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11. 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

smooth

300

12. A block of metal weighing 2 kg is resting on a frictionless plane. It isstruck by a jet releasing water at a rate of 1 kg /s and at a speed of 5m/s. Calculate the initial acceleration of the block.

2 kg

Black

13. Find the relation between the acceleration of rod A and wedge B in the

arrangement shown in the figure. The entire surfaces are smooth.

B

A

14. All the strings and the pullies in the figure shown are masslessand frictionless. Find the relation between the accelerations ofthe block B and mass m as shown in the figure.

a2a1

B

m

15. Two blocks each of mass m are connected by a masslessinextensible string. They are kept on a fixed wedge as shownin the figure. The surface of wedge is frictionless. If thesystem is just released from rest, find the instantaneousacceleration of blocks.

m

ma

16. In the figure shown, find the acceleration of the block B if thatof the A is 0.3 m/s2. A

B

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17. Initially half of the chain’s length ( = 4 m) is overhanging,what will be the speed of the chain when it just slips off thesmooth table ?

/2

/2

18. A body slides down a smooth curved fixed track which is a quadrant ofa circle with radius 10 m. Find the speed of the body at the bottom ofthe 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 thetension 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 betweenthe 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 initialspeed of the particle.

22. A particle moves from position (3 k̂6 ĵ2î ) to position (14 k̂9 ĵ13î ) and a force (constant) of

k̂3 ĵî4 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 thetension in the chain at a distance y from the rigid support.

24. A block of mass M is placed on a smooth inclined surface asshown in the figure. Draw the free body diagram of the massM. Find tension in the string and the normal reaction exertedby 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 themasses is = 0.1.

M m=0.1

F


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26. A body of mass M is moving in+x direction with accelerationa. Find the minimum value ofcoefficient of friction ,

between the block so that abody of mass m can be heldstationary with respect to M onthe vertical side of M.

mM

a

=0

27. (a) Two blocks of mass 2 kg and 1 kg are at rest on a rough surface andare separated by a distance of 3.75 m as shown in the figure. Thecoefficient of friction between each block and surface is 0.20.

1kg 2kg

The 2 kg block is given a velocity of 8 m/s directed towards the second block. It collides with 1kgblock which is at rest. Find the loss of energy of 2 kg block during collision if coefficient of

restitution is ½. (Take g = 10 2m/s ).

(b) Calculate pressure inside a cylindrical liquid drop of diameter d and surface tension S, ifatmospheric 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/s2 )

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 inwhich an automobile can be stopped while travelling at 54 km/hr.

31. Two blocks are kept as shown. A horizontal time varying force isapplied on upper block (F = 20 t). Find the time when the relativemotion between the blocks starts. [Given that mass of each blockis m kg]

20 tm

m

0

= 0

32. A mass m is moving with a constant velocity v0 along a line y = -a and away from the origin. Findthe magnitude of its angular momentum with respect to origin.

33. On a smooth inclined plane, a block of mass m isplaced over the plank. If the plank and the block isreleased from rest. Find the acceleration of block.

Assume friction between block and plank is sufficientto 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 theother end is connected to a particle of mass 0.5 kg lying on a frictionless horizontal table. Thespring remains horizontal. If the mass is made to rotate at an angular velocity of 2 rad/s, find theelongation 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 theother end is connected to a particle of mass 0.5 kg lying on a frictionless horizontal table. Thespring remains horizontal. If the mass is made to rotate at an angular velocity of 2 rad/s, find theelongation of the spring.

36. Find out the magnitude of R1 and R2 contact forces actingbetween the blocks as shown in figure. (take all the surfaces tobe smooth.) 4 kg 3 kg

1 kg

R2

R1

40 N

37. A block of mass 2 kg is kept on a rough inclined surface and it is movingdownward with a constant velocity of v = 1 m/s along the incline. Findout the value of net force applied by the inclined surface on the block.

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.

B (M2)

A (M1)

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/s

2)T1

T2

3 kg

2kg

40. Two blocks of masses 4 kg and 6 kg connected by amassless string are kept on a rough surface havingcoefficient of sliding friction 0.2. Two horizontal forces of 30N and 10 N is applied on 4 kg and 6 kg blocks. Find thefriction force on both the blocks and tension in the string.(Consider the string to be taught initially)

4 kg 6 kg10 N30 N

=0.2 =0.2

41. Two blocks, each of mass m are attached with a massless andinextensible string. One of the blocks is placed on the rough horizontalsurface of a table for which coefficient of friction is . Find out theforce exerted by the table on the block if > 1.

m

m

42. One end of a massless spring of spring constant 100 N/m and natural length 0.5 m is fixed and theother 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 velocityof 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 rodis vertical and rotates with constant angular speed in such a way that the rod AB remainshorizontal. 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 arekept on a smooth horizontal surface. A horizontalforce F = 90N is applied at the first block. Blocks areattached by light strings. Calculate the tensions T1 and T2 acting on the strings.

T2 T1 T1T2m2m3 m1

F = 90N1Kg2Kg3Kg

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 isgiven a velocity of 10 m/s up the slope. Assuming the coefficient of sliding friction between the blockand 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 alight string of length 2L as shown. A continuousforce 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 byax =

22 xL

x

m

F

F

BA

2L

48. A particle rests on top of a hemisphere of radius R. Find the smallest horizontal velocity that mustbe 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 africtional 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. Thecoefficient 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 .

m2

4m

m1


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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.m1

m2

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 connectedto another mass m by means of a massless string passing over a masslessand frictionless pulley as shown in the figure. Find the force exerted by thepulley on the table.

M

m

(ii) A body of mass M is moving withacceleration a in the direction as shown.Find the minimum value of so that abody of mass m can be held stationarywith respect to M on the vertical side of M.

mM

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

m

g cot

55. Two masses ‘m’ and ‘2m’ are connected by a massless string whichpasses over a light frictionless pulley as shown in fig.1. The massesare initially held with equal lengths of the strings on either side of thepulley. Find the velocity of the masses at the instant the lighter massmoves up a distance of 6.54 m. The string is suddenly cut at thatinstant. Calculate the time taken by each to reach the ground. (g =9.81 m/s2)

13.08 m

2mm

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.

4 kg 6 kg10 N30 N

=0.2


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57. 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 crossthe same level.

4m

m

A

B

4m

58. Two identical blocks A and B are placed on a roughinclined plane of inclination 450. The coefficient offriction between block A and incline is 0.2 and that ofbetween B and incline is 0.3. The initial separation

between the two blocks is 2 m. The two blocks arereleased 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 attainingabove motion position.

450

A

B

B =0.3

A =0.2

2 m

59. A man of mass m is moving with a constantacceleration a w.r.t. plank. The plank lies on asmooth horizontal floor. If mass of plank is also mthen calculate acceleration of plank and man w.r.t.ground, and frictional force extended by plank onman.

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 gl6 horizontally. Find the

tension in the string when string deflects through an angle 1200 from the vertical downward.

61. Two blocks of masses m1 and m2 connected by a non-

deformed light spring rest on a horizontal plane. Thecoefficient of friction between the blocks and the surface isequal to . What minimum constant force has to be applied inthe horizontal direction to the block of mass m1 in order to shiftthe other block?

m1 m2

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 2/3 .

(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 2/3 .


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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 projectedup the same plane with an initial speed v0. 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 anda spring. The spring passes over a frictionless pulley as shown

in figure. Block 'B' slides on the horizontal surface of a fixedblock 'C' and the block 'A' slides along the vertical side of 'C',

both with same uniform speed. The coefficient of frictionbetween 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

A

kC

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 thesurface is 0.2.

B

=0.2 A

(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)

B

smoothA


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70. Two blocks are connected by a rod and fixed at Pand 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)

m1m2 12 N

P RQ

m1 = 2kg, m2 = 3 kg,

mass of the rod = 1 kg

71. A body is thrown upward from A with speed ‘u’ m/son an incline plane having inclination the body willcome back to the point A of projection after some

time. All planes are frictionless. Calculate time.

fixed

u

m

B

A

72. Consider the situation shown in the figure. The

10 kg block is tied with a light string on one sideand a force of 10 N is applied on it from the

other side. All contact surfaces are smooth, and

the block remains at rest.

F = 10 Nm = 10 kg

light string

(i) Draw the free body diagram of the 10 kg block.(ii) Find the tension in the string.

73. A block of mass 2kg is placed on a rough surface as shownin the figure. Find the distance travelled by the block in thefirst 2 sec.

60

2 kgk =0.2

s=0.3

6 N

74. In the shown figure a triangular wedge of mass M and asmall cube of mass m is placed. The coefficient of frictionbetween wedge and cube is 1 and between wedge andground is 2. (2 = 1/2). Find the force of friction onwedge due to ground. It is given that 1 > tan .

M

m1

2

75. A particle of mass 1 kg has an initial velocity ) ĵ2î(vi

m/s. It collides with another body and the

impact time is 0.1s, resulting in a velocity )k̂5 ĵ4î6(v f

m/s after impact. Find average force of

impact on the particle.

76. An open elevator is ascending with zero acceleration . The speed v = 10m/sec. A ball is thrownvertically 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 springconstant 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.

B

A

C

78. 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

2

1

F

79. 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 thestring to be light and inextensible. The masses of the blocks are m, 2m

and 3m respectivelyC

B

A

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 uniformacceleration 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 theceiling?(take x from the point where the string is connected to the

ceiling)

m

L

a

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 manhimself. 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 ropeto 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 20kg 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. A chain consisting of 5 links each of mass 0.1 kg is lifted

vertically with a constant acceleration 2.5 m/s2 as shown in

figure. Find the force of interaction between the top link andlink immediately below it.

` 1

2

3

4

5

84. Two masses of 5 kg and 10 kg connected by a massless string

passing over a frictionless pulley are in equilibrium as shown in thefigure. 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. 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 Nis applied on the first block due to which they move with certain

constant acceleration. Calculate the force between the blocks.

2kg4kg12 N

87. A force F is applied from its centre along horizontal direction Finddirection and magnitude of frictional force acting on the cylinder.(No slipping) r

Fm

88. 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. A block of mass M with a semicircular track of radius R rests ona horizontal frictionless surface. A uniform cylinder of radius 'r'and mass 'm' is released from rest at the top point A. Thecylinder slips in the semicircular frictionless track. How fast isthe block moving when the cylinder reaches the bottom of the

track ?

A m

B

M

90. In the shown figure a triangular wedge of mass M and a smallcube of mass m is placed. The coefficient of friction betweenwedge 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 isgiven that 1 > tan .

M

m1

2

91. Two masses m1 = 10 kg and m2 = 5 kg areconnected 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 isreleased from rest, Calculate the velocity when m2 has descended by 4m .

m2

m1

92. Two blocks of masses m1 and m2 connected by a non-deformedlight spring rest on a horizontal plane. The coefficient of frictionbetween the blocks and the surface is equal to . What minimumconstant force has to be applied in the horizontal direction to theblock of mass m1 in order to shift the other block?

m1 m2

93. Two masses m1 = 10 kg and m2 = 5 kg areconnected by an ideal string as shown in the figure.The coefficient of friction between m1 and thesurface is = 0.2. Assuming that the system isreleased from rest, Calculate the velocity when m2 has descended by 4m .

m2

4m

m1

94. See the diagram, The system is in equilibrium. There is no

friction anywhere. Spring, string, pulley are massless. Find

spring compression.

m

M

k


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95. A cabin is moving upwards with a constant acceleration g. Aboy standing in the cabin wants to whirl a particle of mass m

in a vertical circle of radius . (mass is attached to an idealstring) calculate minimum velocity which should be providedat lowermost point (w.r.t cabin) so that particle can just

complete the circle.

g

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 accelerationof block relative to ground. 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 justbefore A hits the ground.

h

A B

98. 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 thatapplication of acceleration a is not enough to support the block inequilibrium.

Ma

m

99. All the strings and the pullies in the figure shown are massless andfrictionless. Find the relation between the accelerations of theblock A, B and C having masses m1, m2 and m3 respectively.

A m1

B m2

C m3a3

a2

a1

100. A pulley fixed to the ceiling of a lift carries a thread whose ends areattached to the masses of 4kg and 2kg. The lift starts going downwith on acceleration of a0 = 4 m/s

2 relative to ground. Calculate(a) The acceleration of the load 4kg relative to the ground andrelative 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


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101. Consider the situation shown in figure. Both thepulleys and the string are light and all the surfacesare frictionless. Calculate (a) the acceleration ofmass m (b) the tension in the string PQ and (c)force exerted by the clamp on the pulley C.

m2

m PQ

C

102. A pulley fixed to the ceiling of a lift carries a thread whose endsare attached to the masses of 4kg and 2kg. The lift starts goingdown with on acceleration of a0 = 4 m/s

2 relative to ground.Calculate(a) The acceleration of the load 4kg relative to the ground andrelative to lift.(b) The force exerted by the pulley on the ceiling of the lift

(Thread and pulley are massless, g=10 m/s2)

a0

4 kg 2 kg

103. In the shown figure a triangular wedge of mass M and a smallcube of mass m is placed. The coefficient of friction betweenwedge 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 isgiven that 1 > tan .

M

m1

2

104. The figure shows L shaped body of mass M placed onsmooth horizontal surface. The block A is connected to thebody by means of an inextensible string, which is passingover a smooth pulley of negligible mass. Another block B ofmass m is placed against a vertical wall of the body. Findthe minimum value of the mass of block A so that block Bremains stationary relative to the L-shape body. Coefficientof friction between the block B and the verticalwall is .

M

B

A

m

105. Figure shows a man of mass 60 kg standing on a light weighingmachine kept in a box of mass 30 kg. The box is hanging from apulley fixed to the ceiling through a light rope, the other end of whichis 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 hasone end fixed to its centre. To the other end of the spring is attached a mass m which is making nrevolutions per second around the centre. Find the radius r of this uniform circular motion and thetension 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 aconstant rate of "n" revolutions per second. The wall of the funnel makes an angle with the


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113. The figure shows an L shaped body of massM is placed on smooth horizontal surface. Theblock A is connected to the body by means ofan inextensible string, which is passing over asmooth pulley of negligible mass. Another

block B of mass m is placed against a verticalwall of the body. Find the minimum value ofthe mass of block A so that block B remainsstationary relative to the wall. Coefficient offriction between the block B and the verticalwall is .

M

B

A

m

114. A plank of mass M is placed on a rough horizontalsurface and a constant horizontal force F is appliedon it. A man of mass m runs on the plank. Find theaccelerations of the man so that the plank does not

move on the surface. Co-efficient of friction betweenthe plank and the surface is Assume that theman does not slip on the plank.

m

FM

115. A smooth fixed wedge has one face inclined at 300 to thehorizontal and a second face at 450 to the horizontal. Thefaces are adjacent to each other at the top of the wedge.Particles of masses 2m and 5m are held on these respectivefaces connected by a taut inelastic string passing over asmooth pulley at the top of the wedge as shown in the figure.Find the acceleration of the system if the particles aresimultaneously released and show that the force acting on

the pulley is7

10 mg (1 + 2) cos (5221 )0.

2m5m4530

116. The figure shows an L shaped body of massM is placed on smooth horizontal surface. Theblock A is connected to the body by means ofan inextensible string, which is passing over asmooth pulley of negligible mass. Anotherblock B of mass m is placed against a verticalwall of the body. Find the minimum value ofthe mass of block A so that block B remainsstationary relative to the wall. Coefficient offriction between the block B and the verticalwall is .

M

B

A

m

117. 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 frictionlesslyon 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 ?

A

B

m

M

v0


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118. 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?m

L

a

x

119. 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 andthe string to be light and inextensible. The

masses of the blocks are m, 2m and 3m

respectively

C

B

A

120. 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 tm

m

0

= 0

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 themiddle 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 ofthe bead relative to the rod.

a

122. A rough inclined plane with inclination = 370 with the horizontal is accelerated horizontally till ablock of mass m originally at rest with respect to the plane just begins to slip up the plane. Thecoefficient of static friction between the surfaces in contact is = 5/9. Find the acceleration of theplane. (tan 370 = 3/4)

123. A time varying force F = 5t N is applied on the upper block asshown in figure. When will the upper block start moving with

respect to lower block? What is the acceleration of lower block atthat instant (here t is in sec.)

m1 = 1kg

m2 =2kg

= 0.5

= 0

F = 5t

124. A rod of length 1m and mass 4 kg , can rotate freelyin a vertical plane around its end A. The rod isinitially held in a horizontal position and thenreleased. At the time the rod makes an angle45with the vertical, calculate(a) its angular acceleration ,(b) its angular velocity.

A

45


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125. A particle of mass m moving with a speed v0 strikes perpendicularly one end of a uniform rod ofmass M and length L initially resting on a smooth horizontal plane. The particle returns in the same

line with speed of 3

v0 . IfM

m =

32

1 find the linear and the angular speed of the rod.

126. A uniform disc of mass m and radius R is projected horizontally with velocity v0 on a roughhorizontal floor having coefficient of friction equal to . Find the time after which it starts purerolling.

127. A uniform solid sphere of mass m and R starts rolling without slipping down an inclined plane oflength 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 aconstant force F as shown in figure. and spills on the flatcar from a stationary hopper. The velocity of loading isconstant and equal to m kg/s. Find the velocity of the carafter 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 otherend stretched to A as shown in figure. Ball is given

velocity V0 = 2 m/s at an angle 30 to OA. Find themaximum elongation of the spring. Given that springconstant is k = 2 N/m and OA = 1m

v0

30

A O

130. Thin threads are tightly wound on the ends of a uniformsolid cylinder of mass m. The free ends of the threads areattached to the ceiling of a lift. The lift starts moving up withan acceleration a0. Find the acceleration of the cylinderrelative to the lift and the force F exerted by the cylinder on

the ceiling.

`

a0

131. Find the acceleration of the body m1 in thearrangement shown in figure with masses ofrespective bodies. The friction is absent.

Also the masses of the pulleys and thethreads are negligible.

m1 m2

m0


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132. What is the minimum and maximum acceleration with which bar A (Fig.) should be shifted horizontally tokeep bodies 1 and 2 stationary relative to the bar?The masses of the bodies are equal , and the co-efficient of friction between the bar and the bodies is

equal to k. The masses of the pulley and the threadsare negligible, the friction in the pulley is absent.

2

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 plankis 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 nofriction anywhere. Calculate resultant acceleration of block Cwhen the system is released. Pulley is massless. A

B

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. A block of mass m is kept over a fixed smooth wedge. Theblock is attached to a sphere of same mass through fixedmassless pullies P1 and P2. Sphere is dipped inside water asshown. If specific gravity of material of sphere is 2, then findthe acceleration of the sphere.

30

P1 P2

m

m

137. A mass of 2.9 kg is suspended from a string of length 50 cm and is at rest. Another body of mass100 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?

138. Two blocks of masses m1 = 2 kg and m2 =4 kg are attachedby light ideal spring of force constant k = 1000 N/m. Thesystem is kept on a smooth inclined plane inclined 30 0 withhorizontal. A force F=15 N is applied on m1 and system isreleased from rest. The block m2 is attached with a lightstring whose another end is connected with a mass m3 = 1kg. Assume initially the spring is at relaxed position and thesystem is released from rest. Find the

(a) maximum extension of the spring.(b) acceleration of the system at this instant

m3

m2

m1F


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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 asshown in figure. A constant force F is applied on the block of mass M due to which the systemaccelerates

(i) find the minimum value of coefficient of frictionbetween the two blocks to prevent sliding

m

FM

(ii) If the coefficient of friction between the blocks is one third that of calculated in part (i) find the workdone by the friction force during the first t seconds. Also find the energy dissipated into heatassuming the length of M is large enough.

140. Consider the arrangement shown in the

figure. If the system is set free at t = 0with the horizontal bar at a height of h as

shown in the figure, obtain

C

BAh

(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. 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 ofnegligible mass. Another block B of mass m is placed

against a vertical wall of the body. Find the minimumvalue of the mass of block A so that block B remains

stationary relative to the wall. Coefficient of frictionbetween the block B and the vertical wall is .

M

B

m

142. In the shown figure. The blocks and pulley

are ideal and force of friction is absentexternal horizontal force F is applied as

shown in figure. Find acceleration of blockC.

m1

B A

M

m2 C F

143. A circle of radius R = 2m is marked on upper of a horizontal board, initially at rest. An insect startsfrom rest along the circle with a tangential acceleration a = 0.25 m/s2. At the same instant boardaccelerates upwards with acceleration b = 2.5 m/s2. If the coefficient of friction between board andinsect 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 boardforms an angle = 370 with horizontal floor of the elevatoraccelerating 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 v2 = 4 m/s down

A

370

a


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the board when it returns to point A. Calculate acceleration 'a'of the elevator and coefficient of friction between the boardand the block.

145. Figure shows a man of mass 60 kg standing on a light weighingmachine kept in a box of mass 30 kg. The box is hanging froma pulley fixed to the ceiling through a light rope, the other endof which is held by the man himself. If he manages to keep thebox at rest (i) what is the weight shown by the machine ? (ii)what force should the man exert on the rope to get his correctweight on the machine ?

146. Consider the situation shown in figure. Both the

pulleys and the string are light and all the surfacesare frictionless. Calculate (a) the acceleration ofmass m (b) the tension in the string PQ and (c)force exerted by the clamp on the pulley C.

m

2

m PQ

C

147. A mass m is released from rest on the incline. The coefficientof friction is = x, where x is the distance travelled along theincline and is a constant. Find(a) the distance travelled by the mass till it stops, and

(b) the maximum velocity over this distance.

m

148. In the shown figure a triangular wedge of mass M and a smallcube of mass m is placed on the wedge. The coefficient offriction between wedge and cube is 1 and between wedge andground is 2. (2 = 1/2). Find the force of friction on wedge dueto ground. It is given that 1 > tan .

M

m1

2

149. Three blocks A, B and C have masses 1kg , 2kg and 3 kg respectivelyare arranged as shown in figure. The pulleys P and Q are light andfrictionless. 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 Palong 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 withground.(iii) the height by which C is raised when B loses contact with ground

P

F = 40 t

A B C

Q


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SOLUTION OF ABOVE PROBLEM

1. For no motion

Fcos60 (mg + Fsin60)

F/2 32

1( 3 g +

2

3F)

F/2 g Fmax = 20 N

F.B.D.Fcos60

mg

f

Fsin60

2. TakingF

as the net force

k̂3

10 ĵ

3

8î

3

4a

m/s2

2ta2

1tus

= 9k̂3

10 ĵ38î

34

210

= k̂15 ĵ12î6

sx = 6m; sy = 12m sz = 15m

3. isinuicosuvi

j)gtsinu(icosuv f

0v.v f i

t =sing

u

j)cos

u

sinu(icosuv f

| uf | = u cot

4. Acceleration is zero

Frictional force = mg sin 300 = 10 N.

5. Tangent acceleration at =dt

dv = 2t +1

normal acceleration an =R

v 2

(at)t=0 = 1 m/s2

(an)t=0 = 2.0

1

R

v2

0 = 5 m/s2


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6. Fx = max

10 2 cos 45 = 10 a 10 = 10 a a = 1 m/s2.

x

100 N

N

y

450

10 2

7. Limiting friction f L = NN – 100 = m 0 N = 100

f L = 0.4 100 = 40 Nf L > 20 N thus block does not move.20 – f = 10 10f = 20 N in backward direction.

20 N

N

f

100 N

8. Acceleration of both blocks will be same = g sin

9. a =7

21 = 3 ms2

f = 3 3 = 9 N

10. 100 = (3 + 7)a

a = 1010

100 m/s2

consider FBD of B

R = ma R = 7 10 = 70 N.

N1

R

FBD of A

100 N

30 N

N2

R

FBD of B

70 N

11. 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/s2 N – mg cos300 = ma

= 0

N = mg cos 30

=2

3100 = 50 3 N

N

mg sin 30

30 mg = 100 N

mg cos 30 30

12. a =2

15

m

dt/vdm

m

F = 2.5 m/s2 .

13. tanx

y a A = aB tan .

14. a2 = 6a1


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15. mg – T = ma . . . (i)T – mg sin = ma . . . (ii)Solving (i) and (ii) mg - mg sin = 2ma

a =g(1 sin )

2

.

T

mg

ma

NTa

mgkx

16. 0.15 m/s2.

17. 30 m / s .

18. 10 2 m/sec.

19. MgL

yL

20.

cosFF

sinF

21

2

21. zero

22. Displacement of the particle, )k̂6 ĵ2î3()k̂9 ĵ13î14(r r s 12

= k̂15 ĵ11î11

Therefore, work = )k̂15 ĵ11î11).(k̂3 ĵî4(S.F

= 44 + 11 + 45 = 100 units.

23. Mg L y

L

24. N = Mg cos T = Mg sin

Mg sin Mg cos

Mg

NT

25. F – N – f 1 = Ma . . . (i)N – f 2 = ma . . . (ii)

N = 2N.M

F

f 1

NM

Nm

f 2

26. N = maf = N = mama > mg a > g/ amin = g/

N

f

mg

ma

Mm


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27. (a) Velocity of 2 kg block when it strikes 1 kg block is v such that2 2v 8 2 2 3.75

v = 7 m/sFrom COM, 14 = 2

1 2v v . . . (1)

For e, 2 1v v1

2 7

2 1v v =7

2 . . . (2)

From (1) and (2)

1

7v

2 , 2v =7

Loss of energy = 21 492 72 4 = 36.75 J

(b) Let us consider the equilibrium of half of the

liquid drop of length .

2S + p0d = pd

p = p0 +2S

d

d

28. mvdv

F 6x 4dx

v dv 3x 2dx

v 2

o o

vdv 3x 2 dx

v 20m / s.

29. f max > mg sin a = 0f = 30 N

6 kg

=3/2

30

30. The friction force retarding the motion of car = N.= mg = 0.2mg

retardation produced a = -0.2mg

m = - 0.2g

If x is the distance of stopping, then


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v2 = u2 + 2ax, as v = 0, u = 54 5

18 = 15 m/s.

hence, x =2(15)

2 0.2 10

x = 56.25 m.

31. Let the time be t0 when the relative motion starts

then, 20 t0 - 0mg = m

mm

t20 0

or, 20 t0 - 0 mg = 10 t0 or, 10 t0 = 0 mg

t0 =

10

mg0 sec.

32. |pr ||L|

= mv0a

33. If you consider plank and block as a system acceleration is g sin . Since there is no tendency ofrelative motion between block and plank, acceleration of both are samei.e. g sin .

34. kx = m2 (0 + x) x 1 cm.

35. kx = m2 (0 + x) x 1 cm.

36. R1 = 20 NR2 = 5 N

37. As it is moving with = v 1m/ s , a = 0

net force acting = 2 2F mg sin cos mg mg cos

mg sin

90o

m

38. (A block)

(B block)

(C block)

N2 m2g

N3

m3g

N2

m1g

N1

N1


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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 tensionmaximum friction force 4 kg block

f max = 8 N30 = T + f max T = 30 – 8 = 22 Nfor 6 kg block22 – 10 = f = 12 N

41. If > 1,f max = mg > mgTherefore friction force acting.f = mgNet force exerted by the table on the block is

2 2N f = 2 mg

f T = mg

f N

42. kx = m2 (0 + x) x 1 cm.

43. (a)

mg

NY

Nx

A B

(b) Nx = M221 . . . (i)

Ny = Mg . . . (ii)Net force by the thin rod is

F =2

y2x NN =

4

L4gM

22

44. Acceleration of the system a =6

90 = 15 m/s2

90 – T1 = 1 15 T1 = 75N

1kg

90NT1

75 – T2 = 2 15T2 = 45N

2kgT1T2


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45. N + F sin30 – Mg = 0N = mg – F sin30

f r = N = (mg – Fsin30) As the block is moving uniformly F cos30 – (mg – Fsin30) = 0F (cos30 + sin30) = mg

F =

30sin30cos

mg =

2

13.0

2

3

101003.0

= 296N 300N

30

mg

N

f r

Fsin30

F

F cos30

46. N

mf

20 20

mgmgsin20

mgcos20

10m/s

f = frictional forceN = mgcos20F = N = 0.2 (mgcos20)mgsin20 + f = ma

mgsin20 + 0.2 (mgcos20) = maa = 3 + 1.8 = 4.8 m/s2 down the plane distance block will travel up the plane;O = 102 – 2 (4.8)s

S = m6.9

100

S= 10.4 m

47. For motion of point C ,

F - 2Tcos = 0T = F/2cos Consider the motion of mass at A towardsB or vice versa . Then as component of Tin the direction of motion will be

Tcos(90 - ) = Tsin So if ax is the acceleration of m along xaxis then from F = ma

C

F

BA

Y-axis

X-axisT

TL

T

T

Tsin = max or ax =T

m

sin (2)

From (1) and (2)ax =

F

F m

F

m

F

m

x

L x2 2 2 2 2cos

sintan

48. The smallest velocity v so that it leaves without sliding down should be such be such that the requiredcentripetal force at that instant must be completelycontributed by weight.

v

R


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mg =R

mv 2

v2 = Rg

v = Rg .

49. Total force exerted by the branch on the monkey

= 22 2048 = 2704

50. For 10 kg massT- 0.2x10g = 10 a . . . (1)

For 5 kg5g-T = 5 a . . . (2)

On adding 1 and 2a=g/5given s=4mso, v= 4m/s


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51. (i) N1 by the inclined plane

(ii) m1g by the gravity

(iii) T by the string

N1

T

m1

(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

T

m2

m2g

90

52. m = 1000 kg, v = 10 m/s, u = 0, a = ? s = 50 ma = 1

502

100

s2

uv 22

m/s2

F = ma = 1000 1 = 1000 N

53. (i) mg - T = maT = Ma

F = 2T =mM

Mmg

M

m

T T

T

T

mg

Mg

N

(ii) N = maf = N = mama > mg a > g/ amin = g/

N

f

mg

ma

Mm

54. In frame of wedgeN + (mg cot ) sin = mg cos N = 0,so block is falling freely. It’s acceleration is g.

g cot

mg cot

mg sin

mg cos


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55.

FBD of m;

mg

T

and T – mg = ma …(1) FBD of 2m;

2mg

T

and 2mg – T = 2ma …(2)

Adding (1) and (2) mg = 3ma

a = g/3Hence velocity of ‘m’ after moving up 6.54 m is;

V2 = 2 (g/3) (6.54) = 2

3

81.9 (6.54) = 2(3.27) (6.54)

V = 6.54 m/s upwardsVelocity 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 groundt = 2.7 seconds‘2m’ falls a distance of 13.08 – 6.54 = 6.54 mt = 0.6 seconds

56. Friction on 4 kg has to be maximum for non zero tension

30 = T + f max

T = 30 – 8 = 22 Nfor 5 kg block

22 – 10 = f = 12 N

57. 4mg – T = 4maT – mg = ma

a =3g

5 ; x = 2

1at

2

4 – x = 21

at2

x = 21

at2

t =4 20 2

a 3g 3 sec.

v A = Bv at 24 m/s

58. a A = g sin 45 – 0.2g cos 45 = 24 2 m/ s

aB = g sin 45 – 0.3 g cos 45= 27

2 m/s2

a AB = 0.5 2 m/s2

s AB =2

AB

1a t

2

t2 =2 2

40.5 2

t = 2 sec.

sB =2

B

1a t 7 2 m

2

s A =2

A

1a t 8 2

2 m


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59. FBD of plank:f

(only horizontal forces are shown)

a1 =f

m … (i) (acceleration of plank w.r.to ground)

For man:ma1 + f = mafrom (i) 2ma1 = ma a1 = a/2

acceleration of man w.r.t. plank = a -a a

2 2

acceleration of man w.r.to ground = aMP + aPG =a

2 + (-

a

2) = 0

f =ma ma

2 2

60. By C.O.E .theorem

22 mv21)120cos1(mglmu

21

gives v = gl3

at point B,

T + mg cos 600 =l

mv 2

by putting v = gl3

we get, T = mg2

5.

1200

l

Tmg

vB

A

u = gl6

61. kx0 = m2g …(i)

Fx0 = m1 gx0 + 20kx2

1 … (ii)

F = m1g +2

gm2

62. For block to move down the plane without accelerationF + mg sin30 – f r = 0F = mg cos30 –mg sin30

=2 10

2

1

2

3

2

3

= 20 [1.06 – 0.5]F= 11.2N

For block to move up the plane without accelerationF – mg sin30 – mg cos30= 0F = mg [sin30 + cos30]

= 20 [1.06 + 0.5] = 31.2N

63. (i) 20 sin 30 = F +2

3 (20 cos 30)

F.B.D. of block ;


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66. As the system is in dynamic equilibrium

For block BT = NB . . . (I)NB = mBg . . . (ii)

For block A

N A = 0 . . . (iii)

N A + T = m Ag . . . (iv)solving (I), (ii), (iii), (iv)

m Ag = mB g

mB = Am =

2.0

2 = 10 kg

FBD

B

A

C

NB

MBg

T T

T

N A

N A

NB

T

from (iv) T = m Ag = 20 N

T = kx Where x is the expansion in the spring

x =kT =

200020 = 0.01 m

Energy stored in the spring

E =2

1kx2 =

2

1 2000 10-4 = 0.1 J.

67. T – mg cos = m(v)2 / T = mg cos + m(v)2 / From COE,

)cos1(mgvm2

1

mv2

1 22

Tv

mg cos mg

68. a A = aB = a

f L = N A = 0.2 50 = 10 NIf A and B both are at restT = f and T = 100 but f L = 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

T

N A

50 N 100 N

T

f

FBD of A FBD of B


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69. Here acceleration of A is twice the

acceleration of B

Let acceleration of B be ‘a’ then

acceleration of A = 2a. thenT = 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.

T

N A

50 50

T

a

T

FBD of A FBD of B

70.a = 2s/m2

kg6

N12

TR = (2 + 1)kg 2 m/s2 = 6 N

TP = (2kg) (2 m/s2

) = 4 NTQ = (2+0.5)kg 2 m/s

2

TQ = 5 N.

m1m2 12 N

2 kg 1 kg

R TR

2kg P TP

2 kg 0.5 kg

P R

Q

TQQ

71. using relation, vf = vi + axt

vi = uax = - g sin

t = sing u

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.

T = 2t =sing

u2

fixed

u

m

vf = 0

A

B

g sin

72. Fx = 0 10 – T = 0 T = 10 N.

N

10 NT

100 N

73. Fapp = 6 cos60 = 3N(f s)max = N = .3 (20 - 33)(f s)max > FappBlock 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 andwedge is zero.


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75. 10 ( k̂5 ĵ6î5 ) N

76. (i) Velocity of ball relative to elevator = 30 m/sVelcoity of ball relative to ground = 10 + 30 = 40 m/s

Maximum height attained by ball =

102

4040

2

= 120m

(ii) When they meet again their displacement is same

10t = 40t -2

1 10 t2

t = 6 sec

77. When the spring has aquired a stable configuration the free body diagrams for the blocks can beshown as follows.

A B

3mg

T

T

2mg

T

mg

T

N1

C

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

5mg = 6ma a = g65 T =

g

65gm3 = mg/2

K(l) =2

mg (l) =

K2

mg (l = change in length of the spring)

78. For limiting case

6 - T - 2 = 1 aT - 4 = 1 a a = 0it will take infinite time.

T

1

F=6

T

2N2N

2N


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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

a1 =10

g9 , a2 =

10

g, a3 =

10

g7

T = mg10

9

F1

T

a1

mg

2mg

2T

a2

T

3mg

a3

80. T - m g = ma … (i)T = (a +g) m

m = m + xLLM

T = (a +g)

L

x1Mm

Tx = 0 = (a +g)(m+M)

Fore exerted on the ceiling = (m + M) (a + g) downward.

aL-x

T

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 aand (ii) becomes 350 +N- T = 35 a

solving (i) and (ii) we getT = 2100 N

82. (i) Let at x = x0 , 5kg starts slipping

F = kx0 = 25 a

m = 5 kg, M = 20 kg


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a =525

5 00 xx

5g = maa = g

5x0 = 0.2 10 x0 = 10 m

mg

ma5 kg

20 kgkx

mg

distance from starting point is 9 m when mass of 5 kg starts slipping.

(ii) Acceleration of 20 kgf = friction force = mgkx - mg = Max = 14 + 1 = 155 15 - 0.2 5 10 = 20 a75 - 10 = 20 a

a =

20

65 = 3.25 m/s2.

f

5 kg

kx20 kg

f

Acceleration of 5 kg

mg = maa = 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 bothT – 5g = 0 …(i)

N + 7 – 10 g …(ii)Solving for N & T

we get T = 50 N & N = 50 Newton

5g

T

5

10g

T

10

N


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85. T1 = T2 sin Mg = T2 cos

tanMg

T1

T1 = Mg tan T2 = Mg sec

Mg

T1

T2

86. a =6

12

42

12

= 2 m/s2

f = 4 a = 8N

4kg

a

87. By Newton's lawF - f = m. aC torque about centre of mass = f. rI = f.r

2

mr 2 = f. r =

mr

f 2

Ff

Since pure rolling takes place ac = .r

mr

f 2

r

ac f =2

mac

F - f = m.

m

f 2

f =3

F

direction of frictional force is opposite to the F.

88. Velocity of ball just before 1st collision = gh2 , downward.

Velocity of ball just after 1st collision = e gh2 , upward.

Velocity of ball just before 2nd collision = e gh2 , downward.

Velocity of ball just after 2nd collision = e2 gh2 , upward.

v2

= e2

gh2 (a) so, gh2ehg2 2 or, h = e4 h = 0.32 cm

(b) Average impulsive force (1st collision) =t

]e1[gh2

= 7589.5 N

Average impulsive force (2nd collision) =t

]e1[gh2e

= 1517.9 N.


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89. Initial energy of the system = mgR

final energy when the cylinder reaches the bottom of the

track B =2

1mv2 +

2

1M 21v

where v is the absolute velocity of 'm'

and v1 is the absolute velocity of 'M'.

mg(R-r) =2

1mv2 +

2

1M 21v . . . (I)

A m

BM

Initial momentum of the system = 0

final momentum when cylinder has reached bottom of the track B

= mv - Mv1 . .. (ii) (assuming track moves towards left)

mv - Mv1 = 0 . .. (2)

v1 =M

mv and mg(R-r) =

2

1[m] [v]2 +

2

1[M]

2

M

mv

=

2

1mv2 +

M

vm

2

1 22

=

M

mm

2

v 22

or v2 =

mM

)r R(gM2

M

m1

)r R(g2

v =

2/1

mM)r R(gM2

and v1 =

2/12/1

)mM(M)r R(g2m

mM)r R(gM2

Mm


91. For 10 kg massT- 0.2x10g = 10 a . . . (1)

For 5 kg5g-T = 5 a . . . (2)

On adding (1) and (2)a=g/5given s=4m ; v2 = 2(g/5)(4) = 16so, v= 4m/s

92. kx0 = m2g …(i)

Fx0 = m1 gx0 + 20kx2

1 … (ii)

F = m1g +2

gm2


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93. 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/5given s=4mso, v= 4m/s

94. FBD of block

NT

mg

mg sin

mg cos

FBD of wedgeT T cos

F

N1

N cos

N sin

T

N

mg

T sin

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 wedgeT + 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

Hence, compression =k

sinmg .

95. At top most point,

T + 2mg =r

mv 2

In critical condition, v2 = 2rg Applying conservation of energy (or work energy theorem) between top and bottom points.

r 2.mgr 2mgmv21mu

21 22

u2 = v2 + 8gr = 10 gr

u = gr 10

96. FBD of block FBD of wedge


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N

mg

T

NT

N

N

T

Mg

N 2T

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)

Here y is constant; - w2

2

adt

xd and b2

2

adt

zd

Hence, differentiating (iv) twice with respect totime; ab = aW (in magnitude)

B A

C

mD

Solving, ab = aw= 2(mM

mg

Required acceleration of the block relative to ground

= 2b2b aa =

)2(mM

mg2

97. Writing constraint relationy A = yB tan differentiate w.r.to. t we getv A = vB tan … (i)using COE,

y A

yB

mgh = 2 A2

B mv2

1)v(m

2

1 … (ii)

putting value v A in equation (ii) and solving we get

vB = gh2 cos , v A = gh2 sin

98.

mg

ma

f rN

(non - I.F.R.)

FBD of block


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a

Mg N

N

(I.F.R.)

FBD of wedge

Therefore, ma cos = mg sin a = g tan

Nmacos

masin

mg cos

mg sin

fr + ma cos = mg sin N = mg cos + ma sin fr max = N min(mg cos + ma sin) = mg sin - ma cos

min =gsin acos

gcos a sin

=gtan a

g a tan

mg

ma

f rN

99. x1 + x4 = 1

x3 – x4 + x2 – x4 = 2

a1 + a4 = 0a3 + a2 – 2a4 = 0a2 + a3 – 2(-a1) = 0 2a1 + a2 + a3 = 0

A m1

B m2

C m3

x1

x2

x3

x4

a1

a2

a3

100. In lift frame, F.B.D of masses 4kg, 2kg, and pulley are

4 kg

4g

T

a1

Block (A)

4a0

2 kg

2g

T

a1

Block (B)

2a0 T1

Pulley- (C)

T T

1a acceleration of 4kg and 2kg w.r.to lift

Equation of block 0 1 A 4g 4a T 4a (1)

Equation of block B 0 12a T 2g 2a (2)

Equation of Pulley T1 = 2T (3)

From (1), (2) & (3)


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21a 2m/ s and 1T 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

101.

T

F1 N1 M2gF2 N2

F3 V1

F1

a/2

F2

F4 V2

N2

M1g

T1 =2T

R M

N1

r M

a

M2

a/2

a/2

Pure rolling is assumed for both rollers

T – F1 – F2 = M2a . . . (i)

1 =a

2R

, 2 =a

2r

M1g – 2T = M1(a/2) . . . (ii)From (i) and (ii)

M1g – 2(F1 + F2) = 1

2

Ma( 2M )

2

a =

1 1 2

1 2

2M g 4(F F )

(M 4M ) . . . (iii)

Taking torque about lower contact point of M

F1 2R = I1 1 =2

2 21

MR 3 aMR MR

2 2 2R

=

3MRa

4

F1 =3

Ma8 . . . (iv)

Taking torque about lower contact point of m

F2 2r = I22 =

22 2

2

mr 3 amr mr .

2 2 2r =

3mr

4

F2 =3

ma8

. . . . (v)

From (iv) and (v),

F1 + F2 = 3 3

a(M m) a(5 2.5)8 8

=

3 75 45

a a80 16

Putting this in (iii)

a =

2 1 10 4 (45 /16)a1 4

a = 1613

m/s2

(a) Acceleration of block M1 =a 8

2 13 m/s2

(b) From FBD of M,F1 – F3 = M(a/2)

F1 =3 3

Ma 58 8

16 30

N(Rightward)13 13


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F3 = 1

Ma8

F3 =10

N13

(rightward)

(c) Acceleration of smaller roller =a 8

2 13 m/s2

.

[4]

102. Free body diagrams :f 1

P

M Ag

f 1 f 2

MBg

f 2

f 2

MCg

Q

f 1

Cylinder A Cylinder B Cylinder C

C A

For the cylinder B

MBg – (f 1 + f 2) = MB aB

f 1 + f 2 = 25 – 2.5aB . . . (i)

Taking torque about point P for cylinder A

P = IP A = f 1 2r 1 + M Ag r

[ 21

2 (0.2)2

+ 2 (0.2)2 ] A = f 1 0.4 + 2 10 0.2

0.1 2 A = 0.4 f 1 + 4 . . . (ii)

Taking torque about point Q for cylinder C

Q = IQ C = f 2 2r 2 + Mc.g r 2 [ 2 2 c 2

11 (0.1) 1 (0.1) ] f 2 0.1 1 10 0.1

2 ]

0.015 c = 0.2 f 2 + 1 . . . (iii)

Also, aB = A 0.4

A = B B1

a 2.5a0.4

and aB = C 0.2

C= 5 aB

Putting values of A C in (ii) and (iii) ,

0.12 2.5 aB = 0.4 f 1 + 4 0.3 aB = 0.4 f 1 + 4 . . . (iv)0.015 5aB = 0.2 f 2 + 1 0.15 aB = 0.4f 2 + 2 . . . (v)

from (iv) and (v)

(0.3 + 0.15)aB = 0.4(f 1 + f 2) + 6

0.45 aB = 0.4 (25 – 2.5 aB) + 6

aB =16

1.45 = 11.03 m/s2


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104. F.B.D. of A

mAg - T = mAa (1)

mA

T

F.B.D. of B relative to L

ma

mg

f

N

N = maf = N Here f = limiting friction

Also f = mg ma = mg

a = g/ (2)F.B.D. of M

T

N

T - N = Ma

T = (M + m)a (3)

From (1) , (2) and (3), m A = 1μ

mM

105. (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 = 300N = 150 N(weight shown by the machine = 15 kg)

T

N

300 N

a

(ii) Now if N should be 600 N.equ (i) becomes N + T – 600 = 60 a . . . (i)and (ii) becomes T – N – 300 = 30 a . . . (ii)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).


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Tension = T = k(r - 0) . . . (i)Linear velocity of the motion = v = 2rn

Centripetal force =r

mv2 =

r

)rn2(m 2 = 42rn2m. . . . (ii)

Equating (I) and (ii), we get,K(r - 0) = 42 rn2 m

or r =mn4k

k22

0

substituting, T =)mn4k(

kmn4220

22

.

107. FBD of m (for maximum frequencyN

v2/r

Fr

mg

m

mr

For maximum value of n the mass will have a tendency to move upwards & so frictional force will beacting downwards.

N - mg cos =r

mv2 sin . . . (I)

Fr + mg si =

r

mv2cos . . . (ii)

Also Fr = N & v = 2n1r

Solving n1 =21

sin(cosr

cos(sing = maximum frequency allowed.

FBD of m (for minimum frequency)N1

v /r

Fr

mg

m

Fr

N1 - mg cos =r

mv2sin . . . (iii)

mg sin - Fr =r

mv2 cos . .. (iv)

Fr = N1, v = 2n2 r.

Solving n2 =21

)sin(cosr )cos(sing

= minimum

frequency allowed.

108. If the real acceleration of the rod perpendicular to the surface of wedge be a, and acceleration ofthe wedge be A.

mgcos - N = ma (1)Nsin = MA (2)

Constraint equation can be written asa = Asin (3)


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Using equations (1) , (2) and (3) we get

(mgcos - ma)sin =

sin

aM mgcossin = masin +

sin

Ma

mgcossin = asin

Msinm

a =

sin

Msinm

sincosmg

and A =

sin

Msinm

cosmg

mgcos

Nsin

A

N

aN

109. 12 – f max = 0 f max = 12 N.

S =40

12 =

10

3 = 0.3

F = 9 (a)a = F/9 m/s2

4kg5kg

12 N

N

12 N

40 N

f max

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 f max = 12 N

F 4

9 (12) = 27 N and resulting acc 3 m/s2

check :- Let F = 28 N

common acceleration =9

28 m/s2

for 4 kg to have this acceleration for force needed = 4 (28/9)

=9

112N

But max. it can get is 12N < 9

112

4 kg block will slip.

110. At equilibrium mg = TLet strings are further elongated by a vertically downward force F. Due to this extra tension F instrings, tension in each spring increase by 2F. Hence increase in elongation of springs is

321 k

F2,

k

F2,

k

F2and

4k

F2 respectively.


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According to geometry of the arrangement, downward displacement of the block from its equilibrium

position is y = 2

4321 k

F2

k

F2

k

F2

k

F2.

If the block is released now, it starts to accelerate upwards due to extra tension F in string.

Hence restoring force on the block

F = -

4321 k

1

k

1

k

1

k

14

y

Restoring acceleration of block =m

F

a = -

4321 k

1

k

1

k

1

k

1m4

y

Hence 2

=

4321 k

1

k

1

k

1

k

1m4

1

Hence T = 2

4321 k

1

k

1

k

1

k

1m4 .

111. After time t, the point A has fallen through

21 gt

2

1x

and, the platform has moved upward by2

02 ta2

1x

x2

A

A

a0

v2

v1

x

x1

x2

The total length of the chain resting on the platform is x = x1 + x2 = 0ag2

1 t2

The normal reaction of the platform consists of weight component and thrust component.N = Nwt + Nthrust

Nwt = x (g + a0) = 22

0 tag2

1

Nthrust = vrel dtdm

vrel = relative velocity of chain w.r.t platformvrel = v1 + v2 = gt + a0t = (g + a0)t

rel

dx v

dt

dm dm

dt dx

2 2

0thrust N g a t

Total reaction is 2 22 2

0 0

1

2 N g a t g a t 220 tag

2

3=


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112.

(1)

a0m

M

arel

arelcos

mg

N

Nsin

N

Let the bar of mass m slides with an acceleration arel in downward direction and prism movestowards right with an acceleration a0 Now arel = ab.p. (acceleration of block w.r.t. prism)acceleration of block w.r.t. ground isab.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 + msin

2)

or a0 =g

m

M

cos sin

sin

2

113. F.B.D. of A

mAg - T = mAa (1)

mA

T

F.B.D. of B relative to L

ma

mg

f

N

N = maf = mg, Here f = limiting friction

ma = mg a = g/ (2)

F.B.D. of M

T

N

T - N = Ma

T = (M + m)a (3)

From (1) , (2) and (3), m A =

M m

1


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114. F.B.D. of man

f 1

mg

N

f 1 = force between man and the plank.

f 1 = ma (i) N = mg (ii)F.B.D. of plank

F

mg

N

f 1

f 2 f 2

f 2 = force of friction between the plank and surface.

f 1 can have any value in the following range. F - f 2max f 1 F + f 2max

F - (m + M)g ma F + (m + M)g

F

m

m M g

ma

F

m

m M g

M

115. The system is illustrated in the figure. Let the tensionin the string be T and the accelerations of the systembe a. The equation of motion for the masses are formass 2m,

T - 2mg sin 300

= 2ma . . . (I)and for mass 5m5mg sin 450 - T = 5ma

2

mg5 - T = 5 ma . . (ii)

2m 5m4530

2mg

TT T

T

5mg

Adding (i) & (ii),2

mg5 - mg= 7ma . . . (iii)

or

1

2

5g = 7a

Hence the acceleration of the system is a =7

1

2

5

g

from (i) T = mg + 2ma

T = mg +7

12

5mg2

=7

)255(mg

=7

)21(mg5

The force on this pulley is the resultant of the tension in the string on the two sides.


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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.

N1

T

a1

mg

2mg

2T

a2

N3

3mg

a3

T

120. Let the time be t0

when the relative motion startsthen,

20 t0 - 0mg = m

mmt20 0

or, 20 t0 - 0 mg = 10 t0 or, 10 t0 = 0 mg

t0 =

10

mg0 sec.

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 directionforming an angle with the rod. Find the acceleration ofthe bead relative to the rod. mg

ar sin

N cos

ar cos N sin

ar

N

( relative acceleration is simply the vector difference between the absolute acceleration)or ay = ar cos + a . . . (i)and ar sin = ax - 0or 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)

mg

sin = mar + ma cos or ar =g sin - a cos


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122. Let the plane moves to the left with acceleration a. According to FBD of the block (mass = m) pseudo forcema 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, whenma cos = mg sin + f

= 370

ma cos

ma

mg cos

ma sin

f

Na

mg sin

Simplifying, we get

a =

sincos

gcos(sin

Substituting the values of and ,a = 22.4 m/s2

123. 5t m1g = 0

t =5

1015.0 = 1 sec

m1g = m2aa =

2

5 = 2.5 m/s2

5t

N

m1g

N

Upper Block

m1g

N

m2g

Lower Block

N

124. (a) For the position of the rod, shown in the figure.Total external torque = Mg[(L/2) sin]as I

We get, Mg[(L/2) sin] =ML d

dt

2

3

or = d dt

gL

32

sin

As = 45 , =3

2 2

g

L

mg

(b)d

d

g

L

3

2sin

d g

Ld

0 0

3

2 sin

2

2

3

2

g

Lcos =

3g

L

cos

for = 45 , =3

2

g

L

= 4.6 rad/s

125. By COM

mv0 = Mv -3

mv0

v =24

v

M3

mv4 00

32

1

M

m

By COAM

mv02

L =

2

L

3

mv

12

ML 02

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