Phy 1 (13)
Short Description
gmh...
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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 4ˆi + 8 ˆj + 10 kˆ . 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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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 ˆi 2ˆj 6kˆ ) to position (14 ˆi 13ˆj 9kˆ ) and a force (constant) of 4ˆi ˆj 3kˆ 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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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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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
http://www.rpmauryascienceblog.com/ 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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(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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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 .
http://www.rpmauryascienceblog.com/ (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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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 ( ˆi 2ˆj ) m/s. It collides with another body and the impact time is 0.1s, resulting in a velocity v f (6ˆi 4ˆj 5kˆ ) 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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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
http://www.rpmauryascienceblog.com/ (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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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
http://www.rpmauryascienceblog.com/ 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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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
http://www.rpmauryascienceblog.com/ 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
http://www.rpmauryascienceblog.com/ 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
http://www.rpmauryascienceblog.com/ 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 45with the vertical, calculate (a) its angular acceleration , (b) its angular velocity.
A 45
= 0.5
m1 = 1kg
=0
m2 =2kg
F = 5t
http://www.rpmauryascienceblog.com/ 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
http://www.rpmauryascienceblog.com/ 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
http://www.rpmauryascienceblog.com/ 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
http://www.rpmauryascienceblog.com/ 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
http://www.rpmauryascienceblog.com/ SOLUTION OF ABOVE PROBLEM
1.
2.
For no motion Fcos60 (mg + Fsin60) F/2
F/2 g Fmax = 20 N
2 3
( 3 g +
F 3 ) 2
sy = 12m
Fcos60
f Fsin60
Taking F as the net force 4 8 10 a ˆi ˆj kˆ m/s 2 3 3 3 1 2 s ut at 2 1 4 8 10 ˆ = 0 ˆi ˆj k 9 2 3 3 3 = 6 ˆi 12ˆj 15kˆ
sx = 6m; 3.
1
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
http://www.rpmauryascienceblog.com/ 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=33=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
http://www.rpmauryascienceblog.com/
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 (14ˆi 13ˆj 9kˆ ) (3ˆi 2ˆj 6kˆ ) = 11ˆi 11ˆj 15kˆ Therefore, work = F.S ( 4ˆi ˆj 3kˆ ).(11ˆi 11ˆj 15kˆ ) = 44 + 11 + 45 = 100 units.
23.
Mg
Ly L
Ly 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
http://www.rpmauryascienceblog.com/
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
http://www.rpmauryascienceblog.com/ 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 mm 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 = m2 (0 + x) x 1 cm.
35.
kx = m2 (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
http://www.rpmauryascienceblog.com/ 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 = m2 (0 + x) x 1 cm.
43. (a) NY
Nx
A
B
mg
1 (b) Nx = M2 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
http://www.rpmauryascienceblog.com/ 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 = Mm
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
http://www.rpmauryascienceblog.com/ 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
http://www.rpmauryascienceblog.com/ 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
http://www.rpmauryascienceblog.com/
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 - 33) (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.
http://www.rpmauryascienceblog.com/
75.
10 ( 5ˆi 6ˆj 5kˆ ) 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=1a T-4=1a a=0 it will take infinite time.
F=6 2N 2N
T 1
2N T
http://www.rpmauryascienceblog.com/ 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 = ma … (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
http://www.rpmauryascienceblog.com/
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 24 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. v2 = 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 Mm 1 M 1/ 2
2gM(R r ) v= Mm
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)
http://www.rpmauryascienceblog.com/
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
http://www.rpmauryascienceblog.com/ 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
http://www.rpmauryascienceblog.com/ 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
http://www.rpmauryascienceblog.com/ 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 = I22 = 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
http://www.rpmauryascienceblog.com/ 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)
http://www.rpmauryascienceblog.com/ 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
http://www.rpmauryascienceblog.com/ Tension = T = k(r - 0) Linear velocity of the motion = v = 2rn
. . . (i)
mv 2 m(2rn)2 = = 42rn2m. r r Equating (I) and (ii), we get, K(r - 0) = 42 rn2 m k 0 or r = k 4 2n 2m
Centripetal force =
42n2m 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 = 2n1r
N - mg cos =
FBD of m (for minimum frequency) N1 Fr
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 Fr = N1, v = 2n2 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)
http://www.rpmauryascienceblog.com/ Using equations (1) , (2) and (3) we get a (mgcos - ma)sin = M sin
and
109.
M mgcossin = 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
mgcossin = 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
http://www.rpmauryascienceblog.com/ 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
http://www.rpmauryascienceblog.com/ 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.cossin = 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)
http://www.rpmauryascienceblog.com/ 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.
http://www.rpmauryascienceblog.com/ 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 = ma 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)
http://www.rpmauryascienceblog.com/ 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 mm 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 =
http://www.rpmauryascienceblog.com/ 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 2g (t) = 0 + t R for pure rolling v(t) = R(t) v v 0 - gt = 2gt t = 0 3g
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
http://www.rpmauryascienceblog.com/ 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
http://www.rpmauryascienceblog.com/ 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.
http://www.rpmauryascienceblog.com/ 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)
http://www.rpmauryascienceblog.com/ 139. (i) For just slipping f = mg F - mg = Ma m f and mg = ma F M a= mM 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
http://www.rpmauryascienceblog.com/ 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)
http://www.rpmauryascienceblog.com/ 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 = (42)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
http://www.rpmauryascienceblog.com/ 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 + 2x 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 + 4a - 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.
http://www.rpmauryascienceblog.com/
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
http://www.rpmauryascienceblog.com/ 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 – 310 = 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
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