Page 1 29. When the spider eats up the moth and travels towards A with velocity v 2 relative to rod. 24 2 48 0 m v v mv R R + æ è ç ö ø ÷ + = [v R = velocity (absolute) of rod] Þ v v R = - 6 Option (c) is correct. 30. Time taken by spider to reach point A starting from point B = + 4 8 2 L v L v/ = = = 20 20 20 L v L L T T / = ´ = 20 4 80 s 31. Form CM not to shift Þ ( ) ( ) x L m x L m ¢ + + ¢ + 8 24 6 48 = ´ 64 48 m i.e., x L ¢ = - 8 3 Option (a) is correct. More than One Cor rect Op tions 1. Along vertical : 2 2 45 mV m v sin sin q = × ° i.e., 2 2 2 V v sin q = â€¦(i) Along horizontal : mv m v mV = ° + 2 45 2 cos cos q i.e., 2 1 1 2 2 V v cos q = - æ è ç ö ø ÷ â€¦(ii) Squaring and adding Eq. (i)and (ii), 8 2 2 2 2 2 2 2 V v v v = æ è ç ö ø ÷ + - æ è ç ö ø ÷ = + + - × × v v v v v 2 2 2 8 7 2 2 2 = - 5 4 2 2 2 v v Dividing Eq. (ii) by Eq. (i) tan q = - 1 2 2 2 2 1 2 2 = - < 1 2 2 1 1 \ q < ° 45 Thus, the divergence angle between the particles will be less than p 2 . Option (b) is correct. Initial KE = 1 2 2 mv Final KE = æ è ç ö ø ÷ + 1 2 2 1 2 2 2 2 m v mV = + × - × é ë ê ù û ú 1 2 1 4 2 5 32 2 1 8 2 2 mv As Final KE < Initial KE Collision is inelastic. Option (d) is correct. 2. v m m m m v 2 2 1 1 2 2 ¢ = - + = - + m m m m v 5 5 2 = - 2 3 2 v = - 2 3 2gl Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 197 48 m 48 m x' 8L M + S v/2 m 2m q 45° L Rest 2m m v v 2 v' 2 v' 1 v = 0 1 m = 5m 1 m = m 2 Page 2 29. When the spider eats up the moth and travels towards A with velocity v 2 relative to rod. 24 2 48 0 m v v mv R R + æ è ç ö ø ÷ + = [v R = velocity (absolute) of rod] Þ v v R = - 6 Option (c) is correct. 30. Time taken by spider to reach point A starting from point B = + 4 8 2 L v L v/ = = = 20 20 20 L v L L T T / = ´ = 20 4 80 s 31. Form CM not to shift Þ ( ) ( ) x L m x L m ¢ + + ¢ + 8 24 6 48 = ´ 64 48 m i.e., x L ¢ = - 8 3 Option (a) is correct. More than One Cor rect Op tions 1. Along vertical : 2 2 45 mV m v sin sin q = × ° i.e., 2 2 2 V v sin q = â€¦(i) Along horizontal : mv m v mV = ° + 2 45 2 cos cos q i.e., 2 1 1 2 2 V v cos q = - æ è ç ö ø ÷ â€¦(ii) Squaring and adding Eq. (i)and (ii), 8 2 2 2 2 2 2 2 V v v v = æ è ç ö ø ÷ + - æ è ç ö ø ÷ = + + - × × v v v v v 2 2 2 8 7 2 2 2 = - 5 4 2 2 2 v v Dividing Eq. (ii) by Eq. (i) tan q = - 1 2 2 2 2 1 2 2 = - < 1 2 2 1 1 \ q < ° 45 Thus, the divergence angle between the particles will be less than p 2 . Option (b) is correct. Initial KE = 1 2 2 mv Final KE = æ è ç ö ø ÷ + 1 2 2 1 2 2 2 2 m v mV = + × - × é ë ê ù û ú 1 2 1 4 2 5 32 2 1 8 2 2 mv As Final KE < Initial KE Collision is inelastic. Option (d) is correct. 2. v m m m m v 2 2 1 1 2 2 ¢ = - + = - + m m m m v 5 5 2 = - 2 3 2 v = - 2 3 2gl Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 197 48 m 48 m x' 8L M + S v/2 m 2m q 45° L Rest 2m m v v 2 v' 2 v' 1 v = 0 1 m = 5m 1 m = m 2 T mv l mg - ¢ = 2 2 or T mv l mg = ¢ + 2 = + m g mg 8 9 = 17 9 mg Option (a) is correct. Velocity of block v m m m v 1 2 1 2 2 2 ¢ = + = + × 2 5 2 m m m gl = 1 3 2gl Option (c) is correct. Maximum height attained by pendulum bob = ¢ = = v g gl g l g 2 2 2 8 9 2 4 / Option (d) is correct. 3. v u cos cos f = q Þ v u = f cos cos q and e v u = f sin sinq = f ´ f cos cos sin sin q q = tan tan f q Option (b) is correct. Change in momentum of particle = - f - + ( sin ) ( sin ) mv mu q \ Impulse delivered by floor to the particle = mv mu sin sin f + q = f + é ë ê ù û ú mv u v sin sin sin q q = + é ë ê ù û ú mv u v e u v sinq = + mu e sin ( ) q 1 Option (d) is correct. u e 1 1 2 2 - - ( ) sin q = - + u e 1 2 2 2 sin sin q q = + u e cos sin 2 2 2 q q = + f u v u cos sin 2 2 2 2 q = + f u v 2 2 2 2 cos sin q = f + f v v 2 2 2 2 cos sin =v Option (c) is correct. cos sin 2 2 2 q q + e = + f cos tan tan sin 2 2 2 2 q q q = + cos ( tan ) 2 2 1 q f = f cos 2 2 q sec = f cos cos 2 2 q = = v u 2 2 FinalKE InitialKE Option (d) is correct. 4. u i j ® - = + ( ) ^ ^ 3 2 1 ms Impulse received by particle of mass m = - + ® ® m m u v = - + + - + m m ( ) ( ) ^ ^ ^ ^ 3 2 2 i j i j = - + m( ) ^ ^ 5 i j unit Option (b) is correct. Impulse received by particle of mass M = - (impulse received by particle of mass m) = + m( ) ^ ^ 5i j Option (d) is correct. 198 | Mechanics-1 v v cos f v sin f u f u sin q q + u cos q m u M v = (â€“2 i + j) m/s ^ ^ ® m Page 3 29. When the spider eats up the moth and travels towards A with velocity v 2 relative to rod. 24 2 48 0 m v v mv R R + æ è ç ö ø ÷ + = [v R = velocity (absolute) of rod] Þ v v R = - 6 Option (c) is correct. 30. Time taken by spider to reach point A starting from point B = + 4 8 2 L v L v/ = = = 20 20 20 L v L L T T / = ´ = 20 4 80 s 31. Form CM not to shift Þ ( ) ( ) x L m x L m ¢ + + ¢ + 8 24 6 48 = ´ 64 48 m i.e., x L ¢ = - 8 3 Option (a) is correct. More than One Cor rect Op tions 1. Along vertical : 2 2 45 mV m v sin sin q = × ° i.e., 2 2 2 V v sin q = â€¦(i) Along horizontal : mv m v mV = ° + 2 45 2 cos cos q i.e., 2 1 1 2 2 V v cos q = - æ è ç ö ø ÷ â€¦(ii) Squaring and adding Eq. (i)and (ii), 8 2 2 2 2 2 2 2 V v v v = æ è ç ö ø ÷ + - æ è ç ö ø ÷ = + + - × × v v v v v 2 2 2 8 7 2 2 2 = - 5 4 2 2 2 v v Dividing Eq. (ii) by Eq. (i) tan q = - 1 2 2 2 2 1 2 2 = - < 1 2 2 1 1 \ q < ° 45 Thus, the divergence angle between the particles will be less than p 2 . Option (b) is correct. Initial KE = 1 2 2 mv Final KE = æ è ç ö ø ÷ + 1 2 2 1 2 2 2 2 m v mV = + × - × é ë ê ù û ú 1 2 1 4 2 5 32 2 1 8 2 2 mv As Final KE < Initial KE Collision is inelastic. Option (d) is correct. 2. v m m m m v 2 2 1 1 2 2 ¢ = - + = - + m m m m v 5 5 2 = - 2 3 2 v = - 2 3 2gl Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 197 48 m 48 m x' 8L M + S v/2 m 2m q 45° L Rest 2m m v v 2 v' 2 v' 1 v = 0 1 m = 5m 1 m = m 2 T mv l mg - ¢ = 2 2 or T mv l mg = ¢ + 2 = + m g mg 8 9 = 17 9 mg Option (a) is correct. Velocity of block v m m m v 1 2 1 2 2 2 ¢ = + = + × 2 5 2 m m m gl = 1 3 2gl Option (c) is correct. Maximum height attained by pendulum bob = ¢ = = v g gl g l g 2 2 2 8 9 2 4 / Option (d) is correct. 3. v u cos cos f = q Þ v u = f cos cos q and e v u = f sin sinq = f ´ f cos cos sin sin q q = tan tan f q Option (b) is correct. Change in momentum of particle = - f - + ( sin ) ( sin ) mv mu q \ Impulse delivered by floor to the particle = mv mu sin sin f + q = f + é ë ê ù û ú mv u v sin sin sin q q = + é ë ê ù û ú mv u v e u v sinq = + mu e sin ( ) q 1 Option (d) is correct. u e 1 1 2 2 - - ( ) sin q = - + u e 1 2 2 2 sin sin q q = + u e cos sin 2 2 2 q q = + f u v u cos sin 2 2 2 2 q = + f u v 2 2 2 2 cos sin q = f + f v v 2 2 2 2 cos sin =v Option (c) is correct. cos sin 2 2 2 q q + e = + f cos tan tan sin 2 2 2 2 q q q = + cos ( tan ) 2 2 1 q f = f cos 2 2 q sec = f cos cos 2 2 q = = v u 2 2 FinalKE InitialKE Option (d) is correct. 4. u i j ® - = + ( ) ^ ^ 3 2 1 ms Impulse received by particle of mass m = - + ® ® m m u v = - + + - + m m ( ) ( ) ^ ^ ^ ^ 3 2 2 i j i j = - + m( ) ^ ^ 5 i j unit Option (b) is correct. Impulse received by particle of mass M = - (impulse received by particle of mass m) = + m( ) ^ ^ 5i j Option (d) is correct. 198 | Mechanics-1 v v cos f v sin f u f u sin q q + u cos q m u M v = (â€“2 i + j) m/s ^ ^ ® m 5. T m a = 1 and m g T m a 2 2 - = Solving, a m m m g = + 2 1 2 ( ) a m a m m m x CM = + + 1 2 1 2 0 = + m m m a 1 1 2 = + m m m m g 1 2 1 2 2 ( ) Option (b) is correct. ( ) ( ) a m m a m m y CM = + + 1 2 1 2 0 = + m m m a 2 1 2 = + æ è ç ö ø ÷ m m m g 1 2 2 Option (c) is correct. 6. As the block comes down, the CM of the system will also come down i.e., it does not remain stationary. a mg m M g CM = + ¹ a CM is downwards and also a g CM < . Option (d) is correct. As no force acts along horizontal direction, the momentum of the system will remain conserved along horizontal direction. Option (c) is correct. 7. Velocity of B after collision : v e v 1 2 1 2 ¢ = + æ è ç ö ø ÷ = 3 4 v [as e = 1 2 and v v 2 = (given)] ¹ v 2 Impulse given by A to B = change in momentum of B = æ è ç ö ø ÷ - × m v m 3 4 0 = 3 4 mv Option (b) is correct. Velocity of A after collision v e v 2 2 1 2 ¢ = - æ è ç ö ø ÷ = v 4 Loss of KE during collision = - ¢ + ¢ 1 2 1 2 2 2 1 2 2 2 mv m v v ( ) = - æ è ç ö ø ÷ - æ è ç ö ø ÷ é ë ê ù û ú 1 2 4 3 4 2 2 2 m v v v = 3 16 2 mv Option (c) is correct. 8. As the mass of the system keeps on decreasing momentum of the system does not remain constant. Thrust force is developed on the rocket due to Newtonâ€™s 3rd law of motion. Option (b) is correct. As, a dv dt v m dm dt g i = = - æ è ç ö ø ÷ - The value of a will remain constant if v i and - dm dt are constant. Option (c) is correct. F F t net = (Thrust force due to gas ejection) - W (weight of rocket) a F m = net Thus, Newtonâ€™s 2nd law is applied. Option (d) is correct. Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 199 m 1 m 2 mg 2 a T T T T x y a v 2 Before collision v' 2 m 2 m 1 m 1 m 2 v = 0 1 A B After collision v' 1 B A Page 4 29. When the spider eats up the moth and travels towards A with velocity v 2 relative to rod. 24 2 48 0 m v v mv R R + æ è ç ö ø ÷ + = [v R = velocity (absolute) of rod] Þ v v R = - 6 Option (c) is correct. 30. Time taken by spider to reach point A starting from point B = + 4 8 2 L v L v/ = = = 20 20 20 L v L L T T / = ´ = 20 4 80 s 31. Form CM not to shift Þ ( ) ( ) x L m x L m ¢ + + ¢ + 8 24 6 48 = ´ 64 48 m i.e., x L ¢ = - 8 3 Option (a) is correct. More than One Cor rect Op tions 1. Along vertical : 2 2 45 mV m v sin sin q = × ° i.e., 2 2 2 V v sin q = â€¦(i) Along horizontal : mv m v mV = ° + 2 45 2 cos cos q i.e., 2 1 1 2 2 V v cos q = - æ è ç ö ø ÷ â€¦(ii) Squaring and adding Eq. (i)and (ii), 8 2 2 2 2 2 2 2 V v v v = æ è ç ö ø ÷ + - æ è ç ö ø ÷ = + + - × × v v v v v 2 2 2 8 7 2 2 2 = - 5 4 2 2 2 v v Dividing Eq. (ii) by Eq. (i) tan q = - 1 2 2 2 2 1 2 2 = - < 1 2 2 1 1 \ q < ° 45 Thus, the divergence angle between the particles will be less than p 2 . Option (b) is correct. Initial KE = 1 2 2 mv Final KE = æ è ç ö ø ÷ + 1 2 2 1 2 2 2 2 m v mV = + × - × é ë ê ù û ú 1 2 1 4 2 5 32 2 1 8 2 2 mv As Final KE < Initial KE Collision is inelastic. Option (d) is correct. 2. v m m m m v 2 2 1 1 2 2 ¢ = - + = - + m m m m v 5 5 2 = - 2 3 2 v = - 2 3 2gl Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 197 48 m 48 m x' 8L M + S v/2 m 2m q 45° L Rest 2m m v v 2 v' 2 v' 1 v = 0 1 m = 5m 1 m = m 2 T mv l mg - ¢ = 2 2 or T mv l mg = ¢ + 2 = + m g mg 8 9 = 17 9 mg Option (a) is correct. Velocity of block v m m m v 1 2 1 2 2 2 ¢ = + = + × 2 5 2 m m m gl = 1 3 2gl Option (c) is correct. Maximum height attained by pendulum bob = ¢ = = v g gl g l g 2 2 2 8 9 2 4 / Option (d) is correct. 3. v u cos cos f = q Þ v u = f cos cos q and e v u = f sin sinq = f ´ f cos cos sin sin q q = tan tan f q Option (b) is correct. Change in momentum of particle = - f - + ( sin ) ( sin ) mv mu q \ Impulse delivered by floor to the particle = mv mu sin sin f + q = f + é ë ê ù û ú mv u v sin sin sin q q = + é ë ê ù û ú mv u v e u v sinq = + mu e sin ( ) q 1 Option (d) is correct. u e 1 1 2 2 - - ( ) sin q = - + u e 1 2 2 2 sin sin q q = + u e cos sin 2 2 2 q q = + f u v u cos sin 2 2 2 2 q = + f u v 2 2 2 2 cos sin q = f + f v v 2 2 2 2 cos sin =v Option (c) is correct. cos sin 2 2 2 q q + e = + f cos tan tan sin 2 2 2 2 q q q = + cos ( tan ) 2 2 1 q f = f cos 2 2 q sec = f cos cos 2 2 q = = v u 2 2 FinalKE InitialKE Option (d) is correct. 4. u i j ® - = + ( ) ^ ^ 3 2 1 ms Impulse received by particle of mass m = - + ® ® m m u v = - + + - + m m ( ) ( ) ^ ^ ^ ^ 3 2 2 i j i j = - + m( ) ^ ^ 5 i j unit Option (b) is correct. Impulse received by particle of mass M = - (impulse received by particle of mass m) = + m( ) ^ ^ 5i j Option (d) is correct. 198 | Mechanics-1 v v cos f v sin f u f u sin q q + u cos q m u M v = (â€“2 i + j) m/s ^ ^ ® m 5. T m a = 1 and m g T m a 2 2 - = Solving, a m m m g = + 2 1 2 ( ) a m a m m m x CM = + + 1 2 1 2 0 = + m m m a 1 1 2 = + m m m m g 1 2 1 2 2 ( ) Option (b) is correct. ( ) ( ) a m m a m m y CM = + + 1 2 1 2 0 = + m m m a 2 1 2 = + æ è ç ö ø ÷ m m m g 1 2 2 Option (c) is correct. 6. As the block comes down, the CM of the system will also come down i.e., it does not remain stationary. a mg m M g CM = + ¹ a CM is downwards and also a g CM < . Option (d) is correct. As no force acts along horizontal direction, the momentum of the system will remain conserved along horizontal direction. Option (c) is correct. 7. Velocity of B after collision : v e v 1 2 1 2 ¢ = + æ è ç ö ø ÷ = 3 4 v [as e = 1 2 and v v 2 = (given)] ¹ v 2 Impulse given by A to B = change in momentum of B = æ è ç ö ø ÷ - × m v m 3 4 0 = 3 4 mv Option (b) is correct. Velocity of A after collision v e v 2 2 1 2 ¢ = - æ è ç ö ø ÷ = v 4 Loss of KE during collision = - ¢ + ¢ 1 2 1 2 2 2 1 2 2 2 mv m v v ( ) = - æ è ç ö ø ÷ - æ è ç ö ø ÷ é ë ê ù û ú 1 2 4 3 4 2 2 2 m v v v = 3 16 2 mv Option (c) is correct. 8. As the mass of the system keeps on decreasing momentum of the system does not remain constant. Thrust force is developed on the rocket due to Newtonâ€™s 3rd law of motion. Option (b) is correct. As, a dv dt v m dm dt g i = = - æ è ç ö ø ÷ - The value of a will remain constant if v i and - dm dt are constant. Option (c) is correct. F F t net = (Thrust force due to gas ejection) - W (weight of rocket) a F m = net Thus, Newtonâ€™s 2nd law is applied. Option (d) is correct. Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 199 m 1 m 2 mg 2 a T T T T x y a v 2 Before collision v' 2 m 2 m 1 m 1 m 2 v = 0 1 A B After collision v' 1 B A Match the Columns 1. If x 0 is the compression made in the spring, the restoring force on B will decrease from kx 0 to zero as the spring regains its original length. Thus, the acceleration of B will also decrease from kx m B 0 to zero. So, the a CM will also decrease from kx m m A B 0 + to zero. \ (a) ® (r) When spring is released after compressing it, the restoring on B will accelerate it towards right while the reaction force on A will apply a force on the wall which in turn will apply equal and opposite force on A and consequently A will travel towards right. As both travel towards right the velocity of CM will be maximum in the beginning. After this A will start compressing the spring and at a certain instant when the spring is compressed to maximum value both the blocks will travel towards right with a constant velocity and then the velocity of CM will become constant. \ (b) ® (q) As the blocks will never move along y-axis, the y-component of the CM of the two blocks will not change. \ (d) ® (p) As the two blocks will keep on moving towards right (surface below being smooth) the x-coordinate of the CM of the blocks will keep on increasing. \ (c) ® (s) 2. Initial a m g m g m m CM = + + + + ( ) ( ) = + g = + 10 SI unit \ (a) ® (q) Initial v m m m m CM = - + ´ + ( ) 20 0 = - 10 \ | | v CM = 10 SI unit \ (b) ® (q) For the time taken by the first particle to return to ground s ut at = + 1 2 2 0 20 5 2 = - + ( ) t t Þ t = 4 s Now, as the collision of the first particle with the ground is perfectly inelastic, the first particle will remain on ground at rest. Now, let us find the position of 2nd particle at t = 5 s s = + ( ) ( ) 0 5 1 2 10 5 2 = 125 m The particle (2nd) will still be in space moving downwards. a m m g m m CM = × + × + 0 = = g 2 5 (SI unit) \ (c) ® (p) Velocity of 2nd particle at t = 5 s v = + ´ 0 10 5 = 50 ms -1 200 | Mechanics-1 B A m B m A k B A x 0 kx 0 kx 0 â€“1 20 ms m 1st particle + 180 m 2nd particle u = 0 Page 5 29. When the spider eats up the moth and travels towards A with velocity v 2 relative to rod. 24 2 48 0 m v v mv R R + æ è ç ö ø ÷ + = [v R = velocity (absolute) of rod] Þ v v R = - 6 Option (c) is correct. 30. Time taken by spider to reach point A starting from point B = + 4 8 2 L v L v/ = = = 20 20 20 L v L L T T / = ´ = 20 4 80 s 31. Form CM not to shift Þ ( ) ( ) x L m x L m ¢ + + ¢ + 8 24 6 48 = ´ 64 48 m i.e., x L ¢ = - 8 3 Option (a) is correct. More than One Cor rect Op tions 1. Along vertical : 2 2 45 mV m v sin sin q = × ° i.e., 2 2 2 V v sin q = â€¦(i) Along horizontal : mv m v mV = ° + 2 45 2 cos cos q i.e., 2 1 1 2 2 V v cos q = - æ è ç ö ø ÷ â€¦(ii) Squaring and adding Eq. (i)and (ii), 8 2 2 2 2 2 2 2 V v v v = æ è ç ö ø ÷ + - æ è ç ö ø ÷ = + + - × × v v v v v 2 2 2 8 7 2 2 2 = - 5 4 2 2 2 v v Dividing Eq. (ii) by Eq. (i) tan q = - 1 2 2 2 2 1 2 2 = - < 1 2 2 1 1 \ q < ° 45 Thus, the divergence angle between the particles will be less than p 2 . Option (b) is correct. Initial KE = 1 2 2 mv Final KE = æ è ç ö ø ÷ + 1 2 2 1 2 2 2 2 m v mV = + × - × é ë ê ù û ú 1 2 1 4 2 5 32 2 1 8 2 2 mv As Final KE < Initial KE Collision is inelastic. Option (d) is correct. 2. v m m m m v 2 2 1 1 2 2 ¢ = - + = - + m m m m v 5 5 2 = - 2 3 2 v = - 2 3 2gl Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 197 48 m 48 m x' 8L M + S v/2 m 2m q 45° L Rest 2m m v v 2 v' 2 v' 1 v = 0 1 m = 5m 1 m = m 2 T mv l mg - ¢ = 2 2 or T mv l mg = ¢ + 2 = + m g mg 8 9 = 17 9 mg Option (a) is correct. Velocity of block v m m m v 1 2 1 2 2 2 ¢ = + = + × 2 5 2 m m m gl = 1 3 2gl Option (c) is correct. Maximum height attained by pendulum bob = ¢ = = v g gl g l g 2 2 2 8 9 2 4 / Option (d) is correct. 3. v u cos cos f = q Þ v u = f cos cos q and e v u = f sin sinq = f ´ f cos cos sin sin q q = tan tan f q Option (b) is correct. Change in momentum of particle = - f - + ( sin ) ( sin ) mv mu q \ Impulse delivered by floor to the particle = mv mu sin sin f + q = f + é ë ê ù û ú mv u v sin sin sin q q = + é ë ê ù û ú mv u v e u v sinq = + mu e sin ( ) q 1 Option (d) is correct. u e 1 1 2 2 - - ( ) sin q = - + u e 1 2 2 2 sin sin q q = + u e cos sin 2 2 2 q q = + f u v u cos sin 2 2 2 2 q = + f u v 2 2 2 2 cos sin q = f + f v v 2 2 2 2 cos sin =v Option (c) is correct. cos sin 2 2 2 q q + e = + f cos tan tan sin 2 2 2 2 q q q = + cos ( tan ) 2 2 1 q f = f cos 2 2 q sec = f cos cos 2 2 q = = v u 2 2 FinalKE InitialKE Option (d) is correct. 4. u i j ® - = + ( ) ^ ^ 3 2 1 ms Impulse received by particle of mass m = - + ® ® m m u v = - + + - + m m ( ) ( ) ^ ^ ^ ^ 3 2 2 i j i j = - + m( ) ^ ^ 5 i j unit Option (b) is correct. Impulse received by particle of mass M = - (impulse received by particle of mass m) = + m( ) ^ ^ 5i j Option (d) is correct. 198 | Mechanics-1 v v cos f v sin f u f u sin q q + u cos q m u M v = (â€“2 i + j) m/s ^ ^ ® m 5. T m a = 1 and m g T m a 2 2 - = Solving, a m m m g = + 2 1 2 ( ) a m a m m m x CM = + + 1 2 1 2 0 = + m m m a 1 1 2 = + m m m m g 1 2 1 2 2 ( ) Option (b) is correct. ( ) ( ) a m m a m m y CM = + + 1 2 1 2 0 = + m m m a 2 1 2 = + æ è ç ö ø ÷ m m m g 1 2 2 Option (c) is correct. 6. As the block comes down, the CM of the system will also come down i.e., it does not remain stationary. a mg m M g CM = + ¹ a CM is downwards and also a g CM < . Option (d) is correct. As no force acts along horizontal direction, the momentum of the system will remain conserved along horizontal direction. Option (c) is correct. 7. Velocity of B after collision : v e v 1 2 1 2 ¢ = + æ è ç ö ø ÷ = 3 4 v [as e = 1 2 and v v 2 = (given)] ¹ v 2 Impulse given by A to B = change in momentum of B = æ è ç ö ø ÷ - × m v m 3 4 0 = 3 4 mv Option (b) is correct. Velocity of A after collision v e v 2 2 1 2 ¢ = - æ è ç ö ø ÷ = v 4 Loss of KE during collision = - ¢ + ¢ 1 2 1 2 2 2 1 2 2 2 mv m v v ( ) = - æ è ç ö ø ÷ - æ è ç ö ø ÷ é ë ê ù û ú 1 2 4 3 4 2 2 2 m v v v = 3 16 2 mv Option (c) is correct. 8. As the mass of the system keeps on decreasing momentum of the system does not remain constant. Thrust force is developed on the rocket due to Newtonâ€™s 3rd law of motion. Option (b) is correct. As, a dv dt v m dm dt g i = = - æ è ç ö ø ÷ - The value of a will remain constant if v i and - dm dt are constant. Option (c) is correct. F F t net = (Thrust force due to gas ejection) - W (weight of rocket) a F m = net Thus, Newtonâ€™s 2nd law is applied. Option (d) is correct. Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 199 m 1 m 2 mg 2 a T T T T x y a v 2 Before collision v' 2 m 2 m 1 m 1 m 2 v = 0 1 A B After collision v' 1 B A Match the Columns 1. If x 0 is the compression made in the spring, the restoring force on B will decrease from kx 0 to zero as the spring regains its original length. Thus, the acceleration of B will also decrease from kx m B 0 to zero. So, the a CM will also decrease from kx m m A B 0 + to zero. \ (a) ® (r) When spring is released after compressing it, the restoring on B will accelerate it towards right while the reaction force on A will apply a force on the wall which in turn will apply equal and opposite force on A and consequently A will travel towards right. As both travel towards right the velocity of CM will be maximum in the beginning. After this A will start compressing the spring and at a certain instant when the spring is compressed to maximum value both the blocks will travel towards right with a constant velocity and then the velocity of CM will become constant. \ (b) ® (q) As the blocks will never move along y-axis, the y-component of the CM of the two blocks will not change. \ (d) ® (p) As the two blocks will keep on moving towards right (surface below being smooth) the x-coordinate of the CM of the blocks will keep on increasing. \ (c) ® (s) 2. Initial a m g m g m m CM = + + + + ( ) ( ) = + g = + 10 SI unit \ (a) ® (q) Initial v m m m m CM = - + ´ + ( ) 20 0 = - 10 \ | | v CM = 10 SI unit \ (b) ® (q) For the time taken by the first particle to return to ground s ut at = + 1 2 2 0 20 5 2 = - + ( ) t t Þ t = 4 s Now, as the collision of the first particle with the ground is perfectly inelastic, the first particle will remain on ground at rest. Now, let us find the position of 2nd particle at t = 5 s s = + ( ) ( ) 0 5 1 2 10 5 2 = 125 m The particle (2nd) will still be in space moving downwards. a m m g m m CM = × + × + 0 = = g 2 5 (SI unit) \ (c) ® (p) Velocity of 2nd particle at t = 5 s v = + ´ 0 10 5 = 50 ms -1 200 | Mechanics-1 B A m B m A k B A x 0 kx 0 kx 0 â€“1 20 ms m 1st particle + 180 m 2nd particle u = 0 \ At t = 5 s v m m m m CM = × + × + 0 50 =25 (SI unit) \ (d) ® (s) 3. Initial KE of block B = 4 J \ 1 2 4 2 ´ ´ = 0.5 u Þ u = 4 ms -1 \ Initial momentum of B = ´ 0.5 4 = - 2 1 kg ms \ (a) ® (r) Initial momentum p p p A B CM = + = + 0 2 = - 2 1 kgms \ (b) ® (r) Velocity given to block B will compress the spring and this will gradually increase the velocity of A. When the spring gets compressed to its maximum both the blocks will have the same velocities i.e., same momentum as both have same mass. p p A B = (at maximum compression of the spring) But, p p A B + = initial momentum of B. \ p p A A + = 2 i.e., p A = 1 kgms -1 \ (c) ® (q) After the maximum compression in the spring, the spring will gradually expand but now the velocity of block A will increase and that of B will decrease and when the spring attains maximum expansion the velocity of B will be zero and so will be its momentum. \ (d) ® (p) 4. If collision is elastic, the two blocks will interchange there velocities (mass of both balls being equal). Thus, velocity of A after collision = v \ (a) ® (r) If collision is perfectly inelastic, the two balls will move together (with velocities V). \ mv m m V = + ( ) Þ V v = 2 \ (b) ® (s) If collision is inelastic with e = 1 2 , v e v 1 2 1 2 ¢ = + × = + × 1 1 2 2 v [Q v v 2 = (given)] = 3 4 v \ (c) ® (p) If collision is inelastic with e = 1 4 , v v v 1 1 1 4 2 5 8 ¢ = + æ è ç ö ø ÷ = \ (d) ® (q). 5. If A moves x towards right Let plank (along with B) move by x¢ to the right. \ x x ´ + ¢ + + + = 30 60 30 30 60 30 0 ( ) ( ) i.e., x x ¢ = - 3 = x 3 , towards left. \ (a) ® (r) If B moves x towards left Let plank (along with A) move x¢ to the left \ x x × + ¢ + + + = 60 30 30 60 30 30 0 ( ) ( ) i.e., x x ¢ = - = x, towards right Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 201 B A m = 0.5 kg m C 30 kg B A 50 kg 60 kg SmoothRead More

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!

### Chapter 9 - Mechanics of Rotational Motion( Part - 1) - Physics, Solution by D C Pandey, NEET

- Doc | 7 pages
### Chapter 9 - Mechanics of Rotational Motion( Part - 2) - Physics, Solution by D C Pandey, NEET

- Doc | 6 pages
### Chapter 9 - Mechanics of Rotational Motion( Part - 3) - Physics, Solution by D C Pandey, NEET

- Doc | 5 pages