http://www.rpauryascienceblog.co/ LWS OF OIO PROBLE D HEIR SOLUIO. What is the axiu value of the force F such that the F block shown in the arrangeent, does not ove? 60 = =3kg 3. particle of ass 3 kg oves under a force of 4 î + 8 ĵ + 0 kˆ. ewton. Calculate the acceleration (as vector) to which the particle is subjected to. If the particle starts fro 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 ove at right angles to the forer direction after tie t = u / g sin when its velocity would be v = u cot. 4. block is placed on an inclined plane of inclination 30 0 fro 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. particle oves in a circle of radius 0 c at a speed given by v = + t + t /s where t is tie in s. Find the initial tangential and noral acceleration. 6. What is the acceleration of 0 kg block in the situation as 0 0 kg shown in the figure 45 0 sooth 7. What is the agnitude and direction of the force of friction acting on the 0 kg block shown in figure? = 0.4 0 kg 0 8. Find the acceleration of kg and 4 kg blocks. 4 kg kg 9. wo blocks of ass 3 kg and 4 kg are kept in contact with each other on a sooth horizontal surface. horizontal force of is applied on the second block due to which they ove with certain acceleration. Calculate the force between the blocks. 3 kg 4 kg 0. For the situation shown in the figure, (a) Draw free body diagras of and B. (b) What is the acceleration of the blocks and B? F = 00 3 kg B 7 kg sooth (c) What is the agnitude of force applied by on B.
http://www.rpauryascienceblog.co/. For the situation, shown in the figure, (a) Draw the free body diagra of 0 kg block (b) Find the acceleration of the block (c) Find the force exerted by the block on the incline. ll contact surfaces are assued to be sooth. 30 0 0 kg sooth. block of etal weighing kg is resting on a frictionless plane. It is struck by a jet releasing water at a rate of kg /s and at a speed of 5 /s. Calculate the initial acceleration of the block. Black kg 3. Find the relation between the acceleration of rod and wedge B in the arrangeent shown in the figure. he entire surfaces are sooth. B 4. ll the strings and the pullies in the figure shown are assless and frictionless. Find the relation between the accelerations of the block B and ass as shown in the figure. B a a 5. wo blocks each of ass are connected by a assless inextensible string. hey are kept on a fixed wedge as shown in the figure. he surface of wedge is frictionless. If the syste is just released fro rest, find the instantaneous acceleration of blocks. a 6. In the figure shown, find the acceleration of the block B if that of the is 0.3 /s. B
http://www.rpauryascienceblog.co/ 7. Initially half of the chain s length ( = 4 ) is overhanging, what will be the speed of the chain when it just slips off the sooth table? / / 8. body slides down a sooth curved fixed track which is a quadrant of a circle with radius 0. Find the speed of the body at the botto of the track. 0 0 9. chain of ass and length L held vertical by fixing its upper end to a rigid support. Find the tension in the chain at a distance y fro the rigid support. 0. wo forces F and F are acting on a particle and the angle between the is. he angle between the resultant force and F is. Find tan in ters of F, F and.. he co-ordinates of a oving particle at any tie t are given by x = ct and y = bt. Find initial speed of the particle.. particle oves fro position (3 î ĵ 6kˆ ) to position (4 î 3ĵ 9kˆ ) and a force (constant) of 4î ĵ 3kˆ acts on it. Calculate the work done by the force. 3. chain of ass and length L held vertical by fixing its upper end to a rigid support. Find the tension in the chain at a distance y fro the rigid support. 4. block of ass is placed on a sooth inclined surface as shown in the figure. Draw the free body diagra of the ass. Find tension in the string and the noral reaction exerted by the inclined surface on the block. 5. Find the contact force between the two asses, if F = 0, = 4 kg, = kg and coefficient of friction between the surface and both the asses is = 0.. F =0.
http://www.rpauryascienceblog.co/ 6. body of ass is oving in +x direction with acceleration a. Find the iniu value of coefficient of friction, between the block so that a body of ass can be held stationary with respect to on the vertical side of. =0 a 7. (a) wo blocks of ass kg and kg are at rest on a rough surface and are separated by a distance of 3.75 as shown in the figure. he coefficient of friction between each block and surface is 0.0. kg kg he kg block is given a velocity of 8 /s directed towards the second block. It collides with kg block which is at rest. Find the loss of energy of kg block during collision if coefficient of restitution is ½. (ake g = 0 /s ). (b) Calculate pressure inside a cylindrical liquid drop of diaeter d and surface tension S, if atospheric pressure is p 0. 8. force F depends on displaceent x as F= 6x + 4, where F is in ewton and x in eter and it acts on a ass = kg which is initially at rest at point x = 0. hen find velocity of ass when x =. ssue that no other force is acting on ass. 9. block of ass 6 kg is kept on rough surface as shown in figure. Find acceleration and friction force acting on the block. (ake g = 0 /s ) 6 kg 30 =3/ 30. If the coefficient of static friction between tyres and the road is 0., what is the shortest distance in which an autoobile can be stopped while travelling at 54 k/hr. 3. wo blocks are kept as shown. horizontal tie varying force is applied on upper block (F = 0 t). Find the tie when the relative otion between the blocks starts. [Given that ass of each block is kg] 0 = 0 0 t 3. ass is oving with a constant velocity v 0 along a line y = -a and away fro the origin. Find the agnitude of its angular oentu with respect to origin. 33. On a sooth inclined plane, a block of ass is placed over the plank. If the plank and the block is released fro rest. Find the acceleration of block. ssue friction between block and plank is sufficient to prevent slipping.
http://www.rpauryascienceblog.co/ 34. One end of a assless spring of spring constant 00 / and natural length 0.5 is fixed and the other end is connected to a particle of ass 0.5 kg lying on a frictionless horizontal table. he spring reains horizontal. If the ass is ade to rotate at an angular velocity of rad/s, find the elongation of the spring. 35. One end of a assless spring of spring constant 00 / and natural length 0.5 is fixed and the other end is connected to a particle of ass 0.5 kg lying on a frictionless horizontal table. he spring reains horizontal. If the ass is ade to rotate at an angular velocity of rad/s, find the elongation of the spring. 36. Find out the agnitude of R and R contact forces acting R between the blocks as shown in figure. (take all the surfaces to 40 be sooth.) 4 kg R 3 kg kg 37. block of ass kg is kept on a rough inclined surface and it is oving downward with a constant velocity of v = /s along the incline. Find out the value of net force applied by the inclined surface on the block. kg 38. hree blocks, B and C are placed one over the other as shown in figure. Draw the free body diagra of all the three blocks. ( ) B ( ) C ( 3) 39. kg and 3 kg blocks are oved up with a coon acceleration of /s. Find the agnitude of tensions ( & ) in the strings. (take g = 0 /s ) kg 3 kg 40. wo blocks of asses 4 kg and 6 kg connected by a assless string are kept on a rough surface having coefficient of sliding friction 0.. wo horizontal forces of 30 and 0 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 0 4 kg 6 kg =0. =0. 4. wo blocks, each of ass are attached with a assless 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 >. 4. One end of a assless spring of spring constant 00 / and natural length 0.5 is fixed and the other end is connected to a particle of ass 0.5 kg lying on a frictionless horizontal table. he
http://www.rpauryascienceblog.co/ spring reains horizontal. If the ass is ade to rotate about the fixed point at an angular velocity of rad/s, find the elongation of the spring. 43. One end of a unifor rod B of length L and ass is hinged to a thin and rigid rod. he thin rod is vertical and rotates with constant angular speed in such a way that the rod B reains horizontal. Position of thin rod does not change. (a) Draw the free body diagra of the rod B. (b) Find the force on the rod B by the thin rod. 44. hree blocks = kg, = kg and 3 = 3kg are kept on a sooth horizontal surface. horizontal force F = 90 is applied at the first block. Blocks are attached by light strings. Calculate the tensions and acting on the strings. 3Kg Kg Kg 3 F = 90 45. 00 kg load is uniforly pulled over a horizontal plane by a force F applied at an angle = 30 0 to the horizontal. Find this force if the coefficient of friction between the load and the plane is 0.3. 46. block of ass 5 kg is placed on a slope which akes an angle of 0 with the horizontal and is given a velocity of 0 /s up the slope. ssuing the coefficient of sliding friction between the block and the slope is 0., find how far the block travels up the slope? ake g = 0 /s. ake cos0 = 0.9 and sin0= 0.3. 47. wo particles each of ass are connected by a light string of length L as shown. continuous force F is applied at the idpoint of the string (x = 0) at right angles to the initial position of the string. Show that acceleration of in the direction at right angles to F is given by F x a x = L x L F B 48. particle rests on top of a heisphere of radius R. Find the sallest horizontal velocity that ust be iparted to the particle if it is to leave the heisphere without sliding down it. 49. onkey is sitting on the branch of a tree. he branch exerts a noral force of 48 and a frictional force of 0. Find the agnitude of the total force exerted by the branch on the onkey? 50. wo asses = 0 kg and = 5 kg are connected by an ideal string as shown in the figure. he coefficient of friction between and the surface is = 0.. ssuing that the syste is released fro rest, Calculate the velocity when has descended by 4. 4
http://www.rpauryascienceblog.co/ 5. (a) Draw the FBD of and. (b) What will the relation between,,,? so that, reains stationary. ll surfaces are sooth. fixed 5. car of ass = 000 kg starting fro rest attains a speed of 0 /s over a distance of 50. ssuing constant acceleration, find the resultant force acting on the car. 53. (i) body of ass is placed in a sooth horizontal table and is connected to another ass by eans of a assless string passing over a assless and frictionless pulley as shown in the figure. Find the force exerted by the pulley on the table. (ii) body of ass is oving with acceleration a in the direction as shown. Find the iniu value of so that a body of ass can be held stationary with respect to on the vertical side of. a 54. sooth wedge of ass is oving with horizontal acceleration g cot. sall block of ass is there on inclined sooth face of wedge. Calculate acceleration of block of ass with respect to ground. g cot 55. wo asses and are connected by a assless string which passes over a light frictionless pulley as shown in fig.. he asses are initially held with equal lengths of the strings on either side of the pulley. Find the velocity of the asses at the instant the lighter ass oves up a distance of 6.54. he string is suddenly cut at that instant. Calculate the tie taken by each to reach the ground. (g = 9.8 /s ) 3.08 ground Fig. 56. wo blocks of asses 4 kg and 6 kg connected by a assless string are kept on a rough surface having coefficient of sliding friction 0.. wo horizontal forces of 30 and 0 is applied on 4 kg and 6 kg blocks. Find the friction force on both the blocks and tension in the string. 30 4 kg 6 kg =0. 0
http://www.rpauryascienceblog.co/ 57. wo blocks and B of ass 4 and respectively, are connected through a assless string passing through an ideal pulley. Initially block is 4 above the block B as shown in the figure. ow syste is released at t = 0 sec. Find the tie after which both blocks cross the sae level. lso find velocity of the blocks at the instant when they cross the sae level. 4 B 4 58. wo identical blocks and B are placed on a rough inclined plane of inclination 45 0. he coefficient of friction between block and incline is 0. and that of between B and incline is 0.3. he initial separation between the two blocks is. he two blocks are released fro rest, then find (a) the tie after which front faces of both blocks coe in sae line and (b) the distance oved by each block for attaining above otion position. B 45 0 B =0.3 =0. 59. an of ass is oving with a constant acceleration a w.r.t. plank. he plank lies on a sooth horizontal floor. If ass of plank is also then calculate acceleration of plank and an w.r.t. ground, and frictional force extended by plank on an. 60. bob of ass '' is suspended by a light inextensible string of length 'l' fro a fixed point. Such that it is free to rotate in a vertical plane. he bob is given a speed of 6gl horizontally. Find the tension in the string when string deflects through an angle 0 0 fro the vertical downward. 6. wo blocks of asses and connected by a nondefored light spring rest on a horizontal plane. he coefficient of friction between the blocks and the surface is equal to. What iniu constant force has to be applied in the horizontal direction to the block of ass in order to shift the other block? 6. block of ass kg slides on an inclined plane which akes an angle of 30 0 with the horizontal. he coefficient of friction between the block and the surface is 3 /. (i) What force should be applied to the block so that the block oves down without any acceleration? (ii) What force should be applied to the block so that it oves up without any acceleration.. 63. block of ass kg slides on an inclined plane which akes an angle of 30 0 with the horizontal. he coefficient of friction between the block and the surface is 3 /.
http://www.rpauryascienceblog.co/ (i) What force should be applied to the block so that the block oves down without any acceleration? (ii) What force should be applied to the block so that it oves up without any acceleration. 64. block slides down an inclined plane of slope angle with constant velocity. It is then projected up the sae plane with an initial speed v 0. How far up the incline will it ove before coing to rest? will it slide down again? 65. onkey of ass is standing on a ladder of ass - which is counter balanced by ass by a string passing over a sooth frictionless pulley. he onkey clibs up the leader by a distance w.r.t. ledder. Find the displaceent of the leader. - 66. wo blocks and B are connected to each other by a string and a spring. he 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 '' slides along the vertical side of 'C', both with sae unifor speed. he coefficient of friction between the surface of the blocks is 0.. Force constant of spring is 000 /. If ass of the block is kg. Calculate the ass of block B and the energy stored in the spring B C k 67. body of point ass is attached with a string of length. he body is under a otion in vertical circle with velocity v at the lowest position. Draw the free body diagra of the body when it akes an angle with the vertical, and find the tension in the string. 68. Find the acceleration of and B in the arrangeent shown in the figure. ass of = 5 kg and ass of B = 0 kg and coefficient of friction between and the surface is 0.. =0. B (ssue pulleys to be light & frictionless and string to be light & inextensible) 69. Find the acceleration of blocks and B. Contact surface are sooth. ass of each of and B is 5 kg. (ssue pulleys to be light & frictionless and string to be light & inextensible) sooth B
http://www.rpauryascienceblog.co/ 70. wo blocks are connected by a rod and fixed at P and R points a force F() is applied on block. So find the tension in rod at P, Q, R points. (Q is the id point of P and R) P Q R = kg, = 3 kg, ass of the rod = kg 7. body is thrown upward fro with speed u /s on an incline plane having inclination the body will coe back to the point of projection after soe tie. ll planes are frictionless. Calculate tie. u B fixed 7. Consider the situation shown in the figure. he 0 kg block is tied with a light string on one side and a force of 0 is applied on it fro the other side. ll contact surfaces are sooth, and the block reains at rest. (i) Draw the free body diagra of the 0 kg block. (ii) Find the tension in the string. light string = 0 kg F = 0 73. block of ass kg is placed on a rough surface as shown in the figure. Find the distance travelled by the block in the first sec. 6 60 k =0. s=0.3 kg 74. In the shown figure a triangular wedge of ass and a sall cube of ass is placed. he coefficient of friction between wedge and cube is and between wedge and ground is. ( = /). Find the force of friction on wedge due to ground. It is given that > tan. 75. particle of ass kg has an initial velocity v i (î ĵ) /s. It collides with another body and the ipact tie is 0.s, resulting in a velocity v f (6î 4ĵ 5kˆ ) /s after ipact. Find average force of ipact on the particle. 76. n open elevator is ascending with zero acceleration. he speed v = 0/sec. ball is thrown vertically up by a boy when he is at a height h = 0 fro the ground. he velocity of projection is v = 30/sec with respect to elevator. Find (i) the axiu height attained by the ball (ii) the tie taken by the ball to eet the elevator again. (iii) tie taken by the ball to reach the ground after crossing the elevator.
http://www.rpauryascienceblog.co/ 77. In the shown arrangeent the asses of blocks, B and C are, and 3 respectively. If all the surfaces are sooth and spring constant of the spring be K, find (i) acceleration of each block (ii) tension in the cord connecting blocks and B. (iii) extension in the spring. B C 78. wo identical cubes placed one over other each of ass '' and side 'a' are placed as shown. Calculate tie the upper block will fall. = 0. = 0. F = 6 = kg F 79. In the pulley block syste shown, find the accelerations of, B, C and the tension in the string. ssue the friction to be negligible and the string to be light and inextensible. he asses of the blocks are, and 3 respectively C B 80. ball of ass is hanging fro a unifor rope of ass and length L whose one end is connected to the ceiling of an elevator as shown in the figure. If the elevator has a unifor acceleration of a /s 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 fro the point where the string is connected to the ceiling) L a 8. Figure shows a an of ass 70 kg standing on a light weighing achine kept in a box of ass 35 kg. he box is hanging fro a pulley fixed to the ceiling through a light rope, the other end of which is held by the an hiself. If he anages to keep the box at rest (a) what is the weight shown, by the achine? (b) what force should the an exert on the rope to get his correct weight on the achine? 8. block of ass 0 kg is lying on a frictionless table. block of 5kg is kept on the block of 0 kg. If a variable force F given by F = kx is applied on the block of ass 0 kg and initially the ass of 0 kg is lying at x = and = 0. and k = 5 /, find (i) the distance after which 5 kg ass starts slipping
http://www.rpauryascienceblog.co/ (ii) Calculate acceleration of asses at 4 fro starting point. 83. chain consisting of 5 links each of ass 0. kg is lifted vertically with a constant acceleration.5 /s as shown in figure. Find the force of interaction between the top link and link iediately below it. ` 3 4 5 84. wo asses of 5 kg and 0 kg connected by a assless string passing over a frictionless pulley are in equilibriu as shown in the figure. One of the ass is resting on the surface. Find (a) ension in the string (b) oral reaction on the 0kg block (take g = 0 /s ) 5 kg 0 kg 85. If a ass is hung with a light inextensible string as shown in the figure. Find the tension in horizontal and inclined strings in ters of, g and. 86. wo blocks of ass kg and 4 kg are kept in contact with each other on a sooth horizontal surface. horizontal force of is applied on the first block due to which they ove with certain constant acceleration. Calculate the force between the blocks. kg 4kg 87. force F is applied fro its centre along horizontal direction Find direction and agnitude of frictional force acting on the cylinder. (o slipping) r F 88. ball is dropped fro a height 00 c on the ground. If the coefficient of restitution is 0., (a) what is the height to which the ball will go up after it rebounds for the nd tie. (b) If duration of each collision is ill sec. hen find the average ipulsive forces during st and nd collision. (eglect any air resistance)
http://www.rpauryascienceblog.co/ 89. block of ass with a seicircular track of radius R rests on a horizontal frictionless surface. unifor cylinder of radius 'r' and ass '' is released fro rest at the top point. he cylinder slips in the seicircular frictionless track. How fast is the block oving when the cylinder reaches the botto of the track? B 90. In the shown figure a triangular wedge of ass and a sall cube of ass is placed. he coefficient of friction between wedge and cube is and between wedge and ground is. ( = /). Find the force of friction on wedge due to ground. It is given that > tan. 9. wo asses = 0 kg and = 5 kg are connected by an ideal string as shown in the figure. he coefficient of friction between and the surface is = 0.. ssuing that the syste is released fro rest, Calculate the velocity when has descended by 4. 9. wo blocks of asses and connected by a non-defored light spring rest on a horizontal plane. he coefficient of friction between the blocks and the surface is equal to. What iniu constant force has to be applied in the horizontal direction to the block of ass in order to shift the other block? 93. wo asses = 0 kg and = 5 kg are connected by an ideal string as shown in the figure. he coefficient of friction between and the surface is = 0.. ssuing that the syste is released fro rest, Calculate the velocity when has descended by 4. 4 94. See the diagra, he syste is in equilibriu. here is no friction anywhere. Spring, string, pulley are assless. Find spring copression. k
http://www.rpauryascienceblog.co/ 95. cabin is oving upwards with a constant acceleration g. boy standing in the cabin wants to whirl a particle of ass in a vertical circle of radius. (ass is attached to an ideal string) calculate iniu velocity which should be provided at lowerost point (w.r.t cabin) so that particle can just coplete the circle. g 96. Friction coefficient between the wedge and the block is. here is no friction between the wedge and horizontal floor. his syste is not in equilibriu. Find acceleration of block relative to ground. 97. wo sooth wedges of equal ass are placed as shown in figure. ll surfaces are sooth. Find the velocities of & B just before hits the ground. B h 98. body of ass is oving in -x direction with acceleration a. Find the iniu value of so that a body of ass can be held stationary with respect to on the inclined side of. Consider that application of acceleration a is not enough to support the block in equilibriu. a 99. ll the strings and the pullies in the figure shown are assless and frictionless. Find the relation between the accelerations of the block, B and C having asses, and 3 respectively. a a B 00. pulley fixed to the ceiling of a lift carries a thread whose ends are attached to the asses of 4kg and kg. he lift starts going down with on acceleration of a 0 = 4 /s relative to ground. Calculate (a) he acceleration of the load 4kg relative to the ground and relative to lift. (b) he force exerted by the pulley on the ceiling on the lift (hread and pulley are assless g=0 /s ) 4 kg kg C 3 a 3 a 0
http://www.rpauryascienceblog.co/ 0. 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 ass (b) the tension in the string PQ and (c) force exerted by the clap on the pulley C. Q P C 0. pulley fixed to the ceiling of a lift carries a thread whose ends are attached to the asses of 4kg and kg. he lift starts going down with on acceleration of a 0 = 4 /s relative to ground. Calculate (a) he acceleration of the load 4kg relative to the ground and relative to lift. (b) he force exerted by the pulley on the ceiling of the lift (hread and pulley are assless, g=0 /s ) 4 kg kg a 0 03. In the shown figure a triangular wedge of ass and a sall cube of ass is placed. he coefficient of friction between wedge and cube is and between wedge and ground is. ( = /). Find the force of friction on wedge due to ground. It is given that > tan. 04. he figure shows L shaped body of ass placed on sooth horizontal surface. he block is connected to the body by eans of an inextensible string, which is passing over a sooth pulley of negligible ass. nother block B of ass is placed against a vertical wall of the body. Find the iniu value of the ass of block so that block B reains stationary relative to the L-shape body. Coefficient of friction between the block B and the vertical wall is. B 05. Figure shows a an of ass 60 kg standing on a light weighing achine kept in a box of ass 30 kg. he box is hanging fro a pulley fixed to the ceiling through a light rope, the other end of which is held by the an hiself. If he anages to keep the box at rest (i) what is the weight shown by the achine? (ii) what force should the an exert on the rope to get his correct weight on the achine? 06. sooth table is placed horizontally and a spring of instretched length 0 and force constant k has one end fixed to its centre. o the other end of the spring is attached a ass which is aking n revolutions per second around the centre. Find the radius r of this unifor circular otion and the tension in the spring. 07. very sall cube of ass is placed on the inside of a forel rotating about a vertical axis at a constant rate of "n" revolutions per second. he wall of the funnel akes an angle with the
http://www.rpauryascienceblog.co/ 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 fro the axis of rotation, what are the largest and sallest values of n for which the block will not ove with respect to the funnel? 08. In the arrangeent shown in the figure, the rod of ass held by two sooth walls, reains always perpendicular to the surface of the wedge of ass. ssuing all the surfaces are frictionless, find the acceleration of the rod and that of the wedge. fixed wall 09. 4 kg block is put on top of 5 kg block. In order to cause the top block to slip on the botto one, a horizontal force of ust be applied to the top block. ssue a frictionless table and find (a) he axiu horizontal force which canbe applied to the lower block so that the blocks will ove together and (b) the resulting acceleration of the blocks. 0. In the arrangeent shown in figure, pulleys are sall and light and springss are ideal. k, k k 3, and k 4 are force constants of the springs. Calculate period of sall vertical oscillations of block of ass. K K4 K K3. chain of length l is held vertical and then released. It falls on a platfor which starts, fro rest and oves vertically upward with a constant acceleration a 0. Deterine the noral force exerted on the platfor by the chain as a function of tie. ass per unit length of the chain is. l a 0. pris of ass and with angle rests on a horizontal surface. bar of ass is placed on the pris. ssuing the friction to be negligible, find the acceleration of the pris.
http://www.rpauryascienceblog.co/ 3. he figure shows an L shaped body of ass is placed on sooth horizontal surface. he block is connected to the body by eans of an inextensible string, which is passing over a sooth pulley of negligible ass. nother block B of ass is placed against a vertical wall of the body. Find the iniu value of the ass of block so that block B reains stationary relative to the wall. Coefficient of friction between the block B and the vertical wall is. B 4. plank of ass is placed on a rough horizontal surface and a constant horizontal force F is applied on it. an of ass runs on the plank. Find the accelerations of the an so that the plank does not ove on the surface. Co-efficient of friction between the plank and the surface is ssue that the an does not slip on the plank. F 5. sooth fixed wedge has one face inclined at 30 0 to the horizontal and a second face at 45 0 to the horizontal. he faces are adjacent to each other at the top of the wedge. Particles of asses and 5 are held on these respective faces connected by a taut inelastic string passing over a sooth pulley at the top of the wedge as shown in the figure. Find the acceleration of the syste if the particles are siultaneously released and show that the force acting on 0 the pulley is g ( + ) cos (5 ) 0. 7 30 0 45 0 5 6. he figure shows an L shaped body of ass is placed on sooth horizontal surface. he block is connected to the body by eans of an inextensible string, which is passing over a sooth pulley of negligible ass. nother block B of ass is placed against a vertical wall of the body. Find the iniu value of the ass of block so that block B reains stationary relative to the wall. Coefficient of friction between the block B and the vertical wall is. B 7. he asses of blocks and B are and. Between and B there is a constant frictional force F, but B can slide frictionlessly on the horizontal surface. is set in otion with velocity v 0 while B is at rest. What is the distance oved by relative to B before they ove with the sae velocity? B v 0
http://www.rpauryascienceblog.co/ 8. ball of ass is hanging fro a unifor rope of ass and length L whose one end is connected to the ceiling of an elevator as shown in the figure. If the elevator has a unifor acceleration of a /s upward, find the tension in the rope as a function of x. What is the force exerted by the rope on the ceiling? L x a 9. In the pulley block syste shown, find the accelerations of, B, C and the tension in the string. ssue the friction to be negligible and the string to be light and inextensible. he asses of the blocks are, and 3 respectively B C 0. wo blocks are kept as shown. horizontal tie varying force is applied on upper block (F = 0 t). Find the tie when the relative otion between the blocks starts. [Given that ass of each block is kg] 0 = 0 0 t. bead of ass '' is fitted on to a rod can slide on it without friction. t the initial oent the bead is in the iddle of the rod. he rod oves translationally in a horizontal plane with an acceleration 'a' in a direction foring an angle with the rod. Find the acceleration of the bead relative to the rod. a. rough inclined plane with inclination = 37 0 with the horizontal is accelerated horizontally till a block of ass originally at rest with respect to the plane just begins to slip up the plane. he coefficient of static friction between the surfaces in contact is = 5/9. Find the acceleration of the plane. (tan 37 0 = 3/4) 3. tie varying force F = 5t is applied on the upper block as shown in figure. When will the upper block start oving with respect to lower block? What is the acceleration of lower block at that instant (here t is in sec.) = 0.5 = 0 = kg =kg F = 5t 4. rod of length and ass 4 kg, can rotate freely in a vertical plane around its end. he rod is initially held in a horizontal position and then released. t the tie the rod akes an angle 45with the vertical, calculate (a) its angular acceleration, (b) its angular velocity. 45
http://www.rpauryascienceblog.co/ 5. particle of ass oving with a speed v 0 strikes perpendicularly one end of a unifor rod of ass and length L initially resting on a sooth horizontal plane. he particle returns in the sae v line with speed of 0. If = find the linear and the angular speed of the rod. 3 3 6. unifor disc of ass and radius R is projected horizontally with velocity v 0 on a rough horizontal floor having coefficient of friction equal to. Find the tie after which it starts pure rolling. 7. unifor solid sphere of ass 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 tie. 8. fiat car of ass starts oving to the right due to a constant force F as shown in figure. and spills on the flat car fro a stationary hopper. he velocity of loading is constant and equal to kg/s. Find the velocity of the car after t seconds. 9. spring is fixed at one end O on a sooth horizontal table. atural length of spring is very sall (tends to zero). ow a ball of ass = kg is attached at other end stretched to as shown in figure. Ball is given velocity V 0 = /s at an angle 30 to O. Find the axiu elongation of the spring. Given that spring constant is k = / and O = O v 0 30 30. hin threads are tightly wound on the ends of a unifor solid cylinder of ass. he free ends of the threads are attached to the ceiling of a lift. he lift starts oving up with an acceleration a 0. Find the acceleration of the cylinder relative to the lift and the force F exerted by the cylinder on the ceiling. ` a 0 3. Find the acceleration of the body in the arrangeent shown in figure with asses of respective bodies. he friction is absent. lso the asses of the pulleys and the threads are negligible. 0
http://www.rpauryascienceblog.co/ 3. What is the iniu and axiu acceleration with which bar (Fig.) should be shifted horizontally to keep bodies and stationary relative to the bar? he asses of the bodies are equal, and the coefficient of friction between the bar and the bodies is equal to k. he asses of the pulley and the threads are negligible, the friction in the pulley is absent. 33. an of ass is oving with a constant acceleration a w.r.t. plank. he plank lies on a sooth horizontal floor. If ass of plank is also then calculate acceleration of plank and an w.r.t. ground, and frictional force extended by plank on an. 34. ass of blocks, B and C is 7.5 kg, 6 kg and kg. here is no friction anywhere. Calculate resultant acceleration of block C when the syste is released. Pulley is assless. C B 35. Friction coefficient between the wedge and the block is. here is no friction between the wedge and horizontal floor. Find acceleration of block relative to ground. 36. block of ass is kept over a fixed sooth wedge. he block is attached to a sphere of sae ass through fixed assless pullies P and P. Sphere is dipped inside water as shown. If specific gravity of aterial of sphere is, then find the acceleration of the sphere. 30 0 P P 37. ass of.9 kg is suspended fro a string of length 50 c and is at rest. nother body of ass 00 g, which is oving horizontally with a velocity of 50 /s strikes and sticks to it. What is the tension in the string when it akes an angle of 60 with the vertical? 38. wo blocks of asses = kg and =4 kg are attached by light ideal spring of force constant k = 000 /. he syste is kept on a sooth inclined plane inclined 30 0 with horizontal. force F=5 is applied on and syste is released fro rest. he block is attached with a light string whose another end is connected with a ass 3 = kg. ssue initially the spring is at relaxed position and the syste is released fro rest. Find the (a) axiu extension of the spring. (b) acceleration of the syste at this instant F 3
http://www.rpauryascienceblog.co/ 39. block of ass is resting on a frictionless surface. second block of ass is placed on it as shown in figure. constant force F is applied on the block of ass due to which the syste accelerates (i) find the iniu value of coefficient of friction between the two blocks to prevent sliding F (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. lso find the energy dissipated into heat assuing the length of is large enough. 40. Consider the arrangeent shown in the figure. If the syste is set free at t = 0 with the horizontal bar at a height of h as shown in the figure, obtain C h B (i) velocities of the wedges 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. (eglect any friction. ass of each wedge is while that of C is ) 4. he figure shows an L shaped body of ass is placed on sooth horizontal surface. he block is connected to the body by eans of an inextensible string, which is passing over a sooth pulley of negligible ass. nother block B of ass is placed against a vertical wall of the body. Find the iniu value of the ass of block so that block B reains stationary relative to the wall. Coefficient of friction between the block B and the vertical wall is. B 4. In the shown figure. he 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. B C F 43. circle of radius R = is arked on upper of a horizontal board, initially at rest. n insect starts fro rest along the circle with a tangential acceleration a = 0.5 /s. t the sae instant board accelerates upwards with acceleration b =.5 /s. If the coefficient of friction between board and insect is = 0., what distance will the insect travel on the board without sliding? 44. board fixed to the floor of an elevator such that the board fors an angle = 37 0 with horizontal floor of the elevator accelerating upwards. block is placed on point of the board as shown. When block is given a velocity v = 4 /sec up the board w.r. to board it coes to rest after oving a distance =.6 relative to the board. Its velocity was v = 4 /s down 37 0 a
http://www.rpauryascienceblog.co/ the board when it returns to point. Calculate acceleration 'a' of the elevator and coefficient of friction between the board and the block. 45. Figure shows a an of ass 60 kg standing on a light weighing achine kept in a box of ass 30 kg. he box is hanging fro a pulley fixed to the ceiling through a light rope, the other end of which is held by the an hiself. If he anages to keep the box at rest (i) what is the weight shown by the achine? (ii) what force should the an exert on the rope to get his correct weight on the achine? 46. 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 ass (b) the tension in the string PQ and (c) force exerted by the clap on the pulley C. Q P C 47. ass is released fro rest on the incline. he 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 ass till it stops, and (b) the axiu velocity over this distance. 48. In the shown figure a triangular wedge of ass and a sall cube of ass is placed on the wedge. he coefficient of friction between wedge and cube is and between wedge and ground is. ( = /). Find the force of friction on wedge due to ground. It is given that > tan. 49. hree blocks, B and C have asses kg, kg and 3 kg respectively are arranged as shown in figure. he pulleys P and Q are light and frictionless. ll the blocks are resting on a horizontal floor and the pulleys are held such that strings reain just taut. t oent t = 0 a force F = 40 t ewton starts acting on pulley P along vertically upward direction as shown in figure. Calculate (i) the ties when the blocks lose contact with ground. (ii) the velocities of when the blocks B and C loses contact with ground. (iii) the height by which C is raised when B loses contact with ground Q F = 40 t P B C
http://www.rpauryascienceblog.co/ SOLUIO OF BOVE PROBLE. For no otion Fcos60 (g + Fsin60) F 3 F/ ( 3 g + ) 3 F/ g F ax = 0. akingf as the net force 4 8 0 a î ĵ kˆ /s 3 3 3 s ut at 4 8 0 = 0 î ĵ kˆ 9 3 3 3 = 6 î ĵ 5kˆ s x = 6; s y = s z = 5 3. v i ucos i usin i v f ucos i (usin gt)j u vi.v f 0 t = gsin u v f ucosi (usin )j cos u f = u cot 4. cceleration is zero Frictional force = g sin 30 0 = 0. F.B.D. f Fsin60 Fcos60 g dv 5. angent acceleration a t = = t + dt noral acceleration a n = R v (a t ) t=0 = /s (a n ) t=0 = v 0 R = 5 /s 0.
http://www.rpauryascienceblog.co/ 6. F x = a x 0 cos 45 = 0 a 0 = 0 a a = /s. y 45 0 0 x 00 7. Liiting friction f L = 00 = 0 = 00 f L = 0.4 00 = 40 f L > 0 thus block does not ove. 0 f = 0 0 f = 0 in backward direction. f 00 0 8. cceleration of both blocks will be sae = g sin 9. a = 7 = 3 s f = 3 3 = 9 0. 00 = (3 + 7)a 00 a = 0 /s 0 consider FBD of B R = a R = 7 0 = 70. 00 30 FBD of R 70 FBD of B R. cceleration perpendicular to the incline = 0 Let acceleration parallel to the incline be a, then g sin 30 = a a = g sin 30 = 5 /s hus resultant acceleration = 5 /s g cos30 0 = a = 0 = g cos 30 = 00 3 = 50 3 g sin 30 0 30 0 g cos 30 0 30 0 g = 00. a = F vd/ dt 5 =.5 /s. y 3. tan x a = a B tan. 4. a = 6a
http://www.rpauryascienceblog.co/ 5. g = a... (i) g sin = a... (ii) Solving (i) and (ii) g - g sin = a g( sin ) a =. g a kx a g 6. 0.5 /s. 7. 30 / s. 8. 0 /sec. 9. g L y L 0. F sin F F cos. zero. Displaceent of the particle, s r r (4î 3ĵ 9kˆ) (3î ĵ 6kˆ ) = î ĵ 5kˆ herefore, work = F.S (4î ĵ 3kˆ).(î ĵ 5kˆ ) = 44 + + 45 = 00 units. 3. g L y L 4. = g cos = g sin g sin g g cos 5. F f = a... (i) f = a... (ii) =. F f f 6. = a f = = a a > g a in = g/ a > g/ a f g
http://www.rpauryascienceblog.co/ 7. (a) Velocity of kg block when it strikes kg block is v such that v 8 3.75 v = 7 /s Fro CO, 4 = v v... () For e, v v 7 v v = 7... () Fro () and () 7 v, v =7 Loss of energy = 49 7 4 = 36.75 J (b) Let us consider the equilibriu of half of the liquid drop of length. S + p 0 d = pd p = p 0 + S d d 8. vdv F 6x 4 dx v dv 3x dx v o vdv 3x dx o v 0 / s. 9. f ax > g sin a = 0 f = 30 6 kg 30 =3/ 30. he friction force retarding the otion of car =. = g = 0.g retardation produced a = - 0.g = - 0.g If x is the distance of stopping, then
http://www.rpauryascienceblog.co/ v = u + ax, as v = 0, u = 54 5 8 (5) hence, x = 0. 0 x = 56.5. = 5 /s. 3. Let the tie be t 0 when the relative otion starts 0t then, 0 t 0-0 g = 0 or, 0 t 0-0 g = 0 t 0 or, 0 t 0 = 0 g t 0 = 0g sec. 0 3. L r p = v 0 a 33. If you consider plank and block as a syste acceleration is g sin. Since there is no tendency of relative otion between block and plank, acceleration of both are sae i.e. g sin. 34. kx = ( 0 + x) x c. 35. kx = ( 0 + x) x c. 36. R = 0 R = 5 37. s it is oving with = v / s, a = 0 net force acting = F g sin cos g g cos 90 o g sin 38. ( block) g (B block) 3g g (C block) 3
http://www.rpauryascienceblog.co/ 39. = 5( 0 + ) = 60. = 3 ( 0 + ) = 36. 40. Friction on 4 kg has to be axiu for non zero tension axiu friction force 4 kg block f ax = 8 30 = + f ax = 30 8 = for 6 kg block 0 = f = 4. If >, f ax = g > g herefore friction force acting. f = g et force exerted by the table on the block is f = g f f = g 4. kx = ( 0 + x) x c. 43. (a) Y x B (b) x = g... (i) y = g et force by the thin rod is F = x = y g 4L 4... (ii) 90 44. cceleration of the syste a = = 5 /s 6 kg 90 90 = 5 = 75 75 = 5 = 45 kg
http://www.rpauryascienceblog.co/ 45. + F sin30 g = 0 = g F sin30 f r = = (g Fsin30) s the block is oving uniforly F cos30 (g Fsin30) = 0 F (cos30 + sin30) = g g 0.300 0 F = = cos 30 sin 30 3 0.3 = 96 300 f r Fsin30 30 g F F cos30 46. gsin0 00 S = 9.6 S= 0.4 0 0/s g 0 f gcos0 f = frictional force = gcos0 F = = 0. (gcos0) gsin0 + f = a gsin0 + 0. (gcos0) = a a = 3 +.8 = 4.8 /s down the plane distance block will travel up the plane; O = 0 (4.8)s 47. For otion of point C, F - cos = 0 = F/cos Consider the otion of ass at towards B or vice versa. hen as coponent of in the direction of otion will be cos(90 - ) = sin So if a x is the acceleration of along x axis then fro F = a L C F Y-axis B X-axis sin = a x or a x = sin () Fro () and () F sin F F x a x = tan Fcos L x 48. he sallest velocity v so that it leaves without sliding down should be such be such that the required centripetal force at that instant ust be copletely contributed by weight. R v
http://www.rpauryascienceblog.co/ v g = R v = Rg v = Rg. 49. otal force exerted by the branch on the onkey = 48 0 = 704 50. For 0 kg ass - 0.x0g = 0 a... () For 5 kg 5g- = 5 a... () On adding and a=g/5 given s=4 so, v= 4/s
http://www.rpauryascienceblog.co/ 5. (i) by the inclined plane (ii) g by the gravity (iii) by the string (i) by the inclined plane (ii) g by the gravity (iii) by the string. For = g sin (i) For = g sin (ii) sin = sin. 5. = 000 kg, v = 0 /s, u = 0, a =? s = 50 v u 00 a = /s s 50 F = a = 000 = 000 90 0 g 53. (i) g - = a = a g F = = g g (ii) = a f = = a a > g a in = g/ a > g/ a f g 54. In frae of wedge + (g cot ) sin = g cos = 0, so block is falling freely. It s acceleration is g. g cot g cot g sin g cos
http://www.rpauryascienceblog.co/ 55. FBD of ; g and g = a () FBD of ; g and g = a () dding () and () g = 3a a = g/3 Hence velocity of after oving up 6.54 is; V 9.8 = (g/3) (6.54) = (6.54) = (3.7) (6.54) 3 V = 6.54 /s upwards Velocity of at that instant = 6.54 /s downwards. When string is cut falls to the ground fro a height of 3.08 + 6.54 = 9.6 9.6 = 6.54 + / (9.8) t where t is the tie taken to reach the ground t =.7 seconds falls a distance of 3.08 6.54 = 6.54 t = 0.6 seconds 56. Friction on 4 kg has to be axiu for non zero tension 30 = + f ax = 30 8 = for 5 kg block 0 = f = 57. 4g = 4a g = a a = 3g 5 ; x = at 4 x = at x = 4 0 t = sec. a 3g 3 v = v B at 4 /s at 58. a = g sin 45 0.g cos 45 = 4 / s 7 a B = g sin 45 0.3 g cos 45= /s a B = 0.5 /s s B = a B t t = 4 0.5 t = sec. s B = a t 7 B s = a t 8
http://www.rpauryascienceblog.co/ 59. FBD of plank: f (only horizontal forces are shown) a = f (i) (acceleration of plank w.r.to ground) For an: a + f = a fro (i) a = a a = a/ acceleration of an w.r.t. plank = a - a a acceleration of an w.r.to ground = a P + a PG = a + (- a ) = 0 f = a a 60. By C.O.E.theore u gl( cos0) v gives v = at point B, 3 gl + g cos 60 0 = by putting v = we get, = 3 gl 5 g. v l 6. kx 0 = g (i) Fx 0 = gx 0 + kx 0 (ii) F = g + g 6. For block to ove down the plane without acceleration F + g sin30 f r = 0 F = g cos30 g sin30 3 3 = 0 = 0 [.06 0.5] F=. For block to ove up the plane without acceleration F g sin30 g cos30= 0 F = g [sin30 + cos30] = 0 [.06 + 0.5] = 3. l 0 0 u = B g 6 gl v 63. (i) 0 sin 30 = F + 3 (0 cos 30) F.B.D. of block ;
http://www.rpauryascienceblog.co/ F = 0-30 = 0(.4) 30 6 = - F = 6 down the plane. F f 3 (ii) F = (0cos30) 0sin30 30 = + 0 g=0 30 0 F.B.D. of block ; F = 44. f 30 0 g=0 64. s it slides down with constant velocity v g sin - f = 0 f = g sin But f = g cos g cos = g sin tan = ow when projected upwards with velocity v 0. f + g sin = a g cos + g sin = a g sin = a 0 = v 0 - (g sin ) (s) g v v 0 s = 4gsin whenever it coes to a halt, the force acting on it are g sin down the plane and f = g sin up the plane it will not slide down further. 65. s the sae string connects the ladder onkey and counterass the centre of asses will ove sae distance. Let's say, ladder oves by a distance x Centre of ass displaceent for = x Centre of ass displaceent for - and L ( ) x ( x) L ( ) 0 = x = - x = x ; x = x L
http://www.rpauryascienceblog.co/ 66. s the syste is in dynaic equilibriu For block B = B... (I) B = B g... (ii) For block = 0... (iii) + = g... (iv) solving (I), (ii), (iii), (iv) g = B g B = = = 0 kg 0. B B B Bg C FBD fro (iv) = g = 0 = kx Where x is the expansion in the spring 0 x = = = 0.0 k 000 Energy stored in the spring E = kx = 000 0-4 = 0. J. 67. g cos = (v) / = g cos + (v) / Fro COE, v v g ( cos) g v g cos 68. a = a B = a f L = = 0. 50 = 0 If and B both are at rest = f and = 00 but f L = 0 thus and B will ove. f For B, 00 = 0 a (i) For, 0 = 5a (ii) fro (i) and (ii) we get a = 6 /s 50 FBD of 00 FBD of B
http://www.rpauryascienceblog.co/ 69. Here acceleration of is twice the acceleration of B Let acceleration of B be a then acceleration of = a. then = 5 a = 0 a (i) 50 = 5a (ii) fro (i) and (ii) 50 = 5a a = /s thus acceleration of = a = 4 /s acceleration of B = a = /s. 50 FBD of 50 a FBD of B 70. a = / s 6kg R = ( + )kg /s = 6 P = (kg) ( /s ) = 4 Q = (+0.5)kg /s Q = 5. P R Q kg R kg R P kg P kg 0.5 kg Q Q 7. using relation, v f = v i + a x t v i = u a x = - g sin u t = gsin s there is no friction, so tie to ove fro point to B and tie to coe back fro B to will be sae. u = t = gsin g sin u v f = 0 B fixed 7. F x = 0 0 = 0 = 0. 0 00 73. F app = 6 cos60 = 3 (f s ) ax = =.3 (0-33) (f s ) ax > F app Block will not ove hence distance travelled is 0. 74. Since > tan so cube will not slip on the wedge. Hence force of friction between the ground and wedge is zero.
http://www.rpauryascienceblog.co/ 75. 0 ( 5î 6 ĵ 5kˆ ) 76. (i) Velocity of ball relative to elevator = 30 /s Velcoity of ball relative to ground = 0 + 30 = 40 /s 40 axiu height attained by ball = 40 = 0 0 (ii) When they eet again their displaceent is sae 0t = 40t - 0 t t = 6 sec 77. When the spring has aquired a stable configuration the free body diagras for the blocks can be shown as follows. B C g g If a be the coon agnitude of their accelerations = a () + g - = a () 3g - = 3a (3) dding all the three equations, we get 5g = 6a 5 5 a = g = 3 g g = g/ 6 6 g g K(l) = (l) = K (l = change in length of the spring) 3g 78. For liiting case 6 - - = a - 4 = a a = 0 it will take infinite tie. F=6
http://www.rpauryascienceblog.co/ 79. he FBD of,b,c are shown = a.() g = a.() 3g = 3a 3.(3) constraint relation : a 3 +a a = 0.(4) solving the equations 9 g g 7g a =, a =, a3 = 0 0 0 = 9 g 0 g g F a a 3g a 3 80. - g = a (i) = (a +g) = + L x L x = (a +g) L x = 0 = (a +g)(+) Fore exerted on the ceiling = ( + ) (a + g) downward. L-x g a 8. (a) FBD of an ; is the tension in the string + = 700...(i) 700 FBD of box, = + 350... (ii) Solving (i) and (ii) = 75 (weight shown by the achine = 7.5 kg) (b) ow if should be 700. equ (i) becoes 700 = 70 a and (ii) becoes 350 +- = 35 a solving (i) and (ii) we get = 00 350 8. (i) Let at x = x 0, 5kg starts slipping F = kx 0 = 5 a = 5 kg, = 0 kg
http://www.rpauryascienceblog.co/ 5x 0 x 0 a = 5 kg 5 5 g 5g = a a = g x 0 = 0. 0 x 0 = 0 5 distance fro starting point is 9 when ass of 5 kg starts slipping. a 0 kg g kx (ii) cceleration of 0 kg f = friction force = g kx - g = a x = 4 + = 5 5 5-0. 5 0 = 0 a 75-0 = 0 a f 5 kg 0 kg f kx 65 a = = 3.5 /s. 0 cceleration of 5 kg g = a a = g = 0. 0 = /s 83. Let f be the upward force. hen F 5 g = 5 a (i) F g F =a (ii) fro (i) and (ii) F = 4.9 84. Drawing the FBD of both blocks writing the equations of otion for both 5g = 0 (i) + 7 0 g (ii) Solving for & we get = 50 & = 50 ewton 5 5g 0 0g
http://www.rpauryascienceblog.co/ 85. = sin g = cos tan g = g tan = g sec g 86. a = = /s 4 6 f = 4 a = 8 4kg a 87. By ewton's law F - f =. a C torque about centre of ass = f. r I = f.r r f = f. r = r Since pure rolling takes place a c =.r a f c f = r r f F - f =. f = 3 F a c direction of frictional force is opposite to the F. f F 88. Velocity of ball just before st collision = gh, downward. Velocity of ball just after st collision = e gh, upward. Velocity of ball just before nd collision = e gh, downward. Velocity of ball just after nd collision = e gh, upward. (a) so, v = e gh gh e gh or, h = e 4 h = 0.3 c (b) verage ipulsive force ( st gh[ e] collision) = = 7589.5 t verage ipulsive force ( nd e gh[ e] collision) = = 57.9. t
http://www.rpauryascienceblog.co/ 89. Initial energy of the syste = gr final energy when the cylinder reaches the botto of the track B = v + v where v is the absolute velocity of '' and v is the absolute velocity of ''. B g(r-r) = v + v... (I) Initial oentu of the syste = 0 final oentu when cylinder has reached botto of the track B = v - v... (ii) (assuing track oves towards left) v - v = 0... () v v = and g(r-r) = [] [v] v + [] = v + v = v or v = g(r r) g(r r) v = g(r r) / and v = g(r r) / g(r r) ( ) / 90. Since > tan so cube will not slip on the wedge. Hence force of friction between the ground and wedge is zero. 9. For 0 kg ass - 0.x0g = 0 a... () For 5 kg 5g- = 5 a... () On adding () and () a=g/5 given s=4 ; v = (g/5)(4) = 6 so, v= 4/s 9. kx 0 = g (i) Fx 0 = gx 0 + kx 0 (ii) F = g + g
http://www.rpauryascienceblog.co/ 93. For 0 kg ass - 0.x0g = 0 a... () For 5 kg 5g- = 5 a... () On adding and a=g/5 given s=4 so, v= 4/s 94. FBD of block FBD of wedge cos g cos g g sin sin F sin cos Equation of otion for the block: g sin = 0 along inclined plane (i) g cos = 0 perpendicular to the inclined plane (ii) For the wedge + sin - cos - F = 0 Horizontally (F is spring force) Fro (i), (ii) and (iii) ; g sin + (g cos ) sin - g sin cos - F = 0 F = g sin gsin Hence, copression =. k g 95. t top ost point, v + g = r In critical condition, v = rg pplying conservation of energy (or work energy theore) between top and botto points. u v gr g.r u = v + 8gr = 0 gr u = 0 gr 96. FBD of block FBD of wedge
http://www.rpauryascienceblog.co/ g g Let accelerations of block in the downward vertical direction be a b and that of wedge in forward direction is a w. Hence, equation of otion, g - = a b (i) = a w (ii) Horizontal otion of wedge : = a w (iii) Constraint equation: Let B = x, BC = y, CD = z Hence length x + y + z = constant (iv) d x d z Here y is constant; - a w and a b dt dt Hence, differentiating (iv) twice with respect to tie; a b = a W (in agnitude) g Solving, a b = a w = ( Required acceleration of the block relative to ground = b b a a = g ( ) D B C 97. Writing constraint relation y = y B tan differentiate w.r.to. t we get v = v B tan (i) using COE, gh = (v B) v (ii) putting value v in equation (ii) and solving we get v B = gh cos, v = gh sin y B y 98. f r a g (non - I.F.R.) FBD of block
http://www.rpauryascienceblog.co/ a g herefore, a cos = g sin a = g tan (I.F.R.) FBD of wedge acos g sin asin g cos fr + a cos = g sin = g cos + a sin fr ax = in (g cos + a sin) = g sin - a cos in = gsin acos gcos a sin = gtan a g a tan g f r a 99. x + x 4 = x 3 x 4 + x x 4 = a + a 4 = 0 a 3 + a a 4 = 0 a + a 3 (-a ) = 0 a + a + a 3 = 0 x a a B x4 x x3 C 3 a3 00. In lift frae, F.B.D of asses 4kg, kg, and pulley are 4a 0 a 0 a 4 kg kg a 4g g Pulley- (C) Block () Block (B) a acceleration of 4kg and kg w.r.t o lift Equation of block 4g 4a0 4a () Equation of block B a0 g a () Equation of Pulley = (3) Fro (), () & (3)
http://www.rpauryascienceblog.co/ a / s and 3 (a) acceleration of 4kg w.r.t 0 lift = /s acceleration of 4kg w.r.t 0 ground = +4=6 /s (b) Force exerted by the pulley on ceiling will be = =3 0. a/ a/ a F g F R F F 3 V Pure rolling is assued for both rollers F F = a = a R, = a r g = (a/) Fro (i) and (ii) g (F + F ) = a( ) g 4(F F ) a =... (iii) ( 4 ) F 4 r F V... (i)... (ii) aking torque about lower contact point of F R = I = R R 3 a R R F = 3 a... (iv) 8 aking torque about lower contact point of F r = I = r 3 a r r. r = 3 r 4 = g = 3Ra 4 F = 3 a.... (v) 8 Fro (iv) and (v), 3 3 3 75 45 F + F = a( ) a(5.5) = a 8 8 80 6 Putting this in (iii) 0 4 (45 / 6)a a = 4 a = 6 3 /s (a) cceleration of block = a 8 /s 3 (b) Fro FBD of, F F 3 = (a/) F = 3 a 3 5 6 30 (Rightward) 8 8 3 3 a a/
http://www.rpauryascienceblog.co/ F 3 = a 8 F 3 = 0 3 (rightward) (c) cceleration of saller roller = a 8 /s. 3 0. Free body diagras : f C f [4] P Q g f f f f Cg Cylinder Cylinder B Cylinder C For the cylinder B Bg B g (f + f ) = B a B f + f = 5.5a B aking torque about point P for cylinder... (i) P = I P = f r + g r [ (0.) + (0.) ] = f 0.4 + 0 0. 0. = 0.4 f + 4... (ii) aking torque about point Q for cylinder C Q = I Q C = f r + c.g r [ (0.) (0.) ] c f 0. 0 0. ] 0.05 c = 0. f +... (iii) lso, a B = 0.4 = a.5a B B 0.4 and a B = C 0. C = 5 a B Putting values of C in (ii) and (iii), 0..5 a B = 0.4 f + 4 0.3 a B = 0.4 f + 4... (iv) 0.05 5a B = 0. f + 0.5 a B = 0.4f +... (v) fro (iv) and (v) (0.3 + 0.5)a B = 0.4(f + f ) + 6 0.45 a B = 0.4 (5.5 a B ) + 6 a B = 6 =.03 /s.45
http://www.rpauryascienceblog.co/ 03. Since > tan so cube will not slip on the wedge. Hence force of friction between the ground and wedge is zero. 04. F.B.D. of g - = a () F.B.D. of B relative to L f a g F.B.D. of = a f = Here f = liiting friction lso f = g a = g a = g/ () - = a = ( + )a (3) Fro (), () and (3), = μ 05. (i) FBD of an ; is the tension in the string + = 600...(i) FBD of box, = + 300... (ii) Solving (i) and (ii) + 300 = 600 = 300 = 50 (weight shown by the achine = 5 kg) 600 a (ii) ow if should be 600. equ (i) becoes + 600 = 60 a and (ii) becoes 300 = 30 a By solving (i) and (ii) = 800 06. Let be the tension in the spring. ow, elongation produced in the spring = (r - 0 ).... (i)... (ii) 300
http://www.rpauryascienceblog.co/ ension = = k(r - 0 ) Linear velocity of the otion = v = rn v Centripetal force = = r Equating (I) and (ii), we get, K(r - 0 ) = 4 rn k0 or r = k 4 n substituting, = 4 n 0k. (k 4 n ) (rn) r... (i) = 4 rn.... (ii) 07. FBD of (for axiu frequency v /r r g F r For axiu value of n the ass will have a tendency to ove upwards & so frictional force will be acting downwards. - g cos = v sin... (I) r v F r + g si = cos r lso F r = & v = n r Solving n = g(sin cos r(cos sin FBD of (for iniu frequency)... (ii) = axiu frequency allowed. - g cos = v sin r... (iii) F r g v /r F r v g sin - Fr = cos... (iv) r F r =, v = n r. g(sin cos) Solving n = r(cos sin) frequency allowed. = iniu 08. If the real acceleration of the rod perpendicular to the surface of wedge be a, and acceleration of the wedge be. gcos - = a () sin = () Constraint equation can be written as a = sin (3)
http://www.rpauryascienceblog.co/ Using equations (), () and (3) we get a (gcos - a)sin = gcossin = asin + sin gcossin = sin a sin gcos sin a a = sin gcos sin gcos and = sin sin a sin sin 09. f ax = 0 f ax =. 3 S = = 40 0 = 0.3 5kg 4kg fax F = 9 (a) a = F/9 /s 40 F 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 ax = F 4 9 () = 7 and resulting acc 3 /s check :- Let F = 8 coon acceleration = 9 8 /s for 4 kg to have this acceleration for force needed = 4 (8/9) 90 = 9 But ax. it can get is < 4 kg block will slip. 9 0. t equilibriu g = 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 F. Hence increase in elongation of springs is F F F F,, and respectively. k k k3 k 4
http://www.rpauryascienceblog.co/ ccording to geoetry of the arrangeent, downward displaceent of the block fro its equilibriu F F F F position is y =. k k k3 k4 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 = - 4 k k k3 k4 F Restoring acceleration of block = a = - 4 k Hence = Hence = k y k 4 3 k 4 k k k 4 3 k 4. k k k3 k4. fter tie t, the point has fallen through x gt and, the platfor has oved upward by x a0t a0 x x v v x x he total length of the chain resting on the platfor is x = x + x = g a 0 t he noral reaction of the platfor consists of weight coponent and thrust coponent. = wt + thrust wt = x (g + a 0 ) = g a 0 t d thrust = v rel dt v rel = relative velocity of chain w.r.t platfor v rel = v + v = gt + a 0 t = (g + a 0 )t d d dx vrel dt dx dt g a t thrust 0 g a t g a t otal reaction is = g a 0 t 0 0 3
http://www.rpauryascienceblog.co/. a 0 a rel cos a rel () g sin Let the bar of ass slides with an acceleration a rel in downward direction and pris oves towards right with an acceleration a 0 ow a rel = a b.p. (acceleration of block w.r.t. pris) acceleration of block w.r.t. ground is a b.g. = a b.p. + a p.g. = -a 0 + a rel cos (in horizontal direction) a b.g. = a rel sin (in vertical direction) ow choosing the x and y axis along horizontal and vertical direction we have by f.b.d. () g - cos = a b.g. = a rel sin () sin = a b.g. = (a rel cos - a 0 ) () ow by f.b.d. () sin =.a 0 (3) By () and () = [gcos - a 0 sin] (4) ow by (3) and (4) (gcos = a 0 sin)sin =.a 0 g.cossin = a 0 ( + sin ) g cos sin or a 0 = sin 3. F.B.D. of g - = a () F.B.D. of B relative to L f a g F.B.D. of Fro (), () and (3), = = a f = g, Here f = liiting friction a = g a = g/ () - = a = ( + )a (3)
http://www.rpauryascienceblog.co/ 4. F.B.D. of an f = force between an and the plank. f f f g f = a (i) = g (ii) F.B.D. of plank F f f = force of friction between the plank and surface. g f can have any value in the following range. F - f ax f F + f ax F - ( + )g a F + ( + )g F g F a g 5. he syste is illustrated in the figure. Let the tension in the string be and the accelerations of the syste be a. he equation of otion for the asses are for ass, - g sin 30 0 = a... (I) and for ass 5 5g sin 45 0 - = 5a 5g - = 5 a.. (ii) 5g dding (i) & (ii), - g= 7a... (iii) or 5 g = 7a Hence the acceleration of the syste is a = fro (i) = g + a 5 g = g + 7 = g(5 5 7 ) 5 g 7 30 0 g 5g( ) = 7 he force on this pulley is the resultant of the tension in the string on the two sides. 45 0 5g 5
http://www.rpauryascienceblog.co/ he angle between the two tensions is (60 0 + 45 0 ) = 05. herefore the force on the pulley is cos (05/) 0 g = cos 5 = 0 ( + ) cos 5. 7 6. F.B.D. of g - = a () F.B.D. of B relative to L f a g F.B.D. of Fro (), () and (3), = μ = a f = g, Here f = liiting friction a = g a = g/ () - = a = ( + )a (3) 7. F = a (a is absolute retardation of ) F = a (a is absolute acceleration of ) Relative acceleration of = a + a 0 s = v = (a + a)s v0 v0 (a a ) F( ) 8. = g = a = (a +g) = + L x L x = (a +g) L x = 0 = (a +g)(+) Fore exerted on the ceiling = (+)(a+g) downward. L-x g a
http://www.rpauryascienceblog.co/ 9. he FBD of,b,c are shown = a.() g = a.() = 3a 3.(3) constraint relation : a 3 + a 3 = a.(4) solving the equations a = (3/5)g, a =(/5)g, a 3 = g/5, = (3/5)g. g a 3 a g a 3 3g 0. Let the tie be t 0 when the relative otion starts then, 0t 0 t 0-0 g = 0 or, 0 t 0-0 g = 0 t 0 or, 0 t 0 = 0 g t 0 = 0g sec. 0. bead of ass '' is fitted on to a rod can slide on it without friction. t the initial oent the bead is in the iddle of the rod. he rod oves translationally in a horizontal plane with an acceleration 'a' in a direction foring an angle with the rod. Find the acceleration of the bead relative to the rod. sin a r cos a r cos g a r sin ( relative acceleration is siply the vector difference between the absolute acceleration) or a y = a r cos + a... (i) and a r sin = a x - 0 or a x = a r sin... (ii) Fro FBD of the bead (projecting forces vertically and horizontally) g - cos = a r sin... () and sin = (a r cos + a)... (B) eliinating between () and (B) g sin = a r + a cos or a r = g sin - a cos
http://www.rpauryascienceblog.co/. Let the plane oves to the left with acceleration a. ccording to FBD of the block (ass = ) pseudo force a acts on the block to the right. oral force = g cos + a sin Force of friction f = (down the plane) he block just begins to slip up the plane, when a cos = g sin + f Siplifying, we get (sin cos g a = cos sin Substituting the values of and, a =.4 /s a g sin f = 37 0 a cos a a sin g cos 3. 5t g = 0 0.50 t = = sec 5 g = a 5 a = =.5 /s Upper Block 5t Lower Block g g g 4. (a) For the position of the rod, shown in the figure. otal external torque = g[(l/) sin] as I L d We get, g[(l/) sin] = 3 dt or = d g 3 dt L sin s 3g = 45, = L (b) d g d 3 sin L 3g d d L sin 0 0 3 g L cos = 3g cos L for = 45, = 5. By CO v v 0 = v - 0 3 4v v v = 0 0 3 4 3 By CO v L L v L 0 = 0 3 3g = 4.6 rad/s L g
http://www.rpauryascienceblog.co/ v 0L 4 L 3 8v v = 0 0 L 4L 6. v(t) = v 0 - gt g (t) = 0 + t R for pure rolling v(t) = R(t) v 0 - gt = gt t = v 0 3g 7. gsin - f = a () fr = I () a = R.(3) fro equations () () and (3) g sin gsin 5 a = gsin I 7 R 5 5 f = gsin 5 7 f = 7 gsin (up the incline) work done by the f is zero since it is static frictional force. v(t) = at = 7 5 g(sin)t a(t) = 7 5 gsin 8. Instantaneous ass of the car = + t d dv d F = ( t)v (vt) dt dt dt Fdt = dv + d(vt) integrating Ft = v + vt Ft v = t 9. ngular oentu will be conserved V 0 O sin30 0 = vr V 0 = Vr fro energy conservation k(o) + v0 = kr + v + v 0 = r + v 4r 0
http://www.rpauryascienceblog.co/ r =.68 /s 30. g + a 0 = a (i). R = I (ii) a = R (iii) fro the above equations a = (g a0 ) and = (g a0 ) 3 3 3. a 0 R a 0 a 0 a 0 a a 0 g = 0 a 0 (i) = (ii) ( + a 0 ) g = a (iii) g ( + a 0 ) = a (iv) 4 0( ) a = g 4 ( ) 0 g g 3. For iniu, f a f a g g a + f = a + kg = = g = a ka + = g g( k) a in = k for axiu (a is ore than ) a f a f g a = + kg g
http://www.rpauryascienceblog.co/ = a = ka + g g( k) a ax =. k 33. FBD of plank: f (only horizontal forces are shown) f a = (i) (acceleration of plank w.r.to ground) For an: a + f = a fro (i) a = a a = a/ acceleration of an w.r.t. plank = a - a acceleration of an w.r.to ground = a P + a PG = a + (- a ) = 0 f = a a a 34. Suppose acceleration of block and B be a (rightward and b(leftward) respectively. hen horizontal block C is b (leftward) and (a+ b). FBD of the blocks: 3 7.5 g C g 6g = 7.5 a g - = (a + b) = (b) - = 6 (b)... (i)... (ii)... (iii)... (iv) Solving: a =.04 /sec ; b =. /s herefore horizontal and vertical coponent of acceleration of block C are b =.9 /s (leftwards) and a + b =.59 /s (downwards) respectively. Hence, resultant acceleration = (.9) (.59) =.43 /s. 35. FBD of block FBD of wedge
http://www.rpauryascienceblog.co/ g g Let accelerations of block in the downward vertical direction be a b and that of wedge in forward direction is a w. Hence, equation of otion, g - = a b (i) = a w (ii) Horizontal otion of wedge : = a w (iii) Constraint equation: Let B = x, BC = y, CD = z Hence length x + y + z = constant (iv) d x d z Here y is constant; - a w and a b dt dt Hence, differentiating (iv) twice with respect to tie; a b = a W (in agnitude) g Solving, a b = a w = ( Required acceleration of the block relative to ground = b b a a = g ( ) D B C 36. Force acting on block along the face of wedge - gsin30 = a () Force acting on sphere Weight (g) - Buoyant force F g = a () Solving we get a = 0 37. oentu before collision = v = 00 0 3 (50) 60 oentu just after collision = ( + )v = [(00 0 3 ) +.9]v 50 c 50 /s Conserving oentu in the horizontal direction v = ( + )v 00 g.9 kg v v = = 5 /s ow let velocity at 60 angle be V; Conserving energy between these two positions ( )v ( ) v + ( + )g [L ( - cos)] v = 4.74 /s.
http://www.rpauryascienceblog.co/ 60 is the tension in the string ( ( + )gcos60 = = 50. )V L ( + )g 38. Let at any instant of tie and is displaced x (down) and x (down) respectively and let x > x elongation of the spring at this instant x = x x Free body diagras : F kx 3 g kx g 3 g Considering the forces parallel to plane g sin + kx F = a i.e. a = + g sin - kx = a i.e. a = as 3 g = 3 a gsin kx F gsin kx 3g 3... (i)... (ii) cceleration of relative to, a = a a gsin kx 3g gsin kx F = ( ) 3 fro (iii) ( 4 0 000x 0) ( 0 000x 5) a = 5 a = 8.5 700 x.... (iv) dv a = v = 8.5 700 x dx.... (iii) 0 vdv 0 0 x ax (8.5 700x)dx Which gives x ax =.4 c Substituting the value of x ax in (iv) a = -8.5 /s Hence syste oves with acceleration 8.5 /s up the plane.
http://www.rpauryascienceblog.co/ 39. (i) For just slipping f = g F - g = a and g = a F a = a F in = = g ( ) g in F (ii) if = 3 3( )g F - g = g = a F(3 ) = 3( ) F a = 3( ) displaceent of block of ass in tie t seconds F(3 ) s = t 3( ) work done by friction force w = gs F F(3 )t = g F 3( )g 6( ) f f F F (3 )t = 8( ) displaceent of block of ass in t seconds F s = t 3( ) Relative displaceent s = s - s = Ft 6( ) Ft ( ) Ft = = 3( ) 3 Heat dissipiation = g s F Ft = g 3( )g 3 3 = F t. 9( )
http://www.rpauryascienceblog.co/ 40. (i) Fro the figure shown y L/ y x L tan O d x dy 0 dt tan dt y/tan x y/tan dy d x tan dt dt If v and u be the agnitudes of the velocities of bar and wedges then v = utan... (i) or h = X tan X = horizontal displaceent of each wedge when the bar coe down by h gh = v + u... (ii) Fro (i) and (ii) Velocity of bar c when it strikes floor = Speed of wedges at the sae instant = (ii) Workdone by sin on the syste of the wedge sin. X = u sin. X = u (Work energy theore) (Work energy theore) gh sin. X =. tan gsec which gives = tan 4. F.B.D. of g - = a () gh tan cos tan tan sin gh F.B.D. of B relative to L f a g F.B.D. of = a f = g, Here f = liiting friction a = g a = g/ () - = a = ( + )a (3)
http://www.rpauryascienceblog.co/ Fro (), () and (3), = 4. F = a 3 = a = a For constraints relation a = 3a + a by solving we get F a = ( 9 ) 4 43. When particle starts sliding, friction on it is equal to liiting friction. ow noral force = (g + b)...(i) Let the particle starts sliding after oving a distance x on the board. Considering tangential otion; u = 0 ; a = 0.5 /s ; v =? v = u + ax v = x/... (ii) ow, resulting horizontal acceleration = v R a / = x R a / his acceleration is provided by liiting friction x R a / = (g + b) x = 4 = 6. 44. Let retardation up the board = relative to the elevator. ow, v = u +as (0) = (4) - (.6) = 0 /s Horizontal coponent of resultant acceleration of the block = cos 37 0 = 8 /s vertical coponent of resultant acceleration of the block = a - sin 37 0 = (a - 6) /s FBD of the block: g
http://www.rpauryascienceblog.co/ g sin 37 0 + cos 37 0 = (8)... (i) cos 37 0 - g - sin 37 0 = (a - 6)... (ii) Solving 40 + 4 a + 3a = 0... (iii) ow considering otion of the block down the plane Let acceleration of the block relative to elevator = this tie. Hence : v = u + x where v = 4 /s, u = 0 Hence = 5 /s x =.6 ow horizontal coponent of resultant acceleration = cos 37 0 = 4 /s vertical coponent of resultant acceleration = a - sin 37 0 = (a - 3) /s Hence : cos 37 0 + sin 37 0 - g = (a - 3)... (iv) sin 37 0 - cos 37 0 = (4)... (v) Hence solving 40 + 4a - 3a = 5 Solving (iii) and (vi) a =.5 /s, = 0.5... (vi) 45. (i) FBD of an ; is the tension in the string + = 600...(i) 600 FBD of box, = + 300... (ii) Solving (i) and (ii) + 300 = 600 = 300 = 50 (weight shown by the achine = 5 kg) (ii) ow if should be 600. equ (i) becoes 600 = 60 a and (ii) becoes 300 +- t = 30 a = -60 a and 900 + 60 a = 30 a or 900 = - 30 a a = - 30 /s (i.e. upwards) and = - 60 (-30) = 800. 300
http://www.rpauryascienceblog.co/ 46. If oves x then / oves x ow if acceleration of is a then acceleration of / a Equation of block = a () Equation of block B () ow () + () g = 3a a = g/3 =0/3 /s (b) fro () = g 5 6 3 tension in thread PQ = 0 3 (c) R 3 5 g a a R / g B.a 47. cceleration of block, a = g sin - g cos x = 0, 0 x vdv g(sin cos.x)dx 0 = g (sin.x - 0 0 tan. otal distance travelled = tan For v to be ax., a = 0, x = tan v ax = sin g cos.. cos.x. ) 48. Since > tan so cube will not slip on the wedge. Hence force of friction between the ground and wedge is zero. 49. (i) When loses contact with ground = g = 0 = 0....(i) = and F = 40 t = = 4 Hence, = 0 t.... (ii) fro (I) and (ii) t = sec, Hence will lose contact at t = sec siilarly for B = B g = 0 = 0 t t = sec, Hence B will lose contact at t = sec. Siilarly, for C = 3 0 = 0 t F = 40 t P Q B C
http://www.rpauryascienceblog.co/ t =.5 sec, hence C will lose contact at t =.5 sec. g = a dv 0 t 0 =... (iii) dt (ii) Velocity of when B loses contact with ground v dv = ( 0t 0 0)dt which gives, v = 5 /s 3 / Velocity of when C lose contact v = ( 0t 0) dt = 5/4 /s (iii) For block C - c g = c a dv 0 t 30 = 3 dt a g = 0t cg 3 dv = 0 t dt 30 dt v t t 0 30 dv tdt dt 3 3 v = 0 H 0 dy 3 / 3 / 0t [ 3 H = 0.4. 3 / 30 0t ] dt 4 0 t 0t 3 30 4