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LAWS OF MOTION PROBLEM AND THEIR SOLUTION
1.
F
What is the maximum value of the force F such that the
block shown in the arrangement, does not move?
60
1
=
m=3kg
2 3
2.
A particle of mass 3 kg moves under a force of 4 î + 8 ĵ + 10 k̂ . Newton. Calculate the
acceleration (as vector) to which the particle is subjected to. If the particle starts from rest and was
at origin initially, what are its new coordinates after 3 seconds ?.
3.
If at any point on the path of a projectile its velocity be u and inclination be , show that particle will
move at right angles to the former direction after time t = u / g sin  when its velocity would be v = u
cot .
4.
A block is placed on an inclined plane of inclination 300 from the horizontal, if the coefficient of
friction between the block and inclined plane is 0.8. Find the acceleration and force of friction on
the block.
5.
A particle moves in a circle of radius 20 cm at a speed given by v = 1+ t + t2 m/s where t is time in s.
Find the initial tangential and normal acceleration.
6.
What is the acceleration of 10 kg block in the situation as
shown in the figure
7.
8.
What is the magnitude and direction of the force of friction
acting on the 10 kg block shown in figure ?
10
10 kg
45
2N
0
smooth
10 kg
 = 0.4
20 N
4 kg
Find the acceleration of 2 kg and 4 kg blocks.
2 kg

9.
10.
Two blocks of mass 3 kg and 4 kg are kept in contact with each
other on a smooth horizontal surface. A horizontal force of 21 N is
applied on the second block due to which they move with certain
acceleration. Calculate the force between the blocks.
For the situation shown in the figure,
(a) Draw free body diagrams of A and B.
(b) What is the acceleration of the blocks A and B ?
(c) What is the magnitude of force applied by A on B.
3 kg
4 kg
21 N
B
A
F = 100 N
3 kg
7 kg
smooth
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11.
12.
13.
smooth
For the situation, shown in the figure,
(a) Draw the free body diagram of 10 kg block
(b) Find the acceleration of the block
(c) Find the force exerted by the block on the incline.
All contact surfaces are assumed to be smooth.
10 kg
300
Black
A block of metal weighing 2 kg is resting on a frictionless plane. It is
struck by a jet releasing water at a rate of 1 kg /s and at a speed of 5
m/s. Calculate the initial acceleration of the block.
2 kg
Find the relation between the acceleration of rod A and wedge B in the
arrangement shown in the figure. The entire surfaces are smooth.
A
B
14.

All the strings and the pullies in the figure shown are massless
and frictionless. Find the relation between the accelerations of
the block B and mass m as shown in the figure.
B
a1
15.
Two blocks each of mass m are connected by a massless
inextensible string. They are kept on a fixed wedge as shown
in the figure. The surface of wedge is frictionless. If the
system is just released from rest, find the instantaneous
acceleration of blocks.
m
a
m

16.
In the figure shown, find the acceleration of the block B if that
of the A is 0.3 m/s2.
A
a2
m

B
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17.
18.
/2
Initially half of the chain’s length (  = 4 m) is overhanging,
what will be the speed of the chain when it just slips off the
smooth table ?
/2
A body slides down a smooth curved fixed track which is a quadrant of
a circle with radius 10 m. Find the speed of the body at the bottom of
the track.
10 m
10 m
19.
A chain of mass M and length L held vertical by fixing its upper end to a rigid support. Find the
tension in the chain at a distance y from the rigid support.
20.
Two forces F1 and F2 are acting on a particle and the angle between them is . The angle between
the resultant force and F1 is  . Find tan  in terms of F1, F2 and .
21.
The co-ordinates of a moving particle at any time t are given by x = ct2 and y = bt2. Find initial
speed of the particle.
22.
A particle moves from position (3 î  2 ĵ  6k̂ ) to position (14 î  13 ĵ  9k̂ ) and a force (constant) of
4 î  ĵ  3k̂ acts on it. Calculate the work done by the force.
23.
A chain of mass M and length L held vertical by fixing its upper end to a rigid support. Find the
tension in the chain at a distance y from the rigid support.
24.
A block of mass M is placed on a smooth inclined surface as
shown in the figure. Draw the free body diagram of the mass
M. Find tension in the string and the normal reaction exerted
by the inclined surface on the block.
M

25.
Find the contact force between the two masses, if F = 10 N, M = 4 kg, m
= 1 kg and coefficient of friction between the surface and both the
masses is  = 0.1.
F
M
m
=0.1
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26.
27.
A body of mass M is moving in
+x direction with acceleration
a. Find the minimum value of
coefficient
of
friction
,
between the block so that a
body of mass m can be held
stationary with respect to M on
the vertical side of M.
M
m
a
=0
(a) Two blocks of mass 2 kg and 1 kg are at rest on a rough surface and
are separated by a distance of 3.75 m as shown in the figure. The
coefficient of friction between each block and surface is 0.20.
2kg
1kg
The 2 kg block is given a velocity of 8 m/s directed towards the second block. It collides with 1kg
block which is at rest. Find the loss of energy of 2 kg block during collision if coefficient of
restitution is ½. (Take g = 10 m/s2 ).
(b) Calculate pressure inside a cylindrical liquid drop of diameter d and surface tension S, if
atmospheric pressure is p0.
28.
A force F depends on displacement x as F= 6x + 4, where F is in Newton and x in meter and it acts
on a mass m = 2 kg which is initially at rest at point x = 0. Then find velocity of mass when x = 2m.
Assume that no other force is acting on mass m.
29.
A block of mass 6 kg is kept on rough surface as shown in figure. Find
acceleration and friction force acting on the block.
(Take g = 10 m/s 2 )
6 kg
 =3/2
30
30.
If the coefficient of static friction between tyres and the road is 0.2, what is the shortest distance in
which an automobile can be stopped while travelling at 54 km/hr.
31.
Two blocks are kept as shown. A horizontal time varying force is
applied on upper block (F = 20 t). Find the time when the relative
motion between the blocks starts. [Given that mass of each block
is m kg]
20 t
0
m
m
 =0
32.
A mass m is moving with a constant velocity v0 along a line y = -a and away from the origin. Find
the magnitude of its angular momentum with respect to origin.
33.
On a smooth inclined plane, a block of mass m is
placed over the plank. If the plank and the block is
released from rest. Find the acceleration of block.
Assume friction between block and plank is sufficient
to prevent slipping.
m
M

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34.
One end of a massless spring of spring constant 100 N / m and natural length 0.5 m is fixed and the
other end is connected to a particle of mass 0.5 kg lying on a frictionless horizontal table. The
spring remains horizontal. If the mass is made to rotate at an angular velocity of 2 rad/s, find the
elongation of the spring.
35.
One end of a massless spring of spring constant 100 N / m and natural length 0.5 m is fixed and the
other end is connected to a particle of mass 0.5 kg lying on a frictionless horizontal table. The
spring remains horizontal. If the mass is made to rotate at an angular velocity of 2 rad/s, find the
elongation of the spring.
36.
Find out the magnitude of R1 and R2 contact forces acting
between the blocks as shown in figure. (take all the surfaces to
be smooth.)
37.
R1
R2
40 N
4 kg
3 kg
A block of mass 2 kg is kept on a rough inclined surface and it is moving
downward with a constant velocity of v = 1 m/s along the incline. Find
out the value of net force applied by the inclined surface on the block.
1 kg
2 kg

38.
Three blocks A, B and C are placed one over the other as shown in figure.
Draw the free body diagram of all the three blocks.
A (M1)
B (M2)
C (M3)
39.
2 kg and 3 kg blocks are moved up with a common acceleration of 2 m/s2.
Find the magnitude of tensions (T1 & T2) in the strings. (take g = 10 m/s2)
T1
2kg
T2
3 kg
40.
41.
42.
Two blocks of masses 4 kg and 6 kg connected by a
massless string are kept on a rough surface having
coefficient of sliding friction 0.2. Two horizontal forces of 30
N and 10 N is applied on 4 kg and 6 kg blocks. Find the
friction force on both the blocks and tension in the string.
(Consider the string to be taught initially)
30 N
Two blocks, each of mass m are attached with a massless and
inextensible string. One of the blocks is placed on the rough horizontal
surface of a table for which coefficient of friction is . Find out the
force exerted by the table on the block if  > 1.
4 kg
6 kg
=0.2
10 N
=0.2
m

m
One end of a massless spring of spring constant 100 N/m and natural length 0.5 m is fixed and the
other end is connected to a particle of mass 0.5 kg lying on a frictionless horizontal table. The
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spring remains horizontal. If the mass is made to rotate about the fixed point at an angular velocity
of 2 rad/s, find the elongation of the spring.
43.
One end of a uniform rod AB of length L and mass M is hinged to a thin and rigid rod. The thin rod
is vertical and rotates with constant angular speed  in such a way that the rod AB remains
horizontal. Position of thin rod does not change.
(a) Draw the free body diagram of the rod AB.
(b) Find the force on the rod AB by the thin rod.
44.
Three blocks m1 = 1kg, m2 = 2kg and m3 = 3kg are
kept on a smooth horizontal surface. A horizontal
force F = 90N is applied at the first block. Blocks are
attached by light strings. Calculate the tensions T1
and T2 acting on the strings.
45.
A 100 kg load is uniformly pulled over a horizontal plane by a force F applied at an angle  = 300 to
the horizontal. Find this force if the coefficient of friction between the load and the plane is 0.3.
46.
A block of mass 5 kg is placed on a slope which makes an angle of 20 with the horizontal and is
given a velocity of 10 m/s up the slope. Assuming the coefficient of sliding friction between the block
and the slope is 0.2, find how far the block travels up the slope? Take g = 10 m/s 2. Take cos20 =
0.9 and sin20= 0.3.
47.
Two particles each of mass m are connected by a
light string of length 2L as shown. A continuous
force F is applied at the midpoint of the string (x = 0)
at right angles to the initial position of the string.
Show that acceleration of m in the direction at right
angles to F is given by
F
x
ax =
m L2  x 2
3Kg
m3
T2
T2
2Kg
T1
m2
T1
1Kg
m1
F = 90N
2L
A
B
F
48.
A particle rests on top of a hemisphere of radius R. Find the smallest horizontal velocity that must
be imparted to the particle if it is to leave the hemisphere without sliding down it.
49.
A monkey is sitting on the branch of a tree. The branch exerts a normal force of 48 N and a
frictional force of 20 N. Find the magnitude of the total force exerted by the branch on the monkey
?
50.
Two masses m1 = 10 kg and m2 = 5 kg are connected
by
an ideal string as shown in the figure. The
coefficient of friction between m1 and the surface is  =
0.2. Assuming that the system is released from rest ,
Calculate the velocity when m2 has descended by 4m .

m1
m2
4m
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51.
(a) Draw the FBD of m1 and m2.
(b) What will the relation between m1, m2 , ,  ? so that m1, m2
remains stationary. All surfaces are smooth.
m2
m1


fixed
52.
A car of mass m = 1000 kg starting from rest attains a speed of 10 m/s over a distance of
50 m. Assuming constant acceleration, find the resultant force acting on the car.
53.
(i) A body of mass M is placed in a smooth horizontal table and is connected
to another mass m by means of a massless string passing over a massless
and frictionless pulley as shown in the figure. Find the force exerted by the
pulley on the table.
M
m
(ii) A body of mass M is moving with
acceleration a in the direction as shown.
Find the minimum value of  so that a
body of mass m can be held stationary
with respect to M on the vertical side of M.
m
M
a
54.
A smooth wedge of mass M is moving with horizontal acceleration g cot .
A small block of mass m is there on inclined smooth face of wedge.
Calculate acceleration of block of mass m with respect to ground.
m
g cot 
M

55.
Two masses ‘m’ and ‘2m’ are connected by a massless string which
passes over a light frictionless pulley as shown in fig.1. The masses
are initially held with equal lengths of the strings on either side of the
pulley. Find the velocity of the masses at the instant the lighter mass
moves up a distance of 6.54 m. The string is suddenly cut at that
instant. Calculate the time taken by each to reach the ground. (g =
9.81 m/s2)
2m
m
13.08 m
ground
Fig. 1
56.
Two blocks of masses 4 kg and 6 kg connected by a
massless string are kept on a rough surface having
coefficient of sliding friction 0.2. Two horizontal forces of 30
N and 10 N is applied on 4 kg and 6 kg blocks. Find the
friction force on both the blocks and tension in the string.
30 N
4 kg
=0.2
6 kg
10 N
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57.
58.
Two blocks A and B of mass 4m and m respectively, are connected
through a massless string passing through an ideal pulley. Initially block
A is 4 m above the block B as shown in the figure. Now system is
released at t = 0 sec. Find the time after which both blocks cross the
same level. Also find velocity of the blocks at the instant when they cross
the same level.
Two identical blocks A and B are placed on a rough
inclined plane of inclination 450. The coefficient of
friction between block A and incline is 0.2 and that of
between B and incline is 0.3. The initial separation
between the two blocks is 2 m. The two blocks are
released from rest, then find
(a) the time after which front faces of both blocks
come in same line and
(b) the distance moved by each block for attaining
above motion position.
A 4m
4m
m
B
A =0.2
A
B
B =0.3
2m
450
59.
A man of mass m is moving with a constant
acceleration a w.r.t. plank. The plank lies on a
smooth horizontal floor. If mass of plank is also m
then calculate acceleration of plank and man w.r.t.
ground, and frictional force extended by plank on
man.
60.
A bob of mass 'm' is suspended by a light inextensible string of length 'l' from a fixed point. Such
that it is free to rotate in a vertical plane. The bob is given a speed of
6gl horizontally. Find the
0
tension in the string when string deflects through an angle 120 from the vertical downward.
61.
Two blocks of masses m1 and m2 connected by a nondeformed light spring rest on a horizontal plane.
The
coefficient of friction between the blocks and the surface is
equal to . What minimum constant force has to be applied in
the horizontal direction to the block of mass m1 in order to shift
the other block?
m2
m1
62.
A block of mass 2 kg slides on an inclined plane which makes an angle of 300 with the horizontal.
The coefficient
of
friction between
the block and
the surface is
3/2 .
(i) What force should be applied to the block so that the block moves down without any acceleration
?
(ii) What force should be applied to the block so that it moves up without any acceleration..
63.
A block of mass 2 kg slides on an inclined plane which makes an angle of 300 with the horizontal.
The
coefficient
of
friction
between
the
block
and
the
surface
is
3/2 .
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(i) What force should be applied to the block so that the block moves down without any acceleration ?
(ii) What force should be applied to the block so that it moves up without any acceleration.
64.
A block slides down an inclined plane of slope angle
with constant velocity. It is then projected
up the same plane with an initial speed v 0. How far up the incline will it move before coming to rest
? will it slide down again ?
65.
A monkey of mass m is standing on a ladder of mass M-m which is counter
balanced by mass M by a string passing over a smooth frictionless pulley. The
monkey climbs up the leader by a distance  w.r.t. ledder. Find the
displacement of the leader.
M-m
M
m
66 .
Two blocks A and B are connected to each other by a string and
a spring. The spring passes over a frictionless pulley as shown
in figure. Block 'B' slides on the horizontal surface of a fixed
block 'C' and the block 'A' slides along the vertical side of 'C',
both with same uniform speed. The coefficient of friction
between the surface of the blocks is 0.2. Force constant of
spring is 2000 N/m. If mass of the block A is 2kg. Calculate the
mass of block B and the energy stored in the spring
B
k
C
A
67.
A body of point mass m is attached with a string of length . The body is under a motion in vertical
circle with velocity v at the lowest position. Draw the free body diagram of the body when it makes
an angle  with the vertical, and find the tension in the string .
68.
Find the acceleration of A and B in the arrangement
shown in the figure. Mass of A = 5 kg and mass of B =
10 kg and coefficient of friction between A and the
surface is 0.2.
A
 =0.2
B
(Assume pulleys to be light & frictionless and string to be light & inextensible)
69.
Find the acceleration of blocks A and B. Contact
surface are smooth. Mass of each of A and B is 5
kg. (Assume pulleys to be light & frictionless and
string to be light & inextensible)
smooth
A
B
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70.
71.
72.
73.
74.
75.
76.
Two blocks are connected by a rod and fixed at P
and R points a force F(12N) is applied on block m2.
So find the tension in rod at P, Q, R points. (Q is
the mid point of P and R)
A body is thrown upward from A with speed ‘u’ m/s
on an incline plane having inclination  the body will
come back to the point A of projection after some
time. All planes are frictionless. Calculate time.
P
m2
m1
12 N
m1 = 2kg, m2 = 3 kg,
mass of the rod = 1 kg
u
B
m
A
fixed

Consider the situation shown in the figure. The
10 kg block is tied with a light string on one side
and a force of 10 N is applied on it from the
other side. All contact surfaces are smooth, and
the block remains at rest.
(i) Draw the free body diagram of the 10 kg block.
(ii) Find the tension in the string.
light string
m = 10 kg
A block of mass 2kg is placed on a rough surface as shown
in the figure. Find the distance travelled by the block in the
first 2 sec.
In the shown figure a triangular wedge of mass M and a
small cube of mass m is placed. The coefficient of friction
between wedge and cube is 1 and between wedge and
ground is 2. (2 = 1/2). Find the force of friction on
wedge due to ground. It is given that 1 > tan .
R
Q
F = 10 N
6N
k =0.2
s=0.3
m
2
60
2 kg
1
M


A particle of mass 1 kg has an initial velocity v i  ( î  2 ĵ ) m/s. It collides with another body and the

impact time is 0.1s, resulting in a velocity v f  (6 î  4 ĵ  5k̂ ) m/s after impact. Find average force of
impact on the particle.
An open elevator is ascending with zero acceleration . The speed v = 10m/sec. A ball is thrown
vertically up by a boy when he is at a height h = 10m from the ground. The velocity of projection is v
= 30m/sec with respect to elevator . Find
(i) the maximum height attained by the ball
(ii) the time taken by the ball to meet the elevator again.
(iii) time taken by the ball to reach the ground after crossing the elevator.
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77.
In the shown arrangement the masses of blocks A, B and C are m,
2m and 3m respectively. If all the surfaces are smooth and spring
constant of the spring be K, find
(i) acceleration of each block
(ii) tension in the cord connecting blocks A and B.
(iii) extension in the spring.
A
B
C
78.
79.
Two identical cubes placed one over other each of
mass 'm' and side 'a' are placed as shown.
Calculate time the upper block will fall.
1 = 0.1 2 = 0.2
F = 6 N m = 1 kg
F
2
1
In the pulley – block system shown, find the accelerations of A, B, C and
the tension in the string. Assume the friction to be negligible and the
string to be light and inextensible. The masses of the blocks are m, 2m
and 3m respectively
A
C
B
80.
A ball of mass m is hanging from a uniform rope of mass M
and length L whose one end is connected to the ceiling of an
elevator as shown in the figure. If the elevator has a uniform
acceleration of a m/s2 upward, find the tension in the rope as a
function of x. What is the force exerted by the rope on the
ceiling?
(take x from the point where the string is connected to the
ceiling)
L
a
m
81.
Figure shows a man of mass 70 kg standing on a light weighing machine
kept in a box of mass 35 kg. The box is hanging from a pulley fixed to the
ceiling through a light rope, the other end of which is held by the man
himself. If he manages to keep the box at rest (a) what is the weight
shown, by the machine ? (b) what force should the man exert on the rope
to get his correct weight on the machine ?
82.
A block of mass 20 kg is lying on a frictionless table. A block of 5kg is kept on the block of 20 kg. If
a variable force F given by F = kx is applied on the block of mass 20 kg and initially the mass of 20
kg is lying at x = 1 m and  = 0.2 and k = 5 N/m, find
(i) the distance after which 5 kg mass starts slipping
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(ii) Calculate acceleration of masses at 14 m from starting point.
83.
84.
`
A chain consisting of 5 links each of mass 0.1 kg is lifted
vertically with a constant acceleration 2.5 m/s 2 as shown in
figure. Find the force of interaction between the top link and
link immediately below it.
1
2
3
4
5
Two masses of 5 kg and 10 kg connected by a massless string
passing over a frictionless pulley are in equilibrium as shown in the
figure. One of the mass is resting on the surface. Find
(a) Tension in the string
(b) Normal reaction on the 10kg block (take g = 10 m/s2)
5 kg
10 kg
85.
If a mass M is hung with a light inextensible
string as shown in the figure. Find the tension
in horizontal and inclined strings in terms of M,
g and .

m
86.
87.
88.
Two blocks of mass 2 kg and 4 kg are kept in contact with each
other on a smooth horizontal surface. A horizontal force of 12 N
is applied on the first block due to which they move with certain
constant acceleration. Calculate the force between the blocks.
A force F is applied from its centre along horizontal direction Find
direction and magnitude of frictional force acting on the cylinder.
(No slipping)
12 N
2kg
m
4kg
r
F
A ball is dropped from a height 200 cm on the ground. If the coefficient of restitution is 0.2,
(a) what is the height to which the ball will go up after it rebounds for the 2nd time.
(b) If duration of each collision is 1 mill sec. Then find the average impulsive forces
during 1st and 2nd collision. (Neglect any air resistance)
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89.
90.
91.
92.
93.
A block of mass M with a semicircular track of radius R rests on
a horizontal frictionless surface. A uniform cylinder of radius 'r'
and mass 'm' is released from rest at the top point A. The
cylinder slips in the semicircular frictionless track. How fast is
the block moving when the cylinder reaches the bottom of the
track ?
In the shown figure a triangular wedge of mass M and a small
cube of mass m is placed. The coefficient of friction between
wedge and cube is 1 and between wedge and ground is 2. (2
= 1/2). Find the force of friction on wedge due to ground. It is
given that 1 > tan .
Two masses m1 = 10 kg and m2 = 5 kg are
connected by an ideal string as shown in the figure.
The coefficient of friction between m1 and the
surface is  = 0.2. Assuming that the system is
released from rest, Calculate the velocity when m2
has descended by 4m .

m
A
B
M
m
2
M

m1
m2
Two blocks of masses m1 and m2 connected by a non-deformed
light spring rest on a horizontal plane. The coefficient of friction
between the blocks and the surface is equal to . What minimum
constant force has to be applied in the horizontal direction to the
block of mass m1 in order to shift the other block?
Two masses m1 = 10 kg and m2 = 5 kg are
connected by an ideal string as shown in the figure.
The coefficient of friction between m1 and the
surface is  = 0.2. Assuming that the system is
released from rest, Calculate the velocity when m2
has descended by 4m .
1

m2
m1
m1
m2
4m
94.
See the diagram, The system is in equilibrium. There is no
friction anywhere. Spring, string, pulley are massless. Find
spring compression.
m
k
M

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95.
A cabin is moving upwards with a constant acceleration g. A
boy standing in the cabin wants to whirl a particle of mass m
in a vertical circle of radius . (mass is attached to an ideal
string) calculate minimum velocity which should be provided
at lowermost point (w.r.t cabin) so that particle can just
complete the circle.
96.
Friction coefficient between the wedge and the block is .
There is no friction between the wedge and horizontal
floor. This system is not in equilibrium. Find acceleration
of block relative to ground.
g
m
M

97.
Two smooth wedges of equal mass m are placed as shown in
figure. All surfaces are smooth. Find the velocities of A & B just
before A hits the ground.
A
B
h

98.
99.
A body of mass M is moving in -x direction with acceleration a.
Find the minimum value of  so that a body of mass m can be held
stationary with respect to M on the inclined side of M. Consider that
application of acceleration a is not enough to support the block in
equilibrium.
m

a
M
All the strings and the pullies in the figure shown are massless and
frictionless. Find the relation between the accelerations of the
block A, B and C having masses m1, m2 and m3 respectively.
a1 A m1
a2
B m2
C m3
100.
A pulley fixed to the ceiling of a lift carries a thread whose ends are
attached to the masses of 4kg and 2kg. The lift starts going down
with on acceleration of a0 = 4 m/s2 relative to ground. Calculate
(a) The acceleration of the load 4kg relative to the ground and
relative to lift.
(b) The force exerted by the pulley on the ceiling on the lift
(Thread and pulley are massless g=10 m/s2)
a0
4 kg
2 kg
a3
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101.
Consider the situation shown in figure. Both the
pulleys and the string are light and all the surfaces
are frictionless. Calculate (a) the acceleration of
mass m (b) the tension in the string PQ and (c)
force exerted by the clamp on the pulley C.
m
P
Q
C
m
2
102.
103.
104.
A pulley fixed to the ceiling of a lift carries a thread whose ends
are attached to the masses of 4kg and 2kg. The lift starts going
down with on acceleration of a0 = 4 m/s2 relative to ground.
Calculate
(a) The acceleration of the load 4kg relative to the ground and
relative to lift.
(b) The force exerted by the pulley on the ceiling of the lift
(Thread and pulley are massless, g=10 m/s 2)
In the shown figure a triangular wedge of mass M and a small
cube of mass m is placed. The coefficient of friction between
wedge and cube is 1 and between wedge and ground is 2. (2
= 1/2). Find the force of friction on wedge due to ground. It is
given that 1 > tan .
The figure shows L shaped body of mass M placed on
smooth horizontal surface. The block A is connected to the
body by means of an inextensible string, which is passing
over a smooth pulley of negligible mass. Another block B of
mass m is placed against a vertical wall of the body. Find
the minimum value of the mass of block A so that block B
remains stationary relative to the L-shape body. Coefficient
of friction between the block B and the vertical
wall is .
a0
4 kg
2 kg
m
1
M
2

m
B
M
A
105.
Figure shows a man of mass 60 kg standing on a light weighing
machine kept in a box of mass 30 kg. The box is hanging from a
pulley fixed to the ceiling through a light rope, the other end of which
is held by the man himself. If he manages to keep the box at rest (i)
what is the weight shown by the machine ? (ii) what force should the
man exert on the rope to get his correct weight on the machine ?
106.
A smooth table is placed horizontally and a spring of instretched length 0 and force constant k has
one end fixed to its centre. To the other end of the spring is attached a mass m which is making n
revolutions per second around the centre. Find the radius r of this uniform circular motion and the
tension T in the spring.
107.
A very small cube of mass m is placed on the inside of a formel rotating about a vertical axis at a
constant rate of "n" revolutions per second. The wall of the funnel makes an angle  with the
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horizontal. If the coefficient of static friction between the cube and the funnel is  and the centre of
the cube is at a distance r from the axis of rotation, what are the largest and smallest values of n for
which the block will not move with respect to the funnel ?
108.
In the arrangement shown in the figure, the
rod of mass m held by two smooth walls ,
remains always perpendicular to the surface of
the wedge of mass M. Assuming all the
surfaces are frictionless, find the acceleration
of the rod and that of the wedge.
fixed wall
m

M
109.
A 4 kg block is put on top of 5 kg block. In order to cause the top block to slip on the bottom one, a
horizontal force of 12 N must be applied to the top block. Assume a frictionless table and find (a)
The maximum horizontal force which canbe applied to the lower block so that the blocks will move
together and (b) the resulting acceleration of the blocks.
110.
In the arrangement shown in figure, pulleys are
small and light and springss are ideal. k1, k2 k3, and
k4 are force constants of the springs. Calculate
period of small vertical oscillations of block of mass
m.
K2
K4
m
K3
K1
111.
112.
A chain of length l is held vertical and then released. It falls on a
platform which starts, from rest and moves vertically upward with a
constant acceleration a0. Determine the normal force exerted on
the platform by the chain as a function of time. Mass per unit length
of the chain is .
A
l
a0
A prism of mass M and with angle  rests
on a horizontal surface. A bar of mass m
is placed on the prism . Assuming the
friction to be negligible, find the
acceleration of the prism.
m
M

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113.
114.
115.
116.
117.
The figure shows an L shaped body of mass
M is placed on smooth horizontal surface. The
block A is connected to the body by means of
an inextensible string, which is passing over a
smooth pulley of negligible mass. Another
block B of mass m is placed against a vertical
wall of the body. Find the minimum value of
the mass of block A so that block B remains
stationary relative to the wall. Coefficient of
friction between the block B and the vertical
wall is .
m
B
M
A
A plank of mass M is placed on a rough horizontal
surface and a constant horizontal force F is applied
on it. A man of mass m runs on the plank. Find the
accelerations of the man so that the plank does not
move on the surface. Co-efficient of friction between
the plank and the surface is
Assume that the
man does not slip on the plank.
A smooth fixed wedge has one face inclined at 300 to the
horizontal and a second face at 450 to the horizontal. The
faces are adjacent to each other at the top of the wedge.
Particles of masses 2m and 5m are held on these respective
faces connected by a taut inelastic string passing over a
smooth pulley at the top of the wedge as shown in the figure.
Find the acceleration of the system if the particles are
simultaneously released and show that the force acting on
10
1
the pulley is
mg (1 + 2) cos (52 )0.
7
2
The figure shows an L shaped body of mass
M is placed on smooth horizontal surface. The
block A is connected to the body by means of
an inextensible string, which is passing over a
smooth pulley of negligible mass. Another
block B of mass m is placed against a vertical
wall of the body. Find the minimum value of
the mass of block A so that block B remains
stationary relative to the wall. Coefficient of
friction between the block B and the vertical
wall is .
m
M
2m
30
0
F
45
0
5m
m
The masses of blocks A and B are m and M. Between A and B
there is a constant frictional force F, but B can slide frictionlessly
on the horizontal surface. A is set in motion with velocity v0
while B is at rest. What is the distance moved by A relative to B
before they move with the same velocity ?
B
M
A
v0
A
B
m
M
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118.
119.
A ball of mass m is hanging from a uniform rope of mass M and
length L whose one end is connected to the ceiling of an elevator
as shown in the figure. If the elevator has a uniform acceleration of
a m/s2 upward, find the tension in the rope as a function of x. What
is the force exerted by the rope on the ceiling?
In the pulley – block system shown, find the
accelerations of A, B, C and the tension in the
string. Assume the friction to be negligible and
the string to be light and inextensible. The
masses of the blocks are m, 2m and 3m
respectively
x
L
a
m
A
C
B
120.
121.
Two blocks are kept as shown. A horizontal time varying
force is applied on upper block (F = 20 t). Find the time
when the relative motion between the blocks starts.
[Given that mass of each block is m kg]
A bead of mass 'm' is fitted on to a rod can slide on it
without friction. At the initial moment the bead is in the
middle of the rod. The rod moves translationally in a
horizontal plane with an acceleration 'a' in a direction
forming an angle  with the rod. Find the acceleration of
the bead relative to the rod.
20 t
0
m
m
=0
a

122.
A rough inclined plane with inclination  = 370 with the horizontal is accelerated horizontally till a
block of mass m originally at rest with respect to the plane just begins to slip up the plane. The
coefficient of static friction between the surfaces in contact is  = 5/9. Find the acceleration of the
plane. (tan 370 = 3/4)
123.
A time varying force F = 5t N is applied on the upper block as
shown in figure. When will the upper block start moving with
respect to lower block? What is the acceleration of lower block at
that instant (here t is in sec.)
124.
A rod of length 1m and mass 4 kg , can rotate freely
in a vertical plane around its end A. The rod is
initially held in a horizontal position and then
released. At the time the rod makes an angle
45with the vertical, calculate
(a) its angular acceleration ,
(b) its angular velocity.
A
45
 = 0.5
m1 = 1kg
=0
m2 =2kg
F = 5t
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125.
A particle of mass m moving with a speed v0 strikes perpendicularly one end of a uniform rod of
mass M and length L initially resting on a smooth horizontal plane. The particle returns in the same
v
m
1
line with speed of 0 . If
=
find the linear and the angular speed of the rod.
3
M
32
126.
A uniform disc of mass m and radius R is projected horizontally with velocity v 0 on a rough
horizontal floor having coefficient of friction equal to . Find the time after which it starts pure
rolling.
127.
A uniform solid sphere of mass m and R starts rolling without slipping down an inclined plane of
length L and inclination 30 to the horizontal. Find
(a) the frictional force and its direction.
(b) work done by the frictional force.
(c) linear speed and linear acceleration of the sphere as a function of time.
128.
A fiat car of mass M starts moving to the right due to a
constant force F as shown in figure. and spills on the flat
car from a stationary hopper. The velocity of loading is
constant and equal to m kg/s. Find the velocity of the car
after t seconds.
129.
A spring is fixed at one end O on a smooth horizontal
table. Natural length of spring is very small (tends to
zero). Now a ball of mass m = 1 kg is attached at other
end stretched to A as shown in figure. Ball is given
velocity V0 = 2 m/s at an angle 30 to OA. Find the
maximum elongation of the spring. Given that spring
constant is k = 2 N/m and OA = 1m
130.
131.
v0
Thin threads are tightly wound on the ends of a uniform
solid cylinder of mass m. The free ends of the threads are
attached to the ceiling of a lift. The lift starts moving up with
an acceleration a0. Find the acceleration of the cylinder
relative to the lift and the force F exerted by the cylinder on
the ceiling.
Find the acceleration of the body m1 in the
arrangement shown in figure with masses of
respective bodies. The friction is absent.
Also the masses of the pulleys and the
threads are negligible.
30

O
A
`
a0
m0
m1
m2
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132.
What is the minimum and maximum acceleration with
which bar A (Fig.) should be shifted horizontally to
keep bodies 1 and 2 stationary relative to the bar?
The masses of the bodies are equal , and the coefficient of friction between the bar and the bodies is
equal to k. The masses of the pulley and the threads
are negligible, the friction in the pulley is absent.
1
A
133.
A man of mass m is moving with a constant acceleration a w.r.t.
plank. The plank lies on a smooth horizontal floor. If mass of plank
is also m then calculate acceleration of plank and man w.r.t. ground,
and frictional force extended by plank on man.
134.
Mass of blocks A, B and C is 7.5 kg, 6 kg and 1 kg. There is no
friction anywhere. Calculate resultant acceleration of block C
when the system is released. Pulley is massless.
2
B
A
C
135.
Friction coefficient between the wedge and the block is .
There is no friction between the wedge and horizontal floor.
Find acceleration of block relative to ground.
m
M

136.
137.
138.
A block of mass m is kept over a fixed smooth wedge. The
block is attached to a sphere of same mass through fixed
massless pullies P1 and P2. Sphere is dipped inside water as
shown. If specific gravity of material of sphere is 2, then find
the acceleration of the sphere.
m
P2
P1
0
m
30
A mass of 2.9 kg is suspended from a string of length 50 cm and is at rest. Another body of mass
100 gm, which is moving horizontally with a velocity of 150 m/s strikes and sticks to it.
What is the tension in the string when it makes an angle of 60 with the vertical?
Two blocks of masses m1 = 2 kg and m2 =4 kg are attached
by light ideal spring of force constant k = 1000 N/m. The
system is kept on a smooth inclined plane inclined 30 0 with
horizontal. A force F=15 N is applied on m1 and system is
released from rest. The block m2 is attached with a light
string whose another end is connected with a mass m3 = 1
kg. Assume initially the spring is at relaxed position and the
system is released from rest. Find the
(a) maximum extension of the spring.
(b) acceleration of the system at this instant
F
m1
m2

m3
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139.
A block of mass M is resting on a frictionless surface. A second block of mass m is placed on it as
shown in figure. A constant force F is applied on the block of mass M due to which the system
accelerates
(i) find the minimum value of coefficient of friction
m
between the two blocks to prevent sliding
F
M
(ii) If the coefficient of friction between the blocks is one third that of calculated in part (i) find the work
done by the friction force during the first t seconds. Also find the energy dissipated into heat
assuming the length of M is large enough.
140.
Consider the arrangement shown in the
figure. If the system is set free at t = 0
with the horizontal bar at a height of h as
shown in the figure, obtain
C
A
B
h


(i) velocities of the wedges A and B at the instant C hits the floor.
(ii) Force exerted by the bar C on each of the wedge. When C hits the floor.
(Neglect any friction. Mass of each wedge is m while that of C is M)
141.
142.
The figure shows an L shaped body of mass M is
placed on smooth horizontal surface. The block A is
connected to the body by means of an inextensible
string, which is passing over a smooth pulley of
negligible mass. Another block B of mass m is placed
against a vertical wall of the body. Find the minimum
value of the mass of block A so that block B remains
stationary relative to the wall. Coefficient of friction
between the block B and the vertical wall is .
In the shown figure. The blocks and pulley
are ideal and force of friction is absent
external horizontal force F is applied as
shown in figure. Find acceleration of block
C.
m
B
M
B
A
M
m1
F
m2
C
143.
A circle of radius R = 2m is marked on upper of a horizontal board, initially at rest. An insect starts
from rest along the circle with a tangential acceleration a = 0.25 m/s2. At the same instant board
accelerates upwards with acceleration b = 2.5 m/s 2. If the coefficient of friction between board and
insect is  = 0.1, what distance will the insect travel on the board without sliding?
144.
A board fixed to the floor of an elevator such that the board
forms an angle  = 370 with horizontal floor of the elevator
accelerating upwards. A block is placed on point A of the board
as shown. When block is given a velocity v1 = 4 2 m/sec up
the board w.r. to board it comes to rest after moving a distance 
= 1.6 m relative to the board. Its velocity was v 2 = 4 m/s down
a
A
0
37
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the board when it returns to point A. Calculate acceleration 'a'
of the elevator and coefficient of friction  between the board
and the block.
145.
Figure shows a man of mass 60 kg standing on a light weighing
machine kept in a box of mass 30 kg. The box is hanging from
a pulley fixed to the ceiling through a light rope, the other end
of which is held by the man himself. If he manages to keep the
box at rest (i) what is the weight shown by the machine ? (ii)
what force should the man exert on the rope to get his correct
weight on the machine ?
146.
Consider the situation shown in figure. Both the
pulleys and the string are light and all the surfaces
are frictionless. Calculate (a) the acceleration of
mass m (b) the tension in the string PQ and (c)
force exerted by the clamp on the pulley C.
m
P
Q
C
m
2
147.
148.
149.
A mass m is released from rest on the incline. The coefficient
of friction is  = x, where x is the distance travelled along the
incline and  is a constant. Find
(a) the distance travelled by the mass till it stops, and
(b) the maximum velocity over this distance.
In the shown figure a triangular wedge of mass M and a small
cube of mass m is placed on the wedge. The coefficient of
friction between wedge and cube is 1 and between wedge and
ground is 2. (2 = 1/2). Find the force of friction on wedge due
to ground. It is given that 1 > tan .
m

m
1
M
2

Three blocks A, B and C have masses 1kg , 2kg and 3 kg respectively
are arranged as shown in figure. The pulleys P and Q are light and
frictionless. All the blocks are resting on a horizontal floor and the
pulleys are held such that strings remain just taut.
At moment t = 0 a force F = 40 t Newton starts acting on pulley P
along vertically upward direction as shown in figure. Calculate
(i) the times when the blocks lose contact with ground.
(ii) the velocities of A when the blocks B and C loses contact with
ground.
(iii) the height by which C is raised when B loses contact with ground
F = 40 t
P
Q
A
B
C
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SOLUTION OF ABOVE PROBLEM
1.
2.
For no motion
Fcos60  (mg + Fsin60)
1

F/2 


F/2  g
Fmax = 20 N
2 3
( 3 g +
Fcos60
f
Fsin60

Taking F as the net force
 4 8 10
a  î  ĵ  k̂ m/s 2
3 3
3
  1 2
s  ut  at
2
1 4
8
10 
= 0   î  ĵ 
k̂  9
2 3
3
3 
= 6 î  12 ĵ  15k̂
sx = 6m;
sy = 12m
3.
F 3
)
2
F.B.D.
sz = 15m

v i  u cos i  u sin i

v f  u cos i  (u sin   gt ) j
 
u
 t=
v i .v f  0
g sin 

u
v f  u cos i  (u sin  
)j
cos 
 | uf | = u cot 
4.
Acceleration is zero
Frictional force = mg sin 300 = 10 N.
5.
Tangent acceleration at =
dv
= 2t +1
dt
v2
normal acceleration an =
R
 (at)t=0 = 1 m/s2
2
v
1
(an)t=0 = 0 
= 5 m/s 2
R
0.2
mg
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6.
Fx = max
 10 2 cos 45 = 10 a
 10 = 10 a
 a = 1 m/s2.
y
10
N
2
0
45
x
100 N
7.
Limiting friction fL = N
N – 100 = m  0
 N = 100
 fL = 0.4  100 = 40 N
fL > 20 N thus block does not move.
20 – f = 10  10
f = 20 N in backward direction.
N
20 N
f
100 N
8.
Acceleration of both blocks will be same = g sin 
9.
a=
10.
100 = (3 + 7)a
100
a=
 10 m/s2
10
consider FBD of B
R = ma
 R = 7  10 = 70 N.
11.
21
= 3 ms2
7
 f=33=9N
N1
13.
y
 tan 
x
14.
a2 = 6a1

R
30 N
70 N
FBD of A
F vdm / dt 5  1
= 2.5 m/s2 .


m
m
2
a=
R
100 N
Acceleration perpendicular to the incline = 0
Let acceleration parallel to the incline be a, then
Mg sin 30 = ma
 a = g sin 30 = 5 m/s2
Thus resultant acceleration = 5 m/s 2
N – mg cos300 = ma = 0
 N = mg cos 30
100  3
=
= 50 3 N
2
12.
N2
aA = aB tan .
FBD of B
N
0
mg sin 30
30
300
0
mg = 100 N
mg cos 300
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15.
mg – T = ma
T – mg sin  = ma
Solving (i) and (ii)
 mg - mg sin  = 2ma
g(1  sin )
a=
.
2
. . . (i)
. . . (ii)
N
T
a
T
ma
kx
mg
mg
16.
0.15 m/s2.
17.
30 m / s .
18.
10 2 m/sec.
19.
Mg
20.
F2 sin 
F1  F2 cos 
21.
zero
22.
  
Displacement of the particle, s  r2  r1  (14 î  13 ĵ  9k̂ )  (3 î  2 ĵ  6k̂ ) = 11î  11ĵ  15k̂

Therefore, work = F.S  ( 4 î  ĵ  3k̂ ).(11î  11ĵ  15k̂ ) = 44 + 11 + 45 = 100 units.
23.
Mg
Ly
L
Ly
L
N
24.
T
 N = Mg cos 
T = Mg sin 
Mg sin 
Mg cos 
Mg

25.
F – N – f1 = Ma
N – f2 = ma

N = 2N.
. . . (i)
. . . (ii)
F
N
M
N
f2
f1
26.
N = ma
f = N = ma
ma > mg
amin = g/
f
a > g/
M
ma
m
m
mg
N
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27.
(a) Velocity of 2 kg block when it strikes 1 kg block is v such that
v 2  82  2  2  3.75

v = 7 m/s
From COM,
14 = 2 v1  v 2
For e,
v  v1
1
 2
2
7
7
v 2  v1 =
2
. . . (1)
. . . (2)
From (1) and (2)
7
v1  , v 2 =7
2
1
49 

Loss of energy =  2 72 
= 36.75 J
2
4 

(b) Let us consider the equilibrium of half of the
liquid drop of length .
2S + p0d = pd
2S
 p = p0 +
d
28.
F  6x  4 
v

d
mvdv
dx
dv
 3x  2
dx
v
2
 vdv 
  3x  2 dx
o

o
v  20m / s.
29.
fmax > mg sin
a=0
f = 30 N
6 kg
 =3/2
30
30.
The friction force retarding the motion of car = N.
= mg = 0.2mg
0.2mg
retardation produced a = = - 0.2g
m
If x is the distance of stopping, then
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v2 = u2 + 2ax, as v = 0, u = 54 
5
= 15 m/s.
18
(15)2
2  0.2  10
x = 56.25 m.
hence, x =
31.
32.
Let the time be t0 when the relative motion starts
 20 t 0 
then, 20 t0 - 0mg = m 

mm
or, 20 t0 - 0 mg = 10 t0
or, 10 t0 = 0 mg
  mg 
t0 =  0
 sec.
 10 

 
| L || r  p | = mv 0a
33.
If you consider plank and block as a system acceleration is g sin . Since there is no tendency of
relative
motion
between
block
and
plank,
acceleration
of
both
are same
i.e. g sin .
34.
kx = m2 (0 + x)
 x  1 cm.
35.
kx = m2 (0 + x)
 x  1 cm.
36.
R1 = 20 N
R2 = 5 N
37.
As it is moving with = v  1m / s , a = 0
net force acting = F  mg sin2   cos2   mg
mg cos
o
90

m
38.
(A block)

m1 g
N2

N1
N1
(B block)

m2 g
N2
m3g (C block)
N3
mg sin
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39.
T1 = 5( 10 + 2) = 60 N.
T2 = 3 ( 10 + 2) = 36 N.
40.
Friction on 4 kg has to be maximum for non zero tension
maximum friction force 4 kg block
fmax = 8 N
30 = T + fmax
T = 30 – 8 = 22 N
for 6 kg block
22 – 10 = f = 12 N
41.
If  > 1,
fmax = mg > mg
Therefore friction force acting.
f = mg
Net force exerted by the table on the block is
N2  f 2 =
42.
T = mg
f
f
2 mg
N
kx = m2 (0 + x)
 x  1 cm.
43. (a)
NY
Nx
A
B
mg
1
(b) Nx = M2
2
Ny = Mg
Net force by the thin rod is
F=
44.
2
N2x  Ny = M g2 
. . . (ii)
 4L2
4
Acceleration of the system
90 – T 1 = 1  15
 T1 = 75N
75 – T 2 = 2  15
T2 = 45N
. . . (i)
a=
90
= 15 m/s 2
6
1kg
T1
90N
2kg
T2
T1
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45.
F
N + F sin30 – Mg = 0
N = mg – F sin30

fr = N =  (mg – Fsin30)
As the block is moving uniformly

F cos30 – (mg – Fsin30) = 0
F (cos30 + sin30) =  mg
mg
0.3 100 10
F=
=
cos 30   sin 30
3
1
 0. 3 
2
2
= 296N  300N
N
46.
10m/s
f
m
20
20
mgsin20
S=
mgcos20
mg
Fsin30
30
F cos30
fr
N
mg
f = frictional force
N = mgcos20
F = N = 0.2 (mgcos20)
mgsin20 + f = ma
mgsin20 + 0.2 (mgcos20) = ma
a = 3 + 1.8 = 4.8 m/s 2 down the plane
 distance block will travel up the plane;
O = 102 – 2 (4.8)s
100
m
9.6
S= 10.4 m
47.
For motion of point C ,
F - 2Tcos = 0
T = F/2cos
Consider the motion of mass at A towards
B or vice versa . Then as component of T
in the direction of motion will be
Tcos(90 - ) = Tsin
So if ax is the acceleration of m along x
axis then from F = ma
Tsin = max
or
ax =
A
B
T
L
T sin 
m
T
X-axis
T

T
C
F
Y-axis
(2)
From (1) and (2)
F
sin 
F
F
x
ax =


tan  
2
2 F cos 
m
2m
2m L  x2
48.
The smallest velocity v so that it leaves without sliding
down should be such be such that the required
centripetal force at that instant must be completely
contributed by weight.
v
R
http://www.rpmauryascienceblog.com/

49.
Total force exerted by the branch on the monkey
=
50.
mv 2
R
v2 = Rg
v = Rg .
mg =
482  202 =
2704
For 10 kg mass
T- 0.2x10g = 10 a
For 5 kg
5g-T = 5 a
On adding 1 and 2
a=g/5
given s=4m
so, v= 4m/s
. . . (1)
. . . (2)
http://www.rpmauryascienceblog.com/
51.
N1
(i) N1 by the inclined plane
(ii) m1g by the gravity
(iii) T by the string
T
m1

T
(i) N2 by the inclined plane
(ii) m2g by the gravity
(iii) T by the string.
For m1
T = m1 g sin  … (i)
For m2
T = m2 g sin  … (ii)
m1 sin  = m2 sin .
N2
m2
0
90

m2g
52.
m = 1000 kg, v = 10 m/s, u = 0, a = ? s = 50 m
v 2  u2
100
a=

 1 m/s2
2s
2  50
F = ma = 1000  1 = 1000 N
53.
(i) mg - T = ma
T = Ma
Mmg
 F = 2T =
Mm
N
T
T
M
T
Mg
T
m
mg
(ii) N = ma
f = N = ma
ma > mg
amin = g/
f
a > g/
M
ma
m
N
mg
54.
In frame of wedge
N + (mg cot ) sin  = mg cos 
 N = 0,
so block is falling freely.
 It’s acceleration is g.
mg cot 

g cot 

mg cos 
mg sin 
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55.
T
T
FBD of m;
and T – mg = ma …(1)
FBD of 2m;
mg
Adding (1) and (2) mg = 3ma
 a = g/3
Hence velocity of ‘m’ after moving up 6.54 m is;
and 2mg – T = 2ma …(2)
2mg
 9.81 
 (6.54) = 2(3.27) (6.54)
 3 
V2 = 2 (g/3) (6.54) = 2 
 V = 6.54 m/s upwards
Velocity of ‘2m’ at that instant = 6.54 m/s downwards.
When string is cut ‘m’ falls to the ground from a height of 13.08 + 6.54 = 19.62 m
 19.62 = – 6.54 + 1/2 (9.81) t2 where ‘t’ is the time taken to reach the ground
t = 2.7 seconds
‘2m’ falls a distance of 13.08 – 6.54 = 6.54 m
t = 0.6 seconds
56.
Friction on 4 kg has to be maximum for non zero tension
30 = T + fmax
T = 30 – 8 = 22 N
for 5 kg block
22 – 10 = f = 12 N
57.
4mg – T = 4ma
T – mg = ma
3g
1
a=
; x = at 2
5
2
1 2
1
4 – x = at 
x = at 2
2
2
4
20
2
t=


sec.
a
3g
3
vA = v B  at  24 m/s
58.
aA = g sin 45 – 0.2g cos 45 = 4 2 m / s 2
aB = g sin 45 – 0.3 g cos 45=
aAB = 0.5
2
2 m/s
1
a AB t 2
2
2 2
t2 =
 4  t = 2 sec.
0.5 2
1
sB = a B t 2  7 2 m
2
1
sA = a A t 2  8 2 m
2
sAB =
7
2 m/s 2
2
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59.
FBD of plank:
f
(only horizontal forces are shown)
 a1 =
f
m
… (i)
For man:
ma1 + f = ma
from (i) 2ma1 = ma
(acceleration of plank w.r.to ground)
 a1 = a/2
acceleration of man w.r.t. plank = a -
a a

2 2
acceleration of man w.r.to ground = aMP + aPG =
f=
60.
a
a
+ (- ) = 0
2
2
ma ma

2
2
By C.O.E .theorem
1
1
mu2  mgl(1  cos120)  mv 2
2
2
gives v = 3gl
B
v
T
120
mg
0
at point B,
T + mg cos 600 =
by putting v =
mv 2
l
l
A
3gl
u=
5
we get, T = mg .
2
61.
kx0 = m2g
6gl
…(i)
1 2
kx 0
2
m2g
F = m1g +
2
Fx0 = m1 gx0 +
… (ii)
62.
For block to move down the plane without acceleration
F + mg sin30 – fr = 0
F =  mg cos30 –mg sin30
 3
3 1
=2  10 

 
 2 2 2 
= 20 [1.06 – 0.5]
F= 11.2N
For block to move up the plane without acceleration
F – mg sin30 –  mg cos30= 0
F = mg [sin30 +  cos30]
= 20 [1.06 + 0.5] = 31.2N
63.
(i) 20 sin 30 = F +
3
(20 cos 30)
2
F.B.D. of block ;
http://www.rpmauryascienceblog.com/
F = 10 -
30
2
F=

=
16
2
10(1.4)  30
2
=-
16
N
F
f
2
down the plane.
Mg=20N
0
30
3
( 20 cos 30)  20 sin 30
2
(ii) F =
=
=
64.
F.B.D. of block ;
30
2
44
2
N
F
+ 10
N.
f
Mg=20N
0
30
As it slides down with constant velocity v

mg sin  - f = 0

f = mg sin 
But f = mg cos 

mg cos  = mg sin 

tan  = 
Now when projected upwards with velocity v0.
f + mg sin  = ma
mg cos  + mg sin  = ma
2g sin  = a

0 = v 20 - (2g sin ) 2(s)
v
mg

v 20
4g sin 
whenever it comes to a halt, the force acting on it are mg sin  down the plane and f = mg sin  up
the plane

it will not slide down further.

65.
s=
As the same string connects the ladder monkey
and countermass the centre of masses will
move same distance.
Let's say, ladder moves by a distance x
Centre of mass displacement for M = x
Centre of mass displacement for M - m and m
L

(M  m)  x   m(  x ) (M  m) L  m  0
2

2
=

M
M
 Mx  m
=
M
m
m - Mx = Mx ;
x=
2M
x
L

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66.
As the system is in dynamic equilibrium
For block B
T = NB
. . . (I)
NB = mBg
. . . (ii)
For block A
NA = 0
. . . (iii)
NA + T = mAg
. . . (iv)
solving (I), (ii), (iii), (iv)
mAg = mB g
2
m
 mB = A =
= 10 kg

0.2
FBD
NB
NB
T
T
B
T
MBg
C
NA
T
A
NA
from (iv) T = mAg = 20 N
T = kx Where x is the expansion in the spring
T
20

x=
=
= 0.01 m
k
2000
Energy stored in the spring
1
1
E = kx2 =  2000  10-4 = 0.1 J.
2
2
67.
68.
T – mg cos  = m(v)2 / 
T = mg cos  + m(v)2 / 
From COE,
1
1
mv 2  mv 2  mg(1  cos )
2
2
T
a A = aB = a
fL = NA = 0.2  50 = 10 N
If A and B both are at rest
T = f and T = 100 but fL = 10 thus A and B will move.
For B,
100 – T = 10 a
… (i)
For A,
T – 10 = 5a
… (ii)
from (i) and (ii) we get
a = 6 m/s2
v

mg
mg cos 
T
NA
T
f
50 N
FBD of A
100 N
FBD of B
http://www.rpmauryascienceblog.com/
69.
70.
Here acceleration of A is twice the
acceleration of B
Let acceleration of B be ‘a’ then
acceleration of A = 2a. then
T = 5  2a = 10 a
… (i)
50 – 2T = 5a
… (ii)
from (i) and (ii) 50 = 5a
 a = 2 m/s2
thus acceleration of A = 2a = 4 m/s2
acceleration of B = a = 2 m/s2.
a=
T
NA
T
T
a
50
50
FBD of A
12N
 2m / s2
6kg
FBD of B
P
R
Q
m2
m1
T R = (2 + 1)kg  2 m/s2 = 6 N
TP = (2kg)  (2 m/s2) = 4 N
T Q = (2+0.5)kg  2 m/s2
T Q = 5 N.
1 kg
R
2 kg
2kg
P
0.5 kg
72.
using relation, vf = vi + axt
vi = u
ax = - g sin 
u
t=
g sin 
As there is no friction, so time to move from point
A to B and time to come back from B to A will be
same.
2u
T = 2t =
g sin 
 Fx = 0  10 – T = 0
 T = 10 N.
TR
TP
Q
2 kg
71.
12 N
TQ
vf = 0
u
B
m
g sin

fixed
A
N
10 N
T
100 N
73.
Fapp = 6 cos60 = 3N
(fs)max = N = .3 (20 - 33)
(fs)max > Fapp
Block will not move hence distance travelled is 0 m.
74.
Since 1 > tan  so cube will not slip on the wedge. Hence force of friction between the ground and
wedge is zero.
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75.
10 ( 5 î  6 ĵ  5k̂ ) N
76.
(i) Velocity of ball relative to elevator = 30 m/s
Velcoity of ball relative to ground = 10 + 30 = 40 m/s
40 2 = 120m
Maximum height attained by ball = 40 
2  10
(ii) When they meet again their displacement is same
1
10t = 40t -  10  t2
2
t = 6 sec
77.
When the spring has aquired a stable configuration the free body diagrams for the blocks can be
shown as follows.
T
N1
T
A
mg
T
C
B
2mg
T
3mg
If a be the common magnitude of their accelerations
T = ma
(1)
T + 2mg - T = 2ma
(2)
3mg - T = 3ma
(3)
Adding all the three equations , we get
5 
5

5mg = 6ma 
a= g

T = 3m g  g = mg/2
6 
6

mg
mg
K(l) =

(l) =
(l = change in length of the spring)
2
2K
78.
For limiting case
6-T-2=1a
T-4=1a

a=0
it will take infinite time.
F=6
2N
2N
T
1
2N
T
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79.
The FBD of A,B,C are shown
T = ma1
… .(1)
2mg – 2T = 2ma2
….(2)
3mg – T = 3ma3
….(3)
constraint relation :
a3+2a2 – a1 = 0
….(4)
solving the equations
9g
g
7g
a1 =
, a2 =
, a3 =
10
10
10
9
T=
mg
10
a1
mg

T
F1
T
2T
a2


2mg
80.
a3
3mg
T - m g = ma
… (i)
T = (a +g) m
M
m = m + L  x 
L

x 

 T = (a +g) m  M 1  
 L 

Tx = 0 = (a +g)(m+M)
 Fore exerted on the ceiling = (m + M) (a + g) downward.
T
L-x
m’g
81.
(a) FBD of man ; T is the tension in the
string
 N + T = 700
. . .(i)
N
T
700 N
FBD of box, T = N + 350
. .. (ii)
Solving (i) and (ii)
N = 175 N
(weight shown by the machine = 17.5 kg)
T
N
350 N
(b) Now if N should be 700 N.
equ (i) becomes 700 – N –T = 70 a
and (ii) becomes 350 +N- T = 35 a
solving (i) and (ii) we get
T = 2100 N
82.
(i) Let at x = x0 , 5kg starts slipping
F = kx0 = 25 a
m = 5 kg, M = 20 kg
a
http://www.rpmauryascienceblog.com/
ma
5x 0
x
5 kg
 0
mg
25
5
20 kg
5g = ma
a = g
x0
= 0.2  10
x0 = 10 m
5
distance from starting point is 9 m when mass of 5 kg starts slipping.
a=
(ii) Acceleration of 20 kg
f = friction force = mg
kx - mg = Ma
x = 14 + 1 = 15
5  15 - 0.2  5  10 = 20 a
75 - 10 = 20 a
65
a=
= 3.25 m/s 2.
20
Acceleration of 5 kg
mg = ma
a = g = 0.2  10 = 2 m/s
83.
Let f be the upward force. Then
F – 5 mg = 5 ma
… (i)
F – mg – F12 =ma
…(ii)
from (i) and (ii) F12 = 4.92 N
84.
Drawing the FBD of both blocks writing the
equations of motion for both
T – 5g = 0
…(i)
N + 7 – 10 g
…(ii)
Solving for N & T
we get T = 50 N & N = 50 Newton
mg
kx
5 kg
f
f
kx
20 kg
N
T
5
T
10
5g
10g
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85.
T2
T1 = T 2 sin 
Mg = T2 cos 
T1

 tan 
Mg

T1 = Mg tan 
T2 = Mg sec 

T1
Mg
a
12
12

= 2 m/s2
24
6
 f = 4  a = 8N
86.
a=
87.
By Newton's law
F - f = m. aC
torque about centre of mass
 = f. r
I  = f.r
4kg
F
f
mr 2
2f
 = f. r   =
2
mr
Since pure rolling takes place
 ac = .r
a
2f
mac
 c 
 f=
r
mr
2
2f
 F - f = m.
m
F
f=
3
 direction of frictional force is opposite to the F.
88.
Velocity of ball just before 1st collision =
2gh , downward.
Velocity of ball just after 1st collision = e 2gh , upward.
Velocity of ball just before 2nd collision = e 2gh , downward.
Velocity of ball just after 2nd collision = e2 2gh , upward.
v2 = e2 2gh
(a) so,
2gh  e2 2gh
or,
h = e4 h = 0.32 cm
2gh [1  e]
= 7589.5 N
t
e 2gh [1  e]
Average impulsive force (2nd collision) =
= 1517.9 N.
t
(b) Average impulsive force (1st collision) =
http://www.rpmauryascienceblog.com/
89.
Initial energy of the system = mgR
A
m
final energy when the cylinder reaches the bottom of the
track B =
1
1
mv 2 + M v12
2
2
B
where v is the absolute velocity of 'm'
M
and v1 is the absolute velocity of 'M'.
1
1
mv2 + M v12
. . . (I)
2
2
Initial momentum of the system = 0
mg(R-r) =

final momentum when cylinder has reached bottom of the track B
= mv - Mv1
. .. (ii) (assuming track moves towards left)

mv - Mv 1 = 0

mv
1
1
 mv 
v1 =
and mg(R-r) = [m] [v]2 + [M] 

M
2
2
M 
=
v2
2
or v2 =
. .. (2)
2
=
1
1 m2 v 2
mv2 +
2
2 M

m2 
m



M 

2g(R  r )  2gM(R  r ) 


 m
 Mm 
1

 M


1/ 2
 2gM(R  r ) 
v= 

 Mm 
m  2gM(R  r ) 
and v 1 =
M  M  m 
1/ 2
 2g(R  r ) 
 m

 M(M  m) 
1/ 2
90.
Since 1 > tan  so cube will not slip on the wedge. Hence force of friction between the ground and
wedge is zero.
91.
For 10 kg mass
T- 0.2x10g = 10 a
. . . (1)
For 5 kg
5g-T = 5 a
. . . (2)
On adding (1) and (2)
a=g/5
given s=4m ; v2 = 2(g/5)(4) = 16
so, v= 4m/s
92.
kx0 = m2g
1 2
kx 0
2
m2g
F = m1g +
2
Fx0 = m1 gx0 +
…(i)
… (ii)
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93.
94.
For 10 kg mass
T- 0.2x10g = 10 a
For 5 kg
5g-T = 5 a
On adding 1 and 2
a=g/5
given s=4m
so, v= 4m/s
FBD of block
T
. . . (1)
. . . (2)
FBD of wedge
T cos 
T
N
T
T sin 
N sin 
N1
mg sin 
N
F
mg cos  mg
N cos 
mg
Equation of motion for the block:
T – mg sin  = 0 along inclined plane
…(i)
N – mg cos  = 0 perpendicular to the inclined plane
…(ii)
For the wedge
T + N sin  - T cos  - F = 0 Horizontally (F is spring force)
From (i), (ii) and (iii) ;
mg sin  + (mg cos ) sin  - mg sin  cos  - F = 0
 F = mg sin 
mg sin 
Hence, compression =
.
k
95.
At top most point,
mv 2
T + 2mg =
r
In critical condition, v 2 = 2rg
Applying conservation of energy (or work energy theorem) between top and bottom points.
1
1
mu2  mv 2  mg2r  mg.2r
2
2
 u2 = v2 + 8gr = 10 gr
 u =
96.
10gr
FBD of block
FBD of wedge
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N
T
N
2T
N
T
N
T
mg
Mg
N
Let accelerations of block in the downward vertical direction be ab and that of wedge in forward
direction is aw.
Hence, equation of motion, mg – T - N = mab
…(i)
N = maw
…(ii)
Horizontal motion of wedge : T – N = Maw
…(iii)
Constraint equation:
Let AB = x, BC = y, CD = z
Hence length x + y + z = constant …(iv)
d2 z
d2 x
Here y is constant; - 2  a w and 2  ab
dt
dt
Hence, differentiating (iv) twice with respect to
time; ab = aW (in magnitude)
mg
Solving, ab = aw=
M  m(2   
C
D
m
B
A
Required acceleration of the block relative to ground
2mg
= a b2  a b2 =
M  m(2  )
97.
Writing constraint relation
yA = yB tan 
differentiate w.r.to. t we get
vA = vB tan 
… (i)
using COE,
1
1
mgh = m( v B )2  mv 2A
… (ii)
2
2
putting value vA in equation (ii) and solving we get
vB =
2gh cos , vA =
98.
N
fr


mg
(non - I.F.R.)
FBD of block
ma
2gh sin 
yA

yB
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N
a
N
Mg
(I.F.R.)
FBD of wedge
N
Therefore, ma cos = mg sin
a = g tan
macos
masin
mg cos
mg sin
N
fr + ma cos = mg sin
N = mg cos + ma sin
frmax = N
 min(mg cos + ma sin) = mg sin - ma cos
gsin   a cos 
g tan   a
 min =
=
gcos   a sin 
g  a tan 
99.
fr
ma

x1 + x4 = 1
x3 – x4 + x2 – x4 = 2
a1 + a4 = 0
a3 + a2 – 2a4 = 0
a2 + a3 – 2(-a1) = 0
 2a1 + a2 + a3 = 0
mg
x1
x4
A m1
a1
x2
a2
B m2
x3
C m3
100.
In lift frame, F.B.D of masses 4kg, 2kg, and pulley are
T
a1
4a0
4 kg
T
T1
2a0
2 kg
a1
T
4g
Block (A)
Block (B)
a1  acceleration of 4kg and 2kg w.r.to lift
Equation of block A  4g  4a0  T  4a1
(1)
Equation of block B 
Equation of Pulley T1 = 2T
(2)
(3)
From (1), (2) & (3)
2a0  T  2g  2a1
T
Pulley- (C)
2g
a3
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a1  2m / s2 and T1  32N
(a) acceleration of 4kg w.r.t0 lift = 2m/s2
acceleration of 4kg w.r.t0 ground = 2+4=6 m/s2
(b) Force exerted by the pulley on ceiling will be =T1 =32 N
a/2
101.
N1
F1
N1
T
F2
M2 g
N2
F1
a
M2
a/2
F2
M
R
r
T1 =2T
M
N2
F3
F4
V1
Pure rolling is assumed for both rollers
T – F1 – F2 = M2a
V2
M1g
. . . (i)
a
a
1 =
, 2 =
2R
2r
M1g – 2T = M1(a/2)
From (i) and (ii)
M1g – 2(F1 + F2) = a(
. . . (ii)
M1
 2M2 )
2
2M1g  4(F1  F2 )
a=
. . . (iii)
(M1  4M2 )
Taking torque about lower contact point of M
 MR2

3
a
F1  2R = I1 1 = 
 MR2  1  MR 2
=

2
2
2R


F1 =
3
Ma
8
3MRa
4
. . . (iv)
Taking torque about lower contact point of m
 mr 2

3
3
a
 mr 2   2  mr 2 . = mr

4
2
2
2r


F2  2r = I22 = 

F2 =
3
ma .
8
. . . (v)
From (iv) and (v),
F1 + F2 =
3
3
a(M  m)  a(5  2.5)
8
8
=
3  75
45
a
a
80
16
Putting this in (iii)
a=
 a=
2  1  10  4  (45 / 16)a
1 4
16
m/s2
13
(a) Acceleration of block M1 =
a
8
m/s2

2 13
(b) From FBD of M,
F1 – F3 = M(a/2)
F1 =
3
3
16 30
Ma   5 

N(Rightward)
8
8
13 13
a/2
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1
8
10
F3 =
N (rightward)
13
F3 =  Ma
(c)
Acceleration of smaller roller =
a
8

m/s2.
2 13
[4]
102.
Free body diagrams :
f 1
f 2
C
A
Q
P
f1
f1
f2
f2
MAg
MCg
Cylinder B
Cylinder A
Cylinder C
MBg
For the cylinder B
MBg – (f1 + f2) = MB aB
f1 + f2 = 25 – 2.5aB
Taking torque about point P for cylinder A
P = IP A = f1  2r1 + MAg  r
. . . (i)
1
2
[  2  (0.2)2 + 2  (0.2)2 ] A = f1  0.4 + 2  10  0.2
0.1 2 A = 0.4 f1 + 4
Taking torque about point Q for cylinder C
Q = IQ C = f2  2r2 + Mc.g  r2
. . . (ii)
1
2
[  1 (0.1)2  1 (0.1)2 ] c  f2  2  0.1  1 10  0.1 ]
0.015 c = 0.2 f2 + 1
Also, aB = A  0.4
A =
. . . (iii)
1
aB  2.5aB
0.4
and aB = C  0.2
C= 5 aB
Putting values of A  C in (ii) and (iii) ,
0.12  2.5 aB = 0.4 f1 + 4
 0.3 aB = 0.4 f1 + 4
0.015  5aB = 0.2 f2 + 1

0.15 aB = 0.4f2 + 2
from (iv) and (v)
(0.3 + 0.15)aB = 0.4(f1 + f2) + 6
0.45 aB = 0.4 (25 – 2.5 aB) + 6
aB =
16
= 11.03 m/s2
1.45
. . . (iv)
. . . (v)
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103.
Since 1 > tan  so cube will not slip on the wedge. Hence force of friction between the ground and
wedge is zero.
104.
F.B.D. of A
T
 mAg - T = mAa
(1)
mA
F.B.D. of B relative to L

N = ma
f = N Here f = limiting friction
Also f = mg

ma = mg

a = g/
(2)
f
N
ma
mg
F.B.D. of M
N

T - N = Ma

T = (M + m)a
(3)
T
From (1) , (2) and (3),
105.
mA =
M  m
μ  1
(i) FBD of man ; T is the tension in the string
 N + T = 600
. . .(i)
N
T
600 N
FBD of box, T = N + 300
. .. (ii)
Solving (i) and (ii) 2N + 300 = 600
2N = 300
N = 150 N
(weight shown by the machine = 15 kg)
T
N
300 N
(ii) Now if N should be 600 N.
equ (i) becomes N + T – 600 = 60 a
and (ii) becomes T – N – 300 = 30 a
By solving (i) and (ii)
T = 1800 N
106.
Let T be the tension in the spring.
Now, elongation produced in the spring = (r - 0).
. . . (i)
. . . (ii)
a
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Tension = T = k(r - 0)
Linear velocity of the motion = v = 2rn
. . . (i)
mv 2
m(2rn)2
=
= 42rn2m.
r
r
Equating (I) and (ii), we get,
K(r - 0) = 42 rn2 m
k 0
or r =
k  4 2n 2m
 Centripetal force =
42n2m 0k
substituting, T =
107.
(k  4 2n2m)
. . . (ii)
.
FBD of m (for maximum frequency
N
m
v2/r
m
r

Fr

mg
For maximum value of n the mass will have a tendency to move upwards & so frictional force will be
acting downwards.
mv 2
sin 
r
mv 2
Fr + mg si  =
cos 
r
Also Fr = N & v = 2n1r
N - mg cos  =
FBD of m (for minimum frequency)
N1
Fr
2
v /r
m
Fr

mg
108.
. . . (ii)
g(sin    cos 
= maximum frequency allowed.
r(cos    sin  
1
2
Solving n1 =
. . . (I)
mv 2
sin
r
mv 2
mg sin  - Fr =
cos 
r
Fr = N1,
v = 2n2 r.
N1 - mg cos =
. . . (iii)
. .. (iv)
g(sin    cos )
1
= minimum
2
r(cos    sin )
frequency allowed.
Solving n2 =
If the real acceleration of the rod perpendicular to the surface of wedge be a, and acceleration of
the wedge be A.
mgcos - N = ma
(1)
Nsin = MA
(2)
Constraint equation can be written as
a = Asin
(3)
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Using equations (1) , (2) and (3) we get
 a 

(mgcos - ma)sin = M

 sin  

and
109.
M 

mgcossin =  m sin  
a
sin  

mg cos  sin 
a=
M 

 m sin  

sin  

mg cos 
A=
M 

 m sin  

sin  

 12 – fmax = 0

fmax = 12 N.
12
3

S =
=
= 0.3
40
10
F = 9 (a)
a = F/9 m/s 2
mgcossin = masin +
Ma
sin 
a
N
mgcos


N
Nsin
A
N
4kg
12 N
5kg
12 N
fmax
40 N
N1
F
90 N
For this ‘a’ to be in 4 kg block force on it to be (4) (F/9) and this force is to be provided by friction
between 5 kg and 4 kg block.
But fmax = 12 N
9

F
(12) = 27 N and resulting
acc  3 m/s2
4
check :Let F = 28 N
28

common acceleration =
m/s2
9
for 4 kg to have this acceleration for force needed = 4 (28/9)
112
=
N
9
112
But max. it can get is 12N <
9

4 kg block will slip.
110.
At equilibrium mg = T
Let strings are further elongated by a vertically downward force F. Due to this extra tension F in
strings, tension in each spring increase by 2F. Hence increase in elongation of springs is
2F 2F 2F
2F
and
respectively.
,
,
k1 k 2 k 3
k4
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According to geometry of the arrangement, downward displacement of the block from its equilibrium
 2F 2F 2F 2F 
position is y = 2 



.
 k1 k 2 k 3 k 4 
If the block is released now, it starts to accelerate upwards due to extra tension F in string.
Hence restoring force on the block
y
F=1
1
1
1
4 

 
 k1 k 2 k 3 k 4 
 Restoring acceleration of block =
F
m
y
a=-
1
1
1
1
4m  

 
 k1 k 2 k 3 k 4 
1
Hence 2 =
1
1
1
1
4m  

 
 k1 k 2 k 3 k 4 
1
1
1
1
4m  

 .
 k1 k 2 k 3 k 4 
Hence T = 2
111.
After time t, the point A has fallen through
1
x1  gt 2
2
and, the platform has moved upward by
1
x 2  a0 t 2
2
A
x1
A
a0
x2
v1
x
v2
x2
1
g  a0  t2
2
The normal reaction of the platform consists of weight component and thrust component.
N = Nwt + Nthrust
1
Nwt =  x (g + a0) =  g  a0 2 t 2
2
dm
Nthrust = vrel
dt
vrel = relative velocity of chain w.r.t platform
vrel = v 1 + v2 = gt + a0t = (g + a0)t
The total length of the chain resting on the platform is x = x1 + x2 =
dm dm dx

  v rel
dt
dx dt
2
N thrust    g  a0  t 2
1
2
2
2
Total reaction is N    g  a0  t 2    g  a0  t 2 =
3
2
 g  a0  t 2
2
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112.
N
N
Nsin
arelcos
m


mg
arel
a0
M


(1)
Let the bar of mass m slides with an acceleration arel in downward direction and prism moves
towards right with an acceleration a0
Now arel = ab.p. (acceleration of block w.r.t. prism)
acceleration of block w.r.t. ground is
ab.g. = ab.p. + ap.g. = -a0 + arelcos
(in horizontal direction)
ab.g. = arelsin
(in vertical direction)
Now choosing the x and y axis along horizontal and vertical direction we have by f.b.d. (1)
mg - Ncos = mab.g. = marel sin
(1)
Nsin = mab.g. = m(arelcos - a0)
(2)
Now by f.b.d. (2)
Nsin = M.a0
(3)
By (1) and (2)
N = m[gcos - a0sin]
(4)
Now by (3) and (4)
m(gcos = a0sin)sin = M.a0
mg.cossin = a0(M + msin2)
g cos  sin 
or a0 =
m
2 
  sin 
M

113.
F.B.D. of A
T
 mAg - T = mAa
(1)
mA
F.B.D. of B relative to L

N = ma
f = mg, Here f = limiting friction

ma = mg

a = g/
(2)
f
N
ma
mg
F.B.D. of M

T - N = Ma

T = (M + m)a
N
T
From (1) , (2) and (3),
mA =
 M  m
  1
(3)
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114.
F.B.D. of man
N
f1 = force between man and the plank.
f1
mg

f1 = ma
F.B.D. of plank
(i)
N = mg
(ii)
N
f2 = force of friction between the plank and surface.
f1
F
f2
f2
mg


115.
f1 can have any value in the following range.

F - f2max  f1  F + f2max
F - (m + M)g  ma  F + (m + M)g
F  m  M  g
F  m  M  g

a 
m
m
m
M
The system is illustrated in the figure. Let the tension
T
in the string be T and the accelerations of the system
T
T
be a. The equation of motion for the masses are for
T
mass 2m,
2m
0
T - 2mg sin 300 = 2ma
. . . (I)
30
5m
450
and for mass 5m
2mg
5mg sin 450 - T = 5ma
5mg
5mg
- T = 5 ma
. . (ii)
2
5mg
Adding (i) & (ii),
- mg= 7ma
. . . (iii)
2
 5

or 
 1 g = 7a
 2

 5


 1
2
g
Hence the acceleration of the system is a = 
7
from (i) T = mg + 2ma
 5

2mg
 1
 2
 = mg(5  5 2 )

T = mg +
7
7
5mg(1  2 )
=
7
The force on this pulley is the resultant of the tension in the string on the two sides.
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The angle between the two tensions is (600 + 450) = 105. Therefore the force on the pulley is 2T
1
mg
1
cos (105/2)0 = 2T cos 52
= 10
(1 + 2) cos 52 .
2
7
2
116.
F.B.D. of A
T
 mAg - T = mAa
(1)
mA
F.B.D. of B relative to L

N = ma
f = mg, Here f = limiting friction

ma = mg

a = g/
(2)
f
N
ma
mg
F.B.D. of M

T - N = Ma

T = (M + m)a
N
T
From (1) , (2) and (3),
117.
M  m
μ  1
F = ma
(a is absolute retardation of m)
F = Ma
(a is absolute acceleration of M)
Relative acceleration of m = a + a’
 v 20 = 2(a + a)s
 s=
118.
mA =
v 02
Mmv 02

2(a  a) 2F(M  m)
T = m  g = ma
T = (a +g) m
M
m = m + L  x 
L

x 

 T = (a +g) m  M 1  
 L 

Tx = 0 = (a +g)(m+M)
 Fore exerted on the
ceiling
=
(+M)(a+g)
downward.
T
L-x
m’g
a
(3)
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119.
The FBD of A,B,C are shown
T = ma1
… .(1)
2mg – 2T = 2ma2
….(2)
T = 3ma3
….(3)
constraint relation :
a3+ a3 = 2a2
….(4)
solving the equations
a1 = (3/5)g, a2 =(2/5)g, a3 = g/5,
T = (3/5)mg.
a1
mg

T
N1
N3
2T
a2

T

a3
2mg
120.
Let the time be t0
when the relative motion starts
then,
 20 t 0 
20 t0 - 0mg = m 

mm
or, 20 t0 - 0 mg = 10 t0
or, 10 t0 = 0 mg
  mg 
t0 =  0
 sec.
 10 
121.
A bead of mass 'm' is fitted on to a rod can slide on it
without friction. At the initial moment the bead is in the
middle of the rod. The rod moves translationally in a
horizontal plane with an acceleration 'a' in a direction
forming an angle  with the rod. Find the acceleration of
the bead relative to the rod.
3mg
N cos 
N
N sin 
ar cos 

ar

ar sin 

mg
( relative acceleration is simply the vector difference between the absolute acceleration)
or
ay = ar cos  + a
. . . (i)
and
ar sin  = ax - 0
or
ax = ar sin 
. . . (ii)
From FBD of the bead (projecting forces vertically and horizontally)
mg - N cos  = m ar sin 
. . . (A)
and
N sin  = m(ar cos  + a)
. . . (B)
eliminating N between (A) and (B)
sin  = mar + ma cos 
g sin  - a cos 
or
mg
ar =
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122.
123.
124.
Let the plane moves to the left with acceleration a.
According to FBD of the block (mass = m) pseudo force
ma acts on the block to the right.
Normal force N = mg cos  + ma sin 
Force of friction f = N (down the plane)
The block just begins to slip up the plane, when
ma cos  = mg sin  + f
Simplifying, we get
(sin    cos g
a=
cos    sin 
Substituting the values of  and ,
a = 22.4 m/s2
Lower
Block
5t  m1g = 0
0.5  1  10
t=
= 1 sec
5
m1g = m2a
5
a=
= 2.5 m/s2
2
Upper
Block
=
5t
N
m1g
3g
As
 = 45 ,  =
(b)
d
3g

sin  
d
2L
2 2 L


for  = 45 ,  =
0
=
3g
2L
By COM
mv 0
3
4mv 0 v 0
1
 m
v=

 

3M
24
 M 32 
By COAM
ML2  mv 0 L
 mv 0 L =

12
3 2
2
 mv 0 = Mv -

 d 
3g
sin d
2 L 0
3g cos
L
= 4.6 rad/s
ma
ma sin 
mg sin 
0
 = 37
mg cos 
N
N
m2g

mg
2
125.

f
m1g
d 3g

sin 
dt 2 L

3g

cos 
2
2L
N
N
(a) For the position of the rod, shown in the figure.
Total external torque = Mg[(L/2) sin]
as
I
 ML2  d
We get, Mg[(L/2) sin] = 

 3  dt
or
ma cos 

a
http://www.rpmauryascienceblog.com/
mv 0L 4 ML2


2 3
12
8mv 0 v 0
=

ML
4L
v(t) = v0 - gt
2g
(t) = 0 +
t
R
for pure rolling v(t) = R(t)
v
 v 0 - gt = 2gt  t = 0
3g

126.
127.
mgsin - f = ma
fR = I
a
 =
R
 from equations (1) (2) and (3)
mg sin  mg sin  5
a=

 g sin 
I
2
7
m 2
m m
5
R
2 5
 f =  m  g sin 
5 7
…(1)
…(2)
….(3)
2
mgsin (up the incline)
7
work done by the f is zero since it is static frictional force.
5
v(t) = at = g(sin)t
7
5
a(t) = gsin
7
f=
128.
Instantaneous mass of the car = M + mt
d
F=
(M  mt )v   M dv  m d ( vt)
dt
dt
dt
Fdt = Mdv + md(vt)
integrating
Ft = Mv + mvt
Ft
v=
M  mt
129.
Angular momentum will be conserved mV0 OA sin300 = mvr
V0
= Vr
2
from energy conservation
1
1
1
1
k(OA)2 + mv 02 = kr2 + mv2
2
2
2
2
2
v
2 + v 20 = 2r2 + 02
4r
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 r = 1.68 m/s
130.
mg + ma0 – 2T = ma
2T. R = I
a = R
from the above equations
2
2m
a = (g  a0 ) and 2T =
(g – a0 )
3
3
… (i)
… (ii)
… (iii)
131.
T1
a0
R
m1a0 T2
a0
T1
m0g
T2
132.
m1g
…(i)
…(ii)
…(iii)
…(iv)
For minimum,
N1
T
f
T
ma
N2
ma
f
mg
mg
ma + f = T

ma + kmg = T
N1 = mg
N2 = ma
kma + T = mg
g(1  k )
amin =
1 k
for maximum (ma is more than T)
N1
T
T
ma
f
mg
ma = T + kmg
N2
ma
f
mg
m2a0
a
T2
T1 = m0a0
T2 = 2T2
(T2 + m1a0) – m1g = m1a
m2g – (T2 + m2a0) = m2a
4m1m2  m0 (m1  m2 )
a=
g
4m1m 2  m0 (m1  m 2 )
T2
a
m2g
http://www.rpmauryascienceblog.com/
N2 = ma
T = kma + mg
g(1  k )
amax =
.
1 k
133.
FBD of plank:
f
(only horizontal forces are shown)
f
 a1 =
… (i) (acceleration of plank w.r.to ground)
m
For man:
ma1 + f = ma
from (i) 2ma1 = ma
 a1 = a/2
a a
acceleration of man w.r.t. plank = a - 
2 2
a
a
acceleration of man w.r.to ground = aMP + aPG =
+ (- ) = 0
2
2
ma ma
f=

2
2
134.
Suppose acceleration of block A and B be a (rightward and b(leftward) respectively. Then
horizontal block C is b (leftward) and (a+ b).
FBD of the blocks:
N1
T
N2
C
A
N3
T
N2
1g
T
7.5 g
T = 7.5 a
. . . (i)
g - T = 1 (a + b)
. . . (ii)
N2 = 1 (b)
. . . (iii)
T - N2 = 6 (b)
. . . (iv)
6g
Solving:
a = 1.04 m/sec2 ; b = 1.11 m/s2
Therefore horizontal and vertical component of acceleration of block C are b = 1.119 m/s2
(leftwards) and a + b = 2.159 m/s (downwards) respectively.
Hence, resultant acceleration
=
135.
(1.119)2  (2.159)2 = 2.43 m/s2.
FBD of block
FBD of wedge
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N
T
N
N
2T
T
N
T
mg
N
Mg
Let accelerations of block in the downward vertical direction be ab and that of wedge in forward
direction is aw.
Hence, equation of motion, mg – T - N = mab
…(i)
N = maw
…(ii)
Horizontal motion of wedge : T – N = Maw
…(iii)
Constraint equation:
Let AB = x, BC = y, CD = z
Hence length x + y + z = constant …(iv)
d2 z
d2 x
D
Here y is constant; - 2  a w and 2  ab
dt
dt
Hence, differentiating (iv) twice with respect to
time; ab = aW (in magnitude)
mg
Solving, ab = aw=
M  m(2   
Required acceleration of the block relative to ground
2mg
= a b2  a b2 =
M  m(2   )
136.
C
m
B
A
Force acting on block along the face of wedge
T- mgsin30 = ma
…(1)
Force acting on sphere

Weight (mg) - Buoyant force  F 

m 
g  –T = ma
2 
…(2)
Solving we get a = 0
137.
60
50 cm
150 m/s
m2
100 gm
m1
2.9 kg
Momentum before collision = m1v1 = 100  10–3 (150)
Momentum just after collision = (m1 + m2)v2
= [(100 10–3) + 2.9]v2
 Conserving momentum in the horizontal direction
m1v1 = (m1 + m2)v 2

v2 =
m1v1
= 5 m/s
m1  m2
Now let velocity at 60 angle be V;
 Conserving energy between these two positions
1
1
(m1  m2 )v 22  (m1  m2 )v 2 + (m1 + m2)g [L (1 - cos)]
2
2
 v = 4.74 m/s.
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T is the tension in the string
T
(m1  m2 )V 2
T – (m1 + m2)gcos60 =
L
60
 T = 150 N.
(m1 + m2)g
138.
Let at any instant of time m1 and m2 is displaced x1 (down) and x2 (down) respectively and let x2 > x1
elongation of the spring at this instant x = x2 – x1
Free body diagrams :
T
F
N1
kx
m1
N2
m3
m2
kx
T
m1g
m3g
m2 g
Considering the forces parallel to plane
m1g sin  + kx – F = m1 a1
i.e.
T + m2g sin  - kx = m2a2
i.e.
m1g sin   kx  F
m1
m 2g sin   kx  m3 g
a2 =
m2  m3
a1 =
as m3g T = m3a2
Acceleration of m2 relative to m1, a = a2 – a1
m g sin   kx  m3 g m1g sin   kx  F
= 2

. . .. (iii)
(m 2  m 3 )
m1
from (iii)
1
1
( 4  10   1000 x  1 10) (2  10   1000x  15)
2
2
a=

5
2
a = 8.5 – 700 x
. . .. (iv)
dv
a=v
= 8.5 – 700 x
dx
0
 vdv  
0
x max
0
(8.5  700 x )dx
Which gives xmax =2.4 cm
Substituting the value of xmax in (iv)
a = -8.5 m/s 2
Hence system moves with acceleration 8.5 m/s 2 up the plane.
. . . (i)
. . . (ii)
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139. (i) For just slipping f = mg
F - mg = Ma
m
f
and mg = ma
F
M
a=
mM
ma
F
min =
=
mg
(m  M)g

F
(ii) if  = min 
3
3(M  m)g
F - mg = MA
mg = ma
F(3M  2m)

A=
3(M  m)M
F
a=
3(M  m)
displacement of block of mass M in time t seconds
1 F(3M  2m) 2
s1 =
t
2 3(M  m)M
work done by friction force
w = mgs 1
F(3M  2m)t 2
F
=
mg F
3(M  m)g
6(M  m)M
=
F2m(3M  2m)t 2
18(M  m)2 M
displacement of block of mass m in t seconds
1
F
s2 =

t2
3(M  m)
2
Relative displacement
Ft 2
 3M  2m 
s = s 1 - s2 =
 1

6(M  m) 
M

Ft 2
(M  m)
Ft 2
=
3(M  m) M
3M
Heat dissipiation = mg s
=
=
F
Ft 2
 mg 
3(M  m)g
3M
=
F 2t 2m
.
9(M  m)M
f
F
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140.
(i) From the figure shown
2y
2x 
L
tan 
dx
dt

1 dy
0
tan  d t
y
L/2


O
y/tan
y/tan
x
dx
dy
  tan 
dt
dt
If v and u be the magnitudes of the velocities of bar and wedges then
v = utan
. . . (i)
or h = X tan 
X = horizontal displacement of each wedge when the bar come down by h
1
1
Mgh = Mv2 + 2  mu2
. . . (ii)
2
2
From (i) and (ii) 

Speed of wedges at the same instant =
(ii) Workdone by N sin  on the system of the
wedge
1
N sin . X =
mu2 (Work energy theorem)
2
1
N sin . X = mu2 (Work energy theorem)
2
1
2Mgh
N sin . X = m.
2
M tan 2   2m
Mmg sec 
which gives N =
M tan 2   2m
141.
2 M tan 2 gh
Velocity of bar c when it strikes floor =
M tan 2   2 m
2Mgh
M tan 2   2 m
N cos 
N
N sin 

F.B.D. of A
T
 mAg - T = mAa
(1)
mA
F.B.D. of B relative to L
f
N
ma

N = ma
f = mg, Here f = limiting friction

ma = mg

a = g/
(2)
mg
F.B.D. of M
N
T

T - N = Ma

T = (M + m)a
(3)
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From (1) , (2) and (3),
mA =
 M  m
  1
142.
F – 2T = Ma
3T = m1a1
T = m2a2
Form constraints relation
2a = 3a1 + a2
by solving we get
2m1F
a2 =
m(m1  9m2 )  4m1m2
143.
When particle starts sliding, friction on it is equal to limiting friction.
Now normal force = m (g + b)
. . .(i)
Let the particle starts sliding after moving a distance x on the board. Considering tangential motion;
u = 0 ; a = 0.25 m/s2 ;
2
2
v = u + 2ax 
v=?
2
v = x/2
. . . (ii)
Now, resulting horizontal acceleration
 v 2 2

=    a 2 
 R 



1/ 2
1/ 2
 x 2

= 
  a2 
 2R 

This acceleration is provided by limiting friction
 x  2

m 
  a2 
 2R 


144.
x=
1/ 2
 N = m (g + b)
24 = 2 6 m.
Let retardation up the board =  relative to the elevator .
Now, v2 = u2 +2as

(0)2 = (42)2 - 2 (1.6)

 = 10 m/s2
Horizontal component of resultant acceleration of the block
=  cos 370 = 8 m/s2
vertical component of resultant acceleration of the block
= a -  sin 370 = (a - 6) m/s2
N1
FBD of the block:
N1
N
mg
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N
N
mg
N sin 370 + N cos 370 = m (8)
. . . (i)
N cos 370 - mg - N sin 370 = m(a - 6)
. . . (ii)
Solving 40  + 4  a + 3a = 20
. . . (iii)
Now considering motion of the block down the plane
Let acceleration of the block relative to elevator =  this time.
Hence :
v2 = u2 + 2x where v = 4 m/s , u = 0
Hence  = 5 m/s2
x = 1.6 m
Now horizontal component of resultant acceleration
=  cos 370 = 4 m/s2
vertical component of resultant acceleration = a -  sin 370
= (a - 3) m/s2
Hence :
N1 cos 370 + N1 sin 370 - mg = m (a - 3)
. . . (iv)
N1 sin 370 - N1 cos 370 = m (4)
. . . (v)
Hence solving 40  + 4a - 3a = 5
. . . (vi)
Solving (iii) and (vi) a = 2.5 m/s 2 ,  = 0.25
145.
(i) FBD of man ; T is the tension in the
string
 N + T = 600
. . .(i)
N
T
600 N
FBD of box, T = N + 300
. .. (ii)
Solving (i) and (ii) 2N + 300 = 600
2N = 300
N = 150 N
(weight shown by the machine = 15 kg)
T
N
300 N
(ii) Now if N should be 600 N.
equ (i) becomes 600 – N –T = 60 a
and (ii) becomes 300 +N- t = 30 a

T = -60 a
and 900 + 60 a = 30 a
or
900 = - 30 a
a = - 30 m/s 2 (i.e. upwards)
and
T = - 60 (-30)
= 1800 N.
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146.
147.
If m1 moves x then m/2 moves 2x
Now if acceleration of m is a then
acceleration of m/2 2a
Equation of block A 2T = ma
(1)
mg
Equation of block B
 T  ma
2
(2)
Now (1) + (2) 2  mg = 3ma
a = g/3 =10/3 m/s2
mg 5m
(b) from (1) T =

N
6
3
10m
tension in thread PQ =
N
3
5m
(c) R  T 2  T 2 
2N
3
a
T
m
R
T
T
m/2
T
.2a
B
mg
2
Acceleration of block, a = g sin  -  g cos 
0


x
vdv  g(sin    cos .x)dx

0
x = 0,
 0 = g (sin .x -
0
2 tan 
.

 Total distance travelled =
For v to be max., a = 0, x =
vmax = sin 
 cos .x 2
)
2
2 tan 
.

tan

g
.
cos .
148.
Since 1 > tan  so cube will not slip on the wedge. Hence force of friction between the ground and
wedge is zero.
149.
(i) When A loses contact with ground
T1 =mA g = 1  10 = 10 N
. . . .(i)
T = 2T1
and F = 40 t = 2T = 4 T1
Hence, T1 = 10 t
. . .. (ii)
from (I) and (ii) t = 1 sec, Hence A will lose contact at t = 1 sec
similarly for B
T1 =MBg = 2  10 = 10 t
t = 2 sec, Hence B will lose contact at t = 2 sec.
Similarly, for C
T = 3  10 = 20 t
F = 40 t
P
T
Q
T1
T
T1
A
B
C
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t = 1.5 sec, hence C will lose contact at t = 1.5 sec.
T1 – mAg = mAa
dv
10 t – 1  10 = 1 
. . . (iii)
dt
(ii) Velocity of A when B loses contact with ground
T1
2
v
 dv =  (10t  10)dt
0
a
1
which gives, v = 5 m/s
3/2
Velocity of A when C lose contact v =
 (10 t  10) dt
mAg
T = 20t
1
= 5/4 m/s
(iii)  For block C
T - mcg = mca
dv
20 t – 310 = 3
dt
3 dv = 20 t dt – 30 dt
v
t
t
20
30
  dv 
tdt

dt
3 3 / 2
3 3 / 2
0
H
2
  dy 
0


3/2
[
mcg

10t 2
30
] dt
 10t 
3
4
H = 0.14 m.
v=
10 2
30
t  10 t 
3
4