systel7! is
zero when measured with -respect to the center of mass. For
this reason, the center of mass reference frame, with its origin ·
the center of mass, also is,known:as a zero momen~m ·
· jre.ference.frame.
·
0
Jat
Physicists often refer to such a .natural frame as the
... (1)
where r'cM is position vector of the center of mass with.·
respect to the origin 0.
Differentiate the position vectors with respect to time to
obtain the relationship between the velocities:
laboratory frame (even if there is no laboratory to be seen!).
The motion of particles within ·a system are often best
. described with respect to the ·system's center of mass-that :
is, in a reference frame with its· origin stationary at the
system's CM. We shall need to describe _systems in this center
of mass reference frame as well as in a .laboratory frame and
to transform physical quantities between the . two
descriptions. .
I.
I,
.
'
•
--+
I
~
--+
·vj=vj,CM+V·tM
•·.
·'·
.-
--+
'!
--+·
--+
... (2)
=VCM+V'i
The kinetic energy of the ith partjcle as measured in the
·
reference frame with its origin at O is
1
2
-m-v ..
2
'
l
The total kinetic energy of the whole system is the scalar
sum ·of the kinetic energies of each of the particles:
'
KE,0 tal
.Concept: Th_evelocityvjofa>particle in the labor'!tory
·!frame equals its velocity ':J},CM in,t11,e center of m_ass frame
tplus thewelocity of the center of mass with respect ·to 'the
·!laboratory frame :
·
·
,\:
--+
Vi
1
= '1"2
~ -m-v 2
ll
-
; ;: t2.
~!m.v t. v.
t
1
Substituting for ll, using equation (2), we o~tain
, .. (1)-"
www.puucho.com
~1 (_,
_,')(_,
-+,)
KE total== ~-mi
vCM+vi
· vCM+vi
.
; 2
Anurag Mishra Mechanics 1 with www.puucho.com
[ IMPU!,SE Aijo fl!J>MENTUM .
But
.343]
--,
--,,) (-+.
-+,)
2
( VCM+vi
· vCM+vi =,VCM
-+
,2
The total kinetic energy thus is
~ 1
2'
,2
-+
-+, '
KE,otal = ""-mlvCM +vi +2VCM·V;)
, 2
1 ""
2
=-."'1lmivCM
2 i
_MOST IMPORTANT CONCEPT
-+,
+vi +2Vci.i·Vi
1. Kin~tic energy of system in centre of mass reference
, frame is given by
J(/,system/CM
1
. '2 +vCM--,
""
'
(3)
+ ""
..£..i-mivi
,£.mi --,
vi···
i
2
1.
•••••••••••
.
0
\-
.
CM
also
Momentum of both the particles has same magnitude in
CM frame.
, Since CM frame is zero momentum frame
X
Locate a give particle with
two positicin vectors.
_,
Now we interpr~t the three differeµt terms in expression
for KE, 0 ta1.
Iri the first terin, the sum of all the masses is the total
mass m,0 ta1 of the system of particles. v CM is fixed for system
and there is only one v CM . The second is th,e kinetic energy
cif the system of particles with respect to the center of mass
or KE of system in CM frame. For the third sum, consider the
following, from definition of center of mass we can see that
Both the particles must have momentum in opposite
.,
direction in· CM frame. ·
4. Consider-· a gas
filled .cylinder kept in a
random
moving train. The gas
motion
molecules move randomly
of gas
molecular
- in container.. If the trains
--,
velocity is v o and a gas
molecule 'has random
Im/;=D
i
•
.
[If we assign origin of a coordinate ·system at center of
mass, what is position vector of center of mass in this
coordinate system?]
differenti!3-te this with respect to time
we get and so the third term in equation (3) is zero. The total
kinetic energy of the system of particle is then
.1
2
""1
,2
,KEtotal = -mtotalvCM + ."-1-mrvi
2
-, 2
Concept: Therefore ,the tot£1!' ki_netic . energy of the
system is'the'srim of:_
(a) the kinetic energy of the center of mass, as if the total
mass of the $JStem,were concentrate~.all at .that point; dnd
(b) the kinetic energy of the particl~ in reference-frame
with its Origin at the center of mass.
+ KEwrtCM
--,
JP1/CMJ= -JP2/CMJ
Fig. 4~ 18
= KEof<;M
-+ ..
-+
JP1/CMJ= µJvreil
........···
.___Th_at_is.a.,_ _ _ _KE_~to.!;l,l
m1
3. Consider a two particle system
0
z
m1m2
+ m2
µ is referred as reduced mass. Reduced mass
does not have any physical significance, its just a
combination of terms.
·
·
·
2. v rel is the relative velocity of blocks. Always
remember that vre1 • is independent of reference
frame.
3
...::'y•;stem
y, .../···
= 2µv 2rel
.µ=
where
i
2
· 1
.
--,
velocity v. The molecules
' .
-+
-+
~
I
rmrrmn7TTT1TTTTT17TTTTfTlltlmrrmr 1
,I
Flg.4.19
I
resultant veloci_ty is v O+ v.
The total kinetic energy of gas is
K,ota1 = Kin, +KcM
.
Kin, is kinetic energy of gas molecules in refe~ence
frame of container and KCM is kinetic energy-of the_ centre of
mass (bulk motion). The internal energy is independent of
motion of the train.
.
5. Consider a wheel rolling on a surface. An observer in
CM frame will see that
each particle of wheel in
pure rotation about CM.
Iii this case two different
motion are involved one
that of particles w.r. t. CM
: • X1and other that of CM w.r. t
Fig.4.20
ground_
Therefore;· KE,ota! = KE system/CM + KECM
'
!
,
www.puucho.com
'
-
Anurag Mishra Mechanics 1 with www.puucho.com
13~··· ++···
·.'
.In rotational mechanics you will learn that KEsystem/CM is
r~tational kinetic energy.
6. Consider a two blocks system connected with a
spring. An impulse acts on
e~· ' f;I"'
. frame block so thai: it acquires
a velocity v 0 • A ground
m2'°*"""!1 m,
.
observer see the sequence_ of _11_11
i~i,.
events as follows.
· m, :- m2, ,· · ,
(a) Spring beings to :..: ·. · Fig. 4:_21 ~ - ' stretch, now spring force
~etards m1 but accelerates m2 • Stretch in spring continues
fill velocity v 1 and v 2 are equal i.e., v,,1 is .zero. At the state
. of maximum extension both blocks have same velocity.
~v:rce
Sp(i~g.forrv, .
. ,mu:=:::mm;Jmm,b;,
.'"'
·
_j
. Fig._ 4:21 (b)
(b) In ground frame system translates forward .while
blocks also oscillate with respect to CM.
r-~: mwl=t~
n
v~,
Tl
;,,~Vo~CM
r
~
.
Max,trilum,corilpression
- "": f
• L
. "
+m 2
to½9B
.
(d) At maximum compression
2
I
At the instant of maximum extension
.
·. 1'P2Vo '_ ~
vCM= m1 +mi:
!'
CM frame
·1
· Fig. 4:21 .(d)
v
In the absence of any'external force CM ·= canst. At the
instant of maximum stretch of spring both the blocks have
same velocity which is equal to velocity of centre of inass.
In CM frame .blocks will be stationary at this moment.
· (d) While translating forwarcj spring begins to return to
tts natural length. When spring regains its natural length,
blocks still have velocity and spring begins to compress. ·
-··t~~~~-j~~",·r· - vi~1
cl .
of spring blocks are stationary in
CM frame
v~=O
,._
.
(c) At relaxed length of spring
m2vo
Vo
= --"--"--m1
~--._,!
.-=.
v2 a0,
(c) Forward translation of system ~ontinues with
.
(iv)
r1,=~=l"if=2.=~.'""""=.· ~~"if=.~-=-1
Fig. 4.21 ('.")
veIocrty v CM
Fig.4.21 (ej
In ground reference frame system moves forward with
spring . getting compressed and rel.peed, then extended
,,
alternatively. .
Now lets see what· happens in CM frame .
(a) Initial state
'Spring i~ fully ¢tr~~Cti~d1and
system haS translated:forward,
.
(Ill),·.
Relaxe_d length
(b) At maximum extension
:ix:1:::;::;:~~v
.
:
-----
"'
Maximum extensicin
(ii)
(i)
l-.:
I
_· r
... -··
-
. ;lnjtlaJ state
.. --- -··· ------ ---
-- --------- -- --- .
·~
- ........ ____J
.
v;~o ·L , . I
'I D-0 L_:M"~i2!
'J.,,;
l~itlal state
q; -~.... ___
.
' :,
,,,
.<
I
l
(e) At the instant of maximum compression velocity of
both blocks i.s same in ground frame.
·
--:---~n-"•.~.
.
,,
ME(HAtflCS-1
~ ". f
-
•
-·. .
•:"
'.'
'
-
½
-- . I
In CM frame the observer sees only oscillatory motion of
blocks
· At the maximum extensiori of spring blocks are at rest in
CM frame.
or
or
or
Xm~
=/gf vo
CONCEPTUAL EXAMPLE: Initially blocks A and Bare
given impulse in 'opposite directions as shown in Fig. 4.22 (a).
Now we have to calculate the
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
·-- . -- --- -
I IMPULSE AND MOMENTUM
345
Most Important Concept:
...,
...,
...,
+ V CM"/ground
V ,.,/ground ::: V ,.,/CM
,.
...,
Fig. 4.22 (a)
(i) Maximum stretch in spring
(ii) Maximum velocity of block A and B in ground frame.
(iii) Minimum velocity of block A and B in ground frame.
Solution:
vCM = (2m)(2vol- mvo = Vo
3m
At initial instant
...,
-,,
V 8/CM
; A/ground
· 7 frame
= V Ajg-VCM/g = Voi
-,,
-,,
/ frame
is maximum when ; ,.,/CM has maximum
7 frame
...,
-,,
a
...,
....
equilibrium position (spring is relaxed) and v A/CM and v CM/g
are opposite to each other.
...,
Fig. 4.22 (b)
Jv Amin I= 0
Thus
From work energy equation in CM frame
We get
W spring ::: AKE system/CM
_ _!,_kX2 = o-.!:. (2m)(m) (3v )2
0
2
max
2 (2m + m)
Minimum. velocity of B is attained at the instant B is
...,
moving toward left (opposite v cMJ and velocity magnitude is
v O (see Fig. 4.23)
-
1(2 )
:-'-7
-kX ~ =- -m (3v 0 ) 2
2 max 2 3
or
-t
Minimum velocity of A is attained when block is at
1--vo
CM frame
1
-t
Jv Afmaxl=Jv /,/CM
J+JvCM/gmundJ=
2vo
frame
Y·frame
...,
...,
...,
Jv B/maxl = Jv 8/CM J+ Jv c,,i/grnund J= 3v 0
r frame
'/ frame
Similarly
At maxirrium stretch
lo + Xm __.
and
Yframe
l
CM frame
4---
,/CM
/ frame
v CM/ground vectors are in opposite direction.
.,
~~Vo
~
Yframe
•7 frame
...,
VA/CM=
Yframe
+ V CM/ground
/ frame
...,
v
a
= V 8 /CM
V 8 /ground
...,
Similarly v ,/gmund is minimum . when v
Initial state
Ve/CM=
/ frame
,magnitude and is in same direction as vector; cM/f::~d.
A
= V B/g - V CM/g = -2v oi
2vo
...,
Note that velocity of any block in ground frame is
:superposition of two velocity vectors, velocity of block in CM
frame and velocity of CM with respect to ground.
...,
...,
V NCM
similarly
~7 frame
lva,cMI = vo
-
'.""7
~
6
Xmax = ~ : Vo
Vo
lvAJcMI =2
f-- Vo
CM frame
Fig. 4.23
Blocks return to
relaxed state
Vs/CM
=2vo
diagrams
representing
situaUon when
block returns
to relaxed
state
Thus
vNCM =_vo
~
Block
I
_.Vo
What appears in
CM frame
I
I\vo block A and B of the same mass are connected to a light
spring and placed on a smooth horizontal swface. B is given
velocityv 0 [as shown in the Fig. 4E.15 (a)] when the spring is"
-in natural. length. In the subsequent motion:
- -
vAJg = 3v 0
lj'~~iimiRJi . Ji57>
v819 = 0
~
What appears in
ground frame
Fig. 4E.15 (a)
Fig. 4.22 (c)
When spring again regains its natural length in CM
frame.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
I346
,,,
--,· -
..-.~--:
"<·"'"~7: '.·· . '
'. ' . -· .
=---:::--=--·=--=---=--===..:====:::;::::::;-----'--
: = :_
: ::
_
::
.:.
:
'(A} the' max/1n.;ir, velocity of B will be v 0,
'(BJ as seen from ground, A can m~ve towards right or{ly__ .
'.CC) the spring will have maximum extension, wlien A and B
i both stop
,
· '(DJ the spring will be at natural length again when B' {s at
L_.n,st,.____ ____
·.
______:;__~
Initial velocity of B = v - .':'. = ~ v = ~ v to right ·,
5 5
5 ·
· ·
Blocks are executing SHM in CM frame with initial
position as equilibrium-position ,
Step III: Velocity variation of B in ground frame,
considering right as +ve(
).
, _
Solution: In CM frame the blocks will perform SHM as
= Vo
V
2
CM
1
-
[
1
I
-~~~
maximum extension.
L~a:r,,if~t~J 16
!Tu,~- bl~cks Aand B.of masses 2m and 3m placed on smoothl
·t·<;.·
lhorizontalswface are connected with a light spring: T/le twoj
blocks are given velocities as shown in Fig. 4E. l 6 when spring
is at natural len"'h.
_____ .
( '"---·!(
l
t~oooooooOoooooooo.F}--+]v
-------- -~...!~~~~__:__ _ __:_ _______
' ",:, :-::;r;: ' _·
(P)
v
(B) maximum magnitude of velocity of A 1 (Q)
v
:
-_< .•.. ~ ..;
'
'
. ~:;:;:-::::,-'."'.f":4
' '"
,·. ·. _
7
e
Jhownl
-~tring
~Lm
...
·i ·: '
.·:. -
"j :..
A2m
-i:t::_~
, Fig. 4E.17 (a) [' . •; ,
I
Solution:
c9;~;,I-so-1""".~--.-th_e_p_·~-o,-bl_e_m-_i_11_C_,'M_._fr_a_m_e_:_In_C_M_1'-r9-rn~
lpilrtides'A. and B mo~e .Cllorig circular-path with same· ci/
.
VCM
:J
(mx O)+ (2m xv) 2v
= ---~--~ = ,
In CM frame
3m
3
m-7
[
· ·
5 5
length) Ire qn a smooth honzontrtl11lane,A zs mven a.yelofit.Yi
of V m,/s alorzg the gr~un;:t perpendicular to line AJ{ds
in Fig, 4E.17 -(a). Find th~, t~.nsipit in
during- thefirl
subsequent_mo_tio~.•·; _. ,
.,"·':..'·L:'.:'J
.~,-,-.;
'~1·~· ~ -~ ·.' ....,.., "'"'"w!l,!"~" w
_ro7:r,-'"t~~Q~Y.!!I~tl.r:;'.f_.:.--·-~t ~!~):~:f'~~~U,f~~~umn ..11
(vAmax /during motion
4v · v
3v
+ = - ~ '·
Th'.9 mas~es,A and B Connec_ted with' a.n ih.e.Jf!~n:ible ~iring; bfJ
I~
I
·
= v to
Thus minimum velocity of A is -v when. spring is at
5
.·._
.
(A) mhtlm~m magnitude of v~locity of A
(vAmin )_during motion
v
5 5
Both blocks can have maximum velocity v 0 /2 towards,
right in CM frame.
Thus, Cvalmax = v 0 in ground frame
(v A ) min = 0 in ground frame when A has.
velocity v 0 /2 towards left in CM frame
Also, in CM frame Va = v 0 /2lefyin the case when spring
is at natural length
Thus
v a = 0 in ground frame at that instant.
-
+
so
lvamaxl=v&lva · l=·o '
Velocity variation A in ground frame mm · • 7
from
(~•~)-;to--~+~~-~-:
Jv~24£J::·;·~11
·
4v
from
., ,
. c:M.1r·
. . e,3:
5
'7
.,
. '.
;
. : 2m · ·
(C) maximum magnitude of yelo_city of B
(v8 ~ )'during motion
· ·'
(R)
O.
(D) 1velocity .of centre of mass· (vCM) of the
(S)
7v
5
isystem comprised of blocks A, B and
spnng
,
1
.
,
·Fig. 4E.17'(b),
,;
. of zm mass =v-2v = -V
ve1oc1ty
3
3
Angular velocity of particle A
.
Solution: Step I: vcM = C3mv)- Zmv = .':'.
Sm
5
Initial velocity of A =
(-v -~) = - ~ to left
V
I
1I 3
l
I
·
2
v2 1
12
or
www.puucho.com
2mv2
T=--
31
3
,
J
m
•
~1M,
3
V/
Ol=--=T=Mm r=2m.-.-
Step II: In COM frame
~
:o;~
03
•
,
2m
CM frame
•
c, ,·;
V
3
3
'i! •
.·
. ;. .
I' ,, Flg.4~.17(cj ;
j
Anurag Mishra Mechanics 1 with www.puucho.com
.JMPUISE AND MOMENTUM
,!4?]
kexarp:el~J~D>
\nvo small balls .A arid B are-i-nt-e-rc_o_n_n-ec-t-ed_b_!)'_a_n_in'--ext,-e_ns_ib_'~le
· !string of length L. Mass of ball .A = m, mass of ball B = m. The
~alls are resti11g·on a frictionless horizontal swface, with the
distance between them = 3[.15. In this position, ball .A is
suddenly given a horizontal velocicy- v o,perpendicular to the
llinejoining the two bal¼,_[see Fig,3E.18 (a)]
B
I
,
I
When string is jerked along length of string velocity
component along string is same for both particle
illustration 7. In the Fig. 4.24, a block of mass m
moves with velocity v O toward a stationary block of mass M
on a· smooth horizontal surface. Find the maximum
compression in the spring of stiffness constant k.
k,, -M --- -!
c·;;,--~,
. . . D~ :2:mO.
j
1
Velocity of C.M .. is unaffected
/'
by . the compression in the
j
spnng.
Il__________
' - _,_ _____.......J'
Fig. 4.24'
I
·,
i
I
Solution : We apply work energy theorem in CM
frame. At maximum compression blocks are at rest in CM
frame.
A
Fig. 4E.18 {a)
(a) Find the speed of ball B just after the string becomes taut.
:(b) Find the impulse of the tension:in ~tring when _the string
I becomes taut.
·
(c) Find the steady tens.ion, in string /7luch after tf,e string h~I
L_becpm_g: tC!.!!t ' ·
· .• ·
•
' ....J
Solution: When string is jerked the motion of two
particle system is super position of translation ahd rotation
of two particles about CM. In CM frame two particle system
will rotate about CM. Fig. 4E.18 (b) and (c) show lab frame
and CM frame.
B
:b..
..
··,;L
···-•• ~ ·
v0 cos'e
· v0 sin0
,, ·
situation before String gets jerked
Ja<:
"';'.
•· ••r-l'v 0oos e
"--\v 0sin8
Lab frame
=
o-½kXiax ·
Llk = 1µv 2 1 -0
2
"
Velocity of centre of mass,
Iriitial velocity of the block m w.r. t. centre of mass,
,
Mv 0
Ul=Vo-Vc=-m+M
Initial velocity of the blockM w.r.t. centre of mass
,
-mv 0
U2=-Vc = - m+M
1
,2
KE system/CM = mu, + lM,2
u2
2
2
2
.Thus.!(
mM )v~=.!kx orx=v 0~.!(·mM)
2 m+M
2
k m+M
0 Vo
••
wspring
=.!(
~ -"';"
mM
)vo2 =.!µv21
2 m+M
2 "
illustration 8. In the Fig. 4.25 shown, if all the
surfaces are smooth, then determine the maximum height h
attained by the block on the wedge, assuming it to be very
large ..
~~sme
-2ln CM frame velocity of B.
(b)
(c)
Fig. 4E.18
;
2
·111
COS0)
.., .
.
.
2
.1.ens1on 1n string :::; - - - - -
(Vo
L/2
Afthe highest position both the block
and _the wedge move together with the
2L
velocity of CM
When string is jerked
Tangential velocity remains
unchanged whereas velocity change
. , B . v sine
along stnng ,or 1s -.- -
~
v,c~se_u
.
· mvsin8
Impulse of tens10n = - - -
F!g, 4E.18 {d)
2
2
'
mvi cos2 8
'
I'
I1
~-----F~lg~._4._25
_____ __,
Solution:
In CM frame block and wedge are at rest when block is
at maximum height.
Ws,avity = LlKE
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
348
MECHll~ICS-1
-mgh
~ o-1( .mM )v5
2 m+M
h_
or
·
v5 (
M )
2g m+M
If wedge is infinitely massive M
~
=
In CM frame, frqm work energy theorem, we get
F'i x1 + F' 2 x 2 = lk(x1 + x 2 ) 2
2
h = Vo
2
2g
F'1 (x1 + x2 ) = l k(x1 + x 2) 2
2
"
2F' .
or
=>
1\vo"blocks of mass m 1 and m2 a" connected by a spring of,
1
!force constant. k. Block of mass m1• ·is pulled by a constant
lforce F1 and other bloc~ is.pulled by a constant force F2 [see
IF,ig. 4E.19 (a)]. Find the maximum elongation in !he spring.
I
i
8::1::::cr
i
I~_ .
I
F 1 =F1 -m1a, =F1 -m1
( Fi _ Fz )
m1 +m2
= (Fi mz + Fzmi ) towards right
m1
+ X2) = __l
k
x
or
= ~(F1m2 +F2m1)
k.
max
lnvo
Solution: We will solve this problem in CM frame
which is accelerated.
(F1 -Fz)
aCM =
m1 + m2
Assuming that F1 > F2 , CM frame is non-inertial, we
have to apply pseudo force on the blocks.
Therefore net external force ~n m1
(X1
1~6~~~J20~
Fig.4E.19(a)
+ m2
m1
+ m2
·------
blocks m1 and m2 ·are connected by a spring o/f6rce
'constant k and are placed on a frictionless horizontal sutfate,
Initially the spring is given extension x 0, _when the system is
released from rest. Find the· distance moved ·by two-blocks
before they again comes to rest.
·
·
im,.,B
L
,---·-
klo
0000~0~~~~~!~,~o~:oo
:f:1 .
·
Fig. 4E.20 (a)
.
.
Solution:
[:b~~:~2~ :~:~fa~=~~-~:~::aJ
0
-
~33··==·~~
I .
I
·"~""""--«•-
--·•
•
"'••,·•
••,·•-·
lo'-Xo,
---
••••-·••-~···-•·-·-·:,··"--
~
•
[flooooo~~o~~o~GJ
r--
::
Fig. 4E.19 (b)
L--------------
---..i
and on m2 ,
....1
.'
Ax2
·"-
·/-"
7
·:-/'
"Fig, 4E.20 (b)
.
. .
( F1 -F2 )
F 2 = F2 - m
2a, = F2 + m 2
.
m1 + m2
.
= (Fi mz + Fzmi ) towards left
m1 + m2
.
In CM frame, the blocks move. in opposite directions
thereby stretching the spring. The spring will have
mazjmum extension when blocks 1re instantaneously at rest
in CM frame of right block displaces distance x 1 and left
displaces a distance x 2 from their initial positions.
In absence of external force blocks again come to rest
when spring is compressed by x O• There is no change in the
position of C.M. of the system, Le., Axcm = 0 of block m1 ·
displaces by Ax1 and m 2 _displaces by Ax 2, then
We have
... (i)
and
'·-= =· m1 Ax1 + m2Ax2 = 0
""
m1 +m2
... (ii)
After release spring block system will execute
oscillation~. •Figure shows five ~tages
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
349
IMPULSE AND MOMENTUM
Illustration 9. Consider a system
of two particles of masses m1 and m 2
separated by a distance r. Suppose they
start to move towards each other due to
their mutual attraction (attractive force
may be electrical, gravitational, etc.). Since
After solving equation (i) and (ii), we get
... _ 2m2Xo , .. _ 2m1Xo.
UA.J L.l.A2 m1 + m2
m1 + m2
[;5_x_q.~g,1~~~.i211>
-···-: ·~1 F CM F-~2'.
: ......... 4--9:
~.~r1-r2-.:'
.
··7····
_,
nvo blocks of equal mass m are connected by an unstretched·
spring and the system is kept at rest on a frictionless
horizontal surface. A constant force Fis applied on one of the
blocks pulling it away from the other as shown in Fig.
,4E.21(a).
system
particles start from rest, and F,n = 0,
_,
It follows that
'(a) Find the position of CM at time t,
(b) If the extension of the spring is x 0 at time t, find the
displacement of the blocks at that instant.
..
Fig. 4.26 (c)
_,
m1 V1+m2V2
m1
... (1)
0
+ m2
... (2)
... (3)
Fig. 4E.21 (a)
... (4)
The centre of mass of such a system remains at rest
Solution:
C.M.
'
:
~
~SJ
_,
y
1.
ill(:
unless acted upon by an external force. In the eqn. (4), A r1
'
:
1~
r
.
'
:
ill(
:
:
:c2M
:
.! [:}o~~~~~::e~~~~o~~fr
Ol4 - - - - ill<,
mr
2
r1 = - - and.
:
Fig. 4E.21 (b)
(a) The acceleration of centre of mass
->
and A 2 are absolute displacements of particles m1 and m2.
If r1 and r2 are displacements of m1 and m 2, then
r1 +r2 = r
and . m1 r1 -m 2r2 = 0
If follows that
F
F
\aCM\=--=m+m 2m
The position of C.M. at time t
1
2
1 F 2 Ft 2
t.xCM = -aCMt = - - t 2
22m
4m
(b) Displacement of C.M. is given by
t.x
- m1f.X1 + m2M2
CM ,
m1 + m2
where t.x1 and t.x 2 are displacements of m1 and m 2
respectively.
Ft 2 mt.x1 + mt.x 2
or
4m
m+m
2
Ft
... (i)
or
M1 +M2 = 2m
The extension of spring is,
t.x2 -t.x1 = x 0
... (ii)
After solving equations (i) and (ii), we get
2
2
t.x1 =I_[Ft -x0 ] and t.x 2 =I_[Ft +x 0 ]
22m
22m
m1
+ m2
which shows that the particles collide at the centre of
mass.
Illustration 10. A projectile of mass mis fired with
an initial velocity v O at an angle 0 to the horizontal. At its
highest point, it explodes into two fragments of equal mass.
One of the fragments falls vertically with zero initial speed.
Since the only external force acting on the system is
gravitational, the motion of centre of mass of the_ system
(the fragments) follows the same parabolic path as the
projectile would have followed if there had been no
explosion. Force of explosion is internal, it cannot change
the trajectory of the system.
---R-
O+--R/2-+- m
CM
----X2---•
X
Fig. 4.27
No external force acts on the sytem in x-direction.
Therefore
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
f~~o_-_--_-- -_--_·________________________
-,
XCM
R
or
=
-,
(xCM)f = (4M)_(x)+M(x-5R) =:fx--R)
4M+M · ·
··
x-coordinate of centre of mass.is fixed.
'
Therefore CxcM J, = CxcM) t
(L+R) = (x-R)
or x = L+ 2R
m 1 X1 + m 2 X2
m1 +m2
= mxR/2+mxx2
2m
or
'x 2 = 3R/2
If we choose origin, at position of centre of mass, then
-,
XcM
= 0=
-,
m1 X1
[=.. e-,.xam"''iJ
• ---·~-
-,
+ m2 X2
V
,
m~N~+mx½=O
or
x 2 = +R/2
+ If one of the fragments lands back at the initial
position of the projectile,
-R- mx0xmxx2
XcM m+m
or
x 2 = 2R
+ If the vertical component of velocity of both the
fragments, after explosion, is zero, they-land at the
same time. If one of the fragments is moving
downward after the explosion, the other fragment
will have upward component of velocity. In this
case, the downward moving fragment strikes the
ground- first. The ground exerts a force on it before
the second fragment reaches the ground, that is an
external force on the system, therefore our analysis
does not apply after this· instant.
mustration 11. A small
22
'
~
I~
l4.fro~ s,;;~n the end of a lo;;-;;:iard of l~~~r'h
m
(
MEftiiti1cs!f!
i: Th~-boa~a.:
Irests on a frictionless horizontal table. The frog wants to jump I
!to the opposite end of the board. What is the minimum;
jtake·off speed i.e., relative to ground v that allows the frog .to[
\do the_gj_ck? The. board and the frog)1ave 1cqual masses. ____ !
.
;..,-
Solution: Taking v for the plank in ground frame and
conserving linear momentum in horizontal, direction
m:: :~~s;s0)
t
: ri~s······-.. _, ._. :,
2usin0
v~
=- - -
1
lifftil/lH/1UUl1Jillf1M,i
fig.
_J
2
L._ -
g
~!:,?? .
.. L = 2u(u cose + u cos0) sine = 2u sin 20
g
g.
U=
~
v~
Minimumu=N
r::-·--·
. .. ·_7
.. ·······-,;7Ysteml
sphere of radius R is released from j Y
rest on the inner surface of a large ,
/
6R
~-. · .
sphere as shown in Fig. 4.28.
:
\
There is no friction between any
\, (L, O)
x
~
surfaces of contact. When the
small sphere reaches the other
·· ........ ·
extreme position, there is internal [___· -~~~-28
.
transfer of mass without any
external force in the x-dire.ction. Therefore the position of
the centre of mass of the system remains fixed in the
horizontal direction. When the small sphere moves to left,
the larger sphere moves to right so that the centre of mass
does not move in the x-direction.
I bExaJm,~p
=/]
Find total W.D. by friction assuming plank is sufficiently long. I
,
. p7.,.." ·----_ ·1
I
_2m
3
-,
-,
or
m1 LI. x1 + m 2 LI. x2 = 0
Let centre of large sphere move through x towards right.
Then M(-10R+x)+4M>:=0
or
X= 2R
Now coordinates of centre of mass are (L + 2R, OJ
Alternatively,
(x ). = (4M)(L) +M(L + SR) (L +R)
CM'
4M+M
:
I
·- - -····~
,_:J
Solution: Where slipping stops both moves with same
speed by momentum conservation
mu= 3mv
u
v=Work done by friction = LI.KE
=Kt -K,
Now,
,
I
·
!
I..__________ -------==--=-=~..,.,,._._
'--.smooth
. Fig. 4E.23 (a)
. : I
=!.2m(~) +!.m(~)
2
2
3
2
3
2
_!_mu 2
2
2
3mu
1
=----mu 2
18
2 .
= -!.mu 2 joule
3
www.puucho.com
.,
Anurag Mishra Mechanics 1 with www.puucho.com
I,. IM~UISE AND MOMENrilM
,
351
,··---
---,
Solution: Ci)
,Find maximum height reached by smaU'.mass m iii Fig. 4E.24[
'(a} and Fig. 4E.24 (b).
. · .,
I
V
I, i1.1777mmmn=_(•=l
I
;
m
m
'7777m7m7777
Fig.4E.24
i
I'
(b)
Solution: Mass of both the blocks =m
. bigger block remains at rest till smaller reaches at
.bottom of circular part. ,
·
· Velocity of smaller bl6tj{ at-lowest point u = ~2gR.
Now bigger block also start moving let smaller block
reaches up to height h.
By momentum conservation
mu =2mv
u
v=-
2
By energy conservation
increase in PE of smaller block= dee. in KE smaller block
+ KE of bigger block
mgh +.!mv 2 = .!m(u 2 -v 2 )
, 2 · 2
·
·mgh =.!mu 2 _,!2mv 2
2
2
u2
mu 2
=-mu --2mx-=---(4-2)
2
2
. 4
8
1
·
2
2mu
2
1
.
(__
Fig.
=> ~ :
4E.2~~-----.J
Initially no momentum along x-axis. So, final
momentum will be zero also and relative velocity' is also
zero. So, no velocity of any object.
By energy conservation,
initial potential energy =final potential energy
Hence,
8 =90°
AXCM = 0
m(2R-x)=Mx
m(2R) =(M + m)x
2mR
2(M/2)R
x=--=
. M + m M + (M/2)
Ir--:-- - -- ---- :
1C:±J ~I
Fig. 4E.25 (c) _ _~
I
(ii) Maximum velocity of wedge will be when the ball is
at the lowest point in the wedge as till this point the
horizontal component of normal on the wedge will be
speeding the wedge.
Pi= 0
Pt =-Mv+mu
Pi =pt
Mv
U=-=2V
m
Ui +Ki= Ut +Kt
mgR+ 0 = o+.!mu 2 +.!Mv 2
2
2gR
mgh = - - =*· mgh = m 8
4
2
2
2mgR = m(2v) +Mv 2
R
2xM xgR=4mv 2 +Mv 2
h=-
2
2
~~;~_1€~
MgR=4xM xv 2 +Mv 2
2
MgR = 2Mv 2 +Mv 2
In the Fig. 4E.25 (a) shown the
~A
RB
we_dge of mass. M has a semicircular_·
.
smooth
··z
.~
M.
M.
/
groove. 'A -parttc
=~
',. e o, mass-m
·
2 u; ,lm\=s===~
w,~fu
released "from Adt slides on -the
Flg .. 4 E.25 (a)
smooth circular track and startsclimbing
th~ rightface.
.
(i) Find the maximum value of 8 which it can subtend with
veftical and also find the distance displaced by we~e at
this position. ,
(ii) Find the m__aximum velocity of wedge during process of,I
on
I
I
[Y
I
MgR
= 3Mv 2
v=~
Concept Review: ( Revisit Concept Review after
studying Collision Theory)
1. Law of conservation of momentum states that in an
inertial reference frame the momentum of system
remains constant if net external force acting on system
is zero.
motion"---"'"'-'---------------~
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
2.
3.
4.
5.
An isolated system implies a system in which there
is no interaction between system particles and bodies
·
external to system.
Law of conservation of momentum is valid in all
inertial referenee frame although momentum of a
particle depends on reference frame.
Law of conservation of momentum is a fundamental
law of nature, its not consequence of Newton's laws.
Centre of mass of a system is not a physical point these
may or may not be any mass present physically at the
centre of mass.
The location of centre of mass depends on choice of
· reference frame.
·
·
Momentum of system is given by
-,
J:.p,
system
conserved. During collision KE is not conserved.
During c~llision deformation of bodies takes place,
fraction of KE converts to deformation energy.
15. In elastic collision total KE of the particles before and
after the collision is same. Momentum of particles in
CM frame changes in direction only and magnitude
remains same.
. 16.
.R
.
--+--+
·•
Ball JI'.
' statlonary
. all I
·I
.
)I:
vrf2
, vrf2
vrf2
Fig. 4.29 (a)
After collision let the deviation of ball I from ~CM is 0.
then .other particle rrioves at 1t - 0 with ;
opposite sense.
A.
.11/4"•••
I......
---t--+
CM
in the
vrf2
Ball I
9. KE of particles in CM frame is
1
II
CM frame
P:\'CM = m 2(v 2 -vcMl =µ(v 2-v 1)
-,
-,
. P1/CM = P:\'CM = Jlv rel
KE,ystem/CM =
.
oblique and elastic collision .
between two particles-in ground frame
•
=m1(v 1-vCM)=µ(v 1-v 2)
-+
..
Vo
= M,otal VCM
6. Equation of motion for centre of mass is
--+
d --+
. -+
Fextemal = M total - (v CM) = m aCM
dx
.
7. The reference frame in which CM is at rest is referred
.as CM frame. Total momentum of system is zero in CM
frame.
8. ·Momentum of particle 1 in CM frame is given by
--+
_;--+
--+--+'
P1/CM
14. In elastic collision KE before and after collision is
· Fig, 4.29;(b) .
·2
2µvrei
10. During collision. the force of interaction between
colliding bodies is large as compared to gravitational
frictional, (i.e., non-impulsive forces) when smooth
bodies collide, force of .interaction acts along line of
impact.
11. Coefficient of restitution is independent of reference
frame. Experimentally it has been found that e
depends on impact velocity material of colliding·
bodies, shape and size of colliding objects.
12. During perfectly in elastic· collision bodies stick
together and move with common velocity·
0
v1 is velocity of ball I in ground frame.
From figure a. = 0/2 After oblique collision ball's line of
motion make right angle with each other.
a.+~= 1t/2;
mass moves with ; CM • Thus in_ CM frame particles are
at rest. Therefore total KE of the particle is converted
into internal energy of system;
'
~ = ~-~
;,u~1
vof2
<~>
KEbefore collision+ KE after collision
KE always decreases during in elastic collision.
13. II) CM frame total momentum of system is zero so
before collision the two particles have equal and
opposite momentum. After collision the combined
a.= 0/2;
r-------- . 2
,Ballll
l
,.,J.I
Fig. 429 (c) ·
From vector diagrams of ball I and II
we get,
'
.
V2 = Vo SIIlCX
www.puucho.com
2
I
Anurag Mishra Mechanics 1 with www.puucho.com
- ------- - ___________ _
I/ -IMPULSE
AND
MOMENTUM
-- --- .. __________ ------- -~---·
---------·
,
353
,
17. Figure shows oblique elastic
collision with a stationa,y ball
of mass 2m. If ball of man m
turns by an angle of 30° in CM
frame, we wish to determine the
angle of divergence between
balls after collision.
·~
0
.m
Ba/11
2m
Ball2'
Fig. 4.30 (a)
J"rdm
'"?
y
rcM = - - -
M
The component of this
equation are
1
XCM = Mf xdm;
Step 1:
YcM
= ~f ydm;
Step 2:
ZCM
= _!_ f Z dm.
z
Fig. 4.31
M
Consider a thin rod of
mass M and length L as shown in Fig. 4.32.
.
.
Ball I
......,----,,.~ ......
P1tcM
·---'a -,-------.....
·T[]
..
y
Ball II
,,
"
,,
h ::
P21CM
•-0
--.,,
r
d:,
"::
..
I \.,=,,,jj
i .·------- ·-·
z
(b)
(a)
Fig. 4.32
-> I ;::,;_
Vo
IV2fcM
3
--->
Vo
lvcMI = 3
Fig. 4.30 (b)
1v'~CMI=~
3
· Now apply trigonometry on vector triangles
sin a= 2sin(0 - it)
0 = 30°;
a= tan-
1
.J3]
1
[
1
+
a+P=75°+tan-
~,~xgm_~J_'T----~>
,-
So angle of divergence is
.
The infinitesimal element in this case is a slice of length
dx. The rod has to be thin enough to ensure that all the
particles of the element are at the same distance from the
origin. If the volume-mass density (mass per unit
volume) of the rod is p (kg/ m 3 ), the mass of the element dV
is dm = p dV = pA dx. If we define 11. = pA, we have
·dm = 11. dx. The quantity A.= M/L is called the linear mass
density (mass per unit length) and is measured in kg/m.
For a disc or a cylinder, the appropriate element is a ring
of width dr and area dA = 21tr dr, which extends through the
body of the solid as shown in Fig. 4.32 ·(b). Its mass is
dm = p dV = ph dA. If we define cr = ph, we have dm = cr dA.
The quantity cr = M/A is called the areal mass density
(mass per unit area) and is measured in kg/m 2 • Note that A
is the cross-~ectional area in a plane of symmetry.
1
.( a) Show that the CM of a uniform thin rod of length L and,
mass M is at its centre. (b) Determine the CM of the rod
assuming its linear mass density A. (its mass per unit length)
:varies linearly from A. = A. 0 at the left end to double th<).t'.
\value, 11. = 211. 0, at th_e right end.
:.
.J3)
1
(
1
+
FINDING THE CENTRE OF
MASS BY INTEGRATION
A continuous body may be regarded as a collection of
point particles. The typical ith element has mas l'.m, and the
position of the centre of mass is given by ·
!
y
0
dm=Adx
x----.i~
dx
X
L,r/1mi
rcM
=~'~-M
Fig. 4E.26
If we take limit l'.m ~ 0, each element shrinks down to
an infinitesimal element of mass dm. In the limit, the centre
of mass of the extended body is expressed as the integral
Solution: (a) Let the rod be placed such that origin of
coordinate system lies at the left end. The rod is assumed to
be thin, soy CM = 0 and zcM = 0. The linear mass density of
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
,3_5_4 :-+·: :_,_1!,_':c_.·'\:; .'
MECHANJC~~I -j
1
the rod is;\,= M/L. We now imagine the rod as divided into
-infinitesimal elements of length dx, each of mass elm = ;\, dx.
So.
,
X
CM
= _!_rx=L xdm = _!_JL 'Axdx =
MO
AfX=O
!::.I
x21
M 2
L
They-coordinate of elements js y = R sin 0. The angle 0
varies from O to 1t. So
1
' 1
=..!..~~;\,]"
M
. o
2M 2
Thus CM of the uniform rod is at its centre.
(b) Now we have ;\, =;\, 0 at. the left e11d, Le., x =.o
and ;\, =2;\, 0 at x =L. Ac~ording to tp.e g,.ven condition, ;\,_
· varies linearly.,So we.write
;\, =A 0 (1+roc)
Atx = L, ;\, = 2A. 0 , so.a= l/L. Now·
0
=_!_2R2;\,
M
Using ;\,
.
,;~ (x: +~)~·=i~L2
. ijf#a~J5iltjjil
and ;\,
:a
.
'
·
x=O
=1- 0
. , - 0 •
2R
= -...
In this case centre of mass of th~ .body is not within the
body of the object.
·
·
0
. ,', M~JL .dm=JL;\,dx
·
= 7tR
M, we have
7t
o ·. ·L ·
Now w~ will determine M in terms of ;\, 0
sin0d0
1
YCM
=I_A.oJL(l+,~)xdx
M
.
=MR 2 ;\, 1·(-cos0) 1·
XCM = _!_ r=L ;\,'xcfx :
.
M x=O
.
M
M
=_!_J• (Rsin0)A.Rd0
Mo
·.
0
;\,L2
L
=~=-
.'
=-Jydm=-JyA.Rd0
YCM
·
1i{1 + f.) dx
2a ~
Determine ;the position• of centre'. offn.ass of an object ofmassJ ·
M in 't~~ ~hape of right. triangle wh?se dimensions_ a~el .
shown in Frg, 4E.2& The ob;ect has-a .uniform mass.per umt, ·
. '., '
area_
.,
dril'
=A.ol(x:~)( =~A.al
. 5 A- 0 2 5
. xCM = - - L
= -L
, ':
6M . 9_
Then . ·
.
.
,.
Fig. 4E.28 .
Fe-·x~~:
1r·etermine
• ••
I
• -
'-
.
tit,~ ce~tre
~. " ; , ,
.
:
~
_
l Solution:, We divide the triangular lamina intci narrow
of width dx and height y as sho_wn in the Fig. 4E.28.
11le mass dm of each, strip_ is ·
.
·
of. mass of~ loop of radius' R, ~US$)M.
Total
mass
of
the
object
f
th
.
dm= ·
. .
xareao estnp
Y dm::::Ad's=I\.Rd8
,"
1
Total area of the object
.
'
"' ~ .
...,.·-:-M-:-cc· (y dx) =
dx
· f '.rips
,. ~
_2M_y
(1/2) ab ,_.
ab ·
.'
... '
· Now x-coordinate of the ~entre of mass _is
'XCM
X
J
. 1 ·x dm=.1
=-
·M.- '·
Flg.4E.27
.
f"
M
J" X(2M)'
·-.- y dx
O
ab
=·xydx
2
· Solution: We choose origiri of .coordinate system at
ab O •
the centre of curvature. and the y-axis on the loop's line of
. To evaluate this integral we must express y in terms !)f
symmetry. Theri x CM =0 because of symmetry. We ch<;>ose a .
x. From similar triangles in the figure we see that
· mass element of length ds =R d0. Since the· total length of
.
y
b '
b
· the loop is 7tR, the mass per.unit length is;\,= M/7tR, where
-=- or
y=-x
x a'
a
M is 'the ,totalinas~ .. The ·mass of .the element is thus
"·
•·
elm= ;\,ds = A.Rde
.Hence,.
·xCM =-f"x - xdx
.
www.puucho.com
.
2· (b)a
ab
0
.·
,.
Anurag Mishra Mechanics 1 with www.puucho.com
_IM_P_ULS-'-E-'-'A_;_:;Nec:l}cc.M:.:cOMccE;::.N:.:.TU:.:.M::.__~-· ---·
c..l
~2Jax2dx=il(x3 Jla
3
a2
o
-
a2
=~a
= __!_fnl•
dmRsin0
M o .
.
and-·.·. YCM
3
0
1 fn/2 M
· ·
·
= -·
· - -2 (21tRfos0Rd0)Rsin0
0
On similar lines we can calculate the y-coordinate to be
.
2
YCM=-b
M
.
2itR
,.
RJ
· ~~amJ,£Je:j~
=
3
k-lsXaroe;~~
0
·
R
. sm ·a cos e·de =·-·
.n/2 :
.
0
2 .
.
!Determine the coo,rdinates of centre of mass of a half disc ofi . Determine thii'CM of a unifo.111}:solid cone of height h' cind1
!mass M and radius. R, assuming uniform mass distribution . .: · . semiangle a as shown in Fig. 4E,31,
·
·
:
'
!
,
1--.- - -
•
'
",
•
; •,.,'
~
i
f
h
-.,J
--1
_!
_F,..;ig;.._4E,_'.;,29;___x._lc.._'----~__J
---s~i~u~~;
'
.
--d,i
"
I
...-t--y-
1
.
/::
w~:se a semicircular strip of radius " · 1· x
_. ·
and thickness dr as differential element .. From, symmetry we ,· L.--~-=""'-_,___F_le_._4_e._3_~,--·_ ..J.1--------~>:
.1
...:
can see that xCM = 0.
·
Solution: We place the apex of the-cone at the origin. .
Mass of the element
is
thus
It
is.clear
that the CM-will lie along they-axis.:we divide the :
.
'
dm = mass per unit area x area of strip
cone into disc of radius x and thickness dy .' The volume.of'.
.
M.
.
.
such a disc is-dV = nx 2 dj, = ic(y tana) 2 dy_. The mass of the
dm=--1trdr
rtR.2/2
.
disc is dm = dV. First we 'win determine the total mass of.
·
the cone.
·
·
The centre of mass of this element is at 2r/1t, from "
. h
2
2
previous problem. So
M= fdm"'.1tptan af~y d.Y, _
p
y CM = __!_fR 2r dm ~ __!_fR 2r( 2M nr_ dr)",
. M o. 7t
M 9 1t 1tR2 .· .
=__±_fRr2 dr =__±_I (cJI~= 4R
ltR2
0
ltR2
3
'
_
2
h3
-1tptan a.-
.
. The position of the CM-is given- by
;YCM. =:__!_fydm
. M.
31t
0
..
.
kli--e-x--gm~: _"-':·-,l
. lt--~-='--i~:-3-o~
... '= __!_ np'tan 2 af h.j ady
M
!Determine the position of centre of mass. of .
}:~if=e;:; :~;;:b~~
0
3
a:--;-~:
~ass M _a~d ·ra~ius_ R, ~u~ingl
Solution : In .this case
element is a circuhµ· strip of ·
thickness ds. The thickness- of ring
subtends angle d0 at the centre of
the hemisphere _as shown in Fig.
.. 4E.30. Radius of ring element is
R cos 0. Mass of the. element is
.
4
.
2 li ·
=·.1
-1tptan a-,-
M
From eqns. (1) and (2), we have
3h
M
21tR2
Then
xCM =· 0-
.
... (2)
4 ;
Yo.i=+
·mustratioil 12. - If some mass of area is r~moved
from a rigid body, then the position of centre of mass of th~
remaining portion i~ obtained from the" following ~ormuiae: '
•
·-+·
,...+
Ai:"r1 -A 2r2
,- A1 -A2'
·. A1 X1 - A2X2
dm -= mass per unit area x area of circular strip
= -·-
.
_o.
,
.A1·-;-A2
", .
x (21tRc6s0)Rd0
~rY1
-A2Y2
A1 -A2.
from symmetry
or ZcoM
www.puucho.com
A1i1 -A2Z2
. A1 .:..A2
'
'
·
Anurag Mishra Mechanics 1 with www.puucho.com
'
.l
'
ME~HANICS-1
.
·1
. = -ita
Area o·
f crrcu
ar ponon
Here m1,A1,r1,X1,Y1 and Z1 are the values for the
.
'
.
-t
.
.
.-
.
J
2
4
· whole mass w~ile m2, A 2 , r2 , x 2 , y 2 and z 2 are the values ·
2
2
an~ ·a~ea of square poriton = [ ar;;] = _a_
for the mass which has been removed. Let us see two
: ·
2v2
8
examples in support of tl):e above. theory.
If G1 and G be the positions of C,G. of the cut squar~
illustration
13. Find
·
portion
and remaining portion.
the position' of centre of mass of
· the uniform lamina shown in
.
1ta2 (0)- ~2
a
Fig. '.(33.
4
84
32
-a
·
ThenOG=
Solution: Here,
1tU2 . a2 .
( 21t8-1)" 4(21t -1)
A 1 = area of complete circle
4
8
(E.)
. = 1ta2
A 2 = area
or small
-
·. The· C.G. · of the remaining portion is .at a
distance of
a
from the centre.
circle,
=1t(%r =!t:2
4(21t- l)
·; Fig. 4.33' .. ·-~
.(x1 , y 1 ) '." coordinates of centre of mass of large circle
= (0, 0)
_and,
(x2 ,y 2)
.
_
_
=coordinates of centre of mass of small circle_.
=(%,o) ,
circular cone has its base cut out in
!conical shape shown in Fig. 4E.32 (a) such that the' holfow is
right circular cone on the sam~ base. Find whauhould be,
,the height of the hollow so that the centre of mass of the'
la
fnm,_;,rtwo•'Ylth-'~""?""""
Using
we get ·
!A uniform solid right
. -¥(%) . -(½)
a
..xco"!= .2.ita"=(~)a=-6
ira
.l
-4 · 4
Fig. 4E.32 (a)
••
,, '
Solution: .
andy~dM = 0asyl andy~ both are zero.
1 2h' y=·h
M=p-1tr
. (Before removal)
Therefore, coordinates of COM of the lamina shown in .
3 . '
3
Flg. 4.33.are
l
2h ,
h1
M1 = p-1tr
i, Y1 = -. (For removed part)
.
.
6.
3
3
illustration 14. ,·A square hole i~ punched out of a
·.
h'
My-M1Y1
YCM = 1 = - ~ - circular lamina, the .diagonal of the square being the radius
.
M-M1 .
· of the circle. If.' a' •be the diameter of the ciicle, find the .
distance ofC.G. of the remainder from the centre of the
p.! 1tr2(!!:., _
circle.
3
4
4
=-..,,...~---~
· Solution : Consider r,:.......,~:::::::·::::::::::::~---,,
1 2
.
p-1tr (h - h1 )
the Fig. 4.34 shown
3
b'elow. · \et 'AB be the .
.., ,
diameter passing through r~
h1 = h1 + h ~ h1 = !!. '
the diagonal OB of the I .
4.
3
square portion where O.is 1·/\,r-:-,:--"<Gl>---<00(.;r--tG>---<9!B
1
the centre of the circle.
33 ~ : .
Portion where O is the
centre of the circle.
ipetermine th~ centre of gravity of a thin ho_mogeneous plate
Mass of the portions
having theform of a rectangle wit!t sides r and ·41' from :which
can be replaced by. their.
a semicircle with a radius r is cutout ofYi&: 4E.33 (a).~-_ _
respective areas .at their.
C.G..
(:-E., a) .. :.
hf)
l.
~~.;;_1ce,,.j
www.puucho.com
...
,
Anurag Mishra Mechanics 1 with www.puucho.com
IMPUISE ANI> MOMENTUM
357·
-- --- ----·1
!
C
\_ r B
Fig. 4E,33 (a)
Solution : We assume the system to consist of three
uniform square plates. All the plates have the same area,
therefore the mass of eaclt plate is m/3. From symmetry, the
centre of mass of elicit plate is at the geometric centre of
eaclt plate. We cltoose origin as shown in the Fig. 4E.34(a).
The position vectors of centre of mass of eaclt of the plates
are
->
Solution:
r1
xCM = m,x, - m 2 x 2
->
r2
->
r3
-m2
m1 ~ mass of the complete plate
m2 ~ mass of the semicircular
plate
Let density of plate is cr
m1
m1
-
cr2r 2
-+ .
7lT2
. ~ .1
2crr2 (~)- 0"7lT . 4r
2
2
31t
XCM ;:::;_--~--c.-cc-2--
2crr2 _ 0"1tr ,
L_ .!'.L(~-.-4E:::,.3_3--l(bi
2
·2r
_
j
=---
3(4-1t) I
illustration · 15. The position' vector of three
particles_ of l!'ass m1 = 1 kg, m2 = 2 kg and m 3 = 3 kg are
r, =Cl+ 4j+fc)m,r2 ':' ci+ j + k)mandr3 = (2i-j-2k)m
respectively. Find the position vector of their centre of mass.
Solution: The position vector of COM of the three
particles will be given by
"-+
-+
-+
m1 r1 + m2 r2 + m3 r3
m1 +m 2 +m3
Substituting the values, ·we get
·
->
(l)(l + 4j + k) + (2)(1 + j + k) + (3)(2i-]-.2k)
rcoM = --~·~-----~-----~--+ - -
rcoM
.
=~~-~~-~~
I.mir1
=-·-M
= ~[m aj + m(Oni) + m ai] = ~(i+ j)
m 3 ·
3·
· · 3
3
·Alternatively, we may use
• - I.mixi
XCM--m
m
_
or
X CM -
and
YCM
=-·-·-·
I:m.y-
and
YCM
m
·m·
m
-x0+-'-xa+-x0
3
3
·3
a
m
3
m
m
m
-x0+-x.a+-x0
a
= _3
3
3
=-
3,
m
whiclt is same as obtained above. ·
. Method 2. The original system can be considered· to
be the remaining portion if a square of side a is removed
froin a full plate of side 2a. The mass bf the large square is
4m/3, while that of the removed portion is m/3. The position ·
vector of the tnetre of mass of the large plate is ·
->
r1
1+ 2+ 3
91 + 3j - 3ic
=-~--
••
= ai
->
rCM
ix
2
•
= (0m)i
The position vector of the centre of mass of the
three-partjcle system, is
'
m2 =cr-.2 .
•
= aj
1 • • ,.
-(3i+j-K)ni
2 .
•
•
= ai+aj
while that of the·smaller plate is
,-,
3· <" 3 s
r2 =-at+-aJ
34 ~
2
· 2
The system· can be consi.dered
to be the .superposition. of a large ·
. fLocate the· ;osition of ce~e of ,;ass of a unifonn.plate of,
plate and a small plate· ofnegative ·
mass m, as shown. in.Fig. 4E.34 (a).
.. inass (- m/3). · Here· we have a
Y a
two-particle' system in which the
'
contribution of the smaller square is to be subtracted.
6
.~Xa!m.\BJjj
1~1 ·. a
-+
· rcM
·3
.
-+
m 2 r2 )
s
·m
(3 •
3·)]
-=(4m_m) 3m(a1+,aJ)-3lzai+2aj
0
.
2a
Fig. 4E.34 (a)
-+ •
1 [4 · :·
2a -----1-----1----,
2
'l
=. -m (m1 r 1 -
www.puucho.com
3
3
5 : 5 s
=-a1+-aJ
6
6
·
·
Anurag Mishra Mechanics 1 with www.puucho.com
..., .
r1 = Ii
..., .,
.·li@·¥f~'~!I~~ .
. 1c~:is~e;. ~ ·disc of radius a wim'· u~~ m~· distrib~i/cm, .
!from whicha,arcular sectionofradiusb has been removed,·as
·. fshown in Fig.' 4E.35., The centre of the removed disc is at.a ·
. fdistan~e c.trom th,e ~ntr~-of thtf Ia,:ge disc. Find the centr;e of,
½;·.: _: .-., ."·,
t:ass~ftheremammgd,sc.; ··
\-'"
'\- ·,·. -L
' "'
MECHANICS-I )
r2
= lj
'The position vector of the centre of mas~ of the systen:i is
given by
.
.
-+
1
-+
-+:--+
-rCM = -(m1 r1+ m2 r2+ m3 r3)
M
.
1
•
•
•
1 • + j)
•
= -[2m1i + mlj + m(O)j] ='-(21
4m
4
§E$i<a~~,Le~
!A rocbt e:g,l~des att:h~topmost point ~fits trajectory, 550 m
Flg.4E.35_
!from the point of projection. One of t~e fragments is found at
te'tcr
Sol~tion:·
be the sud'ac~ d~nsity ~f ~ass; then
·ma~s'.ofentire,;lisc=ia(:1ta 2)
·
,· '. •. ·' i . , iilass of remov~d d~c, = O'(~b 2 )
·. ·.· .. .<,·_-·
· la locatioJt 550 m east and 120 m _north of the launch point.
.The ;ystein ~ be
full
of a
•
•.,
considered: to be the super position
disc and a small disc of negative mass.
t
:~·,'_
•'
•.
•I
.',
•-+'
---......
A'
•
•
Position' .vectoi:
. , _of ftill disc,' .: r1 =. Oi :I' OJ ·
~
;. _'c
.~·, ',.,-',, ),-.·
•
''
-+•,A
· •PositJon vector of reiµoved disc,
r.1
.. A•
= ~ + Oj
r~)-
':-:;o~ition v~ctor of sys;e~•~ l..cm1 r1:.. m,
'.
. .'
M
'/- .
. .. 'cr(ita~) x 0-c;mb 2(cl + OJ)
· >·",
-
' ; .' ·.-.
.
·a(ita 2 J'-a(itb 2 )
= ,.;b
.,. .
2 .
C
.
.
f
- -,
-~.
.
/SeFond fragment is found at a location .550 m east anq 65 m
south.· of.th·e·. l.aunch point Fi~_,t two. fragments are ofeq_ual
· mass m ~nd•third fragment has mass 2 m. If all: tlte ·three
agrnents_ struck the ground ·simultaneously, what is the
· ~ocation~the thirdfr._agment? , : · ·
· '- ' · · ·,
Solution: Force~ generated by explosion are internal
· · forces. Motion takes place under the gravitational force even
. after explosion..Therefore the centre of mass of the system
continues its parabolic trajectory, i.e., centre of .mass strikes
the. gr_ound a_t the-same place where the entire rocket would
have done, at a distance R = 1100 m east of the -launch
''
•i
Po.int.
z(m)
...---,-......... '
',
._, l'c
~e@/AA,ije~
.
(North)
m
.:-- ..;_
~ ~~-~~-~~
m1X1, + ni2X2·+ m3X3
_m1 +m2 ·+m3
c:itoom) = m(SSO) + m(SSO) + 2mx3
y
. XCM ·-
.
,.
.
or
s_linil;p-!y
I
I
.. ·. ;(a)
.·"
· 2m
"'
:• (b)
.. '·, Flg.AE.36.
,.,
'·
. ·..
".
. Fig. 4E.:i7;
m
·:1 :. .m,
~
Y(m). Fragment 2 \ .
,
•·
' Frag111ent 3
550
x(m)
(East)
<;o~ider ~herr;artides connected 6y m~less. rods.. Find the
location .of:thfcentre of mass of the,system of three po.rticl~
· . as shown_in F.:ig, 4E.36.
·
·
··
·
. i -~- ··:,, ), ,•
Fragment 1
. .)!:I;:!
a ~.:: b2
· _The.ef&re centre of mass of system is· to. the left ~f b. ·
,
~-
·..
,,. . .
. '·
.
-.~ Sbl!Jtlon _.: We__introduce ·a coordinate . system as.
-shown in· !'ig. 4E.3~. ·Th~ position vectors of the three
. masses are' ..
·
y CM =· m1Yi + m2Y 2 +_ m3y 3
m1-+ m2 + m3
Because y CM = 0, so . o·= m(120) + m(-65)+.2my3
4
or
y 3 = -27.Sm
The~efore location_ of third fragment is 1100 m east and
27.5 m south of the launch point.
www.puucho.com
'.
4m
x 3 = 1650m
Anurag Mishra Mechanics 1 with www.puucho.com
IM~ULSE ANI> MOMENTUM
c:,/i·,
,
..
•-'
L:ce~amru:el
-----··-- --~ 38Q;>
IA wedge of mass ~ is kept on a sprilJf;- balance. A sniall. block
!of mass ni1 can move. along the ftictionless incline of the
I~--
2
1wedge. What is the reading of the.lialanc~' while the block
lslides? lg!IQ[!Lthe rec~i1 of the w~c/gt!.
-r
eS,·'
'(.
"
"v"'•.-
"
Fig. 4E.39 (a)
Solution: We co nsider wedge and block as our sys tem.
Action-reaction force between block and wedge is'internal
force. N x and NY represent reactions on the wedge by the
·
balance.
351]
Solution: We assign ·the initial position of bead as
origin. There is no external force along the x-axis; hence
position of CM will remain unchanged.
=i
(XCM)initial = 2mx·O+ml
.
. 3m
3
(X l
_ 2mx+ m(x + 1cos8)
CM final. ·
3m · _
~
: . ?.~·-0
... (1)
'.
.
g,sln B
t
g sin2 ,B
I
Fig. 4E,3B
From Newton's second law,
Ny - (m 1g + m2g) = (m1 + m2l(aCMly
where
.
.
m1 a1y
+ m2a2y
3x+ !cos8
... (2) .
3
On equating eqns. (1) and (2}, we get
l
3x+lcos8
-=
3
3
1(1- cos8)
or
x=
(bl
I_,___ _
=
3
.. :(1)
... (2)
(aCM l y = - ~ - - - ~
·
/111 +m2
.,
= m1 ~in 2 8)+ 0
... (3)
,,
m1 +m2
Note that the block has acceleration of magnitude g sin 8
along an incline y-coinponent of this acceleration is g sin 2 8.
From eqns_. (1) and (3),
1
Ny = (m 1g + m2g) + (m 1 + m2l(aCM ly
2
·
(-m
1g sin .8)
=(m
+m2lg+(m
+m2l
=
·
1
1
.
.
m1 +m2 .
= (m1 + m2)g - m1g sin 2 8
Similarly,
Nx = (m1 + m2)(£icM lx
_ m a1x + m2cz2x
(QCM ) x - 1
where
m1 +m2
= m 1 (g siri.8cos8) + 0
Method 2. CM is initially at rest, in the absence of
external fates it will continue to ·be at rest .
(-g
m1
+ m2
....
...,
or
m1AX 1 +m 2AX 2 = 0
....
....
where AX 1 and AX 2 denote absolute displacement of
masses m1 and m 2. Thus we have
m(l - l cos8 + x) + 2mx = 0
1(1- cos8)
or
x=
.3
. Conceptual discussion: A block is released on the
convex-surface of a hemispherical
wedge as shown in Fig. 4E.39(b).
We wish to determine the
displacement of wedge, when the
· block reaches the angular
position 8. There is no _external
force in the x-direction, so
....
G"be~d of mass 2 m can slide on a smooth rod. A ·particle ofi
mass m is attached to "the bead by a light string of length l.
Initially the particle is held horizontally in level with th_e bead
and the string held just taut. Find the distance through:Which
the bead will move when the string has tumed through an
angle 8 witli the horizontal
·
or
www.puucho.com
....
.
.Smooth!
urfaC8l
. .
Fi~::E.39·(b;~
m1AX 1 + m2AX 2 = 0
m(Rsin8-x)-Mx = 0
mRsin8
x=
m+M
Anurag Mishra Mechanics 1 with www.puucho.com
r.
:,=::,=====:,,e,;,===========~============
j 360
~
M_ECHANICS-1]
,.i
•..•• '::
o;
¼.man of mass BO kg is ridi,ii
d trolley"of mass--,ip,,;;
. '!which is rolling along a level surface a,t a-speed of 2 m/s. 'He
1wnps off the bat:k of the trolley so that his speed relative to
l_th!'. ground.is mfs ui tne directiqn,opposite to the motiq,;t pf,j
jthe trolley.• .
.
.
:
. . ·:
(a} What is the speed of the centre of mass of the man-trql/ey
1 : _,system bef9re and after his jwitj,s? ·
'- · '· :·
'
(p) What.is the speed of the trolley after.the manjwnps?
(c) What is the speed·of the centre .<if mass of.the system.after
( · the man hits the ground and comes to rest?
(d} Whatforce is responsiblefor the change in vCM? · ·
(g) I-Iowmucliprgyfid the man%j1end injumping?
;
"
This is equal to the energy expended by the man in
jumping,
Conceptual discussion: If the man jumps with
velocity v 1 relative to the initial ,state of trolley the absolute
velocity of man is
-->
1_
-->
...,
VmG =VmT+VTG
e
1
IvmGI= -Vi+ V
The equation of conservation of momentum is
-m 1 (-v 1 + v) + m 2v = (m1 + m 2)v
Similarly if the man jumps with velocity v 1 relati_ve to
final state of trolley.
lvmGI= .'.vi +v'2
Solution:
The equation of conservation of momentum reduces to
-m1C-v1 +v'2)+m2v'2=·(m1 +m2)v
Ii' j2'.~n
~~-e,.-.~-~----~-~-~L-!,e-.@-r·-4-,1] ~
~ Initial state
,'Final state ·
..
IA,; exp.losive.ofm.'i.zss 6 kg is pTOJ.·e·.~t. il at 35 m/Sat an ang.' ofil
• Fig. 41!._40 i
e_
(a) Velocity of CM of man-trolle_Y· system before the man
jumps,
.m1V1 +m2v2
vCM =~~-~==v
[asv1 =v 2 =v]
m 1 +m?
breaking into two parts, one 'ofwhit:h has twice the mass o.fi
the other. The two fragments land simultaneously. The ·lighter
!fragment. larids back. at the laui1c1i point. Where' dpes the
_other fragment land. What is the energy_~! the explosion?
~
~
Before the man touches the ground, there is no external
force, hence v' CM = v CM = v
- (b) Determine. ,the final vel~cities of' man · and
trolley by._ v''1 and v'. 2 , we have,
cci!lservation
of
. from
.
.
_
/·····r··,=:=:.:-·-...·
40 .
Fig. 4E.41 (a)
=-
. (d) ·Due to the force of friction exerted by the·
ground, on the man; the velocity of CM is changed.
·
' (e) While jumping, the .force'. between man and
· trolley is internal to ·the system. It has ·no influence on the
. motion of CM. However, it changes. the total energy of the
system by the amount·
JIB = KE 1 - KE; .
.
.
= ( 1 m 1v ,21 +·1 m 2v 2 - ·(1cm 1 +- m 2 Jv
2
2
2
,2)
·
=l_·x 80 X1
2
·
=; 1080J
2
2)
~! X 40X 8
2
,,
2
-
.,
oi:,_R/2-R/2__;;, .. X
(c) Tiie man conies to rest after hitting the. ground,
so the speed of the centre of mass of the system is ·
-~'CM·= m1 xO+m 2v\ = 40x8
2 :67 m/s·
· · m1 +m2 .
80+40
/
,, '"•
2
292g·
v 2 =v+m1 (v 1 +t! 1 )=2~ 80 (2+1);=8m/s
·.,.
2 ._, ij
R::; v~sin 28
g
m
m Path ofm 2
Vo ....)........,.; after 13xplos!on
-m1v(+m 2v 2 = (m1 ._+m 2 )v'
m2
1
H=v0 s1rr =~
~
momentum,
or
le.
60° with th~ horizontal _At the.top of its flight it explodes,
!(80+ 40) X 2
2
·
2
Solution: Just J:,efore thJ' ~plosion the projectile has·
velocity components vx = v cos0, Vy = 0 and is at the
topmost point of its trajectory.
.
From conservation of momentum along x-axis, we have
Mv X- = m1v' Ix +m2v' 2x
(as ,J ]y = v' 2y c" 0)
Fragment m1 will land back at the initial launch point if
v'1x
= -Vx·
Therefore
, _ Mv~ -m1v'1x M +m1
. v 2x .
= --~vx
m2
m2
Time ,taken by.m 2 to reach the ground
.
·
t=i
&=
'I~
=• M +m1 Vax
m2·
Voy
g
Distance covered during this time
_ ,
Voy
d. -v~tm2
g_
-(M+m;)
_M+m '.R
--~v~-----~
www.puucho.com
1
m2
2
Anurag Mishra Mechanics 1 with www.puucho.com
IMPULSE AND MOMENTUM
+
Coordinate of second fragment,
)R
x= R +d=R +(M +m 1
=~R
2
2 · m2 . 2 m 2
.
2
= ~ X ( 3 S) ~in 60°
4 9.81
= 162m;
where m1 = 2 kg, m2 = 4kg, v 0 = 35m/s, 9 = 30°
The energy of tlie explosion is
AD
1
,2 1
,2
1
2
ur, = -m1vx +-m 2 v2.x --Mvx
2
2
2
2m1M 2
=--Var
+
.
.
R
Coord mate
of m 2 ·= R += -3R
~
2
2
COLLISIONS
Consider the collision of a bat
n·
with a ball and take the ball to be a
system. The force exerted by the bat
on the ball is considerably greater
than the weight of the ball, ·so that
the total force on the ball just' equals
the force of the bat during collision,
1,·
approximately. The variation of this
Flg.4,35
--=------.J
force is s!Iown in Fig. 4.35. Forces
that act for short intervals are called impulsive forces.
Examples are: a cube ball hitting an object, a pendulum liob .. released when the string is slack, comet deflected by the
sun.
·Toe impul~ive forces are assumed to be greater than
any external forces present. .
+ In a collision, the individual momenta of the
particles do change, but the total momentum of the .
system of colliding particles does not. During
collision the forces of interaction are an
action-reaction pair, internal forces, for a system of
colliding particles.
+ Momentum of an isolated system just before
collision equals the total momentum of the system
just after the collision. ·
+ If the total kinetic energy- of the particles is
conserved, then the collision js called an elastic
collision.
\F;..,~.
a
I
·-··-.
l
+
b._~:xa~~~
IA 'car of mass M = 25001<g ri,~-in_t_o_a-sm-aller car of -;,,~J
m = 1500 kg parked in traffic lane..The sliding wreckage!
leaves marks on the pavement for a distance of d = 5.2 m. If,
the.coefficient of friction between wreckage and pavement isl
,µ k = _0.34, what was the limousine's initial sp~ed (Fig.:
162 m
L
A system is defined
_... --------.. ,,.SYste;;;-J
to include all the , · ·
particles taking part 1
F
~1
·.-._. /F'.
I
.
12 •
in the collision.
'
',
:
'1
m,. ..·.·
Conservation
of
------"½
I
momentum principle
FHa.p
;
is applicable if the
·-. ·"
i
•••>=•.._system j
total external force
on the system is
.
____
:
either zero or can be
neglected compared •
,.. :-:.. ._He
•: F
P- He.
- Total ···--·
to forces present 1
during collision.
I,
Fig. 4.36
,
If the total kinetic ------ ----------~
energy of the particles is not conserved, the collision
is said to be an inelastic collision.
If the particles stick together after the collision, the
. collision is called . a completely inelastic
collision.
--·-\. --~~\.;~:__
+
m2
2x2x6 .
'
=
·
(35cos30°) 2 "'5512J
4
Method 2: The . internal
force of explosion does not.
change of trajectory of CM.
CM lies here·
m1 X1 = m2X2.
i.----X1
X2.,...
2·
R
X2 =-XR =.I
Fig. 4E.41 (b)
4
2
•
I
361
\
~·-
l1!l.,,1..?J1
_ _ _ _ _ ...... __ -·--· _ ---··· ---·
I
Solution : The incident occurs. in two steps: first the
cats collide and then the sliding wreckage is brought to rest
by friction.
Sehlp: We choose the x-axis to be parallel to the
direction of the first cars incoming velocity. Figure .is the
free-body diagram for the wreckage. We apply Newton's
second law
LFY = (m+M)ay, Lf'x = (M + m)ax.
N-(m+M)g=o.·
-A=(m+M)ax.
The magnitude· of the friction force is fk = µ kN.
Combining
these
relations
gives
us
-µk(m+.M)g = (m+Mlax.
After the collision, the wreckage decelerates uniformly
with ax =-µkg.We use eqn. relating the acceleration a,,
distance Ax = d, and change in speed squared:
vj -vr= 2axAx = '-2µkgd,
where v f is zero and vi = u1 is the wreckage speed after
the collision, Thus:
· ·
·
· ·
U1 = ..j2µ kgd.
.. . (i)
Momentum is conserved in: the- perfectly inelastic
collision, s6:
Px (after)= (M + m)ui
= Mv1 + 0 = Px (before)
... (ii)
www.puucho.com
•
Anurag Mishra Mechanics 1 with www.puucho.com
I~:=========--=--=::.::.::.::.::.::.~=-~=,
362
(,IEC!fiyi!ffi
-----------~--:-~---'-----'--'
·
~
~~-
~
•1• .
·
(a)
: ., rn:_
,
?...Ji
"'' •:- •· _ , " ' "" '. ' '
: ~
•
The spring force is a conservative force so total
mechanical energy is conserved. Thus this is a model of an
elastic encounter. Llnear momentum is also conserved.
Let's see what happens in steps?
Step 1. The incoming block approaches the stationary
object (state A).
. Step 2.- When block conies into contact with spring it is
compressed. Compressed spring exerts force··on both blocks
and it slows the incomlng block and accelerates the
block-plus-spring object to the right. When the spring has its
maximum compression· and both objects have the same
velocity. Spring continues to be compressed till both blocks
·attain common velocity i.e., v rel becomes zero. ·
·
•
~ . wfbru
~
----' :W WM NJ2# Mif
1
1
11 -I
· fsl1
,(b)
! ~~ ~
"'""'!.'li·,lA5'2 'i\,~
(c),
(d)
;; ·
:Z.
+Y
'
L.
.
(m + M)g x
L.._ _ _ _ _ _ _ _ _..,_,Fig, 4E.
42 _ __
. Substituting relation (i) for u 1 into (ii), the initial speed
of the limousine is: ·
·
_ m + M ,,,;;-::;;d
v,-_ (m + M)u1 ---v-'!'kgu
M
M
--------' (4.0 x 103 kg)~2(Q.34)(9.8 m/s 2 )(5.2m)
=
2.Sx 103 kg
=9.4m/s.
·Step 3. The spring begins to regain its natural state it
exerting a leftward force on one block and a rightward force
on the other. Once the block on the left leaves the sppng,
forces stops actings and the collision is over (state C).
We choose the x-axis to be along the direction of motion.
Momentum is conserved throughout the encounter, and the
v CM of system remains constant.At, the end of step 2, state
(B), both objects move at the·cM·velocity; ·
State A
State B
= (m+M)vCM
MODELS FOR ELASTIC AND
INELASTIC COLt.lSIONS
=
The kinetic energy of the two bl\)cks moving together isThe terms ela.stic and inela.stic describe the' result, of a . ·
less than· the original energy of the· smaller block.· Some
collisiJl>n without giying any details of the interaction
energy has been converted to elastic .energy stored in the
between the particles. When objects collide, they exert
spring:
forceJ on each other depending on the objects' geometrical
State (A)
State (B)
structure. The nature of those forces determine the changes
1
2.
1
2
in energy that occur.
-mv 0
Kinetic energy
-(m+M)vcM
2
.2
. A block of mass m and initial speed v O collides with a
_!ks2
spring attacheµ to a stationary block of mass M. (Fig. 4.37).
Elastic energy
0
J\
2
Describe the state of the system when the spring reaches
1 1
2
2
1 2
1
maximum compression and when it has .re-expanded.
-mv 0 =-(m+M)vCM
+-ks
2
2,
2
+Y_
.,,.
L.
1
ks2
1
2
1
2
·
or
= -mv 0 --(m+M)vCM.
.x
X
2 2
2
___,.______ --------···1
e:.
r-e-...;
~
---=,---+
VCM
(a)
.
or
or
(b)
2[
~1
m(m + M)]-1 ks2 --mv
0 1
2
2
(m+M) 2
. 1 mM 2
=---Vo,
2m+M
After the spring regains its natural state (C), the two
objects have velocities umi'and uMi We compare the linear
momentum and energy in states (A) and (C).
The spring is uncompressed in each state and there is no
elastic potential energy.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
)
IMPULSE ANI> MOMENTUM
State ("-)
State (C)
Mass
m
M
m
x-component.ofmomentum mv 0 , none
1
2 none
Kinetic energy_
2mvo
M
Set quantities equal:
Momentum
mv 0 = mum +MuM.
1 2 1
2
1
2
Energy
-mv 0 =-mu +7MuM
2
· 2
m 2
These equations are the same as· those for the elastic
collision and have the same solution. With m2 = M. and
~ 1 = m in eqn., we have :
m-M
um =---Vo
· M+m.
2m
and
UM= ---Vom+M
Analyze
If M. > m,- then um-< 0 and the incoming
block returns to the left.
For ·an elastic collision to occut, the particles have to
exert conservative forces on each other that is, have the
ability to transfonn and store potential energy. .
k,~~!elm~~~
surface. Disc ~ ,is projected towards aisc 2 with velocity ~ 0•
· After collisiqn disc,2 moves .at an angle e~ 45° with x<cixis.
Cal~ulate the magnitude of outgoing velocities of disc 1 and
. ___
·
· disc 2 .
"
Y 2 .-\~~:······
(!lll
"x
1
~fore2
C!m-.- • •
m
~
~50 ,
•---.=;-:
m
'"=
1 ;;'~ ,
.,
Solution: The incoming particles are free to slide on
the frictionless surface. Disc 2 is knocked at an angle to the
direction of disc 1's incident velocity so the problem is two
dimensional. Choose the x-axis to be along the direction of
disc is initial velocity and the y-axis as shown in Fig. 4E.4;!.
We are give disc l's initial and the direction of disc 2's
·outgoing velocity.
Kinetic energy:
1
2
Before
-E=-mv 0 +0
2
After
1
2
+
= 1+2sin9cos9 ·
Fig. 4'/:.43 shows two discs kept 07! a smooth horizontal
I
x-component of momentum:
Before
Px = mv 0 + 0
After
Px = mu 1 cose + mu 2 cos45°
y-component of momentum:
Before
Py=O+O.
After
PY = -mu1 sine+ mu 2 sin45°
From conservation of energy, we get
v~ = uf l!i
... (i)
From conservation of moinentum, we get ·
x-component of momentum:
v O = u 1 cose + u 2 / ./2
... (ii)
y-.tomponent of momentum:
0 = -u1 sin0+u 2 /J2
... (iii)
From eqn. (iii):
... (iv)
u 2 = C.J2)u1 sine
Then, substituting eqn. (iv), into eqn. (i),
v~ = uf (l + 2sin 2 9)
... (v)
and eqn·. (iv) into eqn. (ii),
v 0 =. u1(cose + sin9)
... (vi)
Now, equating v~ in eqns. (v) and (vi),
(1+2'sin 2 9) = (cos0 + sin9) 2
2
1
2
2
E = -mu1 .+-mu 2
(Remember: cos 9+sin 2 9=1.)' Thus sin8= cose, so
9 = 45°. It follows from eqns. (iv) and (v) that:
u1 =u2·=vo/J2:
.
.
2
Two-Dimensional Collisions ·
The momentum of a
system of two particles during
collisions is constant for· an
isolated system. This result is
applicable in each of the
direction x, y and z, as
momentum is · a vector.
Consider a collision between
two particles with particle 2
at rest initially. After the
collision, particle 1 and
particle 2 move at angle e and
<j> w.r:t, horizontal respeGctvely.
Conservation
of
(b) After colliSion
momentum can: be applied. in
,-_ _ _..c.Fc.l,g._
4.38
·.:
_·,$_
the component form.
x-momentum:
ni1Vux + m2v2ix = m1v1fx + m2v.2fx
·... (1)
y-momentum:
'
... (2)
m_1V19' + m2V2iy = m1V1.6' + m2V2ty
In our case, the above equations reduce to
m1vu = m1v11 cos9 + m 2v21 cos<j>
... (3)
O= m1ti 11 sin.9-m 2v 21 sin<j> ' ... (4)
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
, 364
where n_egative sign in eqn.''(4) appears due to the fact . two parts: At 1, period of deformation; At 2 , petiocj of
that after the collision particle 2 has a ,y-component of
recovery.
'
velocity pointed downwards.
. The
impulsive
force
If the collision is elastic, we can apply conservation of
increases to a maximum value
kinetic energy.
'
at the end of the deformation ·
2
1 · 2
1 · 2 -1
period and then decreases to
2m1vi; = 2m1V11 +2m2v2J
zero during the recovery period:
,.,_t.t,----Aj,,~ /
In Fig. 4.40, P and R represent
Oblique Collision : Common normal to the colliding
·
I··
,
· ·c1 ·'
,
Period of, Periodo',
un.pu
,orces
unng ·r '•deIormation recoveli)'•1. •
surfaces is called the line of . !_;
, s1ve.
d
de,ormatmn
an
recovery
.
, . _,
impact. If centre of mass of ,
periods respectively.
1
-Fig. 4.40- L.~
colliding bodies :lies oµ line of I
The coefficient 'of restitution (e) is defined as
impact, the impact is central
Impulse ofrecovery
_impact, otherwise. the impact is
e=
called eccentric impact. In a
Impulse of deformation
head on collision (direct
,
•
= (v's ).- (v'A )~
impact) the · velocities of the
, Direct centralimpacf , J
(v A)n -(vs)~
colliding bodies are along the ' f
Fig. ~.39 (a)
j
= Velocity of separation along the line of impact
line of impact. If the velocities
Velocity of approach along the line of imp~ct
of one or both particles are at an angle with the line of
impact the impact is said to be an, oblique impact or glancing
~. .... - ·0'·m.)t1'~
collision.
•+, ~ : - - -
Let .~s .consider an oblique collisi!]n betwe~n two
particles. We assign different axis for oblique collision,
, normal axis (n-axis) along line of impact and tangential axis
(t-axis) along tangent to surfaces in contact. We assume the
particles to' be _smooth and frictionless, so that the impulsive
force exerted by particles on.each other is internal force fpr
the system. These impulsive forces act along line of impact
(n-axis).
· -----;..,,
·
·•
,1
:·.'.···:.t~)~
· ..: Peiiodo(deform
..ation. /·. __,,]_ ·
(p"'+ -~·::~····
'·
Periodbf/e·covEtry ."
.d,
:\; .
.°
Fig. 4,41_'." '
.·
:< ..-------=---'-'-'-"·-------'·
+
·"-
Note that in fonnula· for' e, velocity components.
· along n°axis are substituted.
•
t~i.ix1S
illustration 16. Consider collision between a block
>,~ .
~. ,' Vg
A and ball B; the ,block is
constrained to move along ..,.. "'-·--,,,----,-1,---~
'" , ,_.
a horizontal surface. Ignore
,,
friction- at· any of the
· • ,, v'A
,
v·
•
.
. ,. A
surfaces.
Impulsive.
force
.... VA
Obli_Que cel'ltra't ·1_tnpact ·
between block and ball is
...:...,+-~+
··'4•. a 1
~:-.w...-''---F!!![4,3~]'...._J_ _ _ __,
along the · line of impact j .....'.:.:J:TITI:IIl:tl:JI__'. ·~
and the impulse ofreacti<;m · 1·
, • •
.
,
• ':' :·•
. We can form 'the following equations.
exerted · by ground is 'in I: '· ··
Fig; 4.42 . ~ ;
1. Since no force acts along t-axis on each particle'.
vertical direction, which is an external force on the system.
considered separately, the component or momentum along
.
following equations can be formed for the system of .
t-axis is· conserved; hence t component of the velocity of· 'ball The
arid block:, ,
·
·
·
·.
each particle remains unchanged. Therefore, we obtain .
1. (v~), = (v's ),
{v A), = (v'A ),; {vs),= (v's ),
... (1).
2:
mAVA_+ms(Vs)x = mAv·'A+(msV's·ix,
_2. 1'otal. momentum of the two particles is conserved
Note that ·we -have. applied l~w of conservation of
along n-axis.
momentum
on ·our systeiµ along x-axis only. Momentiun of
. mA(vA)~ +ms(Vs)n = m,i(v'A )n +ms(V's )n , .. (2)
·
system
is
not
coµserved in the clirection in which a ·body is
3. From definition of. coefficient of restitution,
prevented from motion. . ·
·
·
·
, ..
1
\.
~.v'e n•axis
.:--·;·.
,'Line of
s: , ' impact
a
1·
'',
(v's ). -(v'A ln
• "
,
:-.
·
j"
'.,I:'
= e[(v A) •. -.:" (us).]
.... (3)
Coefficient, of restitution : In a typical collision
between .two bodies the force versus time graph is shown in·
the Fig. 4.40. We can divide the,total period of impact into
I
·.'.;
3. (v'B );- (v'A ln .=,e_[(v Aln -(vs)~]'
. .'
www.puucho.com
;
)
Anurag Mishra Mechanics 1 with www.puucho.com
[i~PULSE A~ll_l\\O~ENTU~ - ---- -
365[
---
OBLIQUE IMPACT
Oblique impact is the impact in which relative velocity
of approach of the colliding bodies is not along the line of
impact.
:~'--0-- .cC)axisG--C?S~-:
Before collision
' Ncrmal axis
After collision
r--i
44
1,__>
i,,}E~~~E?J~J
An elastic collision takes place betwee~ two masses m1 and
m,., moving on a frictionless surface, as shown in Fig. 4E.44.
The-spring constant is k = 600 N / m
' 6 :::B
v2i = 2.5 mis
v 1 i = 4 mis
l
m 1 =1.60kg
m2 =2.10kg
Fig. 4.43
Fig. 4E.44
Procedure for Solving The Problems
Step 1. Drawn-axis and t-axis at the point of impact.
Step 2. Conserve the momentum of the system, along
and perpendicular to the line of impact, i.e., along n-axis and
t-axis.
We obtain
m1u1 cos0 1 + m2u·2 cos0 2 = m1v 1 cosp 1 + m 2 v 2 cosp 2
and
m1u1 sin0 1 + m2u 2 sin0 2 = m1v 1 sinp 1 + m 2v 2 sinp 2 ... (1)
If the colliding bodies are smooth, no force is acting on
m1 and m 2 along the tangent; the momentum of m1 and m2
remains conserved along t-axis;
m,u, sine, = m1V1 sinP1
... (2)
i.e.,
and
m 2u 2 sin0 2 = m 2v 2 sinP 2
... (3)
Coefficient of restitution is defined along line of impact
(a) What is the velocity of the block 2 at the instant.block 1 is'
only.
e = v 1 cosp, -v 2 cosP 2
u 1 cos0 1 - u 2 cos0 2
... (4)
Now we have four equations and four unknowns
v,., v 2,P 1 and p 2. Solving the four equations for the four
unknowns; we obtain
A
(m1 + m2)u 1 cos0 1 + m 2(1 + e)u 2 cos0 2
V1 COSp1 = -~-~~-~-~--~-~
... (5)
m1 + mz
A
m1(1 + e) cos0 1 + (m 2 - em1 )u 2 cos0 2
Vz COSpz =
m1 +m 2
moving to the right with a velocity 3.00m/s?
·(b) What is_ compression of the spring at that instant? .
Solution: (a) From momentum conservation, we
obtain
m1vli + m2 v 21 :;:: m1vv + m2 v 21
(1.60)(4.00) + (2.10)(-250) = (1.60)(3.00) + (2.10)v 2f
·
v 21 = -1.74 m/s
Negative sign implies that block 2 is still continuing in
the same direction.
(b) Because no friction or non-conservative force acts
on the system, we can Use conservation of energy equation.
We obtain
1
21
21
2·1
212
2m1V1; +2m2V2; = 2m1V1/ +2m2V2J
:"zkx
On substituting numerical values, we obtain
X= 0.173m.
[J:~~Q._t;QjB~c:~ ,i45li>
Consider two particles that undergo an elastic collision on a
frictionless surface as shown in Fig. 4E.45. One particle of
mass m2 is at rest initially.
~I~
Before collision
~
0----;,
After collision
tan Pi
= ~ = Final tangential component of velocity
v 1n
•
final normal component of velocity
Similarly we can find v 2 and p 2.
,Remark: . .. -- - - -.
Impulse= m{rl 2 (1+e)(u1 cos8 1 -u2cos8 2)
m1+m2
Energy loss=
m{rl 2 (1-e 2)(u1 cos8 1 -u2 cos8 2)2
_2(m,-i:m 2 ) _ _ ,
Fig. 4E.45
(a) Find the velocity components v 11 , v 21 of the particles
after the collision. Discuss the results.
(b) if m2 » m1, (c) if m1 » m,., (d) if m1 = m2.
Solution: (a) In this case, both the momentum and
kinetic energy are conserved; therefore we have
m 1vi, + m 2 v 2, = m 1v 11 + m 2v 21
... (1)
1
21
2l-2l·2
m1 vu + m2 v 2, = m1 vv + m 2v 21 •.. (2)
2
www.puucho.com
2
2
2
Anurag Mishra Mechanics 1 with www.puucho.com
.MECHANICS"!
j
From eqn. (2),
m,(Vii-Vi1)=m2(V~f -vt)
or m1(vli -vlf )(vli + vlf) = m 2(v 21 -:v2i)(v 21 +v2i) ·
... (3)
From ecjn. (1),
.
... (4)
m1 (vli:-vlf)=m 2(v 21 -v2i)
We divide eqh. (3) by eqn. (4) and obtain
. vli + vlf = V21 + v2i
(vii -v2i)
= -(v,1 ·-v 21 )
... (5)
Eqn. (5) shows that relative speed of the two particles
before collision (vli -v2i) equals the negative of their
relative speed after the collision, -(vl/ -v 21 ). Now we can
· ·
solve eqns. (1) and (5) to obtain ·
4m m
1 2
----~c--=~-2
(m, -m 2) +4m1 m2
The transfer of KE will be maximum when denominator
is minimum,
m1 = m2
. i.e.,
v,1 =·(m, - m2 )vli + ( . 2m2
m1 +m 2
. m1 +m 2
)v2i
: .. (6)
· ... (7)
The ·above · results are very important in solving
problems of one-dimensional collision.
If particle 2 is initially at rest, _then v 2i = O in eqri. (6)
and eqn.. (7). ·
v,1 =(m' -1'.'2)vli
. . m, + !112
·
V2f
=
.
(b) If m2 '>> m1,
v 11 = -vli
V2j
= V2i
When a very heavy particle collides head on-with a very
· light particle that" is initially at rest, tlie heavy particle
cqntinues its motion unaffected after the collision and the
.light particle rebounds with a speed equal to about twice
. the initial speed of the heavy particle. This would .happen
when a moving heavy. atom, such as uranium, with -a light
-ato!]l, such as hydrogen.
(d) If m 1 "'. m 2 , .
vlf =_v 2i
V2J
= Vti
cent.·
Consider bvo particles m 1 and m2 · that und;rg~ p"e,t;~tly
inelastic collision.
. (a)' What is. loss in kinetic energy during" collision? .
:(b) Whatis:frictionless changetn,.kjrtetic energy? ·
(c)0)iscuss _the resultform 2»:mJ.Jmd v2i - O<--·----'
· .·Sohiti~n: After collision two particle's stick together
and move with common velocity v I after collision,
· From conservation of momentum,
(. lm ) ' ·
V'l
m1.+m2
.
and
v 2/ = v2i. ".' 0
When fl very light particle collides head on with a very
heavy particle that is initially at rest, the light particle has
velocity reversed and heavy particle stays .at rest.
(c) if m1 >> m2 ,
vlf ": ·v,i .
.
1
~ = 1; tra~sfer of en~rgy is hundred :per .
If m1 = m 2 ;
v 2~ =(_ .2m,. )vli+(m 2 _-m1 )v2i
m1 +m2
· m1 +m2
.
1
m1 v 11 +m 2v2i
.
or
-
= (m1' +m 2 )v1
..:_ !111:VH
v1 - .
+ m2V2i.
.
m1 -:t-m2··
. Initial kinetic energy of the two-particle system is .
.
., 1
~i =
1
2
2
Final kinetic energy of system is' .
.
.
1.
.
E1 _~ (m; +m 2 )vJ
2
1.,
.. Loss in kinetic energy is ·
... .
KEi-KE/ =
_
[(l
2
1_ [
- -
2
.
· · When particles of equal mass collide, they exchange
velocities. ··
· ·
(e) Kinetic energy transferred from projectile· tb
target:
·2
m1 vli + 2m2v2i
2
m1vli+
.
2
m1vli
1·
2
2)
m2v_2i -
1 • ;]
2
(m1 +m 2 )v1
· 2 ' (m1v 1;.+ m2v 2;)2_].
+ m2V21
:.
(m1
+ m2 )
c' _![m1m2(vii +vt-2vliv2i)]
2
=
Ifv2i=O,
www.puucho.com
m1 +m 2
m11(1 2
2
-~~-(vli-V2i)
2 (m1 +m 2)- .
1·
•
Anurag Mishra Mechanics 1 with www.puucho.com
-----"-=------------~-~---··_·•;::as1 I
"-EE-: .......,.,_;-:--ra;i4s
·:--,.,..
i.;.i!-:~--~~!B!#'iiih">~\,
~
_1_MP_u_1S_E_AN~il_.m~o~.fn_E_N_T_U_M_ _~-·~·
and
~-- -- ----"AC,-------~~~---~ ___ ...._ -----~-----,~·
'"
IA .bullet of ,mass m1 is fired into .a .large block o/mass m 2
which implies that if a light projectile strikes a heavy
target, the entire kinetic energy is lost in the collision.
~~~~~~
iCo,;;;de~. a.
~ne.C.dirnensional elastic; C.ollision b~tween ; ;;-.e~
incoming bodjJ dnd body 2, initially at rest.Bow would you
choose the mass ofB 'in comparison to the mass of A in order
that B · shoµld '.re.co. il w.ith ~. a) gr.·lat~'st speed, (b) grearestj'
momentum, (cJgreiltest kinetic energy?
____ .
-~·
'
-----
""""'
Solution: Since the collision is elasti~, energy as well
as momentum are conserved.
m1vli = m1V11 + m2V21
.
riI2
V1; = V1J +-V2J
1111
For the sake of simplicity we take a.parameter
.·k= m2
. m1
Eqn. (1) reduces to
and
1
2
2
=
1
+ kv22/
-,=.=71 ...............
·JL
~
I ~~m_,~· ·:;;·_-····-+
.
---~--·~l,~~~~==-- -~-·4···~·
•
•
'
(m1 + m2)
Vli =~---VJ
m1
, .. (2)
Now we ~ubstitute.the expression for v11 in eq~. (2) to
obtain
· ·
.
I m
, "
2
2
V1J
'T~,... ·;:;:~1
· Solution: (a) There are two parts in the problem:.
(i) Collision
between
bullet
·and
... (1) ' block: Impulsive force exerted by bullet and block is very
. large, so that we can neglect all the external forces on ·the
system of block and bullet.
·
,.
· ·Froin conservation of.~omenti.tm, ·, ·
··'
,m1v;, =.(m 1 + m 2 )vf
2 1
2' 1
2
m1vli = m 1vv + m 2v 21
2
V1;
. ~.a..
!
= vlf + kv 21
· vli
isuspended fro.m .wires. 'The coUision. is perfectly inelas((c, so
!that the, combined system swings through a height h, · ··• . · ·
ra) What is .the initial speed of th.e bullet?
'(b) If the ballet emerged from the.block with half of its initial
·. 2V1·
+
(ii) The combined · system (bullet
block)
rises to a height h: We will apply·conservation of ei1ergy
after collision.
1 Cm1 +.m_ )vf2
2
2
V21=--'
2km1Vli
P2 = m2v21 =-~~
.
l+k.
Particle 2 will recoil with maximum momentum when
,denominator is· minimum, whici1 is possible if k-, oo or
·
= (m1 +m 2 )gh
= .J2gh
Vf
l+k
(a) Particle .2 will recoil with maxi1,11um speed·when k.
_is minimum, i.e., ,
k """?. 0 or
m 2 <<m1
(b) Momentum of particle 2' is·
, :.(2)
From eqns. (1) and (2), ·
. vi,= (m1 + m2 ). .J2gh .
. ,·
m1
. .
(b) On the pattern ~f pari: (a) .;ye may , write
conservation of momentum equation (for block and .bullet
system) and conservation of energy (for; block) ,after
collision. ,
·
/= mv mi.(½"vli) ..
m1~ 1
.m2 >> m1.
2 2;_;
. 1 . 2
m 2gh= m 2v 2t·
2
from eqns. (3) and (4),
(c) Kinetic energy of particle 2 is
.
1 , 2 , .f · . ( 2vli ) 2
K2 =2m2v21 =_2km1 l+k
... (1)
. , .
.
and
.
_ 2m1v~,k
4{KE 1 )k
2
- Cl+k) . Cl-k) 2:+4k
~
.
www.puucho.com
,·
•, ..-(4)
2 ~
· Vzt
m1 Vu
2g
Bmfa
h=-=,--
Particle 2 will recoil with maximum kin~tic energy when
denominator is minimum.
·
i.e.,
k = 1 or ·
mB = in A
,
~
,.(3)
Anurag Mishra Mechanics 1 with www.puucho.com
Solution :
'
, ,; ,
,•1
~·
.,,
illustration 19. A ball of mass ,m moving at a speed
v makes a head on collision with an identical ball at rest. The
'
khte'tic' energy
of the balls after the collision is 3/4th of the
origin~l. Find tlie coefficient of restitution.
Solution ·, As we have seen in the above discussion,
thaturider the given conditions:
;·,_,, .ri1
rI, ,
(a) By momentum conseiyation,
2(4)-4{2)"' 2c.::.2) + 4{v2)
=*·
v 2 ,= lm/s
'' velocity of separation
(b) - e = - -velocity of approach
1- (-2)
4-(-2)
(c) At maximum deformed state, by conservati_on of
momentum, COffimOn Velocity Js V = p, , •;:; '
· ·
J 0 -=m1 (v-u 1 )=m 2 (v-u 2 )
_=-2(0-'--'4),=-8-N-'s ''
1-
I~,
- , ,·_;,,
= 4(0-2) = -·8N-s
or
=; 4{0-2) = ~8 N-~' (d) Potential energy_1 maidrriufu deformed state;
-u ,; lo;;~ iii kinetic -eiiei-gy during deformation
or
U;, (:! rri1u,2 't- .!__m 2u~ ).,-- _! (in~ +m 2 )v 2
l: ~- .. ·_. 2
' ;:?-
.,,,.
,:~';;- ·'2'2_(4) +24{2)(-, -2(f+4)(0)
' .'ll-
i'
,1
'.
. , 1',d it ,.,;
1
2 )'
11 ' ,- , ,,
r
. '
,,
·
-
2
,,,:·
iv)
or
,'·l·
'..
l
.
. \ .,· ....{D'
:,
,•,'
.= (0.5)(-8) =-_11'1-s, ·.:
., ,
.,
8
·,'
illustration.'18: A ball,is --- . JJ -,.,---n,.:,,?'71"~=,,,,1
•••• ~t,,.¥-,-'t
moving , with velocity 2 m/s . '· ,. '. \ · ,, ·,c
:
'towards\ 'a 'lieayy wall moving. . ,, ,-:·~-.::,:-:;;;:}
'.J
fowards the_ball with speed 1 mh, t ~5 , ::••1'ri\1s
t_
as, shown in Fig. 4.45 (a): F ~ ·: _:, "
~
Assuming collision to be elastic, ,.-"·
;. -.:·,: ;,
,
I''
·1
•
I 11
' ,,-J
'i::
find·. the velocity of the ,o(llr· h. ·
',,,:';.,,±,.<el
immediateiy'after the collision.
t,'·:- .,Flg, 4:4~1'.'L'i:l.'[J
'
,
•
1
1
l
··, ' -
,
;
.
-
1· -, '
Solution : The speed of wall will not change after. the
collision, So; let:UJ,e the velocity.~fthe ball after collision'in
the direction shown in Fig;'4.45 (b), Since collision is elas):ic
(e = l)/,1- ·, ,_ ,1 ~ , :,:,~c , -1·,
'. ,r;-~~,il~l;'r.:-::::,,,c:-:';=c;::;;;~r~1:-:,:,:--bt"c<;o~~lfJ~.:J;Cl71
·,..,
e)-
Substituting the v~.1!~, w~ g_et ( :
(1 + e) 2
or
1
JR_.~-eJ~-.-~-
2'
1
~
', .......=>
_e1= ~R
~
, ._
=2(-2-0)=-4N-s.
)
) _.).ll.: , ' - :, •
'
•••.• =, 4{0-1)
= -4
N-s - J
•• :
,
,_ I '
,; 1...or1I
Given that
,_,.) ·r1 .."1"::• ,,
or __ _[f = 24j6tile ". ·: ·,
(e}, J~_;,,_,ri}1('!_;
~m2(V '.'.:,~2)
or
and
or
2+2e 2 =3
__ or
e.,
2
-= -1
.' ' 2
1
+( ;
+ (1- e) 2 = 3
' 1
· · or
2
e) = ¾
e=-
F2
. Th~ Velocity of the Center of
. ' Mass for
. Collisions
. .
- ·
When various masses collide in-a complicated manner,
the velocity of the center of mass' unchanged, if the external
forces on ,the _system. of colliding particles during the
collisic5n are negligible: compared with the internal colli;ion
forces._ .
·
• .
.
.
,·,
I
Consider a system of two,particles of masses m1 and m2
.
'
.
moving.· with· velocities '.111 and v~/ , ·respectively, and
subjected -to zero total external force. Some times the
external forces are very small, they can be neglected·,
The · position vector of. the. center of . mass of the
· ·
two-particle system is,
~
. ..·,1 . --+
~
..
rcM
.
.
= m1 +m2
-- - (m,
r1 + m2 r2)
,.
-
The velocity of the center .of inass is
, '
or
or
)-
.'
/'.'.
-
separation
speed
= appro~tjl
.
'
,. - -~ - . speed ,
/"
: o/
~,
.
·
I"
v-1=2+1
,V_
''.':'T m/c5:
C
-
.
www.puucho.com
---·
__________
.,..,
----
.
,_.,
., ...:,........,...... __
_,...
.
____
-
.
------·-
Anurag Mishra Mechanics 1 with www.puucho.com
IMPULSE AND MOMENTUM
371
y
m,
m,
X
Fig. 4.48
m, ,
For two-particle collisions (Fig. 4.48), the particles' total
momentum in the CM frame is zero, so their incoming
momentum vectors are equal and opposjte. The particles'
outgoing momentum vectors are also equal and opposite.
Fig. 4.47
Rearrange this slightly to find
( m1
->
->
->
+m 2 )vcM =m 1 vli+m 2 v:,;
... (2)
The terms on the right-hand side of Equation (2) are the
momenta of the individual particles; the term on the
left-hand side 9f the equation is the momentum of the center
of mass, if we imagine the entire mass of the system to be
concentrated at that po'nt.
If the two particles undergo a collision and have velocity
vectors v,f and Vzf, the total momentum after the collision.
is
->
__,
m1 vlf+ m 2 v
-+
-+·
-,
-+
-
__,
->
Iu 2 1=Iv 2 ~ IU 1 I=Iv 1 [ In a two-dimensional elastic collision,
;only the angle 0. - is unchanged by the collision between
;incoming and outgoing velocities is not fixed by the incoming
.
velocities. In Fig. v 1 and v 2 are approach velocities of
2f
But we know that momentum is conserved in all
collisions where external forces are negligible compared
with the internal fores arising from the collision.,Therefore
--)
-+
Vu+V2i
Concept: Conservation of kinetic energy in an elastic
,collision ensures that the magnitude of each particles'
momentum of the CM frame is unchanged by the collision.
In case of two-particle elastic collision in the GM reference
frame of the particles total momentum of a system in its CM
frame is always zero and kinetic energy is conserwd, each
particle has the same speed after the collision as it did before:
---+
-+
= ml V11+ m2 V21
Using this equality in Equation (2), we find
i_ncoming particles and
u and. u are outgoing particles.
1
2
CONCEPTUAL EXAMPLE
2. Show that the
magnitude of each particle's momentum in the CM reference
frame is unchanged when the two particles collide elastically.
Solution: Can•be expresses in terms of its momentum:
1
l
K = -mv 2 = -(mv) 2 /m = p 2 /(2m). The total momentum
->
->
= m1 vlf+m 2 v 2f
This means that the velocity of the center of mass is the
same before and after the collision.
Although we derived this result for two-particle
collision, the result is true for any number of colliding
particles because conservation of momentum, is true for all
collisions (with negligible external forces).
Elastic Collisions ·in The CM Reference Frame
The incoming velocities completely determine the
outcome of a one-dimensional elastic collision. We obtained
an interesting result for two-dimensional collisions, the
angle between the outgoing velocities is fixed, while the
absolute direction of either velocity is not.
2
2
in the CM frame is zero, so the particles have equal and
opposite momenta and both before and after the collision;
-+
-+
-+
IP1_,l=IP,,;I= Pi
and
-+
IP1,f l=IP:1,fl= Pf
... (i)
The total kinetic energy K of the system remains
constant.
Before:
After:
K
K
P1,;2
Pzi2
= 2m + 2m
1
2
P, ( 1
1 )
= ; 2m1 + 2m 2
= P;'j +. P{j = Pf ( -1- + · 2m 1
2m 2
2m1
...
C')
11
1-) ... (iii)
2m 2
Equation (i) was used to similarly eqns. (ii) and (iii).
Eqnating K in eqns. (ii) and (iii) gives P; = Pf, as required.
Since the magnitude of each particle's momentum does
not charge, the only unknown is its direction. Which can not
be fo,md using the conservation laws alone.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
[_@ ____
Inelastic Collisions in CM Reference Frame
An elastic collision is an ideal model. Even when things
like billiard balls interact, a small amount of kinetic energy
is converted to other forms. The opposite ideal case is a
perfectly inelastic collision, in which the colliding objects
stick together. The CM reference frame allows us to give a
~~~:_pr~?se.~e~i.tion:
'
.
Concept: 1. In a perfectly inelastic collision, all the
1
kinetic energy of the incoming objects in their CM reference
frame is transformed to internal energy within the outgoing
:objects.
I
;
2. After · a perfectly inelastic collision, the outgoing
:objects, viewed as particles, have no kinetic energy in the CM,
'frame. They are all stationary in the CM frame-stuck
:together. In the lab frame the system still has its original,
:nonzero linear momentum and thus has some kinetic energy.
1
'f11~.J>g_tgqjr1g_o]Ji,ects_ move together with their CM velocity.
(a) Find velocity of centre of mass
(b) Maximum extension in the spring.
·v+--[ni} 000000~000000 i2m]--+2v
A
Fig. 4E.53
.
4mv-mv
Solution: Velocity of CM vcM = - - - - = v
3m
In COM frame. Initial momentum =0
at the time of maximum elongation both the masses will
be moving in same direction with same speed.
Initial relative velocity v ,.1 =3v
Decreasing in KE = Increase in PE of spring
1
1 2
= -kx
2
-mvrel
2
2
1 mx2m( 3v) 2
2 3m
=_!kx 2
3mv 2
=_!:_ kx 2
[;Ex;ca,t:ri,~le I- 521 }I-~
">,
§.- ---,.:::.=~~l!f~:--,...,-~.:.:.-.,
2
r - - - . ··--
:nvo particles _of mass mi, m 2 moving with initial velocity u1
!and u 2 colli~d,-head-on. Find minimum kinetic energy during,
:collision. ·Thus prove that maximum kinetic energy is lost in
iP!!.rfe~tly_i11elas_tis; ,o_llision.
Solution:
~ u 1· · ~ u2
cv·········~ .
Fig: 4E.52
In C·frame initial kinetic energy of system.! µ(v 2
.
where µ
=
2
-
v 1 )'
2
Concept: 7\vo identical blocks of mass m, each are
connected by a spring as shown in the Fig. 4.49 At any instant
of time t = 0, one block is given a velocity v 1 and other is given
a velocity v 2 ( v 1 > v 2 ) in the same direction simultaneously
as shown in the Fig. 4.49. The maximum energy stored in the
. rs
. given
.
b!}' -1 m( v -v )2 •
'
spnrzg
1
2
4
v,
mi m, . During collision at the instant of
m1 +m2
·,mJ::l::=::~v2
--
maximum deformation we get minimum kinetic energy in
C-frame as they attain same velocity thus no relative
velocity. When system have minimum kinetic energy in
c.frame it also has minimum kinetic energy in ground frame
·
·as velocity of CM is constant.
K0
B
k
smooth
Fig. 4.49
=.!:.µv~.1 + :!.m,v~
at maximum deformation. Thus
2
2
minimum kinetic energy during collision is .!cm,+ m 2 )v:,
w h erevc
=
(m 1u 1
m1
2
+ m 2u 2 )
+ m2
In inelastic collision final kinetic energy is
A smooth ball is dropped from a height hon a smooth incline,
,as shown in the Fig. 4E.54 (a). After collision the velocity of
the ball is_ directed horizontally.
~
.!:.2 m,. v ,.2 of CM is
h
constant.
~--- -
,··..-ii
·r--.., __
~~~~jt~,:i 53
p
0
/;1fo blo;ks-A a~d B. ~f musses ,11 and 2m placed on smooth
Ihorizontal surface are .:onnec'.ed with a light spring. The two
iblock§ are given velocities as .;!wwn when spring is ~t natural·
l~11gth~ --
ucose
u
Fig. 4E.54 (a)
(a) Find the coefficient of restitution.
(b) Ifthe collision is elastic, what is the impulse on the ball?
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
-~-----,
t tMPUISE AND MO~lN_T_UM_~_
Solution: (a) Normal axis and tangential axis are
shown in the figure. Reaction of the incline is along n-axis
and in the absence of friction there is no force along t-axis;
therefore velocity along t-axis remains unchanged,
i.e.,
V COS8
= U sin0
... (1)
From the definition of coefficient of restitution,
vsinB
... (2)
e=
-ucose
... (3)
Or
V = eucot0
From eqns. (1) and (3),
(eu cotB) cosB = u sine
or
e = tan 2 0
(b) When the collision is elastic, the component of
velocity along n-axis is reversed in direction. Therefore the
change in velocity
t.v = 2v cose
- -·
~-
Time taken to return to the point of projection after
d
impact
=---ev 0 cosa
Note that x-component of velocity after impact is
ev 0 cosa.
d
d
Total time of flight = - - - + - - = ;~::a[~v:;ra
There is no change in the vertical component of the
velocity after impact, therefore total time of flight remains
unchanged.
·
2v 0 sina
sina
2./ifi
T = -~-- = ~---
g
d
cosa
.JiFi
or
(1 +e e) = 2./ifi sin a
g
(1:e)= hsi~2a
d
or
u
0
Fl~. 4E.54 (b)
There is no change in velocity along the t-axis, therefore
no impulse along t-axis.
. Change in momentum of the ball
= mt.v
= 2mvcos0
Velocity of the ball when it strik~s the plane= ~2gh.
Thus impulse = 2m~2gh cose .
•
__
_
--;· --
•
u
•
f
al
A pa~cle ~ throw~fr~m-~h;~,;~-h-horizon~cill~-;,~-;,~
·vertical wall movingmvay with a speed.v shown in the Fig.'
4F..56 (a) ..If the particle returrrs:to· ihe point' of projection
after suffering _two ·elastic collisions, one with, the waWand
,another with the ground, find. the total time of fi.ight· ahd/
initial
:
- --- ·'· --· - ---· __ ::_
as
~
!
•
V
. '
--.v/4
_
I
•
.
----1.._
k~~~'Ji'J:>.~~:J _~~~;.>
-
g
X
h
•
,,
--- • - , I
-An inelastic ball is projected with velocity v 0 =..fiFi, at an
'angle a to the horiz@tal, towards a wall distant d from the,
:point of projection. After collision the ball retums to the point'
!of projection. \'\!hat is th_e ~qefficie1JUJf_restiiution?
Solution: Time taken to reach the wall
d
=--v 0 cosa
• ______ Fig. 4E. 56 (a)_____
/
Solution: While colling with wall x component of
velocity gets charged while component remains same. We
have
v0 cosa=vx
Velocity before impact
.... .. :.·b
'..:J"y
ev0 cos-a= Vx
:o..,_d ______..
h ,14··_ _.,,x'---I,
Velocity after impact
Fig. 4E.55
·,
SeparqtiQ1J _xJ,_e~ee11. tl!e_partic(~ C!tlrl th.e_ wall~~-,_;1J_.::._
www.puucho.com
-1
F;g,
4E: 5~.(~) , '
·
- •••. ~- - ,, __ ,,._ ·-
"
I
j
Anurag Mishra Mechanics 1 with www.puucho.com
.----- ---- -------- - - - - - - - - - - - - - - - -
rai4 -
It should be noted that time of flight will be 'it', where t
is time of ball to the ground and first collision must be
occurred with wall and second with grouod. Whereas if. wall
was moving towards the ball then, first collision must occur
at ground and second with wall.
Time of flight= 2
x.ff .
- --- - -
~~=~~!ieI~ .(5al>
:A ball is projected with velocity v 0, at an angle a to the
horizontal, towards a smooth wall which approaches the ball·
:with velocity u. After collision the ball retraces its path to the'
:point of projection. What is the time t taken by the ball from'
!the instant of projection_ to ]?Dint of impact?
Let separate between point A and wall is x 1 when ball
hits the wall
X1
V
-- - - -MECHANICS-I :
- - - --------------·--· - ---- - -~
u
+~=T=2fg{2h
V/2
time taken by ball to cover this distance
t1
=; =%Hi
Initial separate in x = x1
-
v 0 cos a
Fig. 4E.58
~ t1
4
Solution: Since the wall is smooth, the vertical
component of velocity will not change, as explained earlier,
the tangential component, remains unchanged. Collision is
elastic, therefore coefficient of restitution is 1.
Relative velocity of separation
e=
Relative velocity of approach
=(%-~x¾}~=vHi·
V
l=
~:~=:~;J':
• ---
,A ball is shot in a long hall having a roof at a height of 15 mi
M<h ,p,ro
(v 0 cosa+u)
1
<heflOOS ~"
Thus relative velocity of separation v = -(v 0 cosa+u)
Velocity relative to ground = (-v 0 cos a+ u) + u
Time of flight depends on the vertical component of
velocity which is unchanged.
T = 2v 0 sina
g
X
.
.
-
-~-~~~
!
ball lands on the floor at a distan'f'shown x =__ mfr<;>m!
,the point of projection. (A§sume_cq(lisions as elastic ifpri:y)_ ,J
2
Solution: y = xtan0-
gx
2u 2 cos 2 0
Let time taken before impact be t, the distance covered
before and after impact are same,
v 0 cosat = (T-t)(v 0 cosa + 2u)
t(v 0 cos a+ v 0 cosa + 2u) = T(v 0 cosa + 2u)
t = (v 0 cos a+ 2u)v 0 sin a
or
g(v 0 cosa+u)
-- -- - --· - -,--i-
r-· --
ts'o~R=~''.it\M3t!!irJ
I - - --- - -x=15
Fig. 4E.57 (b)
1Oxx 2 x25
3 2X (25) 2 X 9
2
=> x - 6Ox-t-15 x 45 = 0
=> x = 15, 45 (45 is rejected)
Tora!= 30 m
4
=> 15=x·-
59
- --
1i>
--
. .
-
,A smooth ring is kept on a smooth horizontal surface. From a
:point P of the ring a particle is projected at an angle a to the
:radius vector at P. If e is the coefficient of restitution between'
',the ring and the particle, show that the particle will return to
;the point of projection after two reflections if
:
2
1 1
1
cot a=-+-+L
-
-
-
- .
.
e
e2
e3
Solution : Let u be the velocity of projection at P. We
can find the velocity of rebound at point Q from Fig. 4E.59
(b).
www.puucho.com
tanp
usina.
tana
=-eu cosa
e
Anurag Mishra Mechanics 1 with www.puucho.com
! IMPULSE-AND MOMENTUM
L... ____ - - · - - - - -
·--
-
-- --
-- -
--
v = ~u 2 sin 2 a+ eu 2 cos 2 a.
and
Note: that friction between A and B will be
non-impulsive therefore we ignore it.
e =1
=> v=v 2 -v 1
Also
"' (2)
2v
vsinp
tanp tano:
tany = - ~ - = - - = - ev cosp
e
e2
Similarly at R
(1)
w = ~v 2 sin 2 a.+ev 2 sin 2 a
and
R
+
(2)
=3
Vz
Using COM for (A + B)
2
2m( ; ) = 3mv 1
4v
VJ=9
2v
1
42
(-4v )2 =-mv
( 3 )2--(3m)
2
9
27
1
2
(b) Af.=-(2m) -
······-- 'le..'J, 611
(-----.Exam·
_ _____- - ~P•c¥;
61 }:.->'
1·---:-,_.
(a)
u*s~a
5
"·
u
'
,:·····
a)/(v
u sin
A small ball i5 projected from point P on floor towards a wall'
,as shOVfn in Fig. 4E.61 (a). It hits the wall when its velocity i5
;horizontaL Ball reaches point P after on bounce on the floor. If
Ithe coefficient, of restitution i5 the same for the two colli.ions, !
'(i-nd its value_
u eu cos a
··..
~
~/
','Q
,/a,
·····._t-axis
,
I
,'
uCos a
Velocity components
before impact
Velocity components
after impact
Fig, 4E,59 (b)
Since the particle returns to the point of projection,
o: + p+ y = it/2
or
tan(o: + P+ y) = =
or
1-[tano: tanp + tanp tan y + tan ytano:] = 0
Fig. 4E.61 (a)
Solution: We have
2vxvy
2vy
R=---, T=--
g
[1 1 1]
After first collision
.1 = tan 2 o: - + +e e 2 e3
or
2
1
1
1
e
e2
e3
v'x
-
--~
A block of mass m i5 projected with velocity v as shown in Fig. i
'4E.60_ The ground i5 smooth but there i5 friction between A;
and B. If colli.ion i5 elastic
Distance covered before 2 nd
collision
.
T
T evxvy
d1 = v~.- = evx.- = --2
vn
_ 2v" X v" yY
d2 -
[mlµ
V
2m
Is
µ=O
-1
I
(a) Find the final common velocity of A and B.
(b) Find total energy di.sipated in friction.
Assume that A does not fall off 13_._
conservation
mv = mv_ 1 + 2mv 2
Collision between B and C is elastic
g
but
dz
d1
Fig. 4E.61 (b)
= ev'
Zeb
g
X
ev y
g
evxvy
Fig, 4E.60.
Solution: (a) Using
between B and C
2
After second collision
v; = v~
A
[ml--- I
= evx
v'y=Vy
cot o:=-+-+-
or
g
.
,
d1 +d2 =--(1+2e)
g
R 2vxvy
d1 +dz=-=-2
2g .
2e 2 + e -1 = 0
-1±./I+s -i ± 3
e (1 + 2e) = 1;
of momentum
e=----=--
4
... (1)
4
Rejecting -ve value
www.puucho.com
e = 1/2
u
....
til'
Anurag Mishra Mechanics 1 with www.puucho.com
fTh'.;f~~uctz:·au~ §1lna1~ at r~:~~~'/nt.~ontg_ct;<m ~ _table. ;t]
((~. ~ef d¼c/o.,j,s..9me·m.CIJS b.ut, o.,f'".ifou,b.•.le !ad1us ,st;nk.es ,t~e.1!1'
lsjrrlrriemcally, iind '.'itself comes' to . rest .after impact. The
'
Alternative
•'
2./2
cos0=--.
3
----- -·-·~-·-- ---.
·v;
!
F"
I~
,
..
. i ~ : :,'.;_J~!~t~l,___ j_ _ .,___'., .. ·-··
u
Solution:
Fig. 4E.62 (d)
Assume. initial velocity of big block =u
and final velocity of small ball is. v
Conserving momentum
=>
mil = 2mv cos8
V2 -V1
-e=~-~
=>
... (1)
U2 - U1
.v-0
=>
... (1)
-e=---O-1Lcos8
V
=>
e=--
... (2) .
U"COS8
From eqns. (1) and (2)
1
9
e=---~2cos2 8 16
Passage: (Example 63-65)
;
.
. --
-
-
- .
'
··~-·- ---·- . -
~-:-'""·:~7
,Two smooth balis A and B, each of mass m and radius R,'.havei
!their centres at (0, 0,R)and at.(511.,-R,R) respectively) in,~
;coordinate 'system as shown'. Ball A,- moving .along· po_sitivel
'.x:axis, collides with ball B. Just before .the collision,_speed
.ball A is 4 m/s and ball B is stationary. The collision betll/een
!the balis is elastic.
_ ·_ __
· · ·· \ _ .
o/i
!
y
l
' ,,,·
'l
+--+-"+-....,..-sc---x(m) '.
•
I
i
B
I
Fig, 4E.63 (a)
J..,.,,....-~-----
I
,
~~-·-----""· · - - -
.
__l,'
!Velocity of'the ball A just after the' collision is:
(a) (t+.flj)m/s ·
(b) (i---.13i)m/s /.
(c) (2I+.--.13])m/s
(d)
(2i+2])m/s
• I
.
, .
- . '
. ---·- - ---,·-- ._,_ --~J
www.puucho.com
n
Anurag Mishra Mechanics 1 with www.puucho.com
IMPULSE AND MOMENTUM
Solution: (a)
Solution: (b)
_,,. ... 4 sin 30"
A
A
..
Before Collision
__.,. 0
Before Collision
(a)
(a)
4 sin 30'
4 sin 30°
.,. 0
.. ···
B
After Collision
After Collision
(b)
Fig. 4E.63
vA
(b)
Fig. 4E.65
= 4sin 30° [cos 60i + sin 60JJ
1
-(v 2 -v 1 )
---~
-~
(1)
2
vA=i+.f:3]m/s
- ----- \
: _E;x9_rQple :_ ~~-
v
Vz
(2)
Impulse of the force exerted by A on B during the collision, is
equal to:
(a) (./3mi+3mJ)kg-m/s
(b) (
1;
.
=(
-
-,
(_,F 2 Atl}J>le . 65
(a)-(3-v3i+9j)m/s
(c) (6i+3.f:3J)m/s
-
4
4
1--.
- I- .
;,J~~9:~ij)fr:..~
-,
sin 30° JJ - OJ
I,->
Coefficient of restitution during the collision is changed to
1/2, keeping all other parameters unchanged. What is the
velocity of the ball B after the collision ?
A
-
.
9, 3./3,)
-1---J
JAon·B;:::; mVBf-VBr
= (3mi-~3mJ) kg-m/s
r;c:
2
2
= m[ 4cos30° (cos 30° i -
1
= mv 1 + mv 2
3
v 2 = ./3 m/s[cos30° i+sin30° (-J)J
(d) (2v'3mi + 3mJ) kg-m/s
2
(0-4cos30°)
= 2./3
3./3
v 2 = --m/s
(c) (3mi- .f:3mJ) kg-m/s
-,
'
+ V1
mi- 3m]) kg-m/s
Solution: (c)
-
;Jt = ./3
m-2
Vz
- --
1
A
,-;'
(b)-(9i-3-v3j)m/s
. 4
(d) (6i-3./3J)m/s
Two spheres. are moving towards each other. Both have same'
radius but their masses are 2 kg and 4 kg. If the velocities are
4 m/s and. 2 m/s respectively and coefficient of restitution is
e = 1/ 3, find,
(a) The common velocity along the line of impact.
(b) Final velocities along line of.impact.
(c) Impulse of deformation.
(d) Impulse of reformation.
(e) Maximum potential energy of deformation.
(f) Loss in kinetic energy due to collision.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
378
-,---~-- - _ - : ..•:=.-::....-=-:.~-::.---:--_=.--:-.....=::..·_----
J
From the above two equations,
.
2
,
v 1 =--m/s
-J3
1
v 2 =-m/s.
and
.
./"~,
·--~~~-Jf~ ...
~ --------~~~:.?! motion
R... .. R:
R·· •• :f!kg
2m1s a ··-...
Line of motion
-J3
(c)
4kg
= m1 (v-u1 )
· = 2(0-4cos30°) = -4-,J3 N-s
JR= e.fv = .!(-4-./3) = _ _±__ N-s
(d)
-./3
3
I
···-.Line of impact
J0
I
1
j
__ L~~---~g:4E.6~------lJ·
•AB C sm0
.
S o I u t.10n.. I nu
= -BC = -R = -1
AB 2R 2
0=3~
(a) By conservation of momentum along line of
impact.
·-- -------1
• LOI
M
(e) Maximum potential energy of deformation is
equal to loss in kinetic energy during deformation upto
maximum deformed state.
-
U= 1 11!1 (u 1 cos0) 2 + 1 m2 (u 2 cos0) 2 - 1 (m1 +m 2 }v 2
2
2
2
1
1
1
= -2(4cos30° ) 2 +-4(-2cos30° ) 2 --(2+ 4)(0) 2
2
2
2
·
or
U = 18 joule.
(f) Loss in kinetic energy,
uAK£1(
= - m1 u 1 cos0) 2+1
- m2 (u 2 cos0) 2
2
2
4 sin 30°
··...
-(½m,v:
2~!j~4m/s
4 cos 30
1
•,
~cos30°
-
2
+½m v~)
2
1
= -2(4cos30°) +-4(-2cos30°)
2
·2
'.·-xn30°
2
-784kg
2 sin.30°
-
2 sin 30°
Just Before ColHsion Along LOI
-(½2(lJ2 +½4(lJ2)
•••
~ - - - - - - · · F i g . 4E.66 (b) _, _ _ ___ __
]
2(4cos30°)- 4(2cos30°) = (2+ 4)v
or
v = 0 (common velocity along LOI)
(b) Let v 1 and v 2 be the final velocity of A and B .
respectively then, by conservation of momentum along line
of impact.
'ii; /4sin30°
21<~~'.
6-exam,~,;~
. '
'.lwo bl~~ks of mass 2 kg and Mare at rest on an incli~~dpla~e]
and are separat~d by a d~t~nce 6.0 as shown in Fig. 4E._671
~a) .. The coeffi~1ent of friction ·between each block mi.d• the1
inclined plane IS 0.25. The 2 kg block is given a velocify ofr
10.011!/s up the inclined plane. It collides with M, comes batk
and has a velocity of 1.011!/s when it reaches its initial
position . .The other block M after the collision moves 0.5 in up
land tomes :10 rest. Calculate the coefficient of restitution
between the blocks and the mass of the block M. ·
[ Take sin0 = tan0 = 0.()5 and g = 10 m/s 2 .]
,
I
4~2
2 sin 30°
AKE = 16 joule
-Maximum Deformed State
·
~--i
Just After Collision Along LOI
!
,i
L__ Fig. 4E.66 (c)
·-------
I
, "'E
2(4cos:30° )- 4(2cos30°) = 2(v 1 ) + 4(v 2 )
or
0=v 1 +2v 2
... (1)
By coefficient of restitution,
e = velocity of separation along LOI
velocity of approach along LOI
1
V2 -v 1
or
-=--~~~-3 4cos30°+ 2COS30°
or
V2-V1 =-./3
... (2)
www.puucho.com
' C U')
: iii C\!
!
LO 0
, ••••••••••• II II
,E'
:h 1 ;; 6 sin 8
: =0.3m
Re°ferepce level
Fig. 4E.67 (a)
I,
Anurag Mishra Mechanics 1 with www.puucho.com
~I_M_PU_~_E_AN_D~,M_O_ME_N_TU_M_ _ _~----~----~-'-----s=-------'-679j
Solution: This problem can be divided into the
following steps:
1. Block of mass 2 kg moves up the incline collides with
block M with velocity v 1 •
· ·
From the conservation of momentum for collision,
(2)v 1 = Mv 3 - (2)v 2
2. After collision the velocity of the 2 kg block is v 2 • The
2 kg block returns to the original position with velocity 1.0
k~~m.i,;1 68 ~
----···-·-- - - - - - - ,
m/s.
3. B.lock M moves up with velocity v 3 and comes to rest
after moving a distance 0.5 m up the incline.
First step: From the work~energy theorem we may
calculate v 1 as follows :
W friction
= ,iKE + ,iUg
M
= 2(v1 +v2) = 2[8+5]
V3
15.12kg
1.72
Fig. 4K68 (a) shows a smooth spherical ball of mass m
striking two .identical· equilateral triangular wedges of mass
M; At the ill$tant of impact velocity ofthe ball is v 0. TakiJJg
coefficient of restitution e, determine the velocities of the
sphere and the wedges iUSLaftf!r collision....
1
-6µmgcos8 = -m[v; -(10)2] + mgh1
2
= -2[f.j.lg cos8 + ghi] +(100)
where cos8 = .J1- sin 2 8 = ~1- (0.05) 2 = 0.99
= (100)- 2 [(6)(0.25)(10)(0.99) + (10)(0.3)]
v;
v;
v 1 ~ 8m/s
Fig. 4E.68 (a)
Second step: We may apply work-energy ·theorem
on return journey of the 2 kg block.
W friction
= ,il{E + llUg
1
2
2
-6µmgcos8=-m[(l)
-(v 2 ) ]-mgh 1
2
-12(0.25)(10)(0.99) = [(1) 2 -(v 2 ) 2]-2x (10)(0.3) 2
v 2 ~ 5m/s
Third step: We may apply work-energy theorem on
the upward journey of M.
or
·---'------"----'
Solution: Let J be intpulse between ball and wedges
and v 1 and v 2 be the velocities of the ball and the wedge.
From intpulse-momentum equation on the ball,
2J sin 30° = mv 1 - Emvo)
J = mv 1 + mv 0
... (1)
From the wedge, J cos30°= Mv 2
... (2)
On eliminating J from eqns. (1) and (2), we have
2
...(3)
../3Mv 2 = mv 1 +mv 0
After collision
Before colllslon
Fig. 4E.67(b)
·- ... ·--· --·. ---''-------'-'--------"--'
W friction = ,iKE + ,iUg
.
1
-(0.5)µMg cos8 = -M[O- (v 3 ) 2] + Mgh 2
2
-v~ = -µg cos8- 2gh 2
v~ = (0.25)(10)(0.99) + 2(10)(0.025)
v 3 ~ 1.72 m/s
From the definition of coefficient of restitution,
e= Relative velocity of separation
Relative velocity of approach
=
V2 +V3
From the definition of coefficient of restitution,
.
V1COS60°+V2COS30° V1+../JV2
e=~-------=---v0cos600
v0
or
ev 0 =v 1 +../3v 2
... (4)
On solving eqns. (3) and (4) for v 1 and v 2 , we get
(2eM -3m)v 0
V -----l 2M+3m
VI
e=
5+1.72
8
V
0.84
www.puucho.com
../3(1 + e)mv 0
----~
22M+3m
Anurag Mishra Mechanics 1 with www.puucho.com
~-ai
...-c-
/;
. ,r
>.e~:.;;.:.......-'-"-·-"'"-'---~-'.--'-C..-----'"--"·-'--;====~·-":::;' .
_·_-_-_-_-_-,...:.-·· -----·-
MECHA~~f-fj
.
'k qXc(~i~J;~~~
.~--~!~
~------,···-·-· ...,_,, __ ,=
---··-~,.-------·
=
·--
- - --~
I_
,
. .
,
,
,
I
M = 0.25 kg
·1,'.
S/ooth wire
.
0.8m
.
1I
·I'
1
I
d= 0.4m
j
''.
Fig. 4E.69
-------·
·'
Soh,1tion : The velocity of the ring when the string is
vertical can be determined from the energy conservation
equation .
mg(0.8- 0.4) = IMv~
.
2_ _ _ _ _ .
..
v0
1w
=~2mg,
--(0.8-0.4) =--m/s
M
5
Note that the velocity of mass mis zero and that of ring
is v O in the horizontal direction.
When the string makes the maximum angle 0 with the
v~rtical subsequently, we have
(M+m)v=Mv 0
(from conservation cif momentum)
And .from energy conservation, we have
1
2
1
2
mgl(l-cos0)+-.(M+m)v
=-Mv
0
·
-2
.
2
or
.
Kinetic energy before impact= tmv 2
2
=Im
(v + 2u)~
2
.
''
~
Solution:, Velocity of ball relative to _wall before
collision is (V'+ u). After elastic collision the velocity of ball
relative to wall will be -(v + u). The velocity of the ball
relative to ground will be
-(v+u)-u = -(v + 2u)
Kinetic energy after impact
'
or
4E.70
_____ ,,,...._._..____Fig."'"""'""-~·"'""""''""·~·---
,.
I
m = 0.75 kg;
'
.i
'
'I
! i_
i
::--·---:;
IA 'ting.of mass .M. 0.25 kg free to slide on a fixed smooth/
· horizontal wire is attached to a particle of mass m = 0:75 kg1
· by astring of l~ngth 1 m which pas§es over a fixed smooth,peg
Pat a depth-d-= OA m belo~ the wire anq in the same,vertical
plane. The system is released from rest when the ring is 0.8 m
\from the peg. }'ind th~ maximum angle' the ring will make
lwith the verticaLafter it loses contact with th peg.
·
I
, ·-- I
.. i
mg1(1-cos0)=I[M-~]v~
2
M+m
The change in kinetic energy i~ equal to 2mu(u + v ).
Now we calculate the work of reaction forces actir g on
the ball during the _impact. Let the collision continue for t
seconds. Assume the reaction force to be constant· (the result
does not depend on this assumption). Since the impact
changes the momentum by 2m(v + u), the force of reaction is
· ·
2m(v +u)
·
·
F
t
The work of this force is
W =rs= F(ut) = 2m(v-t,u)ut
t
= 2m(v +u)u
So we can see that this work is equal to change in kinetic
energy.
SYSTEM OF VARIABLE MASS;
ROCKET PROPULSION
Fig. 4.50 (a) shows a system of mass M and momen'.tum
Mv at same time t. A tiny infinitesimal mass dM travelling
with velocity u combines with the system in an infinitesimal
time dt; so the mass is M + dM and velocity-is v + dv.
.., u
..,dM
. t; Mv+
Tota1 momentum at time
Total momentum at time t x dt: (M + dM) (v+ dv)
1-cos0= 1 Mv~
2 (M + m)gl
..,
Ixix 196x3
2 4 25 X 9.8
So the change in moment)lm dP is
=0.3
or
cose = 0. 7,
--+
e = cos- (0.7)
~
-._,
-·---- ...
-
- _,_ - ---... ---·
--+
--+
From Ne'\\'.ton's second law, we have
--··1
·A bdll moving with a velocity v strikes a wall moving towa~d·
!the wall with a velocity u. An elastic-impact occurs. Determine'
ithe velocity of the bdll after the impact. What is the cause bf;
lthe change in the kinetic energy of;.t~e. ball? Consider the mass:
)of the wall td be infinitely great. ,
l.:-------··-
--+
=Mdv+vdM +dM dv-udM
~~m~~>
r-· ---·--~·.-.. -:. -.----- -~-_- -_--_--..--------· .....
--+
dP = (M + dM)(v+ dv)-(Mv+_udM)
1
.. ···~""
r.i
....
=dP =Mdv+vdM-udM
d!_ .... ·- __ dt
____ ':",t__
,-~v1
dM
-~
.
~;
!
(a)
(b)
Fig. 4.50
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
381/
- - - -----·------·-dM
=v-
We ,bave neglected the term dM dv/ dt , in the limit of
infinitesimals it is zero. Thus we get
dV
--i
-+
"""?
dM
LFexr =M--(u-v)-·
dt
Note that the quantity
Vrel,
... (1)
dt
(ti- v)
is the relative velocity,
dt
dt
•
where L Fext denotes the external force on the mass M
(for a rocket it would include the force of gravity and air
resistance).
The force exerted by dM on M is
v'rel
dM, which
dt
represents the rate at which momentum is being transferred
to (or from) the mass M (for a rocket this term is called the
thrust).
The equation (2) has application in rocket propulsion. It
propels itself forward by the ejection of burnt gases. The
mass M of the rocket decreases during the process, so
dM/dt < 0. Another application is the dropping of material
(gravel) onto a conveyer belt. In this case the mass of the
loaded conveyor belt increases, so dM/dt > 0.
-
.., -
-
-:-
!--
Note that the direction of force Fext is same
as the velocity of the belt.
--, •
L-~~~~pJ E;:,.l!~J/
A hopper drops gravel at a rate of 75.0kg/s onto a conveyor
,belt moving at a constant speed v = 2.20 m/s.
(a) Detennine the force needed to keep the conveyor belt.
moving.
,
(b) What power output must the motor have that drives the:
conveyor belt?
••
Fig. 4E.71 (a)
Solution : We assume that horizontal component of
velocity of gravel at the moment it lands on the conveyor
belt is zero.
->
= V gra\"el -
->
V belt
= 0-v
As
Fext
dv -, dM
=M--vrel dt
dt
Fig. 4E.71 _(b)_
dw ->->
2dM = 363 W
= Fext' V = V dt
dt
which is the power output required of the motor.
The rate at which gravel is gaining energy is
dK
dt
=j_(.:1cMv2)=_:lcdMv2
dt2
_,
2dt
which is only half the work done by Fex,. The other half
of the external work done goes into thermal energy
produced by friction between the gravel and the belt (the
same friction force that accelerates the gravel).
L:cE~i,ii;ppJ,~,j-72[>
-
-
'
A rocket has a mass of 21000 kg of which 15000 kg is fuel.!
The rocket engine can exhaust fuel at the rate ofl90kg/swith,
an exhaust velocit;y of 2800 m/ s relative to the rocket. If the
rocket is fired vertical(,- upward, calculate:
( a) the thrust of the rocket;
(b) the net force on the rocket at blast-off and just when all
the fuel has been used up;
(c) the rocket's velocity as a function of time, and
,
(d) its final velocity at the bum-out. Assume that·
acceleration due to gravity is constant at g = 9.8m/s2
and there.4. no. air r~istan~ .. .. _
i
Solution: (a) The thrust of the rocket is
_,
= V rel
dM
-
Vrocket
dt
= (-2800)(-190)
= 5.3xl0 5 N
,~Mff(I
->
V rel
,/-"/
-
Fthrust
(a)
Uy
(b) Rate of work done by Fext,
... (2)
->
r
I
->
->
of dM w.r.t. M. So we can rearrange eqn. (1) as
dv
..,
.., dM
M-=LFex,+V«J -
Llx,~-07
dt
= (2.20) (75)
= 165N
We have taken upward positive, so v rel
is negative because exhaust velocity. is
downward, and dM/dt is negative because
the rocket's mass is decreasing.
(b) F,x, is blast-off
=Mg= (2.1 X 1Q4 )(9.80)
;, 2.lxl0 5 N
Fext
at bum-out = M rocke,g
= (6xl0 3 )(9.80)
= 5.9xl0 4 N
=0-(-v)(:)
www.puucho.com
,~M
!
~gasa,
Flg.4E.72
Anurag Mishra Mechanics 1 with www.puucho.com
3::::8~2---'~-----";.;__----'-"-..:.;._--_;;_---·....._.----'-----M-EC..;...HANIC,S~IJ
dv
dM
so,
F••, = (5,3 x 105 - 2.1 x 10 5 ) (blast- off)
So
Fext =M-+vdt
dt
= 3.2x 105
.
dv
4
5
(P-µkpgx) = (px)-+pv 2
F0 . , = (5.3 x 10 - 5.9 x 10 ) (burn- out)
dt
After bum-out only the force of gravity remains, i.e.,
On
rearranging
the
equation,
we
have .
'
-5.9x10 4 N.
2
. !
2
dv P-µkpgx-pv
P
v
-~~---=--µkg-dv
dM
(c) As
M-=Fen +v 1 dt
px
px
x
dt
re dt
!...'.I
dv
= Fen
dt+v 1 dM
M
,. M
where Fext ; ~Mg and M is the mass of the rocket as a
function of time.
So,.
Jvdv=-J'gdt+v,. 1 JM dM
v0
o ·
Mo m
Since v rel is constant we have taken it out of the
integral.
M
Thus
v(t) = v 0 -gt+ vre1 InMo
where v(t) is the rocket's velocity and M its mass at any
time t.
Note that vre1 is negative (-2800m/s) because it is
opposite to the motion and that ln(M/M 0 ) is also negative
because M O > M. Therefore the thrust on the rocket is
positive and acts to increase the velocity.
·
(d) Time taken to use up all the fuel (15000 kg) at a
rate of 190 kg/s, so, at bum-out
If we take
t = l.50x104
190
v 0 = 0,
iciE~x@.ml~Lt,'~J74l;>
IA chain oflength L and ·;ass per ~-n-it-1-en_g_t_h_p._is_p_il_le_d_o_n ~J
horizontal swface. Orie end of the chain is lifted vertically!
with a .constant velocity v by a variable force P. Detennine: I
( a)' P as a/unction of the height xof the end above the"surface.
(b) the energy. lost during tne lifting of the chain.
·
'
l
l
I
X
Flg.4E.74
__
...,_
Solution: (a) Let x be the displacement of the end of
the chain above the surface.
Fext =·P-pgx
= 0-V
dM
-=pv
dt
dv = 0
dt
dv
dM
From the equation, M - = Fext + v rel dt
dt
or
O = (P-pgx) + (0-v)pv
6000
v=-(9.8)(79)+(-2800)(ln
)
.
21000
= 2730m/s
~~~•:~tei@);;>
p =p(gx+v2)
a
IA pile of loose-link chain, mass per ·unit length A lies on
1rough surface with coefficient of kinetic friction µ k· Of!e end
of the chain is 'being pulled horizontally along the surface bya
constant force P. Detennine the acceleration of the chain in
dx
.
'
tenns of x and- = v.
'dt
(b) From work-energy theorem;
fPdx-M:=AK+t.U
f
=--:-=-p•
Fig.4E.73
· dv ·
dM
M-=F +v 1 dt
ext
re dt
here
vrel ::::;Q-v
and
dM
dM dx
·-=--=pv
dt
dx dt
Also
Fen =P-.µkpgx
.
where JP dx is work done by external force P, M: is loss
in energy.
Pdx =
(pgx+pv 2)dx
Ji
1
2
2
=-pgL
+pvL
1- . _,
Solution: ·As
f
Vre1
= 79s
;us ;;e~
,----p-·-
2
On substituting in the work-energy equation, we get
12
2
1212
-pgL +pv L-M: = -pLv +-pgL
2
2
2
1
2
M: =-pLv
2
The link at rest on the platform acquires its velocity
abruptly through an impact with the link above it. Work
done by internal non-elastic forces during impact is
converted into heat and acoustic energy.
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
IMPULSE AND MOMENTUM
3531
. I
---- -,
Only One Alternative is Correct
1. A set ofn identical cubical blocks lies at rest parallel to
each other along a line on a smooth horizontal surface.
The separation between the near surface of any two
adjacent blocks in L. The block at one end is given a
speed v towards the next one at time t = 0, all
collisions are elastic then:
.
(a) The last block starts movmg at t
(b) The last block starts moving at t
(n + l)L
= -'---~-
v
n(n- l)L
2v
(c) The center of mass of the system will have the
final speed v
(d) The center of mass of the system will have the
V
final speed n
2. A boy of mass m is standing on a block of mass M kept
on a rough surface. When the boy walks from left to
right on the block, the centre of mass (boy + block) of
the system:
(a) remains stationary
(b) shifts towards left
(c) shifts towards right
(d) shifts toward right if M > m and toward left if
M<m
3. A uniform sphere is placed on a smooth hori~ontal
surface and a horizontal force F is applied on it at a
distance h above the surface. The acceleration of the
centre:
(a) is maximum when h = 0
(b) is maximum when h = R
(c) is maximum when h = 2R
(d) is independent of h
4. An open water tight railway wagon of mass 5 x 103 kg
coasts at initial velocity of 1.2 m/s without friction on
a railway track. Rain falls vertically downwards into
the wagon. What change then occurred in the kinetic
energy of the wagon, when it has collected 103 kg of
water:
(a) 1200 J
(c) 600 J
(b)° 300 J
(d) 900 J
5. A body falling vertically downwards under gravity
breaks in two parts of unequal masses. The centre of
mass of the two parts taken together shifts
horizontally towards:
(a) heavier piece
(b) lighter piece
(c) does not shift horizontally
(d) depends on the vertical velocity at the time of
breaking
6. A block of mass Mis placed on the top of
a bigger block of mass 10 Mas shown in
figure. All the surfaces are frictionless.
10M '
The system is released from rest, then ~.2.2~::d
the distance moved by the bigger block
at the instant the smaller block reaches the ground:
(a) 0.22 m
(b) 0.20 m
(c) zero
(d) 0.24 m
7. In the figure shown, the two
•,
identical balls of mass M and
radius R each, are placed in
.,
contact with each other on the
1
frictionless horizontal surface. 1
The third ball of mass M and ,
R
;<f_.\
radius -, is coming down
::~·;I
-! :
t
2
\\\
\\\\!
vertically and has a velocity
· -··· __ J
= v O when it simultaneously hits the two balls and
itself comes to rest. Then, each of the two bigger balls
will move after collision with a speed equal to :
(a) 4v 0
(b) 2v 0
./s
(c) ~
www.puucho.com
./s
./s
(d) none .
Anurag Mishra Mechanics 1 with www.puucho.com
384·
- - - ---------·-------· _,._ ·- ·- · MECHAN1cs,TJ
~~----'--------'-------------------~·~·---8. A ball kept in a closed box moves in the box making
collisions with the walls. The box is kept on a smooth
surface. The velocity of the centre of mass:
(a) of the box remains constant
{b) of the box plus the ball system remains constant
(c) of the ball remains constant
(d) of the ball relative to the box remains constant
9. Two identical billiard balls are in contact on a table. A
third identical ball strikes them symmetrically and
comes to rest after impact. The coefficient of
restitution is:
(a)'~
3
(b)
~
(d)
(c)
.!.
3
:./3'
2
6
10. A ball is projected from ground 'with a velocity v at an
angle 8 to the vertical: On its path it makes an elastic
collision with a vertical wall and returns to ground.
The total time of flight of the ball is:
(a) 2vsin8
(b) 2vcos8
g
(c) v sin 28
g
g
(d) v case g
11. A sphere moving with velocityv strikes elastically with
a wall moving towards the sphere with a velocity u. If
the mass of the wall is infinitely large, the work done
by the wall during collision wiHbe:
(a) mu(u + v)
(c) 2mv(u+v)
(b) 2mu(u + v)
(d) 2m(u+v)
12. On a horizontal smooth surface a disc is placed'at rest.
Another disc of same mass is coming with impact
parameter equal to its own radius. First disc is of
radius r. What'should be the radius of coming disc so
that after collision first disc moves at an angle 45° to
the direction of motion of incoming disc ?
(a) 2r
(b) r(-.J2-1)
r
(c)
(d)
r../2
(-.J2-1)
13. A ball is thrown vertically downwards with velocity
2gh from a h_eight h. After colliding with the ground
it just teaches the starting point. Coefficient of
restitution is:
1
.J
(a) -
--
.
(a) (l+e)u2sin28
g
2
(c) (l-e)u sin28
g
15. A ball is dropped from a height h. As it bounces off the
floor, its speed is 80 per cent of what it was just before
it hit the floor. The ball will then rise to a height of
most nearly:
(a) 0.80 h
(b) 0.75 h
(c) 0.64 h
(d) 0.50 h
16. Internal forces can change:
(a) the linear momentum but not the kinetic energy
(b) the kinetic energy but not the linear, momentum
(c) linear momentum as well as kinetic energy
(d) neither the linear momentum nor the kinetic
energy
17. In an elastic collision of two billiard balls which' of the
following quantities is not conserved during the short
time of collision:
(a) Momentum
(b) Total mechanical energy
(c) Kinetic energy
(d) None
18. A block of mass M lying on a smooth· horizontal
surface is rigidly attached ·to a light horizontal ·spring
of spring constant k. The other end of the spring is
rigidly connected to a fixed wall. A stationary gun fires
·bullets of mass m each in horizontal direction with
speed v 0 one after other. The bullets hit the block and
get embedded to it. The first bullet hits the block at
t = 0, the second bullet hits at t = ;7t~M: m, the third
. att
bu IIet hits
=21t~M+m
- k - + 21t~M+2m
k
and so on.
The maximum compression in· the spring after the n th
bullet hits is:
nmv 0 .Jk ·
(b) _(M_+_n_m_)~31_2
(a) ----'C.:.:C,.,,312
(M +nm)
nm11 0 .Jk
(c)
~
(M + nm)
312
(d)
nmv 0
.jk(M +nm)
19. A boy hits a baseball with a bat and imparts an impulse
J to the ball. The boy hits the ball again with the same
../2
(c) 1
14. A ball is projected with initial
.
r---- -- - ---
velocity u at an angle 8 to the i ~ \
horizontal. Then horizontal . ~ x ~ _ J
.
displacement covered by ball -as
it collides third time to the ground would be, if
coefficient of restitution is e: ·
.,,
force, except that the ball and the bat are in contact for
twice the amount of time as in the first hit. The new
impulse equals :
(a) half the original impulse
(b) the original impulse
(c) twice. the original impulse
(d) four times the original impulse
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
_____·_-·3asj
[-,M~UlSE AND M~~~N--'T-=-UM:.c:____ _-.c_ _ _ _ _ _ _ _ _ _ _ _·_-_ - _ · · - - - - - ' - - - - - 20. A shell is fired from a cannon with a velocity v at an
angle 8 with the horizontal direction. At the highest
point in its path, it explodes into two pieces of equal
masses. One of the pieces retraces to the cannon. The
speed of the other piece immediately after the
explosion is:
(a) 3v cos8
(b) 2v cos8
3
(c) -vcos8
2
(d) vcos8
21. A disc of radius R i;cut out from a larger disc ofradius
2R in such a way that the edge of·the hole touches .the
edge of the disc. Then center of mass for the residual
disc is:
(a) at 2R from center of the original disc away from
3
the center of the hole
(b) at R from center of the original disc away from the
3
center of the hole
(c) at the center of the original disc
(d) at the center of the hole
22. There are some passengers inside a stationary railway
compartment. The track is frictionless. The centre of
mass of the compartment itself (without the
passengers) is C1 , while the centre of mass of the
'compartment plus passengers' system is C 2 • If the
passengers move about inside the compartment along .
the track.
(a) Both C1 and C 2 will move with respect to the
ground
(b) Neither C1 nor C 2 will move with respect to the
ground
(c) C1 will move but C 2 will be stationary with respect
to the ground
(d) C2 will move butC 1 will be stationary with respect
to the ground
23. All the particles of a body are situated at a distance R
from the origin. The distance of the center of mass
·
from the origin is:
(a) = R
(b),,; R
(c) > R
(d) e: R
24. Two trains A and B are running in the same direction
on the parallel rails such that A is faster than B. Packets
of equal weight are transferred from A to B. What will
happen due to this:
(a) A will be accelerated and B will be retarded
(b) B will be accelerated and A will be retarded
(c) There will be no change in A but B will be
accelerated
(d) There will be no change in B but A will be
accelerated
25. Three blocks are
initially placed as
shown in the
figure. Block A
has mass m and
initial velocity v to the right. Block B with mass m and
block C with mass 4m are both initially at rest. Neglect
friction. All collisions are elastic. The final velocity of
block A is:
(a) 0.6v to the left
(b) 1.4v to the left
(c) v to the left
(d) 0.4v to the right
26. A square plate of edged and a I w
- . --·- . -circular disc of diameter dare : _1
Placed touching each other at the l&\\\\M~\\\\M
t,.-d ...... d-1 :
midpoint of an edge of plate as
··-- ---· -·- -- --·-·
shown. Then center of mass of the combination will be
(assume same mass per unit area for the two plates):
"2d
(a) - - left to the center of the disc
2+ 1t
.
·
~ right to the center of the disc
(b)
2+7t
(c) _.±!_ right to the center of the disc
4+1t
(d) _.±!_ left to the center of the disc
4+ 1t
27. A rocket of mass 4000 kg is setfor vertical firing. How
much gas must·b~ ejected per second so that the rocket
may have initial upwards acceleration of magnitude
19.6 m/s 2 ? [Exhaust speed of fuel = 980 m/s]
(b) 60 kg s- 1
(a) 240 kg ss1
1
(c) 120 kg s(d) None
28. Select the graph(s) which best represent the graph of
bouncing ball. Assume ball dropped from height and it
y~;;: ffi-~·~-
l=f
(1) ' ~
1
I
.2
~
~ac~ im_p_acr
,
: : •
•
•
:
time
'
I _ ~-·-" __ _
(3)
;t
time~
(a) l, 2, 3, 4
(c) 3, 2.
,
__
' . ~ ~ -:
:i
-·- - -
'
----l
'E
'
(2) ' ~ •.••......... I
time__,.
f
: fr
'
, 'i5
______ ----·
!,gF--
(4) _; ............ II
'8
•u
._a:i,.
~time
~
_1
•
!
'
----- -
(b) 1, 2, 4
(d) 3, 4
·r --- - .
29. A ball of mass m falls
j
vertically from a height h and •, ~
collides with a block of equal : µk= 0,2 /
mass m moving horizontally , .....:--~ v / '
I
m
with a velocity v on a surface. •illi/77777771!~7
:
The coefficient of kinetic ,. - - .---- I
friction between the block and the surface is 0.2, while
the coefficient of restitution e between the ball and the
www.puucho.com
'7n
Anurag Mishra Mechanics 1 with www.puucho.com
block is 0.5. There is no friction acting between the
ball and the block. The velocity of the block decreases
by:
(a) 0 ·
(b) 0.1~2.gh
. (c) o.3-fiih
(d) can't be said
30. A particle of mass m; ·ihitially
at rest, is acted upon by a
variable force F for a brief
interval of. time T. It begins
to move with a velocity u
o<....~"'um_e_.~Ti-...
after theforce stops acting. F
is· shown in the graph as ·a
function of time. The curve is a semicircle, then
.
itR2
~T2
0
(a) u = 2m
(b)u=Sm
(~) u = ~oT
(d) u = FoT
4m
. 2m
31. A particle strikes a horizontal
frictionless floor with a speed u, at
an angle 0 with the vertical, and
rebounds .with a speed v, at· an
angle cj> with the vertical. The
coefficient of restitution between the particle· and the
· floor is e. The angle cj> is equal to:
(a) 0
(b) tan-1 [e tan0]
(c) tan-1[~tan0]
(d)(l+'e)0
32. Two masses A and B
of mass M and 2M
respectively
are
-j
.'
- '
connected
by
a
~
I'\" , ~
compressed
ideal
' '~- '
·spring .. The system is
placed on a horizontal
frictionless table and
given a velocity uti in
the z-directicin as shown in the figure. The spring is
then released. In the subsequent motion the line from
B to A always points along the i unit vector. At some
instant of time mass B has a X'component of velocity as
·L~·· .
v xi The velocity
(a)
v,i+uti
(c) -2vxi+uk
v
A
·
of as A at that instant is:
(b) .:.v,f+uii
(d)2v,I+uk
33. A particle A of mass 100 g moving along +ve x-axis
with 10 m/sec, collides at origin, with particle B ·of
mass 200 gm moving along +ve y-axis with 10 m/sec.
After collision the particle B moves along line
4x.- 3y = 0 with speed 5.m/sec. The equation of line
along which A moves _after collision.
(b) 3y-x=O
(a) y-3x=O
(c) 4y ~· 3x :i O
(d) None
3
34. An open water tight railway wagon of mass 5 x. 10 kg
coasts at an ihitial velocity 1.2 m/s without friction on
a railway track. .Rain drops fall vertically downwards
into the wagon. The velocity of the wagon after it has
collected 10 3 kg of water will be:
(a) 0.5 m/S
(b) 2 m/S
(c) 1 m/s
(d) 1.5 m/s
35. Two identical balls A and B lie
on a smooth horizontal
whjch
gradually
surface,
merges into a curve to a height
3.. 2 m. Ball A is given a velocity 10 m/sec to collide
head on with ball B, which then takes up the curved
path. The minimum coefficient of Restitution 'C for
the collision between A and B, in order that B reaches
the highest point C of curve. (g = 10 m/ sec 2 ) ·
-,-.. -.~. .}.=-i
b_0:+B ·/23:3
(a)
.!.
(b) ~
(c)
.!.
(d)
2
4
5
~
4
36. if collision takes place between 2 particles then which
of the following statement is/are true:
{a) kinetic energy is conser:ved during collision
(b) momentum is conserved during collision .
(c) momentum is conserved only before and after
collision
(d) conservation of momentum during cqllisio,n
depends on the type of collision
·
· ··
3 7. On a smooth horizontal plane, a uniform string of
mass M and length Lis lying in the state of rest. A man
of the same mass M is standing next to one end, of the
string. Now, the man starts collecting the ; string.
Finally the man collects all the string and puts it in his
pocket. What is the displacement of the man with
respect to earth in the process of collection?·
.· ·. · ·. .·. ·.·. . . t I
ii/: I I I 1. I I I I I I I I I I I I: I I I I I_, I I ii
(a) -
(b) ~
L
(c) 8
(d) none
L
2.
4
I
38. In the figure shown surface
r;;;,?.· = =_=. '.: m.,
is frictionless and spring is _ ~ ~ .
in natural condition. If
x 1, x 2 and x 3 are the maximum compression in spring
for elastic, completely inelastic and inelastic (e = 0.5)
respectively then:
(b) X2 > X3 > X1
(a) x 1 > X2 > X3
(d) X2 > X1 > X3
(c) XI > X3 > X2
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
IMPUISE AND MPMENTUM
,,
39. In
a
smooth
stationary cart. of
length d, a small
block is projected
along it's length with
velocity V towards
front, Coefficient of
V
restitution for each
collision is e, The cart
rests on a smooth
____,, __ ..
ground and can move
.freely. The time taken by block to come to rest w,r, t,
cart is :
(a)
ed
(b)
ed
~-J
~1--y----·---
(1 + e)v
(l- e)v
-.----1
43. In the · anangements shown in
V
£
J
figure masses of each ball is 1 kg
1kg ·,
and mass of trolley is 4 kg, In the
kg
'
figure shell of mass 1 kg moving
horizontally with velocity v = 6
m/sec collides with the ball and get stuck to it then its
· maximum deflection of the thread (length 1.5 m) with
vertical is:
(a) 53°
(b) 37°
(c) 30°
(d) 60°
44. Centre of mass of two thin uniform rods of same
length but made up of different materials and kept as
shown, can be, if the meeting point is the origin of
coordinates:
d
(c) -
(d) infinite
e
40. In the shown figure both blocks are
in equilibrium m = 1 kg, a bullet of
mass m moves with velocity 10 m/s
and get embedded into the block A,
then just after collision:
. -1 :•
[ll~i
•
L
I.
t•o;,:10
1
L__rt)_,...;c,_J
Vo
( a) vA =-,Va= 0
2
I
(b)vA=va=~
2
(d) VA =v.
I
'+==========~I X
.__,_ _ _ _ _L_--': :.__ ___J
= Vo
3
(!:2'2!:)
(!:3'3!:)
(a)
41. In the previous question the string will be tight again
after 't' seconds from collision then t =
(a)
I
..
.! sec
2
(b) 1 sec
(c) 2 sec
(c)
I
(d) string .will never be tight again
42. The inclined surfaces of two movable wedges of same
mass M are smoothly conjugated with the horizontal
plane as shown in figure, A washer of mass m slides
down the left wedge from· a height h. To what
maximum height will the washer rise along the right
wedge? Neglect friction.
r·
"'
I (. ·..
45. A is a fixed point at a height h
above a perfectly inelastic
A
h.
. •
smooth surface. A light
extensible string of length
l (l h) has one end connected
to A and other to a heavy particle as shown in figure.
The particle is held at the level of A with the string
tight and released from rest, The height above the
plane, where particle is again instantaneously at rest is
, S, then which is incorrect?
(a) velocity of particle on the surface is = ~2,gh cos8
(b) vel~ of particle when it leaves surface
= -J2,gh cos 2 8
hs
, "" n)I,·
>
(c) S = -
z4
h
(a)
(M
(c)
(b)
+ m) 2
h(~)
M+m
hM
(M+m) 2
2
(d)h(~)
.M+m
(d) None of these
46. A hemisphere of mass 3m and radius R is free to slide
with its base on a smooth horizontal table. A particle
of mass m is placed on the top of the hemisphere. If
particle is displaced with a negligible velocity, then
find the angular velocity of the particle relative to the
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
centre of the hemisphere at an anguT.ar displacement 0,
when velocity of hemisphere is v. ·
(a) ~
--~,-r-:;,r
Rcose
1
-.
~
(b)~
Rcose
5v ·
Cc)
Rcose
.(d)
-1
~
Rcose
47. 1\vo balls with masses in the ratio of 1 : 2 moving in
opposite direction have a head-on elastic collision, .If
their velocities before impact were in the ratio of 3 : 1,
then velocities after impact will have the ratio:
(a) 5 : 3 .
(b) 7 : 5
(c) 4 : 5
(d) 2 : 3
48. A projectile is fired at a speed of 100 m/s at an apgle of
37° ·above the horizontal. At the highest point, the
projectile breaks into two parts of mass ratio 1 : 3, the
smaller coming to rest. Then· the distance between
launching point and the point where the heavier piece
lands:
· (b) 960 m
(a) 480 ni
(d) 640 m
(c) 1120 m
49. 1\vo identical spheres A and B lie on a smooth
horizontal circular groove at opposite ends of a
.diameter: A is. projected along the groove and at the
end of .tinie t, impinges on B. If e is the coefficient of
·' restitution, the second impact will occurs after a time:
' . (a) 2t
(b) !.
, . . e;.
(c) 7tt .
(ii) The kinetic energy of a particle, is independent of
the frame of reference ·, '
(a) Both (i) and (ii) are true·
(b) (i) is true but (ii) -is false
(cj (i) is false but (ii) is true
(d) both (i) and (ii) are false
53. ABC is a part of ring having radius - . · B ,;:]~
R 2 and ADC is a part of disc having : ~
!, ~.;
2
inner radius R1 and outer R2. Part A ....... :, ..... _.•
ABC and ADC have same mass.
R
.e
.
(d) 21tt.
e
e
50. 1\vo blocks of masse·s 10 kg and 4 kg are connected by
a string of negligible mass and placed at a frictionless
h~rizontal surface. An impulse gives a velocity of 14
m/s to the heavier block in the direction of t lighter
block. The velocity of the center of mass is: ·
·ca) 30 m/s
(b) 20 m/s
(c) 10 m/s
·. · (d) 5 m/s
51. A stationary pulley carries a ropes one end of which
supports a ladder with a man and the other a counter
weight of mass M.The man of mass m· climbs 'up a
distance 1c w.r.t. the ladder ·and then stops. The
displacement of the centre of mass of this system is:
(a) ~
(b) ml
M+m
2M
(c)
ml
(d)
ml .
M+2m
2M+m
52. Consider the following two ~tatements:
(i) The linear momentu)Il of a particle is independent
of the frame of reference
r~:
cj
:;::n':!tass will be located,
(a) (R2 -R1)C2R1 +Rz) (above).
31t(R1 +R2)
·
(b) (R2 -R1lC2R1 +R2) (beiow)
3it(R1 +R2 )
2R1 +R 2
(above)
(c) ·
3it
(d) 2Ri + Rz (below)
'3it
54.
.·
__~ J
,__
.
, •• ?.
,,,_,,""'-""'""':""
P}vr:,;~mis. :. ~,~!9~~1
~dtv~l~ci:~ of:~ti:~
particles are as shown in l,1kgn · ~ ·~ A1kg:
the figure. They are kept
on a smooth surface and tc...::,.· • ·
' · . ':::,_,;j
being mutually attracted by gravitational force. Then
position of .the center of mass at t = 2 sec:
(a) X;, Sm
(b) X = 7m
· ·
(c) X = 3m
(d) X = 2tn
.
55. 1\vo identical ba1ls are dropped from the same height
onto a hard surface, the second ball being released
exactly when the first ball collides with the surf~ce. If
the first ball.has made two more collisions by the time
the second one collides; Then the- coefficient of
restitution· between the ball and the surface satisfies:
(a) e > 0.5
(b) e = 0.5 .
(c) e = -.J3 - l ·
(d) e ;< -.J3 - l
2
2_
56. A bullet of mass 20 g traveling
'.lli
horizontally with a speed of 500
,--,;,"'" ·
m/s passes' through a wooden·
.i\\W\\\\\\\\\\\\\\\\1.\1.\.,_\}~~l
block of mass 10 kg initially at
rest on a level -surface. The bullet emerges with a
speed of ,100 m/s and the block slides 20 cm on the
surface before coming to rest. Then coefficient of
friction between the block and the surface is:
(a) 0.8
(b) 0.16
(c) 0.32
(d) 0.24
www.puucho.com
t \~ ~!\: :1
E:&n,Ji~. c':l:Jll
Anurag Mishra Mechanics 1 with www.puucho.com
~-~-~,~= .,
Initially spring is at it's ![iii}+•o~I
_·_IM_PU_LS_E_AN_._D_M_OM_E_N_TU_M_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
5 7. A gun is mounted on a railroad car. The mass of the
car, the gun, the shells and the operator is SO·m where
m is the mass of the one shell. If the muzzle velocity of
shell is 200 m/s, what is recoil speed of car after
second shot?
200
(a)
m/s
(b) 200
+
m/s
49
48 48
(J_ J_)
(J_48 + J_)
m/s
49
m/S
(d) 200 (J_ +
48 48x49
63.
natural length and collision • ~ ~
- ·"'· ·- . -~
is elastic. Then find maximum compression of spring
during motion:
(a) ~ v 0
(c)
(c) 200
l
)
58. A man of mass 60 kg can throw a stone of mass 1 kg up
to a height 5 m. If he is standing on ice skates of
negligible mass, the maximum velocity that he can
generate in same stone if he throws it with same force
in the horizontal direction:
(a) Vmax = 9.9 m/s
(b) Vmax = 12 m/s
(c) Vmax =7 mis
(d) Vmax =10 m/S
59; The density of a linear rod of length L varies are
p = A + Bx where x is the distance from the left end ..
Then, the position of the center of mass from the left
end is: 1
2
2AL·+ 3BL2
(b) AL+ 3BL
(a) 3(2A + BL)
(2A +BL)
2
(c) 3AL + 2BL
(2A +BL)
2
(d) 3AL + 2BL
3(2A +BL)
-· ----·1
60. Three identical balls each of j
+vo=9
I
mass 5 kg are connected with
I
each other as shown in figure,
'~--~
cI
and rests over a smooth
horizontal table at moment t = 0 ball B is given
velocity 9 m/sec then velocity of A in direction of
velocity of B just before collision is:
(a) 9m/sec
(b) zero
(c) 3m/sec
(d) 6m/sec
61. In above question 60 the velocity with which A collides
with C is:
'
(a) 6m/sec
(b) 9m/sec
o--------o----o i
(c) 3m/sec
(d) 2-m/sec
(b) µmt-J2gh
(d) µm(gt +-J2gh)
0
64. In the shown figure, if all the
surfaces are smooth and the two
masses are allowed to move then
centre of mass of the system will
move:
(a) upwards
(b) downwards ·
(c) leftwards
(d) rightwards
65. Four identical rods of mass M 1---· .. "' ·---·1
and length L are placed on one ,
!
another on the table so as to i·
j'
produce
the
maximum r ·
.
--- ~
overhang as shown in figure. '---~
The maximum ·total overhang will be:
(b) 24L
(a) 3L
4
25
(c) 25L
(d) 4L
24
3
66. A 20 g 'bullet passes through a
plate of-mass 1 kg and finally
~
comes to rest inside another : 20 r;' "" ·-··· · ·
plate of mass 2980 g. It makes
1 kg
2980 g
the plates move from rest to '--- - - ----- same velocity. The percentage loss in velocity of bullet
between the plates is:
(a) 0
(b) 50%
(c) 75%
(d) 25%
67. 1\vo particles of mass m1 and m 2 in projectile motion
,
.=-~-· .m·~
I"
have velocities
v
1
<
v
2
respectively at time t
They collide at time t 0 • Their velocities become
v at time 2t
2
0
= 0.
v1 and
w~still moving in air. _The value of
--+--+
--+
--+\.IS:.
\(m, V1+m2V2)-(m,
V1+m2V2)
2
62. A box is put on a scale which is adjusted to read zero ,
when box is empty. A stream of pebbles is poured into
the box from a height h above it's bottom at a rate µ
pebbles/sec. each pebble has a mass m. Consider the
collision between pebble and box to be completely
inelastic. Find the re'ading of scale after t seconds of
falling of pebbles.
·
(a) µtmg
(c) µm(gt--J2gh)
/mv
V2k
(a) zero
(b) (rri, +.m 2 )gt 0
(c) 2(m1 + m 2 )gt 0
(d) :! (m 1 + m 2 )gt 0
2
68. A small disc of mass m slides
:~:
I
down a smooth hill of height h i
from rest and gets onto a plank ih
• i
M
I
of mass M lying on the I)
horizontal plane at the hill. Due
to friction between the disc and the plank the disc
slows down and after a certain moment, moves in one
www.puucho.com
__________,
Anurag Mishra Mechanics 1 with www.puucho.com
[390 ,.·
. MECHANICS,!
. pi,;ce with the plank. Then the work performed by the
friction force in this process is:
(a) zero
mM
(b) - - g h
M
(c) -gh
(d)--gh
m
m+M
mM
M-m
69. In the figure one fourth part of a
uniform disc 9f radius R is shown.
The distance of the centre of mass
of this object from centre 'O' is :
(a) 4R
(b) 2R
3it
3it
-./2. 4R
-./2. 2R
(c)
3it
(d)
(a) 3(it - l)a
(c) 3(it -1)
3it
~
~ .
(b) (it - l)a
6
4
(c) SR ·
(d) 12 cm
(a) (0, O)
(b) (0, R)
(o,~Rr.
(d) none of these
7 4. A truck moving on horizontal road towards east with
velocity 20 ms-1 collides elastically with a light ball
moving with velocity 25 ms- 1 along west. The velocity
of the ball just after collision:
(d) 3(/+ 1)
76. A particle of mass 3 m is projected from the ground at
some ang\e with horizontal. The horizontal range is R .
At the highest point of its path it breaks into two pieces
m and 2m. The smaller mass comes to rest and larger
mass finally falls at a distance x from the ,point of
projection where x is equal to :
(a) 3R
71. When.the momentum of a body increases by 100%, its
K.E. increases by :
(a) 400%
(b) 100%
(c) 300%
(d) none
72. A ·small' sphere is moving at a constant speed in a
vertical circle. Below is a list of quantities that could be
used to describe some aspect of the motion of the
sphere.
I - kinetic energy
II - gravitational potential energy
III - momentum
Which of these quantities will change as this sphere
moves around the circle? ·
(a) I and II only
(b) I and III only
(c) Ill only
(d) II and III only
73. From a uniform disc of radius R,
an equilateral triangle of side
../3 R is cut, as shown in the
figure. The new position of
·
centre of mass is :.
(c)
(a) 65 ms-1 towards east
(b) 25 ms-1 towards west
(c) 65 ms-1 towards west
(d) 20 ins-1 towards east
75. From a circle of radius a, an isosceles right angled
triangle with the hypote.nuse as the diameter of the
circle is .removed. The distance of the centre of gravity
. of the remaining position from the centre of the circle
is :
a
70. In the figure shown a hole of
radius 2 cm is made in a
semicircular disc of radius 6it
at a distance 8 cm from the
centre C of the disc. The ·
distance of the centre of mass
of this system from point C is :
(a). 4 cm
(b) 8 cm
(c) 6 cm
I
(b) 3R
2
(d) 3R
4
77. 1\vo balls A and B having masses 1 kg and 2 kg,
moving with spe~ds 21 m/s and 4 mis respec~vely in
opposite direction, collide head on. After colli/,ion A
moves with a speed of 1 m/s in· the same direction,
then the coefficient of restitution is :
(a) 0.1
(b} 0.2
(c) 0.4
(d) none
,.
78. 1\vo massless string of
length 5 m hang from
the ceiling very near to
each other as shown in
. the figure. 1\vo balls A
and B of masses 0.25 kg
and 0.5 kg are attabched
:.,o. AS .;;___f'
to the string. The a11 A
•.'-~-~'-. , 8
is released from rest at
'
a height 0.45 m as shown in the figure. The collision
between two balls is completely elastic. Immediately
after the collision, the kinetic energy of ball B is 1 J.
The velocity of ball A just after the collision is:
(a) 5 ms-1 to the right
(b) 5 ms-1 to the left
(c) 1 ·ms-1 to the right
(d) ·1 ms-1 to the left ·
ldml
79. An ice block is melting at; constant rate
= µ. Its
.
dt
initial mass is m0 and it is moving with velocity on a
frictionless hori.zontal surface. The distance travelled
by it till it melts completely is :
•
www.puucho.com
Anurag Mishra Mechanics 1 with www.puucho.com
~
(a) 2m 0v
µ
(c) mov
(a)
83. The system of tbe wedge
and tbe block connected by
a massless spring as shown
in tbe figure is released
:with tbe spring in its
M
natural lengtb. Friction is f_1:__ _....:..c~L:,,~I
absent
-maximum
'elongation in tbe spring will be:
5k
(c) 4Mg
5k
.
'
'
2
80. A ball strikes a smootb horizontal ground at an angle
of 45° with tbe vertical. What cannot be tbe possible
angle of its velocity witb tbe vertical after tbe
collision? (Assume e S: 1)
(a) 45°
(b} 30°
(c) 53°
(d) 60°
81. Two identical balls A
and B are released from
tbe positions shown in
figure. They collide
elastically on horizontal
portion MN. All surfaces
are· smootb. The ratio of
heights attained by A and B after collision will be:
(neglect energy loss at M and N)
(a) 1 : 4
(b) 2 : 1
(c) 4 : 13
(d) 2 : 5
82. As shown in tbe figure a
body of mass m moving
vertically witb speed 3
m/s hits a smootb fixed
inclined plane and
rebounds
witb
a
velocity v f
in the
horizontal direction. If
L. of inclined is 30°, tbe velocity v f will be:
(b) -.J3 m/s
(a) 3 m/s
'
1
(d) this is not possible
,(C) -.J3 m/S
(a) 3Mg
.
after it hits tbe wall is :
(d) can't be said
2µ
:.,~
sphere and tbe wall is e =.!. The velocity oftbe sphere
(b) mov
µ
.,,
(b) 6Mg .
5k
(d) 8Mg
5k
i- j
(c) -i-J
(d)
21-J
3
relative to ground is :
(b) ~
(a) 3L
4
(c) 4L
5
4
(d)
~
3
86. Two particles of equal ma~s haye velocities 2ims-1
and 2j ms-1 • First particle has an acceleration·
ci + ms-2 while tbe acceleration of tbe second
. particle is zero. The centre of mass of tbe two particles
moves in?
(a) circle
(b) parabola
(c) ellipse
(d) straight line
87. A man weighing 80 kg is standing at tbe centre of a flat
boat and he is 20 m from tbe shore. He walks 8 m on
tbe boat towards tbe shore and tben halts. The boat
weight 200 kg. How far is he from tbe shore at tbe end
of tbis time?
(a) 11.2 m
(b) 13.8 m
... (d).1-'i~'!:_m
(c) 14.3 _m
88. · A sphere strikes a wall and rebounds witb coefficient
of restitution !. If it rebounds witb a velocity of 0.1
J)
3
m/sec at an angle of 60° to the normal to tbe wall, tbe
loss of kinetic energy ls:
(a) 50%
(b) 33,! %
(c) 40%
(d) 66~ %
3.
3
89. A spaceship of speed. vi, tra~elling ,aloni;:)t,y axis
suddenly shots out one fourtb of its !ffii't ~ speed
2v 0 along +x-axis. ;,;y .axes are fixed witb respect to
ground. The velocity of tbe remaining part is:
,
3
before it hit a vertical wall. The wall is parallel to
vector J and coefficient of restitution between tbe
-i + 2j
85. A man of mass M stands at one end of a plank of
lengtb L which lies at rest on a frictionless surface. The ·
man walks to otber end