Page 1 Ob jec tive Ques tions (Level 1) Single Correct Option 1. Momentum remains conserved. Decrease in momentum of the ball is transferred to sand while KE does not remain conserved as it gets used up in doing work against friction. 2. F M a ext net CM = ´ \ If F ext net = 0, a CM = 0 i.e., d dt v CM = 0 or v CM = constant Option (a) is correct. 3. The forces acting on the blocks would be equal and opposite as per Newton's 3rd law of motion. Acceleration of the blocks will depend upon their masses as per Newton’s 2nd law of motion. Accelerations being different velocities will be unequal. Option (c) is correct. 4. While colliding the balls will apply equal and opposite impulsive force on each other. Impulsive forces will change the momentum of the balls but the total momentum of the system of 2 balls will remain conserved impulsive forces being internal ones. Change in momentum of the system will definitely be due to external gravitational forces on the balls but as the time of impact shall be very less the impulsive force will over shadow the weak gravitational force. 5. External force acting on the cannon shell before explosion is the gravitational force. Now, as no extra net external force would be act on the shell during collision the momentum of the system shall remain conserved and the CM of the system (now broken in pieces) will also keep on following the path which the shell would have followed had the explosion not taken place. Further, as the explosion would never be superelastic, the KE of the system can’t increase after explosion. Option (d) is correct. 6. Velocity of separation Velocity of approach = e As in an elastic collision e < 1 Velocity of separation < velocity of approach (when e = 0, the velocity of separation in zero and the colliding bodies do not separate from each other.) Further, whether the collision in elastic or inelastic the law of conservation of momentum always hold gord. \ Option (d) is correct. 7. p v ® ® = M CM , Option (a) is correct. p p p p 3 ® ® ® ® = + + + 1 2 ...., Option (b) is correct. Further, we define momentum for every type of motion. \ Option (d) is correct. 8. Let us consider a system of two masses as shown in figure. Momentum of system about CM =  +  ® ® ® ® m m 1 1 2 2 ( ) ( ) v v v v CM CM = +  + ® ® ® m m m m 1 1 2 2 1 2 v v v ( ) CM = +  + ® ® ( ) ( ) m m m m 1 2 1 2 v v CM CM = ® 0 Option (c) is correct. 9. Option (a) If collision is inelastic. Option (b) If collision is perfectly inelastic Option (c) If the dimensions of the particles ®0 \ Option (d) would be the answer. 180  Mechanics1 M 1 M 2 F F v CM ® v 1 ® v 2 ® m 1 m 2 CM m A v v v m m B C Page 2 Ob jec tive Ques tions (Level 1) Single Correct Option 1. Momentum remains conserved. Decrease in momentum of the ball is transferred to sand while KE does not remain conserved as it gets used up in doing work against friction. 2. F M a ext net CM = ´ \ If F ext net = 0, a CM = 0 i.e., d dt v CM = 0 or v CM = constant Option (a) is correct. 3. The forces acting on the blocks would be equal and opposite as per Newton's 3rd law of motion. Acceleration of the blocks will depend upon their masses as per Newton’s 2nd law of motion. Accelerations being different velocities will be unequal. Option (c) is correct. 4. While colliding the balls will apply equal and opposite impulsive force on each other. Impulsive forces will change the momentum of the balls but the total momentum of the system of 2 balls will remain conserved impulsive forces being internal ones. Change in momentum of the system will definitely be due to external gravitational forces on the balls but as the time of impact shall be very less the impulsive force will over shadow the weak gravitational force. 5. External force acting on the cannon shell before explosion is the gravitational force. Now, as no extra net external force would be act on the shell during collision the momentum of the system shall remain conserved and the CM of the system (now broken in pieces) will also keep on following the path which the shell would have followed had the explosion not taken place. Further, as the explosion would never be superelastic, the KE of the system can’t increase after explosion. Option (d) is correct. 6. Velocity of separation Velocity of approach = e As in an elastic collision e < 1 Velocity of separation < velocity of approach (when e = 0, the velocity of separation in zero and the colliding bodies do not separate from each other.) Further, whether the collision in elastic or inelastic the law of conservation of momentum always hold gord. \ Option (d) is correct. 7. p v ® ® = M CM , Option (a) is correct. p p p p 3 ® ® ® ® = + + + 1 2 ...., Option (b) is correct. Further, we define momentum for every type of motion. \ Option (d) is correct. 8. Let us consider a system of two masses as shown in figure. Momentum of system about CM =  +  ® ® ® ® m m 1 1 2 2 ( ) ( ) v v v v CM CM = +  + ® ® ® m m m m 1 1 2 2 1 2 v v v ( ) CM = +  + ® ® ( ) ( ) m m m m 1 2 1 2 v v CM CM = ® 0 Option (c) is correct. 9. Option (a) If collision is inelastic. Option (b) If collision is perfectly inelastic Option (c) If the dimensions of the particles ®0 \ Option (d) would be the answer. 180  Mechanics1 M 1 M 2 F F v CM ® v 1 ® v 2 ® m 1 m 2 CM m A v v v m m B C 10. F t m v D D = Þ F m v t = D D = ´  5 65 15 2 ( ) =125N 11. m d m d d c o ´ = ´  1 1 ( ) Þ d m m m d o o c 1 = + = + 8 8 6 d = 4 7 d = ´ ´  4 7 10 10 (1.2 m) = ´  0.64 m 10 10 12. M mv M m V ´ = +  0 ( ) Þ V m m m v =   ( ) =   ´ 3 9 3 16 =   8 1 ms \ KE of 6 kg mass = ´ ´  1 2 6 8 2 ( ) =192J 13. v m m m v 1 2 1 2 2 2 ¢ = + =2 2 v (as m m 1 2 < < ) Option (b) is correct. 14. Horizontal velocity of the leaving coal : v i 1 ® = + v ^ Horizontal velocity of the system v i 2 ® = + v ^ \ U v v ® ® ® ® =  = rel 1 2 1 0 \ F U th rel ® ® ® = = dm dt 0 As, the leaving coal does not exert any thrust force on the wagon, the speed of the wagon won’t change. Option (a) is correct. 15. If n be the number of bullet shots per second n´ [change in momentum per second] £ F i.e., n 40 100 1200 0 144 ´  é ë ê ù û ú £ ( ) or n £ 144 48 or n £ 3 \ Option (a) is correct. 16. Change in momentum along xaxis = m (v v cos cos q q  ) = 0 \ Net change in momentum = Change in momentum along yaxis = +   m v v [( sin ) ( sin )] q q =2mvsinq = mv 2 (as q = ° 45 ) Option (a) is correct. 17. Velocity of ball before first impact i.e., when it reaches point Q of the horizontal plane Centre of Mass, Conservation of Linear Momentum, Impulse and Collision  181 d m o m c d 1 O C CM M – m m M Rest V v 3 kg 6 kg 9 kg v 2 m 2 m 1 v' 2 m 2 v' 1 m 1 v = 0 1 After collision Before collision x O v i ^ system v cos q v sin q v cos q q v y P + – q v cos q v Q x Page 3 Ob jec tive Ques tions (Level 1) Single Correct Option 1. Momentum remains conserved. Decrease in momentum of the ball is transferred to sand while KE does not remain conserved as it gets used up in doing work against friction. 2. F M a ext net CM = ´ \ If F ext net = 0, a CM = 0 i.e., d dt v CM = 0 or v CM = constant Option (a) is correct. 3. The forces acting on the blocks would be equal and opposite as per Newton's 3rd law of motion. Acceleration of the blocks will depend upon their masses as per Newton’s 2nd law of motion. Accelerations being different velocities will be unequal. Option (c) is correct. 4. While colliding the balls will apply equal and opposite impulsive force on each other. Impulsive forces will change the momentum of the balls but the total momentum of the system of 2 balls will remain conserved impulsive forces being internal ones. Change in momentum of the system will definitely be due to external gravitational forces on the balls but as the time of impact shall be very less the impulsive force will over shadow the weak gravitational force. 5. External force acting on the cannon shell before explosion is the gravitational force. Now, as no extra net external force would be act on the shell during collision the momentum of the system shall remain conserved and the CM of the system (now broken in pieces) will also keep on following the path which the shell would have followed had the explosion not taken place. Further, as the explosion would never be superelastic, the KE of the system can’t increase after explosion. Option (d) is correct. 6. Velocity of separation Velocity of approach = e As in an elastic collision e < 1 Velocity of separation < velocity of approach (when e = 0, the velocity of separation in zero and the colliding bodies do not separate from each other.) Further, whether the collision in elastic or inelastic the law of conservation of momentum always hold gord. \ Option (d) is correct. 7. p v ® ® = M CM , Option (a) is correct. p p p p 3 ® ® ® ® = + + + 1 2 ...., Option (b) is correct. Further, we define momentum for every type of motion. \ Option (d) is correct. 8. Let us consider a system of two masses as shown in figure. Momentum of system about CM =  +  ® ® ® ® m m 1 1 2 2 ( ) ( ) v v v v CM CM = +  + ® ® ® m m m m 1 1 2 2 1 2 v v v ( ) CM = +  + ® ® ( ) ( ) m m m m 1 2 1 2 v v CM CM = ® 0 Option (c) is correct. 9. Option (a) If collision is inelastic. Option (b) If collision is perfectly inelastic Option (c) If the dimensions of the particles ®0 \ Option (d) would be the answer. 180  Mechanics1 M 1 M 2 F F v CM ® v 1 ® v 2 ® m 1 m 2 CM m A v v v m m B C 10. F t m v D D = Þ F m v t = D D = ´  5 65 15 2 ( ) =125N 11. m d m d d c o ´ = ´  1 1 ( ) Þ d m m m d o o c 1 = + = + 8 8 6 d = 4 7 d = ´ ´  4 7 10 10 (1.2 m) = ´  0.64 m 10 10 12. M mv M m V ´ = +  0 ( ) Þ V m m m v =   ( ) =   ´ 3 9 3 16 =   8 1 ms \ KE of 6 kg mass = ´ ´  1 2 6 8 2 ( ) =192J 13. v m m m v 1 2 1 2 2 2 ¢ = + =2 2 v (as m m 1 2 < < ) Option (b) is correct. 14. Horizontal velocity of the leaving coal : v i 1 ® = + v ^ Horizontal velocity of the system v i 2 ® = + v ^ \ U v v ® ® ® ® =  = rel 1 2 1 0 \ F U th rel ® ® ® = = dm dt 0 As, the leaving coal does not exert any thrust force on the wagon, the speed of the wagon won’t change. Option (a) is correct. 15. If n be the number of bullet shots per second n´ [change in momentum per second] £ F i.e., n 40 100 1200 0 144 ´  é ë ê ù û ú £ ( ) or n £ 144 48 or n £ 3 \ Option (a) is correct. 16. Change in momentum along xaxis = m (v v cos cos q q  ) = 0 \ Net change in momentum = Change in momentum along yaxis = +   m v v [( sin ) ( sin )] q q =2mvsinq = mv 2 (as q = ° 45 ) Option (a) is correct. 17. Velocity of ball before first impact i.e., when it reaches point Q of the horizontal plane Centre of Mass, Conservation of Linear Momentum, Impulse and Collision  181 d m o m c d 1 O C CM M – m m M Rest V v 3 kg 6 kg 9 kg v 2 m 2 m 1 v' 2 m 2 v' 1 m 1 v = 0 1 After collision Before collision x O v i ^ system v cos q v sin q v cos q q v y P + – q v cos q v Q x v g = + × 0 1 ( ) = g \ Velocity of ball after 1st impact = = ev eg time elapsed between 1st and 2nd impact with the horizontal plane ( ) ( ) ( ) + = + +  e g e g g t Þ t e =2 = 4 3 s (as, e = 2 3 L) Option (c) is correct. 18. x x dm dm x m dx m dx CM = = ò ò ò ò (where m = mass per unit length) = ò ò x Ax L dx Ax L dx 2 2 Q m Ax L = æ è ç ö ø ÷ 2 = ò ò x dx x dx L L 3 0 2 0 = ´ L L 4 3 4 3 = 3 4 L Option (a) is correct. 19. As there is no net external force acting on the system in the horizontal direction, the CM of the system shall not shift along xaxis. \ 50 10 450 5 50 450 50 450 5 50 450 ´ + ´ + = ´ + ´ + + ( ) ( ) x x (Initially) (Finally) Þ x = 1 m Option (b) is correct. 20. As discussed in the answer to previous question no. 19. Mx M x L M M ML M L M M + + æ è ç ö ø ÷ + = + æ è ç ö ø ÷ + 3 2 3 3 2 3 (Initially) (Finally) Þ x M M ML + é ë ê ù û ú = 3 x L = 3 4 i.e., the distance that plank moves = 3 4 L \ The distance that the man moves =  L L 3 4 = L 4 Option (b) is correct. 21. mv m m mu + × = × + 3 0 0 3 182  Mechanics1 dx x L 0 x 450 kg CM (boat) 10 m O 50 kg 50 kg x 450 kg CM (boat) X X M/3 L Smooth X M/3 L Smooth X x O M M O m 3m m 3m Rest Rest v u Before collision After collision P Q + eg 2nd – eg 1st u = 0 Page 4 Ob jec tive Ques tions (Level 1) Single Correct Option 1. Momentum remains conserved. Decrease in momentum of the ball is transferred to sand while KE does not remain conserved as it gets used up in doing work against friction. 2. F M a ext net CM = ´ \ If F ext net = 0, a CM = 0 i.e., d dt v CM = 0 or v CM = constant Option (a) is correct. 3. The forces acting on the blocks would be equal and opposite as per Newton's 3rd law of motion. Acceleration of the blocks will depend upon their masses as per Newton’s 2nd law of motion. Accelerations being different velocities will be unequal. Option (c) is correct. 4. While colliding the balls will apply equal and opposite impulsive force on each other. Impulsive forces will change the momentum of the balls but the total momentum of the system of 2 balls will remain conserved impulsive forces being internal ones. Change in momentum of the system will definitely be due to external gravitational forces on the balls but as the time of impact shall be very less the impulsive force will over shadow the weak gravitational force. 5. External force acting on the cannon shell before explosion is the gravitational force. Now, as no extra net external force would be act on the shell during collision the momentum of the system shall remain conserved and the CM of the system (now broken in pieces) will also keep on following the path which the shell would have followed had the explosion not taken place. Further, as the explosion would never be superelastic, the KE of the system can’t increase after explosion. Option (d) is correct. 6. Velocity of separation Velocity of approach = e As in an elastic collision e < 1 Velocity of separation < velocity of approach (when e = 0, the velocity of separation in zero and the colliding bodies do not separate from each other.) Further, whether the collision in elastic or inelastic the law of conservation of momentum always hold gord. \ Option (d) is correct. 7. p v ® ® = M CM , Option (a) is correct. p p p p 3 ® ® ® ® = + + + 1 2 ...., Option (b) is correct. Further, we define momentum for every type of motion. \ Option (d) is correct. 8. Let us consider a system of two masses as shown in figure. Momentum of system about CM =  +  ® ® ® ® m m 1 1 2 2 ( ) ( ) v v v v CM CM = +  + ® ® ® m m m m 1 1 2 2 1 2 v v v ( ) CM = +  + ® ® ( ) ( ) m m m m 1 2 1 2 v v CM CM = ® 0 Option (c) is correct. 9. Option (a) If collision is inelastic. Option (b) If collision is perfectly inelastic Option (c) If the dimensions of the particles ®0 \ Option (d) would be the answer. 180  Mechanics1 M 1 M 2 F F v CM ® v 1 ® v 2 ® m 1 m 2 CM m A v v v m m B C 10. F t m v D D = Þ F m v t = D D = ´  5 65 15 2 ( ) =125N 11. m d m d d c o ´ = ´  1 1 ( ) Þ d m m m d o o c 1 = + = + 8 8 6 d = 4 7 d = ´ ´  4 7 10 10 (1.2 m) = ´  0.64 m 10 10 12. M mv M m V ´ = +  0 ( ) Þ V m m m v =   ( ) =   ´ 3 9 3 16 =   8 1 ms \ KE of 6 kg mass = ´ ´  1 2 6 8 2 ( ) =192J 13. v m m m v 1 2 1 2 2 2 ¢ = + =2 2 v (as m m 1 2 < < ) Option (b) is correct. 14. Horizontal velocity of the leaving coal : v i 1 ® = + v ^ Horizontal velocity of the system v i 2 ® = + v ^ \ U v v ® ® ® ® =  = rel 1 2 1 0 \ F U th rel ® ® ® = = dm dt 0 As, the leaving coal does not exert any thrust force on the wagon, the speed of the wagon won’t change. Option (a) is correct. 15. If n be the number of bullet shots per second n´ [change in momentum per second] £ F i.e., n 40 100 1200 0 144 ´  é ë ê ù û ú £ ( ) or n £ 144 48 or n £ 3 \ Option (a) is correct. 16. Change in momentum along xaxis = m (v v cos cos q q  ) = 0 \ Net change in momentum = Change in momentum along yaxis = +   m v v [( sin ) ( sin )] q q =2mvsinq = mv 2 (as q = ° 45 ) Option (a) is correct. 17. Velocity of ball before first impact i.e., when it reaches point Q of the horizontal plane Centre of Mass, Conservation of Linear Momentum, Impulse and Collision  181 d m o m c d 1 O C CM M – m m M Rest V v 3 kg 6 kg 9 kg v 2 m 2 m 1 v' 2 m 2 v' 1 m 1 v = 0 1 After collision Before collision x O v i ^ system v cos q v sin q v cos q q v y P + – q v cos q v Q x v g = + × 0 1 ( ) = g \ Velocity of ball after 1st impact = = ev eg time elapsed between 1st and 2nd impact with the horizontal plane ( ) ( ) ( ) + = + +  e g e g g t Þ t e =2 = 4 3 s (as, e = 2 3 L) Option (c) is correct. 18. x x dm dm x m dx m dx CM = = ò ò ò ò (where m = mass per unit length) = ò ò x Ax L dx Ax L dx 2 2 Q m Ax L = æ è ç ö ø ÷ 2 = ò ò x dx x dx L L 3 0 2 0 = ´ L L 4 3 4 3 = 3 4 L Option (a) is correct. 19. As there is no net external force acting on the system in the horizontal direction, the CM of the system shall not shift along xaxis. \ 50 10 450 5 50 450 50 450 5 50 450 ´ + ´ + = ´ + ´ + + ( ) ( ) x x (Initially) (Finally) Þ x = 1 m Option (b) is correct. 20. As discussed in the answer to previous question no. 19. Mx M x L M M ML M L M M + + æ è ç ö ø ÷ + = + æ è ç ö ø ÷ + 3 2 3 3 2 3 (Initially) (Finally) Þ x M M ML + é ë ê ù û ú = 3 x L = 3 4 i.e., the distance that plank moves = 3 4 L \ The distance that the man moves =  L L 3 4 = L 4 Option (b) is correct. 21. mv m m mu + × = × + 3 0 0 3 182  Mechanics1 dx x L 0 x 450 kg CM (boat) 10 m O 50 kg 50 kg x 450 kg CM (boat) X X M/3 L Smooth X M/3 L Smooth X x O M M O m 3m m 3m Rest Rest v u Before collision After collision P Q + eg 2nd – eg 1st u = 0 \ u v = 3 e = Velocity of separation Velocity of approach =   = u v u v 0 0 = 1 3 Option (d) is correct. 22. Change in momentum of A or B = mu (As collision is elastic) Impulse = Change in momentum Ft mu 0 2 = Þ F mu t = 2 0 Option (b) is correct. 23. Acceleration of block A a F m 0 = ¢ Þ F ma ¢ = 0 Acceleration of block B : a F F m B =  ¢ =  F ma m 0 =  F m a 0 24. Option (c) is correct. Impulse on ball = Change in momentum of ball =  æ è ç ö ø ÷  + æ è ç ö ø ÷ 9 20 4 5 0 0 mv mv =  5 4 0 mv 25. If a ball dropped from height h rebounds to a height h¢, then speed of ball just before 1st impact, u gh = 2 Just after 1st impact u gh ¢ = ¢ 2 \ e u u h h = ¢ = ¢ = 64 100 =0.8 i.e., h e h ¢ = 2 Height attained after nth impact = e h n 2 = × ( ) 0.8 2 1 n (as h=1 m) =( ) 0.8 2n Option (d) is correct. 26. Momentum of car (+ block) before throwing block = Momentum of car after throwing block + Momentum of block 500 1 500 25 25 20 ´ =  + ® i v k ^ ^ ( ) ( ) or 475 500 v i k ® =  ( ) ^ ^ or v i k ® =  20 19 ( ) ^ ^ Option (c) is correct. 27. While force is increasing with time F kt = (where k is + ive constant) or ma kt = or dv dt k m t = \ v k m t C = + 2 2 or v k m t = × 2 2 (If at t = 0, v = 0) Centre of Mass, Conservation of Linear Momentum, Impulse and Collision  183 37° 53° 3 5 4 37° 53° v 0 X Y + v cos 37° = v 00 4 5 3 4 v cos 53° 0 v 0 9 20 v 0 3 4 = u m m Rest m u m After collision Before collision Rest A B F' F' m m F Page 5 Ob jec tive Ques tions (Level 1) Single Correct Option 1. Momentum remains conserved. Decrease in momentum of the ball is transferred to sand while KE does not remain conserved as it gets used up in doing work against friction. 2. F M a ext net CM = ´ \ If F ext net = 0, a CM = 0 i.e., d dt v CM = 0 or v CM = constant Option (a) is correct. 3. The forces acting on the blocks would be equal and opposite as per Newton's 3rd law of motion. Acceleration of the blocks will depend upon their masses as per Newton’s 2nd law of motion. Accelerations being different velocities will be unequal. Option (c) is correct. 4. While colliding the balls will apply equal and opposite impulsive force on each other. Impulsive forces will change the momentum of the balls but the total momentum of the system of 2 balls will remain conserved impulsive forces being internal ones. Change in momentum of the system will definitely be due to external gravitational forces on the balls but as the time of impact shall be very less the impulsive force will over shadow the weak gravitational force. 5. External force acting on the cannon shell before explosion is the gravitational force. Now, as no extra net external force would be act on the shell during collision the momentum of the system shall remain conserved and the CM of the system (now broken in pieces) will also keep on following the path which the shell would have followed had the explosion not taken place. Further, as the explosion would never be superelastic, the KE of the system can’t increase after explosion. Option (d) is correct. 6. Velocity of separation Velocity of approach = e As in an elastic collision e < 1 Velocity of separation < velocity of approach (when e = 0, the velocity of separation in zero and the colliding bodies do not separate from each other.) Further, whether the collision in elastic or inelastic the law of conservation of momentum always hold gord. \ Option (d) is correct. 7. p v ® ® = M CM , Option (a) is correct. p p p p 3 ® ® ® ® = + + + 1 2 ...., Option (b) is correct. Further, we define momentum for every type of motion. \ Option (d) is correct. 8. Let us consider a system of two masses as shown in figure. Momentum of system about CM =  +  ® ® ® ® m m 1 1 2 2 ( ) ( ) v v v v CM CM = +  + ® ® ® m m m m 1 1 2 2 1 2 v v v ( ) CM = +  + ® ® ( ) ( ) m m m m 1 2 1 2 v v CM CM = ® 0 Option (c) is correct. 9. Option (a) If collision is inelastic. Option (b) If collision is perfectly inelastic Option (c) If the dimensions of the particles ®0 \ Option (d) would be the answer. 180  Mechanics1 M 1 M 2 F F v CM ® v 1 ® v 2 ® m 1 m 2 CM m A v v v m m B C 10. F t m v D D = Þ F m v t = D D = ´  5 65 15 2 ( ) =125N 11. m d m d d c o ´ = ´  1 1 ( ) Þ d m m m d o o c 1 = + = + 8 8 6 d = 4 7 d = ´ ´  4 7 10 10 (1.2 m) = ´  0.64 m 10 10 12. M mv M m V ´ = +  0 ( ) Þ V m m m v =   ( ) =   ´ 3 9 3 16 =   8 1 ms \ KE of 6 kg mass = ´ ´  1 2 6 8 2 ( ) =192J 13. v m m m v 1 2 1 2 2 2 ¢ = + =2 2 v (as m m 1 2 < < ) Option (b) is correct. 14. Horizontal velocity of the leaving coal : v i 1 ® = + v ^ Horizontal velocity of the system v i 2 ® = + v ^ \ U v v ® ® ® ® =  = rel 1 2 1 0 \ F U th rel ® ® ® = = dm dt 0 As, the leaving coal does not exert any thrust force on the wagon, the speed of the wagon won’t change. Option (a) is correct. 15. If n be the number of bullet shots per second n´ [change in momentum per second] £ F i.e., n 40 100 1200 0 144 ´  é ë ê ù û ú £ ( ) or n £ 144 48 or n £ 3 \ Option (a) is correct. 16. Change in momentum along xaxis = m (v v cos cos q q  ) = 0 \ Net change in momentum = Change in momentum along yaxis = +   m v v [( sin ) ( sin )] q q =2mvsinq = mv 2 (as q = ° 45 ) Option (a) is correct. 17. Velocity of ball before first impact i.e., when it reaches point Q of the horizontal plane Centre of Mass, Conservation of Linear Momentum, Impulse and Collision  181 d m o m c d 1 O C CM M – m m M Rest V v 3 kg 6 kg 9 kg v 2 m 2 m 1 v' 2 m 2 v' 1 m 1 v = 0 1 After collision Before collision x O v i ^ system v cos q v sin q v cos q q v y P + – q v cos q v Q x v g = + × 0 1 ( ) = g \ Velocity of ball after 1st impact = = ev eg time elapsed between 1st and 2nd impact with the horizontal plane ( ) ( ) ( ) + = + +  e g e g g t Þ t e =2 = 4 3 s (as, e = 2 3 L) Option (c) is correct. 18. x x dm dm x m dx m dx CM = = ò ò ò ò (where m = mass per unit length) = ò ò x Ax L dx Ax L dx 2 2 Q m Ax L = æ è ç ö ø ÷ 2 = ò ò x dx x dx L L 3 0 2 0 = ´ L L 4 3 4 3 = 3 4 L Option (a) is correct. 19. As there is no net external force acting on the system in the horizontal direction, the CM of the system shall not shift along xaxis. \ 50 10 450 5 50 450 50 450 5 50 450 ´ + ´ + = ´ + ´ + + ( ) ( ) x x (Initially) (Finally) Þ x = 1 m Option (b) is correct. 20. As discussed in the answer to previous question no. 19. Mx M x L M M ML M L M M + + æ è ç ö ø ÷ + = + æ è ç ö ø ÷ + 3 2 3 3 2 3 (Initially) (Finally) Þ x M M ML + é ë ê ù û ú = 3 x L = 3 4 i.e., the distance that plank moves = 3 4 L \ The distance that the man moves =  L L 3 4 = L 4 Option (b) is correct. 21. mv m m mu + × = × + 3 0 0 3 182  Mechanics1 dx x L 0 x 450 kg CM (boat) 10 m O 50 kg 50 kg x 450 kg CM (boat) X X M/3 L Smooth X M/3 L Smooth X x O M M O m 3m m 3m Rest Rest v u Before collision After collision P Q + eg 2nd – eg 1st u = 0 \ u v = 3 e = Velocity of separation Velocity of approach =   = u v u v 0 0 = 1 3 Option (d) is correct. 22. Change in momentum of A or B = mu (As collision is elastic) Impulse = Change in momentum Ft mu 0 2 = Þ F mu t = 2 0 Option (b) is correct. 23. Acceleration of block A a F m 0 = ¢ Þ F ma ¢ = 0 Acceleration of block B : a F F m B =  ¢ =  F ma m 0 =  F m a 0 24. Option (c) is correct. Impulse on ball = Change in momentum of ball =  æ è ç ö ø ÷  + æ è ç ö ø ÷ 9 20 4 5 0 0 mv mv =  5 4 0 mv 25. If a ball dropped from height h rebounds to a height h¢, then speed of ball just before 1st impact, u gh = 2 Just after 1st impact u gh ¢ = ¢ 2 \ e u u h h = ¢ = ¢ = 64 100 =0.8 i.e., h e h ¢ = 2 Height attained after nth impact = e h n 2 = × ( ) 0.8 2 1 n (as h=1 m) =( ) 0.8 2n Option (d) is correct. 26. Momentum of car (+ block) before throwing block = Momentum of car after throwing block + Momentum of block 500 1 500 25 25 20 ´ =  + ® i v k ^ ^ ( ) ( ) or 475 500 v i k ® =  ( ) ^ ^ or v i k ® =  20 19 ( ) ^ ^ Option (c) is correct. 27. While force is increasing with time F kt = (where k is + ive constant) or ma kt = or dv dt k m t = \ v k m t C = + 2 2 or v k m t = × 2 2 (If at t = 0, v = 0) Centre of Mass, Conservation of Linear Momentum, Impulse and Collision  183 37° 53° 3 5 4 37° 53° v 0 X Y + v cos 37° = v 00 4 5 3 4 v cos 53° 0 v 0 9 20 v 0 3 4 = u m m Rest m u m After collision Before collision Rest A B F' F' m m F Thus, graph between v and t would be While force is decreasing with time. F kt =  (where k is + ive constant) Þ v k m t C =  + ¢ 2 2 At t = 0, v v = 0 \ C v k m t ¢ = + 0 0 2 2 Thus, v v k m t t = +  0 0 2 2 2 ( ) Thus, graph between v and t would be \ Option (a) is correct. 28. x A x A x A A CM = + + 1 1 2 2 1 2 0 3 2 3 2 2 1 2 2 2 =  +   [ ( ) ] [ ] ( ) p p p p p R R x R R R R or 8 2 2 1 2 p p R x R R = or 8 2 1 x R = or x R 1 4 = Option (c) is correct. 29. x A x A x A x A A A CM = + + + + 1 1 2 2 3 3 1 2 3 \ 0 4 2 2 2 1 =   [ ( ) ] p p p R R R x +  +  × [ ] [ ] p p R R R 2 2 3 0 or 14 3 1 x R = or x R 1 3 14 = Option (d) is correct. 30. As no external force is acting along the horizontal direction on the system (wedge + block). The CM of the system shall not change along horizontal when the block moves over the wedge but would change along vertical as net force (= gravitational force) is acting on the block. Further, as no nonconservative force is acting on the system, its total energy will not change. Option (d) is correct. 31. mv M m v = + ¢ ( ) …(i) From final position, v gh ¢ = 2 From Eq. (i), or m M m v gh + = 2 or v M m gh = + æ è ç ö ø ÷ 1 2 Option (c) is correct. 32. As no net extra external force is acting on the system the CM of the gun and the bullet system remains at rest. The force exerted by the trigger of the gun on the bullet is an interval one. 33. m g T m a 1 1  = 184  Mechanics1 T T T T mg 1 mg 2 a v t 0 t Parabola v 0 M Rest v m M v' m m h Final position Just after collision Before collision Rest v t 0 t Graph would be parabolic in nature v 0Read More
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