The mass:
• The mass of the body is a scalar quantity directly proportional to the weight of this body.
In condition that all weights are measured in the same place of the earth and previously its
known as the amount of material contained in the body.
• The mass is denoted by m and it is measured by ton, kg. or gm.
• One ton
=
1000 kilogram.. one kilogram
= 1000 gram,
one gram
=
1000 milligram.
• The mass of the body may change from moment to another moment.
For example :
1. The
mass of a rocket decreases due to the exhaust of burning gasses.
2. The
mass of a raindrop increases during its falling down due to the accumulation of
moisture on its surface.
ExampleO
A rocket of mass 13
i
tons ,it throws off fuel at a constant rate of 150 kg. per second.
Find the mass of the rocket after 50 seconds from the instant of firing.
The mass of thrown out fuel
= ] 50 x 50 = 7500
:. The mass of the rocket after 50 seconds
A falling raindrop
13
=7
i -7
i tons.
i = 6 tons.
is of mass 0.4 gm. at a certain instant.
If vapour accumulates
on its surface during its fall at the rate 5 milligram/sec.
Find the mass of the raindrop
44
=
kg.
after 2 minutes from this instant.
Newton's Laws of motion
The acquired mass = 5 x 120 = 600 milligrams. = 0.6 gram.
:. The total mass = 0.6 + 0.4 = 1 gram.
~ momentum
II Definition
The momentum vector of a particle ,denoted by H is defined as the product of the
(
mass of the particle a~d its ~elocit~ vec~~~._.
..
.__ .__ _
-
..
~
!
..__ __ i
• If the mass of the particle is m ,the velocity vector at an instant = v
...•..
, then the momentum of the particle at this instant is H = m v
• It is clear to us now from the definition that the momentum of a particle at an instant is
a vector having the same direction of its velocity at this instant.
• If the motion is on a straight line ,then each of H and v is parallel to the straight line of
motion , according to this we can express each of these two vectors in terms of the
/\
algebraic measure of them related to a constant unit vector c parallel to this straight line
, then we write
/\.......
/\
i.e. The
/\
:. H c
H=Hc,v=vc
/\
=mvc
Le.H=mv
algebraic measure of the momentum vector = the mass x the algebraic measure
of the velocity vector.
• When the mass is constant then H is proportion to v and the relation between them
becomes linear, that is why the momentum in this case is linear momentum.
We can notice the effect of momentum in many of the surrounding appearences as
an example:
1. If you
hold grain of sand on your hand, you will not feel it , but this particle of tiny
mass can scratch the glass of a rapidly moving car in a sandy storm, this reason is that
the grain of sand had acquired momentum relative to the car (It is equal to the product of
its mass into its velocity relative to the car).
2. If you throw
a big stone on a solid wall it will not penetrate the wall but if you fire
a bullet (although its tiny mass) on the same wall it will be imbedded in the wall ,
the difference here is that the velocity of the bullet is much greater than that of the stone.
45
••. Units of measure of the magnitude of momentum
'.: Momentum H
= m.v
:. The unit of measure momentum
= the unit of measure
mass x the unit of measure
velocity's magnitude as gm.cm./sec. , kg.m./sec. - kg.m zhr.
••. The change in momentum
If HI is the momentum vector of a particle of mass ml and with velocity vI at a certain
~
instant (t1) and H2 is the momentum vector of the same particle after its mass became m2
and its velocity became v 2 ,at the instant (t2) then:
The change in momentum vector ~ H = H2 - HI = m2 v 2 - m 1 vI
If the mass is constant, then:
~
II= m (~
- vI)
I2
~ H = madJ
If a (t) is the acceleration of a moving body then:
• Magnitude of the change in momentum vector
I,
= II H; -
t
HI II
If : v I is the norm of v I -v 2 is the norm of v 2 ,then the change of momentum
,
-.".. ,•.•.•.,~.¥,..<
Example~
A car of mass 2 tons moves with velocity 90 kmJh. Find its momentum in kg.mJsec.
·:H=m.v
8)
:. H = (2 x 1000) x (90 x 15 = 50000 kg.m./sec.
Example
fJ
A ball of mass 15 gm. moves horizontally
with velocity 2.52 kmJhr.
If it hit a racket at rest to rebound with velocity 40 cmJsec.
Find the change in the momentum
46
of the ball.
Newton's Laws of motion
-
Solution
1\
C
First solution using the velocity "ector:
/\
Consider c a unit vector in the direction of
the motion of the ball after collision :
VI
= 2.52 x
250
9
= 70 cm./sec.
/\
'.: vI is in the opposite direction of the unit vector c
/\
:.
VI
=-70c
/\
/\
.: v 2 is in the direction of the unit vector c
:.v2=40c
:. The change in the momentum vector of the ball
~
~
~
~
(/\
/\
= H2 - HI = m (v2 - vI) = 15 40 c - (- 70 c)
) = 15 x 110 c = 1650 c
/\
.', The magnitude of the change in momentum of the ball = II
/\
H;"- H711 = 1650 gm.cm .!sec.
Second solution using the algebraic measure of the velocity:
/\
Consider c as a unit vector in the direction of the motion of the ball after hitting the racket
:. vI (the algebraic measure of the velocity before hitting the racket) = - 70 cm.!sec.
, v 2 (the algebraic measure of the velocity after hitting the racket) = 40 cm./sec.
:. The algebraic measure of the change in the momentum
= m (v2
-
vI) = 15 (40 - (-70») = 1650 gm.cm./sec.
:. The magnitude of the change of momentum of the ball = ] 650 gm.cm./sec.
Third solution using the norm of the velocity vector:
'.: VI
= 70 cm./sec. ,v2 = 40 cm./sec .
.: v I and v2 are in two opposite directions
.'. The change of the momentum of the ball = m (v2 + VI)
= 15 (40 + 70) = 1650 gm.cm.!sec.
:. The magnitude of change of the momentum = 1650 gm.cm./sec.
Example~
A body of mass 20 kg. , moves in a straight line such that
X = (2 t2 -
3 t + 5) ~ , where
/\
unit vector parallels
-.c isisametre.
the direction of motion of the body. If the unit of measuring
X
Find the magnitude of the momentum of the body at the beginning of motion and after
4 seconds from the beginning of motion.
47
~
dX
/\
:. v=crt=(4t-3)c
=0
, v
/\
=-
• At the beginning of motion
:. t
3c
,.: H = m v
:. H = 20 x (- 3 c) = - 60 c
/\
/\
:. The magnitude of the momentum = 60 kg.m./sec.
/\
f!
At t = 4 sec.
/\
, then v = (4 x 4 - 3) c = 13 c
/\
, .: H = m v
/\
:. H = 20 x (13 c) = 260 c
:. The magnitude of the momentum = 260 kg.m./sec.
Example~
A body of mass 10 kg. , moves in a straight line such that its acceleration (a) is given as
a function of time (t) by the relation a
= 6 t2 -14 t
where a is measured in mJsec: ,time
in second. Calculate the change of the momentum of the body in the time interval [1 ,5]
.: ~ H = m
11
(2 ad
t = 10 lS5 (6 t2
-
14 t) d t
5
= 10 [2 t3 - 7 t2] 1 = 10 [(250 - 175) - (2 - 7)] = 800 kg.m./sec.
Example6
A body of variable mass moves in a straight line and its mass at any instant (t) is given
~ 1
/\
/\
by m (5 t + 2) gm. and its displacement vector is given by s 2" (t2 - 8 t) c wher-:..c
=
=
is a unit vector parallels the direction of motion of the body and the magnitude of s in
metre -find :
1. The momentum of the body at any instant (t).
2. The change of momentum during the time interval [2 , 6].
~
'.: s
1
= 2"
-
(t2
~
L H= m v
/\
-
= [(5 t + 2) (t =(5t2-18t-8)~
-.. ~H = [5 t2 - 18 t - 8] ~
= 88 gm. m./sec.
48
:. v
8 t) c
/\
4)] c
= (t -
/\
4) c
Newton's Laws of motion
rrbm
A body fell vertically down
, its momentum
rest ,froID.'a>'lieight;16:9'metre:abov-ellie'gfortnd
became 5460 gm.mJsec. before hltting.the ground pb·~tly
:
Calculate the mass of the body. If this, body rebounded
_;
.:}
':.'
c~ _
-".
',;
-
:
_,"
_"
.;
i'; i '
after hitting t~~grou~d
r
_
i
I.
.-
'.
~- _
j •.••..
>
_•.
:
..
";"
-:
,.
i
I
r
,,"
~!
_~- ~ 'F~
vertically upwards at a height 4.9 m. Calculate the magnitude of the change of the
momentum of the body due to this iriipaci. "'- '
",' " ," -
Solution
/\
Consider c a unit vector in the vertical direction upwards.
• Calculate the velocity at which the body reaches to the ground (v 1)
.: v2 = v~ + 2 gs
:. vi = 0 + 2 x 9.8 x 16.9 = 331.24
.:
= 18.2 m./sec.
HI = m vI where
:.
-r-
:. vI
HI
' .. , ".,
18.2 c
/\
=-
5460 c and it is in the direction of v I
.:
5460 = m x - 18.2
/\
=-
:. vI
'.;.
".
'"
m'= 300
.. i- ..
,
,.>._.
';,
gin:' "; '"
.'
"
• ..':
: : "
,'.
~•
• ~
•~ •- .'
!
'j
)":.:
."
!,
•
-<
.:
.'
(
• Calculate the velocity at which ~he bqdy' rebounds; (v2)
2
2
"
= Vo - 2 gs
v2 = 9$m.lsec.
, ,': v
:.
....
:.0
-..
2";'
= v2 -:-,2x 9.8
..
,";:A'--
:. v~~~}
..,,'
x4.9
~
~.,
,,""
" .
:. H2 = 300 x 9.8 = 2940 gm.m./sec. ,
,
,
:".t'.·
'
".,
:. The algebraic measure of the change of the. momentum' '
.." ..: ',' <>,
i
.'" r:
= H2 - HI = 2940 + 5460 = 8400
:..... -:
!",). ': '.:
-" ._;~ ; _', .
,
:. The magnitude of the change of the momentum
;.1"
.:
'.
,"
.J
'':.'
.'
",
r
:: r;
(.~-~:r.:
.:"!~ -S'-,::'-;j;)
= 8400 gm.cm./sec.
j"'C:~~::
'~i."_·
~~i
Example~
A sphere of mass ~ kg. is projected vertically upwards with velocity 7 mJsec.
from a point below the ceiling of the room at a distance (1.6) metre.
It impinges the ceiling, then it rebounded
downwards.
If the magnitude of change in the magnitude'nfits
fuoniehtuit{dti~'tihhlsTimpah
with the ceiling equals 2400 gm.mJsec.'"
Calculate the velocity of rebounding
Solution
.
.•••
~
,
1\
. .'
.-,
.-
I',
.-,
~
ii,,<
,::,':,
of the sphere.
- '..
-~"
"
-'. ~ ~.-,~
,
.
-,.:
"
~.'
a"!\
.•
~ .
- -
~ . '.
-'
.-.. -,'
. --:-.
1."_
-.
.,
.
Consider c a uhlf'veaor in theveftica1<alr~2tron d6wnw'ards: ,', "','. ",
• Calculate v'i(the velocity of the sphere before impact directl~) .
.: vi = v~ - 2 gs
',<"
:. vi = (7)2 - 2 x 9Jf ~ '1'.6;;
1+.64
. ,":
;
/\
:. VI :::: 4.2
m.lsec.
:.
VI ::::~
4.2 c
.: The magnitude of the change of momentum = I H2 - HI I = I m (v 2 - vI)
:.2400=
~
x
103
I
[v2-(-4.2)]
where V 2 is the algebraic measure of the velocity of rebounding
:.
V2
+ 4.2 = 6.4
Example
:. v 2 = 2.2 m.lsec.
lI!J
A bullet of mass 120 gm, , is fired with velocity 390 mJsec. towards a wooden target
of mass 3 kg. at rest, to stay inside it ,then the system moved afterwards with
a certain velocity. If the momentum does not change due to collision.
Find the velocity of the system after collision.
Solution
/\
-
Consider c a unit vector in the direction of the motion of the bullet
,\
C
/\
:. VI
= 390 c ,ml = 120 gm .
.: m2 (the mass of the system) = 120 + 3000 = 3120 gm.
'.: The momentum does not change due to collision
ffi)
=120gm.
ffi2=3120gm.
/\
/\
:. 120 x 390 c = 3120
X
v2
:. Yz
= 15 c
:. The system moves after collision with velocity of magnitude 15 m.lsec.
in the same direction of the bullet.
A bullet of mass 3 kg. is fired with velocity 300 mJsec. from a cannon in the
direction of a target moving with velocity 60 kmJh. to hit it.
Find the magnitude
of the momentum
of the bullet relative to the target if.
1. The target moves away from the cannon.
2. The target moves towards the cannon.
/\
fonsider c a unit vector in the directi0.E.of the motion of the bullet and suppose that
v A is the velocity vector of the bullet , vB is the velocity vector of the target.
'.' vB in the same direction of v A
50
A
C
3kg.
Newton's Laws of motion
•
•
A
I. e. in direction of c
. ~
" ..•...
5"
50"
.. : A 300 c , vB = 60 x 18 c = 3"" c
=
:. vAB(the velocity vector of the bullet relative to the target)
= ~v A - ~v B = 300 "50"
c - -3
850 "
c =-3
C
:. The momentum vector of the bullet relative to the target
H
..•...
850 "
= m v AB= 3 x - 3
c
"
= 850 C
:. The magnitude of the momentum relative to the target
= 850 kg.m./sec.
Second: The target moves towards the cannon:
1\
C
.,' vB is in the opposite direction of vA
,,50 "
.: V
300
c
,
v:
-'-3··
c
A
B
=
=-
~
~
:. vAB(the velocity vector of the bullet relative to the target)
..•.....•...
"(
50 ) " 950 r.
= vA - vB = 300 c - - 3"" c = 3 c
950 "
.', The momentum vector of the bullet relative to the target H = m vAB= 3 x -3c = 950 "c
:. The magnitude of momentum of the bullet relative to the target = 950 kg.m./sec.
A cloud is at rest at a height 1102.5 m. above the earth surface, water drops fall from
it ,the mass of each drop is 0.1125 gm. If the vabour accumulates on the surfaces of
drops of water during its falling with rate 12.5 milligram per second.
Find the momentum of one drop of water at its reaching to the earth surface
measured in gm. cmJsec.
Consider "
c a unit vector in the vertical direction downwards
Calculate the velocity at which the water's drop reaches to the
earth surface and the time elapsed for that
.,' v2
= v~ + 2 gs
:. v = 147 m./sec.
.: v
= vo + a t
:. v2
= 0 + 2 x 9.8 x
1102.5
:. v = 147 "
c
:. 147
= 9.8 t
:. t =
-147 =
9.8
15 sec .
51
~----~
-------------
:. The increase in the mass of the water's drop
= the rate
x time
= 12.5 x 15 = 187.5 milligram = 0.1875 gm. _
. -, !
:. The mass of the water's; drop when it reachesthe.earth's.surface
m
= 0.1125
+ 0.l875
= 0.3 gm.
A
A
:. The momentum H = m v = 0.3 x 147 c= 44.lc
:. The magnitude of the momentum
= 44.1 gm.m./see. = 4410 .gm.cm.zsec.
mmr£ID] ...
.
,
A wooden block of mass 5 kg. fell down from a tower of jumping on a swimming
pool. The tower is of height 19.6 metre above the water surface in the pool.
After
it collides
do~nwards
with the water's surface it moved inside the water vertically
with uniform retardation
of magnitude 4 mjsec~ ,then .it covered
a distance 4.5 m. before it starts returning
..
Find the magnitude
,:
back to float on the water.
~
_.
...
of change of momentum
.-
'.
-,
"
of the wooden block due to its impact
with the water surface in the basin of swimming. ~~
Solution
. 5·kg.;
;;...
'.
.
~ ~
. ..
;
.'
Consider ca unit vector in the vertical direction downwards
U=O··j
~.
• With respect to the motion of the wooden block before
hitting the water's surface
.: v2 = v~ + 2 gs
--
:. VI
:. vi = 0 + 2 x 9.8 x 19.6 = 384.l6
= 19.6 e
A
• With respect to the motion of the wooden block after hitting the water surface and
moving inside water
.: v2
= v~ + 2 as
:. 0
:. v~
= 36
:. v2 = 6 e
= v~ -
2 x 4 x 4.5
1\
:. The, change of the mon;t~~~u~~e~t()~ = I:Iz~, HI ~,m,>v,2:-;-IT,l Yl
-
•.•••
I'
I
~
-
'.
[-'
,"
• ,
,
.'
~
t
;
:. The magnitude of the change of momentum
.i;'.=
= rn,.(v2 "- ';'1);
-
A
~
A
5 (6·~:-19,.69)'7
'.
=·1- ~SJ..::;,68kg.rn./sec
-.
~
"
A
- ..68 .c,
..
t.,
,': ',::' ,.
.. . :-:-,
!
D
\\~:Eierdse..
QJ
On momentum
Complete the following :
, '
,,'.,
."
( 1) A carof mass 3 tons moves on asttaight,line'withvelocity'72ktn.llT.,;;'lhenthe;'
momentum of the car equals ,
kg.m.lsec.
y.? .:,. i1
r
( 2 ) A force F acted on a body of mass ~) to move it with velocity V for ~ time t , then
"the momentumvector of the body H-~='r.; .. ;.... ::..... .. ., '-'_.
i ; ',.
~-;, .~.~
30 irrt)~~c.,:th~n
( 3 ) A body of ~ass ~'kg.;.moves·with velocity
=
body
j
the in()Il1enf~m'6nhe
gm.cm.zsec.
~;
"
..
>.
"i'
•
) I:
_.
"
'
:.j
_
( 4}A carof mass 18,0.0. kg. .moves with velocity.l 0.0..kmzh. ,then themomentumof
-
car =
-
.
•
-,
"
'.
"
"_
•
~
•.
>
gm.m.ls~c...
_
• • •
. :"
i
",,;'
'....
l
" '. ~.
i ,,"
the
i:
",:'p,'F":~
',c-';(
,i ..•. ~.:.,.,:.
•
A
( 5 ) A rocket of mass 13 tons throws out fuel at a constant rate 9fJ~D kg. p~r second
.
~
'_ i " ~
-
~
'.;.
~
".
-.'
".;'-.
_
, then the mass of the rocket after 40. seconds from the instant of shooting will be
............... tons.
.'
".'
' .. :;,~
( 6) Water v~pq~ condenseson thesurfafe.o{ WC;lte~Ar?p~,wbi.1e.
t~ey,a~~Ja!liHg.~t,fate
2 milligram Isec, the.mass of one. falling ~ctte~.drop was 0.',f; gm. ~t a mOl1)~I)J"then
. .
....
-.
1·····-
the mass of the fallmg water drop after 14~!l!ltes
o
.....
..
. . ....
I,-....
fromthl~ 1l1:9ll}et1t,7";;"~:.:·;">
grams.
Calc)Jlatt(Jh)e J;Il()rnentllIl19f.:QPljIlw;Dqse}massi,&AOtgl).~:m.Q:v<~ngt9,"Xarqs;,tp'~,~orth
with a uniformspeedof
O'
....
a magnitude 72 ~I11.1P.: "'.'"
'.;,
:.;'
2~:~9}S~1kru. Ih,.,
Calculate the momenturnofa carwhosemass is:;800ikgiinovirtg:u:fwardstPle Western
South with a uniform speed of a magnitude 126 km.lh.
,-,
,
"
.,./,,:
,
o Compare between the momentumofacannon.car
i ', "("
•. ' •. l
i
,
.
.
;".'
"""',Y
·r·
'r·.·
H·· )'" "
OEll:lf)(. kg. krn. /hr.:
·\'::1.::·~.:;-;-::-"'··'<."f~
"of,m~ss4. tOljl~ whenit WQ.ves;"Voiti,l
,'.i
!
velocity 2.7 km.lh. and the momentum of a bullet of mass 1 kg. moving with velocity
2400thJsec'.' ':>,0<'::.
' .. [ ,~.;-:;., ':';' "<,j'.. .:
i··~·":?;::«'Hi;h:H:'.='3GD{f.kgjI{lsec.>:,
Ql A body of mass:75granis.
-movesfrom-rest.ina
stl;aighUmewithuniform:!a~~eler.atjon
9 cm.lsec. inthe direction of its motion. Calculate i~smoment~~ after ~inute from
.'. the momentof thebeginning of inciti'on'..\.)ii ':"fG ., ~>:'." ;,.... :'/102'S-grtl'.
cm.zsec.>
i
2
[It A 'body .of mass J 1 gm.. ~moved :t'forrttfes~in ar smight' Hq,e with
'unifor:m 'acceleration
12 cm.lsec~ in the direction of motion. Calculate its':lnQmentumafter coveringdistance
24 metre.
o: 2640 gm.cm.rsec.
QfCalculate
the,momentumofa.stonelofmass25o.
vertically downwards:
.:«:
"
,'.
»
gill. when.it falls a.distanee.a.s.m»:
«2450(')0 gJH;,cm)sec.»
".'
'/
QJ A body
of mass 20 gm. -fell down vertically. What is its momentum after 3 seconds
from the moment of falling.
OJ A cannon
«
5HHOO gm.cm.zsec.
,)
fires 120 bullets per minute. If the mass of the bullet is 12 gm. and its velocity
at the opening of the tube of the cannon is 115 metre ./sec. Find the momentum generated
in the second.
" 2760()O gm.cm.z-e,
.,
tl A body fell down vertically from rest at a height 1.849 metre , its momentum became
3010 gm. cm./sec. at the end point of this distance. Calculate the mass of the body.
"S
gill. "
IijJj A body of mass 1.5 kg. ,is moving at a certain moment with velocity 15 km./h. after
a period of time, it is noticed that the body moves with velocity 11 metre./sec. Find the
magnitude of the change of the momentum of the body if :
( 1 ) The two velocities are in the same direction.
" lfL25 kg.m.'-'cc. "
(2) The two velocities are in opposite directions.
" 22.75 kgm
'Cl:.
mJ A body starts its motion from rest with uniform acceleration
on a straight line.
It covered 27 metres through half a minute. While the change in the momentum
is 5.4 kg.m./sec. Find the mass of this body.
WI A car of mass 2 tons moves from rest with a uniform
" .i
kg ..•
acceleration 150 cm./sec~ Find:
( 1) Its momentum after 10 seconds from the moment of beginning motion.
( 2 ) The change of its momentum during the fifth second.
" 30000
,30U() kg.mzsec .
»
A train wagon of a mass 15 tons moves horizontally with velocity of a magnitude
40 m./sec to collide with a barrier at the end of the railroad - then it rebounds back with
velocity 30 m./sec. Calculate the change of its momentum.
« 1()500()() '''g.ll1.!scc.
.,211 A ping-pong
"
ball of mass 50 gm. is projected horizontally with velocity 500 cm./sec.
It impinges with a bat at rest and it rebounds in the opposite direction with velocity 300 cm./sec.
Find the norm of the change in momentum due to the impact with the bat. '( 40000 gm. cm.rsec.
m!,.,
A rubber body ~f mas~
ioo kg. moves
»
horizontally with velocity 120 cm./se. When
it collides with a vertical wall and rebounds in a perpendicular direction on the wall ,it
loses two thirds of the magnitude of its velocity. Calculate the change of momentum of
the rubber body due to the impact.
« 1600n gm.cm hcc.
.~
»
, A ball of mass 200 gm. moves horizontally at a uniform velocity of magnitude 40 m./sec.
to collide with a vertical wall. If the magnitude of the change of momentum as a result of
impact is 12 kg.m./sec. , calculate the velocity of rebounding the ball.
54
«
20 m.zsec "
Newton's Laws of motion
WI Find the height
at which a body of mass -~kg. should fall so that the magnitude of its
momentum as its impact with the ground is equal to the momentum of a body of mass 70 gm.
moving with velocity of 432 km.lh.
« 14.4 mctrc »
{OJ .'.
A rubber ball of mas 200 gm. is let to fall on a horizontal surface from a height of
90 em. to rebound up to a height of 40 em. Calculate using kg.m.lsec. unit the magnitude
of the change of momentum of the ball as a result of impact.
« J A kg.m i-ce, »
w'J ....
A rubber ball of mass ~ kg. is let to fall from a height of 8.1 meters on horizontal ground
, then it vertically rebounds upwards to a height of 3.6 meters after it impings with the
ground. Calculate the change of the momentum of the ball due to the impact with the ground.
«
10.5 kg.mJsc<.:.»
ID.l A rubber
ball fell vertically from a height of 2.5 m. from the ground. it rebounded to 1.6 m.
height. If the change of momentum is 630 kg.cm.lsec. then calculate its mass.
«
kg. »
+
[ill A ball of mass
125 gm. is projected vertically upwards with velocity 7 m.lsec. from a point
at a distance 1.6 from a ceiling, then it rebounds downwards with velocity 1.8 m.lsec.
Find the magnitude of change of its momentum due to its impact by the ceiling.
" 750 grn. mJscc.
»
£ill ._. From a point under the ceiling of a room of a distance
240 em. ,a ball of mass 40 gm.
is thrown with velocity 980 em/second vertically upwards to collide with the ceiling and
in turn, its momentum changes by a magnitude 0.4 kg.m.lsec. Find the velocity of
rebounding the ball.
«
300 ern.Ace.
»
A body of mass ~ kg. is projected vertically upwards from a point on the ground surface
with velocity 29.4 m.lsec. Calculate the momentum of the body:
( 1 ) After one second.
( 2 ) After two seconds.
( 3 ) When the body becomes at a height of 44.1 metre from the point of projection.
«
9.R kg. rn.rscc. ,4.9 kg. m.rscc. - zero
»
a Choose the correct answer from the given ones:
( 1) The momentum of a bullet whose mass is 100 gm. moving at velocity 240 m.lsec. is
(a) 24 x 10- 3 gm.m.lsec.
(b) 24 kg.m.lsec.
(c) 24 x 104 gm.m.lsec.
(d) 24 x 103 kg.m.lsec.
( 2 ) The momentum of a car whose mass is 2 tons moving in a straight line at velocity
54 km.1h is
.
(a) 108 tons.m.lsec.
(b) 3000 kg.m.lsec.
(c) 30000 kg.m.lsec.
(d) 108000 kg.m.lsec.
.
GlJ' A:bOPY:9f~lmass.:500 gIJ1.~sIet.to fall from a heightof 4.9 meters en ~begrqllnd
it comesthe ground i~
;-:,~llrfac;ei.,.th~l"\.itsmomentumas
.
(a) 2.45 kg.m./sec.(b9r4Lg.kg:m.lsec>_
,n
(c) 2450 kg.m.zsec,
d'~,,:L c; f 'n~-i":-':
lJr~-
~")!,n
r:s ,
J'!
(d) 4.~QQkg.m.zsec.
",>? \_;·"/-.,~,,i~rn~;.'}_~I::,';
L:.I: ,j..:.
!i;
i,l;
c.i(~)~~;fO(?~et{)f m~ss4. t.Qnsjn~luQ.inKt)1efuel i~laul!l,ch~pat yelpci~y ;79() .rp.js~c, ? and
it throws out the fuel at~c()nstallt;r~t~ o{ ~magniWd~Jpo~g.p~,r: ~~p.u4.J.tit~tt
momentum is constant, then the velocity of the rocket after 10 seconds in km./h .
.Liunit is ;';r;'''':;:''''''
;,'
(by600
(~fr80Q
:'rf·. ;'i:.,·<3f~;'J"
:'-_~JC!~t
,.
'(6,800
lit
-:~:;~~,"'-';-
~··(d)960
~
-~,
L;
_~ ,-
( 5 ) A missile
of mass 1 kg. is launched at velocity 720 km./h. towards a tank of mass 50
tons moving towards the mortar at velocity 20 mlsec. ,then
.
,
1
('"
,.~,:_,'
::1:
"
,:-_,;,,'~
'>
'_
,-'
-
"t", _,"., _,:,~
l~"c'
i'
,":
','.
_:_:,
First: 'The magnitude' of the' momentum of the mlssle with respect to the tank is
/
:-~; ,
.,"
",.;
.'~
,"
-
'
,~
-
(a) 200 kg.m./sec.
'I
,,',.'_, ~>~:'~i>:
/)
_,jl')"
_
~_
.
,_
"
\ :
;.
:i~ /~«~~' ,'Ie
Cd) l.TxlOrkg.m./sec.
i
·:);,i~. _~~~_:/;:_: '
",,""j
r':{'" -:_-;!.
:~_';'~i:'l:j-:::
;
,L
,
Second ; Tfi..~IJ}flgnj1\lq~
pf mOlJ\ep~\Unof
thetank, )yitti resPe9tJo the. missile i~
_.,_>.~_~-,,:_.!.,
~ '::,
.-.:-,
•. ~,~'
•.r .• :
'::"':
.
-.
(b) 220 kg.m./sec.
(c)lOJ kg.mJsec,
,t
I ~._
.
->1""
'_,",'.,'
~.
'j';'.".
',__,, __,;
-
,
'r
--i,'_.><,<_,'
-'_,'
'.
-,
-
(a) 200 kg.m./sec.
(b) 220 kg.m./sec.
(c) 107 kg.m./sec.
(d) 1.1 x 107 kg.mlsec.
-,_;
.
""""'0(''''
.
~
WJ.Qjmpletifleach\'ofth~fonoWitig:
( Ii ry A b6dyOf[tha:s~"2 'K~f\rrY6ve~
'on; a';sth\igtrt lit1ef,l the~p6~itiBti\fed6r- ofihe 66dy'at the
i
1i_12,
time t is determined from the relation -; = ( t3 + 1) + (t2
vector
after 2 seconds = ..,....,........
L~Cfij•r~::-:
... '-;:(! Dt) iL:(:";!.~ r-~:F{-;
L:;l
":'~ //:1
.r:',
i
t}r'tha·lthltribihe~tum
-
( 2 ) A body is moving ~here its displacement vector is -;-=
,t
'
j"';,",\f
6t:l lff,8rYarrolU-;-U.is
measured in m~g-~;.j;iJ;ll\~ecggP.r.Ifthe momentum of the b.9Ayj~.3;}}g.P?fs~c.,then
.its mass = ,
gm.
,nou~;<-jL01{~:(
J:'{-;'k'
' s . .-~;
!
•
~,-
~
f"i·~
( 3 ) A body fell down from a height h a bove the ground surface. If the mass of the body
equals 500 gm. and its momentum at the instant of collsion with the ground surface
equals 8400 gm.m./sec. -then h =
metre .
.{ 4 )'l(dPThe 'thortlerittirh i6fa 'bbay
with velocity-of-magnitade
as
rrtass i700gti1.'nloving'ina
sthiightlirte"siarting
m./sec. and with a uniform,a(,:celtfration2.5 :m./sec~ in
(jf
the sam~,;~Iire,cJ;~~nrP.fit~jnjDal
velocity after 12 see .J~Op1~tJ;1ebeginlli9g
of:
motion is
-..
,
~.'
.
-
".)
.w!~Xc.
..
-
.
,
.~qu~qq
',':',':.';'~'.'~:'"
~,g.
.."
.L '~.' ~ ,;_, •._
.."",'_
. '__
r~~~;~ __
~
4:.
,,1
(5)
If a body of mass one kg. moves in a straight line such that !h,e!,a~~~l~rationof
the body isgive~by: tp~Jelation a = 4 t + 2 where a is m~asuredin w..ise~.~
, t in
second ,th~'h'th~.6~ang~· o{momentum of the body in tli~'ti"rri~"i~ter~~i
6] is
1 ~',
1
equal to ,......... "kg:m.lse'C·.
j,..\j.,
~_
.,j
f
t~',
::;_ ~:f _: /
'_)
l
;;.'
.
Newton's Laws of motion
...
.._u ki-'::J.FAcarofmass
m
-
--where C
2: t()rTs)movbsjih:~f~nl"igatHrie·suctFthat.x~<Q3tL:'"'4't +;nC
:,is the unit vector: in.the direetion.ofcthe.mction.of the car ;lfX is measuredin metre, ,!find
the magnitude 'ofthe 'momentutrixlf the car when it starts
to. move. '.'.then after
its motion.
~J
.
Q A body m6~es
rifi a straight
:3.seconds of
<<<::8QoO'.28000kg:mJsec
»
line s\i~h that the acceleration of its motioka' is given as
a function of time t by the relation a = 2 t - 6 where a is measured in m./sec~ unit and
timet in .second. Catculate ,tl;lec~~n~e~?~.~?,merytlJm of the body. in ~~.~.
ti~~ ipt~rval
3 :5t:5 5'if the mass of the body' is 8 kg .. ""
« 32»
,L
,'_.
_;- ~_~
._ ..~_,._ .. ~-
,.-
fllJ [;Q] A'bodyof-mass
~~2~~_~
~_t_,_':"'_~'Y
•.
·
.
,
'
I
.•.
,.
,
.'
..,',
_
48 :gill! ,¥novesili:;r sttaight:Iitle 'sudithat a':i::i(3't''-''i2) htJfsec~;
Calculate the change of momentum in the following time intervals': .
( 1 ) [1 , 3]
~
.)(;2 }'[3~5lc':
:::-2 A car of mas
1.5~ons.:,'iri~v~slri·~;straight
..
., " ..
'«'i~~;;k:g.m.!
sec, zero
Iirie ~uch that a(t)is:'given~y'the
a ~12.t-,-t 2 ~her~ a ,i:~.rn~;asut:edinm/s~c~ unit.andtime tin see, , find:
: ::
(2 }The.change of momentum, of the Garduring .thetimeinterVal,[2
,14]-"
«
~g:.91q~~~iI\astraight
-h.
216 , 360 ton.m.! sec.
»
~n~G,,\>Vp,ereC
line.such t4at;;=·:(~t2;c
"
relation
, ..
( 1) The-change-of mome,B.turm,oLtllecar during the first six seconds,
Wl: ~~ Ab.odY:9f,ma~~)6
»
.
is the, unit vector in the direction parallel
to the motion
.If
the magnitude
of
.,
.
..
,
.' ...
~ S·is in meter
unit ,',t in s~co,ncl r. find t~e, changeof the, mqmentllmJ)fthebody
in the followin~: time
..
-.
,~
'.:
,
"
.
.
.
'.
'
.. ,-
.
,".'
"
-;
intervals:
1 \;
(1)[2,4]
~-T--'7-'
~
'.
128 ,3696 kg.m.! sec.
«
»
~ ........•...
~-,-.
:J"Abqqy
, ••••
~
(2)[5,8]
m~~es ,in a straight.lj~~~;ith a ~mit~rmac~e~efation
.--
•
•
.._
•
•
,
_.
_',
_
"
)
,
•
-',
•
_,
<
•
'.
3111/S~C~and with
~.=;-
_
.'
,"
F
~.'.....,
~
-;.
" .initia] v~loci~ 5, Il}.!~e9~Ifthe massof
the body is, 18 kg. ,.Hnd.the magnitude of the
~~:,-:,.!~.'_.,
change of the momentum in JI;1~f()~I9.wingtime inl~I:vals : i..."
.... ..•.' . -;,',
.l..' .:~,:;
P'.;'
J:!:"_:'-
__
~;
L:~~~'
.s:»::
. j.n
..,..'.....
s ..
' \
"
.}.;'.~
..,~:
( 1 ) [0 ,3]
.s v
~.I:!i .'.'
'J
_
-.::.
~- ...••...
;
...••••
;
i'~
'.J~'
..•.
,<,"'
. "
::.';,
«
..lq~:•.i~'~nr~~~~,~rtiC(llly·~p~~q~.lwithvelo9it)'
.
.-,--;.
:: ...
,.--
.ff;
~~..-
:~~-~~
IT,l~~;
'.:~
rj
~:'~';-;'l
!
,..:;... ..:••.
.:
l.:i
s,
,,~,,~,,\
.•...
:::_:-._-': .. ,:
•
162 ,54 kg,m.! sec.
:, c,,'
(2)[4,8]
»
.; .•.
r.
(3)[7,11]
«-29.4
En,(~ ~}~¥'~f
.'
~8:8nlj~,Cal~palte
t~~':f~~~g~i9f'i~S:Il1o~e~t;~wl~~t~F:,f~il?~~n8jti~~"t?t~~v~ls :",,';,','
(1)[2,5]
'
~.J"~,,,.:
( 2 ) [1 ,2]
ij)j' r.JA;bOdyqf,~~S~.·'l·
.•.
r+ ..
_ -.. __
":"- -:':1"
!! .::
;;.1.
_
sr.»
,-39.2,-39.2kg.m./sec.»
:,",
.. ~". -: -
----:-.~7::-O---,.---
'12Q()'
k'i::ip~{~s~~:~'st;~ght ~iin~:~y~h
that s ';:i 3' ~ .i~'
t'2, ~h~re"S.:is'measured
J ,-;'
:i '
'..•
<.:>.~\
Ll.!'._,'
~~!lf
~.~I
I.,' .....
";'
.•
t .
~:.:
::'.,l., ~ ."'~"'~~/'
I~~ .~:;
{',"
",,~~?
",3
t , :',!; .,
in meter, find the momentumof thecar after ~ sec..from the beginning of themotion. '.
-'.~~l,:~·..:
«
",;r.):,.
-~~Uj~ ';':;.,'.~
'::'
->
i,:,
'.'.,.1
i~~·.;~:";'-~_..;'
)'3f,.
j, .•
';i:\
._' .--,'-:~ '
~ :.,'
'.
57600 kg.m./sec in opposite direction to the direction of beginning motion»
WJ ..."
A body of mass 12 kg. moves in a straight line such that
~
--'"-
--'"-
S is given as a function
of
time t by the relation S = t (6 - t) C where C is the unit vector parallels the direction of
~
the motion. If the magnitude of S is in meter unit and t in second, find the change of
momentum of the body in the following time intervals:
( 1) [1 ,2]
( 2 ) [2 ,5]
( 3 ) [4 ,6]
(, - 2.:1,- T2 ,- 48 kg.m/sec.
t.rlJ , : .,.A body whose mass at any time t in second is equal to
:.straight ~e
1
(t
+
»
52 kg. , moves in
and its displacement at any time t is given in the form S =
i
(t 2 - 4 t
+ 3)
C where C is the unit vector parallels the direction of the motion of the body and the
~
magnitude of S is given in meter:
( 1 ) Find the momentum of the body at any time t.
"
*(
{2 + 3 t - 10)
( 2 ) Find the momentum of the body during the time interval [2 ,5]
ffil ,
m
«
C
6 kg.m/sec.
»
»
A body of variable mass moves in a straight line and its mass at any instant t is
= (4 t + 1) gm. and its displacement
second,
IIS II in em. Find the change
vector is given by the relation
S = (t 2 -
of its momentum in the time interval [3 ,5]
«
WJ A rubber ball of mass
2 t) ; ,t in
116 kg.mJsec.»
100 gm. is let to fall from a height of 40 cm. on a horizontal ground.
If the ball rebounds to one quarter of the height from which it falls after each impact.
Find the magnitude of the change in its momentum. Just before and after the second
impact, measured in units of gm.cm./sec.
«21000 grn ':Il1.1scc."
A body of mass 7 kg. is left to fall vertically downwards. It takes 2 seconds to reach the water
surface. If the body, after the impact with the water, moves in it with a uniform velocity
vertically downwards and covers a distance of 6 metres in 1.5 seconds. Find the magnitude of
the change of its momentum due to the impact with the water.
tnl .,.,A stone
,< JOlJ.2 kgm.rsec.
»
of mass 800 gm. is let to fall from rest for two seconds then it impings with
a surface of a pond and sinks down in a uniform velocity to travel 12 meters is three
seconds. Find the momentum of the stone as a result of impact against water surface.
,,-12.48
fjJJ
»
"A body of mass 90 gm. is let to fall and after three seconds of falling -the body collides
•
with a viscous liquid surface to imbed in it with a uniform velocity to travel 2.2 meters in
half a second. Calculate the change of momentum due to the impact.
58
kg.rnzscc.
«
-2.25 kg.rn.rsec.
»
Newton's Laws of motion
[ill A bullet
of mass 100 gm. ,is fired with velocity 350 m./sec. towards a wooden body of
mass 2 kg. which is at rest. If the 'bullet is imbedded in it and the system moves after that
with a certain velocity. Find the velocity given that the momentum of the system does not
change due to the impact.
"
16
t
III Ise«,
n
[])I A bullet of mass 50 gm. is fired with velocity 810 m./sec. towards a wooden body of
mass 4 kg. which is at rest. If the bullet is imbedded in it and the system moves after that
with a certain velocity. Find this velocity given that the momentum of the system did not
change due to the impact.
«
10
1l1I~el'. "
WJ A fixed cannon fired a projectile
of mass 5 kg. with a velocity of 350 m./sec.
in a horizontal direction towards a tank moving with a velocity of 45 km./h. and it hit it.
Find the absolute value of the momentum of the projectile,
then calculate the magnitude
of the momentum of the projectile relative to the tank if :
( 1 ) The tank is moving towards to the cannon.
"
175() , I
( 2 ) The tank is moving away of the cannon.
IDl A cannon
"
x 12.5 kg.
Ill.l~ec. "
I ()x7.5 kg. m.rxcc . "
fired a projectile of mass 1 kg. with velocity 300 m./sec. towards a tank of
mass one ton moving towards the cannon with velocity 54 km.lh. Find:
a
t~
( 1) The momentum of the projectile relative to the tank.
,. 315
( 2 ) The momentum of the tank relative to the projectile.
.. 3 j :'iOO!)
kg.I1l.!~c<:."
kg.nl.i~l'c.
"
Water vapour condenses on the surface of water drops -while it is falling at rate
10.5 milligram/sec. The mass of one falling water drop is 0.25 gm. Find the momentum
in gm.cm./sec. of one water drop when it reaches the surface of the ground from a height
of 1000 metres.
'.' :'i()OO gl11.1'1n.i~L'l: ,.
A car moves along a straight road with velocity 45 km./hr. Facing a sandy storm in an opposite
direction to that of the car motion with velocity 36 km./hr. Find the momentum of the grain
of the sand relative to the car. Knowing that the mass of a grain of sand is 7.5 milligrams.
7
" lilt
a
glll.llll
.. 'L'l'
"
The dust accumulates on the surface of moving bodies at the rate 14 gm.per minute.
A ball of mass 46 gm. falls from a height of 1440 metres in air saturated with dust.
Find the magnitude of its momentum in units of gm.m./sec. when it reaches the ground.
" X ..U)() g 111.
m./ ~el'. ••
(g}
A particle whose mass is constant is projected vertically upwards with a velocity v0 - write
down the law which gives its velocity in terms of the time, and hence deduce that the rate of
change of the momentum with respect to the time is a constant vector and find its magnitude.
59
Newton's laws of motion
lflIercise ill
a
(1) 60000
(2)mv
(3) 1500000
( 4) 50000000
(5)7
(6)0.35
~
V2= v~ + 2 gs = 0 + 2 x 980 x 184.9
:. v = 602 cmisec.
.: H=mv
:.3010=602m
:. m = 3010 = 5 gm.
602
Q
The momentum
~
of the train
= 40 x 72 = 2880 ton.kmIh
Let ~ be a unit vector in the direction of the first
in the north direction
velocity
• The two velocities are in the same direction
Q
The momentum
:'~1=15Xfs=1j~
of the car
--
:.v2=1l~
--
25
1\
H2-HI=I.5(11
= 800 x 126 = 100800 kgm. kmJhr
C-
6
A)e
=1O.25e
A
in direction of western south
:. The magnitude of the momentum = 10.25 kg.m/sec .
Oi
• The two velocities are in opposite directions.
The velocity v of the cannon's car =
:. HI =
.: H2 =
t x 4000
t
Fa-
x
-& = t misec .
:. HI = 3000 kg.m.!see.
x 2400 = 3000 kg.misec.
= H2
:. HI
(Ii
. ~_25A
...•.._
.. v'-6c,
v2--11c
,--_
/\ 25/\_
A
.. H2 - HI - 1.5 (- 11 c - 6 e) - - 22.75 e
:. The magnitude of the momentum = 22.75 kg.misec.
VlJ
t
.: s = vot +
v = v + at = 0 + 9 x 30 = 270 em.isec.
o
H= ~
:. a = 0.06
v2 = v2o + 2 as = 0 + 2 x 12 x 2400
:. v = 240 cmisee.
:. 27 = 0 +
a[2
:. 27 = 450 a
x 270 = 2025 gm.em.!see.
Q
1\
:. a =
i
t
a x (30)2
m.!sec.2
m.!sec2
.: v = v o + at = 0+ 0.06 x 30
:. v
:. m (1.8 -0) = 5.4
= 1.8 m.!sec.
:. m = 3 kg.
H = 11 x 240 = 2640 gm.em.!see.
Q
WJ
(1)
v2 = v~ + 2 gs = 0 + 2 x 980 x 490
.: v =
Vo
+ at = 150 x 10
= 1500 emisec.
:. V = 980 emisec.
= 15 m.!sec.
H = 250 x 980 = 245000 = 245000 gm.cmisec.
:. The momentum
Q;
v = v0 + gt = 0 + 980 x 3 = 2940 cm.!sec.
:. H = 20 x 2940 = 58800 gm.cmisec.
= 2 x 1000 x 15
= 30000 kg.m.!sec.
(2)
.: v (after 4 seconds) = 150 x 4 = 600 em.!sec.
= 6 misee.
, v (after 5 seconds) = 150 x 5 = 750 cmisec.
The mass of bullets fired in one second
= 120 x 12 = 24 gram
60
:. H = 24 x 11500 = 276000 gm.cm.!sec.
= 7.5 m.!sec.
:. The change in momentum during the fifth second
= 2000 (7.5 - 6) = 3000 kg.m.!sec.
III
0=v~-2x9.8x3.6
The change of momentum = m (v + vJ
:. v 0 = 8.4 mJsec.
= 15000 (30 + 40) = 1050000 kg.misec.
:. The change of momentum
~;
HI 1= 150 (300 + 5(0) 1=40000
gm.cmisee.
ml
• The velocity before collision with the ground directly (v)
III
The rebounded
t = 40 cmisec.
velocity = 120 x
:. The change of momentum = 100 (40 + 120)
= 16000 grn.cmlsec.
m,
.: 12=0.2(4O+v)
y2 = v~ + 2 gs = 0 + 2 x 9.8 x 2.5 = 49
:. v
= 7 mJsec.
• The velocity after rebounding
:. zero = v~ - 2 x 9.8 x 1.6
:. u = 5.6 mJsec.
:. 630000 = m (700 + 560)
WJ
Wi
• The velocity before impact with the ceiling directly (v)
• The second body :
v = 432 x
v2 = v2 _ 2 gs = 49 - 2 x 9.8 x 1.6 = 17.64
fs = 120 mlsec.
H of the second body = 0.07 x 120 = 8.4 kg.mlsec.
• The first body:
H = 8.4 =
t
v
:. v = 4.2 mlsec.
The change of momentum
.: v2=v~+2gs
:. (16.8)2=0+2x9.8s
= (16.8)2 = 14.4 metre
19.6
v2=v~_2gs
:. v2 = (980)2 - 2 x 980 x 240 = 490000
• velocity just belor collision with the ground
= v~ + 2 gs = 0 + 2
~
The velocity before impact with the ceiling directly (v)
Wi
2
= 125 (1.8 + 4.2)
= 125 x 6 = 750 gm.misec.
:. v = 16.8 mlsec .
S
directly (v0)
.: v2 = v~ - 2 gs
.
630000
I
.. m = 1260 = 500 gm. = "2 kg.
:. v = 20 mlsec.
v
[12.6+8.4]
= 10.5 kg.m1sec.
I H2 -
:.
t
=
x
9.8 x 0.9
:. v = 700 cmisec.
:. 40000 = 40
:.1000=v+700
:. v = 300 cmisee.
(v
+ 7(0)
v = 4.2 m1sec.
• The velocity after rebounding
2
v
= v~ - 2 gs
~
( 1 ) After one second :
:. zero = v~ - 2 x 9.8 x 0.4
v=vo-gt=29.4-9.8=
vo = 2.8 m1sec.
H = mv =
the change of momeutum = 0.2 [4.2 + 2.8]
• The velocity just before collision with the ground:
19.6mlsec.
x 19.6 = 9.8 kg.misec.
( 2 ) After two seconds :
= 1.4 kg.mJsec.
fIll!
t
v = 29.4 - 2 x 9.8 = 9.8 mJsec.
H=
(3)
t
:. v2
v2 = v~ + 2 gs = 0 + 2 x 9.8 x 8.1
x 9.8 = 4.9 kg.mlsec.
v2 = u2 - 2 gs = (29.4)2 - 2 x 9.8 x 44.1
= zero
:. v = zero
:. H = zero
v = 12.6 m.lsec.
tJl}J
• The velocity after rebounding
(1)
(b)
(2)
(c)
(3 )(b)
v2=v~-2gs
(4)
(d)
(5)
I": (b)
2nd: (d)
Newton's laws of motion
fJ!Ij
(1)
A
A
24 i+6
j
(2)6.H=
(2)
(4)31.5
(3)
300
=-54
:. the magmlude of the change = 54 kg.m./sec.
(5) 72
~
UIJ_
~
2
18 [-3tL
14.4
dX
5
=z f
(1) 6.H
~
v=Tt=(6t-4)e
5
-9.8 d t= [-9.8 tlz =-29.4
kg.misec.
8
(2) 6. H = [- 9.8 tl. = - 39.2 kg.m.lsec.
vo =-4c
:. The magnitude
of the momentum
in the beginning
= 4 x 2000 = 8000 kg.m./sec.
at t = 3 sec.
Wi
:. v = 14 c
The mementum
II
(3) 6. H = [- 9.8 tl7 = - 39.2 kg.m.lsec.
after 3 see = 14 x 2000
= 28000 kg.m./sec.
v=~=3tZ-24t
dt
:. The momentum after 4 sec. = 1200 [3 (4)2_24
WJ
6.H=83
5
f
(2t-6)dt=8
[t2-6tl3
5
i.e. 57600 kg.m.lsec. in opposite direction of the
direction at the beginning.
= 32 kg.m./sec.
~
~
J
(1 ) 6. H = 0.048
-; = (6 t_t2)
5
( 2 ) 6. H = 0.048)
~
(3 t - 12) d t
3 2
]3 - 72
= 0.048 [ "2 t - 12 t I = ill kg.m./sec.
(3 t - 12) d t
v=~=6-2t
dt
2
(1) 6. H = 12 [6 - 2 tll =- 24 kg.m.lsec.
5
(2)
6. H = 12 [6 - 2 tlz =-72
= 0.048 [ ~ t2 - 12 t]: = zero
6
of
6. H = 1.5
(12 t_t2)
r
(12t-t2)dt
~ =~
dt
= .1 (2 t - 4) ~ = (t - 2) c ,.:
2
II= t (t + 5) (t-2)
6. H = [
= 1 5 [6 t2 _ .1 t3] 14
.
3
2
t
~=
t (t
2
H =m v
+ 3 t-IO) -; kg.m.lsec.
(t2 + 3 t - IO)]~ = 6 kg.m.lsec.
ell
= 360 ton.m.lsec.
~ = d-; = (2 t - 2) ~
~
( 1 ) 6. H = 16
.r
= 16
(2)
12 [6-2tl4=-48kg.m'/sec.
~~
dt
1 t 3]60 = 216 ton.m./sec.
= 1.5 [ 6 t 2 - 3"
(2)6.H=1.52
kg.m.lsec.
6
(3)6.H=
~
(1)
(4)l
= - 57600
dt
II= m ~ = (4 t + 1)(2
(3 t2 - 8 t) d t
[t3 -
4
4 t2l2 = 128 kg.m./sec.
-
6. H = [8
t - 2) ~ = (8 t2 - 6 t - 2) ~
5
t2_6
t- 213 = 116 kg.m./sec.
8
6. H = 16 [t3 - 4 t2l5 = 3696 kg.m./sec.
~
.: The ball after the first impact , it rebounds to
~
3
(l)6.H=
180
f
3
-3dt=
18 [-3tlo=-162
:. the magnitvde of the change = 162 kg.m.lsec.
a height =
t
x 40 = 10 cm.
:. The velocity (v) before the second impact directly
where:
v2 = v~ + 2 gs = 0 + 2 x 980 x 10 = 19600
WJ
The momentum
:. v = 140 cm.!sec.
of the bullet before collision with
the wooden body directly
.: The ball after the second impact,
a height =
t
it rebounds to
= 0.1 x 350 = 35 kg.m.!sec.
x 10 = 2.5 ern.
• After collision,
• The velocity (v 0) after rebounding
directly where:
v2=v~-2gs
the system moved as one body with
the same momentum,
and its mass = 0.1 + 2 = 2.1 kg.
:. 2.1 v = 35
:. v = ~
= 161 m.!sec.
2.1
:. zero = v2 - 2 x 980 x 2.5
o
WI
:. v2 = 5 x 980 = 4900
o
The momentum
:. v 0 = 70 cm.!sec.
:. The change in momentum
of the bullet before its impact with
the target directly = 0.05 x 810 = 40.5 kg.m.!sec.
= 100 (70 + 140)
= 21000 gm.cm.!sec.
• The velocity before impact with the water surface
.: After impact,
the system moved together with the
same momentum,
and its mass = 4 + 0.05 = 4.05
:. 4.05 v = 40.5
:. v = 40.5 = 10 m.!sec.
4.05
WI
The absolute momentum
directly.
The velocity inside the water v = ~
1.5
:. The change of the momentum
of the projectile
= 5 x 350 = 1750 kg.m.!sec.
v = Vo+ gt = 0 + 2 x 9.8 = 19.6 m.!sec.
= 4 m.!sec.
= 7 (4 - 19.6)
v (for the tank) = 45 x
:. The velocity of the projectile relative to the
= - 109.2 kg.m.!sec.
tank = 350 + 12.5 = 362.5
:. The magnitude of the change of momentum
= 109.2 kg.m.!sec.
fs = 12.5 m.!sec.
( 1 ) When the tank is moving towards the cannon
The momentum
= 5 x 362.5 = 1812.5 kg.m.!sec.
( 2 ) When the tank moves away from the cannon
:. The velocity of the projectile relative to the
tank = 350 -12.5
• The velocity before collision the surface
= 337.5 m.!sec.
:. The momentum = 337.5 x 5 = 1687.5 kg.mJsec.
:. v=vo+gt
~
= 9.8 x 2 = 19.6 m.!sec.
v for the tank = 54 x
• The velocity after collision the surface "start of the
dive"
v=
If
fs = 15 m.!sec.
:. The magnitude of the velocity of the projectile
relative to the tank = 300 + 15 = 315
=4m.!sec.
:. The momentum
of the projectile relative to the
tank = 315 x I = 315 kg.m.!sec.
:. i\ H = 0.8 [4 - 19.6] = - 12.48 kg.m.!sec.
:. The magnitude of the velocity of the tank relative
to the projectile = 15 + 300 = 315
• The velocity before impact with the liquid directly
:. The momentum
• The velocity inside the liquid
v=~
= 4.4 m.!sec.
0.5
:. The change of momentum
~
S=vot+
= 0.09 (4.4 - 29.4)
= - 2.25 kg.m.!sec.
of the tank relative to the
projectile = 315 x 1000 = 315000 kg.m.!sec.
v = 0 + 3 x 9.8 = 29.4 m.!sec.
tgt2
:.1000=0+4.9t2
:. t2 = 1000 = 10000 :. t = 100 sec.
7
4.9
49
Newton's laws of motion
The mass of the drop of water at the instant of
reaching the ground's surface
= 0.25 + ~
V=V
o
+gt=O+
H = mv =
( 1 ) .: The body moves by uniform velocity
x~
= 0.4 gm.
1000
7
-to
98
10 x~=7
:. F = 20 + 64 = 84 N
14Om/sec
..
K= 18N
( 2) .: The body moves by uniform velocity
x 14000 = 5600 gm.cmJsec.
:. F + 30 = 75
~
K=20N
v (for the car) = 45 x 2~0 = 1250 cm.!sec.
v (for the sand grain) = 36 x 2~0 = 1000 cm.!sec.
( 3 ) .: The body moves by uniform velocity
The velocity of the sand grain relative to the car
= 1000 + 1250 = 2250 cm.!sec.
H = mv =
~
S = vot +
l~io
x 2250 =
-t
16i
:. 2 K + 70 = 320 + 110
:.K=180N
:.160+F=340
:.F=180N
( 4 ) .: The body moves by uniform velocity
gm.cm.!sec.
:.3F=84
,':
gt2
:. 1440 = zero + 4.9 t2
:. t2= 1440
:. t=~sec.
4.9
7
The rate of accumulating dust =
The increase of the mass = ~
:. F=45N
(5)
, F=28N
:. K=22N
2K=44
.: The body moves by uniform velocity
:. 3 F + 70 = 340
~ri= ~
gm.!sec.
x 1;0 = 4 gm.
:. The mass of the ball at the moment of reaching to
:.3 F = 270
:. F=90N
:. K + 350 = 560
:. K=2ION
( 6 ) The body moves by uniform velocity
:. F+30=90
the ground's surface = 46 + 4 = 50 gm .
,F=60N
:. K=70N
.: v= u + gt=O + 9.8 x~
:. V= 168 m.!sec.
7
:. H = mv = 50 x 168 = 8400 gm.m.!sec.
.: The car moves unifonnJy
v=vo-gt
:.24+2K=68
:. F=R
:. The force of the motor = 5 x 35 = 175 kg.wt.
:. H=mv=mvo-m.gt
d H = _ mg which is a constant vector its norm = mg
dt
lEKercise ill
.~=~
.. R2
Q
(1)225
(2)20
(3)50
(4)
130
(5)85
:. R2 = F = 6t
x 1000 = 6250 kg.wt.
2
:. 45 x 25 = ~
6250
Q
vi
.: The velocity v 2 is uniform
vi
:. v2 = 25V km.lh.
( 1 ) .: body at rest:
F = 32 + 48 = 80 N
(2)':
body at rest :
:. F = 25 N
(3)
, K = 49 N
.: body at rest:
:. F = 40 N
, K = 50 + 70 = 120 N
.: v2 is the maximum velocity
:. R2 = the force = 60 kg.wl.
0: R.« v
Rl_ vl
... R;-V;-
:. v2 = 135 km.lh.