Page 1 Objective Questions (Level 2) Single Correct Option 1. m v m m g l m v 1 1 1 1 0 5 3 + ´ = + m v m g l 1 1 2 3 5 × = \ v m m g l 1 1 3 2 5 = Option (b) is correct. 2. mu mv = 2 Þ v u = 2 0 2 2 2 = + - v g H ( ) Þ H v g = 2 2 = ( / ) u g 2 2 2 = u g 2 8 or H gl g = 2 8 or l l ( cos ) 1 4 - = a or 1 1 4 - = cos a or a = - cos 1 3 4 Option (c) is correct. 3. Here | | | | v v 2 1 ® ® = = v say Net momentum of the two elements as shown in figure = = ® dp | | dp =2vdmsinq = æ è ç ö ø ÷ 2v M R Rd p q q ( )sin = 2Mv d p q q sin p Mv d = ò 2 0 2 p q q p / sin = - 2 0 2 Mv p q p [cos ] / = - - 2 0 1 Mv p [ ] = 2Mv p Option (b) is correct. 4. 1 2 2 1 2 2 0 2 ( ) m v kx = or v k m x = 2 0 or 2 2 0 mv mk x = or F t mkx av ´ = D 2 0 or F mk t x av = 2 0 D Option (b) is correct. 190 | Mechanics-1 l v 1 m 1 m Rest u = Ö5gl v /3 1 m H l m a 2m m v m m Rest u = Ö2gl R q q d q d q 2q dm v 2 v 1 dp dm v 1 ® ® ® ® dl kx 0 kx 0 Page 2 Objective Questions (Level 2) Single Correct Option 1. m v m m g l m v 1 1 1 1 0 5 3 + ´ = + m v m g l 1 1 2 3 5 × = \ v m m g l 1 1 3 2 5 = Option (b) is correct. 2. mu mv = 2 Þ v u = 2 0 2 2 2 = + - v g H ( ) Þ H v g = 2 2 = ( / ) u g 2 2 2 = u g 2 8 or H gl g = 2 8 or l l ( cos ) 1 4 - = a or 1 1 4 - = cos a or a = - cos 1 3 4 Option (c) is correct. 3. Here | | | | v v 2 1 ® ® = = v say Net momentum of the two elements as shown in figure = = ® dp | | dp =2vdmsinq = æ è ç ö ø ÷ 2v M R Rd p q q ( )sin = 2Mv d p q q sin p Mv d = ò 2 0 2 p q q p / sin = - 2 0 2 Mv p q p [cos ] / = - - 2 0 1 Mv p [ ] = 2Mv p Option (b) is correct. 4. 1 2 2 1 2 2 0 2 ( ) m v kx = or v k m x = 2 0 or 2 2 0 mv mk x = or F t mkx av ´ = D 2 0 or F mk t x av = 2 0 D Option (b) is correct. 190 | Mechanics-1 l v 1 m 1 m Rest u = Ö5gl v /3 1 m H l m a 2m m v m m Rest u = Ö2gl R q q d q d q 2q dm v 2 v 1 dp dm v 1 ® ® ® ® dl kx 0 kx 0 5. As the collisions of the striker and the walls of the carrom are perfectly elastic the striker will follow the path OPQROP ... Change in KE = work done against friction 1 2 0 2 mv mgs - = m (m = mass of striker, s = displacement of the striker) Þ s v g = 2 2m = ´ ´ ( ) 2 2 10 2 0.2 =1 m PQ OP OA = = 2 = æ è ç ö ø ÷ = 1 2 2 2 1 2 m OP PQ + = + = 1 2 1 2 1 m \ Striker will stop at point Q where co-ordinates are 1 2 2 1 2 , æ è ç ö ø ÷ . Option (a) is correct. 6. As no force is acting on the system along horizontal, the CM of the system will not shift horizontally. 4 1 1 p p = - ( ) Þ p = 1 5 m Displacement ( ) x of bar when pendulum becomes vertical x p =sinq x p = sinq = ° 1 5 30 sin = 1 10 m = 0.1 m When the ball reaches the other extreme end the bar will further shift to the left by distance x and as such the net displacement of the bar will be 2x i.e., 0.2 m. Option (b) is correct. 7. Momentum imparted to the floor in 1st collision = - - = + p ep p e ( ) ( ) 1 2nd collision = - - = + ep e p ep e ( ) ( ) 2 1 3rd collision = - - = + e p e p e p e 2 3 2 1 ( ) ( ) As theoretically there will be infinite collision, total momentum imparted to floor = + + + + + + ¥ p e ep e e p e ( ) ( ) ( ) .... 1 1 1 2 = + + + + ¥ p e e e ( )[ ] 1 1 2 K = + - p e e ( ) 1 1 1 = + - æ è ç ö ø ÷ p e e 1 1 Option (d) is correct. 8. Let F be the frictional force applied by plate when bullet enters into it \ 1 2 2 mu Fh = â€¦(i) If plate was free to move mu M m u + = + ¢ 0 ( ) \ u m M m u ¢ = + New KE of bullet = - + ¢ 1 2 1 2 2 2 mu M m u ( ) (entering plate) Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 191 x q p 4 kg 4 kg x 1 kg CM 1 â€“ p x CM ep p During first collision Floor 2 ep ep 3 ep 2 ep Floor During Second collision During third collision 45° P R Q O A 45° Page 3 Objective Questions (Level 2) Single Correct Option 1. m v m m g l m v 1 1 1 1 0 5 3 + ´ = + m v m g l 1 1 2 3 5 × = \ v m m g l 1 1 3 2 5 = Option (b) is correct. 2. mu mv = 2 Þ v u = 2 0 2 2 2 = + - v g H ( ) Þ H v g = 2 2 = ( / ) u g 2 2 2 = u g 2 8 or H gl g = 2 8 or l l ( cos ) 1 4 - = a or 1 1 4 - = cos a or a = - cos 1 3 4 Option (c) is correct. 3. Here | | | | v v 2 1 ® ® = = v say Net momentum of the two elements as shown in figure = = ® dp | | dp =2vdmsinq = æ è ç ö ø ÷ 2v M R Rd p q q ( )sin = 2Mv d p q q sin p Mv d = ò 2 0 2 p q q p / sin = - 2 0 2 Mv p q p [cos ] / = - - 2 0 1 Mv p [ ] = 2Mv p Option (b) is correct. 4. 1 2 2 1 2 2 0 2 ( ) m v kx = or v k m x = 2 0 or 2 2 0 mv mk x = or F t mkx av ´ = D 2 0 or F mk t x av = 2 0 D Option (b) is correct. 190 | Mechanics-1 l v 1 m 1 m Rest u = Ö5gl v /3 1 m H l m a 2m m v m m Rest u = Ö2gl R q q d q d q 2q dm v 2 v 1 dp dm v 1 ® ® ® ® dl kx 0 kx 0 5. As the collisions of the striker and the walls of the carrom are perfectly elastic the striker will follow the path OPQROP ... Change in KE = work done against friction 1 2 0 2 mv mgs - = m (m = mass of striker, s = displacement of the striker) Þ s v g = 2 2m = ´ ´ ( ) 2 2 10 2 0.2 =1 m PQ OP OA = = 2 = æ è ç ö ø ÷ = 1 2 2 2 1 2 m OP PQ + = + = 1 2 1 2 1 m \ Striker will stop at point Q where co-ordinates are 1 2 2 1 2 , æ è ç ö ø ÷ . Option (a) is correct. 6. As no force is acting on the system along horizontal, the CM of the system will not shift horizontally. 4 1 1 p p = - ( ) Þ p = 1 5 m Displacement ( ) x of bar when pendulum becomes vertical x p =sinq x p = sinq = ° 1 5 30 sin = 1 10 m = 0.1 m When the ball reaches the other extreme end the bar will further shift to the left by distance x and as such the net displacement of the bar will be 2x i.e., 0.2 m. Option (b) is correct. 7. Momentum imparted to the floor in 1st collision = - - = + p ep p e ( ) ( ) 1 2nd collision = - - = + ep e p ep e ( ) ( ) 2 1 3rd collision = - - = + e p e p e p e 2 3 2 1 ( ) ( ) As theoretically there will be infinite collision, total momentum imparted to floor = + + + + + + ¥ p e ep e e p e ( ) ( ) ( ) .... 1 1 1 2 = + + + + ¥ p e e e ( )[ ] 1 1 2 K = + - p e e ( ) 1 1 1 = + - æ è ç ö ø ÷ p e e 1 1 Option (d) is correct. 8. Let F be the frictional force applied by plate when bullet enters into it \ 1 2 2 mu Fh = â€¦(i) If plate was free to move mu M m u + = + ¢ 0 ( ) \ u m M m u ¢ = + New KE of bullet = - + ¢ 1 2 1 2 2 2 mu M m u ( ) (entering plate) Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 191 x q p 4 kg 4 kg x 1 kg CM 1 â€“ p x CM ep p During first collision Floor 2 ep ep 3 ep 2 ep Floor During Second collision During third collision 45° P R Q O A 45° = - + + æ è ç ö ø ÷ 1 2 1 2 2 2 mu M m m M m u ( ) = - + é ë ê ù û ú 1 2 1 2 mu m M m = + æ è ç ö ø ÷ 1 2 2 mu M M m = ¢ Fh â€¦(ii) Dividing Eq. (ii) by Eq. (i), h h M M m ¢ = + i.e., h M M m h ¢ = + æ è ç ö ø ÷ Option (a) is correct. 9. Now, v e v 1 2 1 2 ¢ = + (entering plase) \ 10 1 2 16 = + × e i.e., e= 1 4 Option (b) is correct. 10. y l b l L b l L l b L b CM = × + × æ è ç ö ø ÷ + æ è ç ö ø ÷ × + × æ è ç ö ø ÷ ( ) ( ) 2 1 2 3 1 2 l bl bL l L b bl bL = + + + 2 2 2 2 6 2 y l CM according toquestion = é ë ê ê ê ù û ú ú ú or l l L l L l L + é ë ê ù û ú = + + 2 2 2 6 2 2 or l L 2 2 2 6 = or l L = 3 11. mv m v mv 1 1 2 2 1 2 = - â€¦(i) e =1 \ 2 2 1 v v = \ m v m v m v 1 2 2 2 1 2 2 ( ) = - \ 3 1 2 m m = or m m 1 2 1 3 = 12. As the resultant of the velocities of 1st and 2nd are just opposite to that of 3rd, the 4th particle will travel in the line in which 3rd is travelling. Let the velocity of 4th particle is u as shown in figure. \ u v v cos cos 45 45 ° + ° = i.e., u v = - ( ) 2 1 Total energy released = + + + - 1 2 1 2 1 2 1 2 2 1 2 2 2 2 mv mv mv m v [ ( )] = + - 1 2 3 2 1 2 2 mv [ ( ) ] = - mv 2 3 2 [ ] Option (a) is correct. 13. 192 | Mechanics-1 1st 2nd 45° 4th u 3rd v v v m m A B A B Rest â€“1 v = 16 ms 2 2 â€“1 v = 2gs = 10 ms 1 ' v 2 mu mv = 2 1 i.e., v u 1 2 = C will increase the tension in the string. mu mv = 2 2 i.e., v u 2 2 = C will also increase the weight of B when collision takes place. Thus, mu m mv = + ( ) 2 3 i.e., v u 3 3 = A m 1 v 1 B m 1 Þ A v 2 B v 2 A B C u 2 mg mg A B C u 2 mg mg A B C u mg mg mg mg mg mg 2mg 2mg 2mg 2mg Page 4 Objective Questions (Level 2) Single Correct Option 1. m v m m g l m v 1 1 1 1 0 5 3 + ´ = + m v m g l 1 1 2 3 5 × = \ v m m g l 1 1 3 2 5 = Option (b) is correct. 2. mu mv = 2 Þ v u = 2 0 2 2 2 = + - v g H ( ) Þ H v g = 2 2 = ( / ) u g 2 2 2 = u g 2 8 or H gl g = 2 8 or l l ( cos ) 1 4 - = a or 1 1 4 - = cos a or a = - cos 1 3 4 Option (c) is correct. 3. Here | | | | v v 2 1 ® ® = = v say Net momentum of the two elements as shown in figure = = ® dp | | dp =2vdmsinq = æ è ç ö ø ÷ 2v M R Rd p q q ( )sin = 2Mv d p q q sin p Mv d = ò 2 0 2 p q q p / sin = - 2 0 2 Mv p q p [cos ] / = - - 2 0 1 Mv p [ ] = 2Mv p Option (b) is correct. 4. 1 2 2 1 2 2 0 2 ( ) m v kx = or v k m x = 2 0 or 2 2 0 mv mk x = or F t mkx av ´ = D 2 0 or F mk t x av = 2 0 D Option (b) is correct. 190 | Mechanics-1 l v 1 m 1 m Rest u = Ö5gl v /3 1 m H l m a 2m m v m m Rest u = Ö2gl R q q d q d q 2q dm v 2 v 1 dp dm v 1 ® ® ® ® dl kx 0 kx 0 5. As the collisions of the striker and the walls of the carrom are perfectly elastic the striker will follow the path OPQROP ... Change in KE = work done against friction 1 2 0 2 mv mgs - = m (m = mass of striker, s = displacement of the striker) Þ s v g = 2 2m = ´ ´ ( ) 2 2 10 2 0.2 =1 m PQ OP OA = = 2 = æ è ç ö ø ÷ = 1 2 2 2 1 2 m OP PQ + = + = 1 2 1 2 1 m \ Striker will stop at point Q where co-ordinates are 1 2 2 1 2 , æ è ç ö ø ÷ . Option (a) is correct. 6. As no force is acting on the system along horizontal, the CM of the system will not shift horizontally. 4 1 1 p p = - ( ) Þ p = 1 5 m Displacement ( ) x of bar when pendulum becomes vertical x p =sinq x p = sinq = ° 1 5 30 sin = 1 10 m = 0.1 m When the ball reaches the other extreme end the bar will further shift to the left by distance x and as such the net displacement of the bar will be 2x i.e., 0.2 m. Option (b) is correct. 7. Momentum imparted to the floor in 1st collision = - - = + p ep p e ( ) ( ) 1 2nd collision = - - = + ep e p ep e ( ) ( ) 2 1 3rd collision = - - = + e p e p e p e 2 3 2 1 ( ) ( ) As theoretically there will be infinite collision, total momentum imparted to floor = + + + + + + ¥ p e ep e e p e ( ) ( ) ( ) .... 1 1 1 2 = + + + + ¥ p e e e ( )[ ] 1 1 2 K = + - p e e ( ) 1 1 1 = + - æ è ç ö ø ÷ p e e 1 1 Option (d) is correct. 8. Let F be the frictional force applied by plate when bullet enters into it \ 1 2 2 mu Fh = â€¦(i) If plate was free to move mu M m u + = + ¢ 0 ( ) \ u m M m u ¢ = + New KE of bullet = - + ¢ 1 2 1 2 2 2 mu M m u ( ) (entering plate) Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 191 x q p 4 kg 4 kg x 1 kg CM 1 â€“ p x CM ep p During first collision Floor 2 ep ep 3 ep 2 ep Floor During Second collision During third collision 45° P R Q O A 45° = - + + æ è ç ö ø ÷ 1 2 1 2 2 2 mu M m m M m u ( ) = - + é ë ê ù û ú 1 2 1 2 mu m M m = + æ è ç ö ø ÷ 1 2 2 mu M M m = ¢ Fh â€¦(ii) Dividing Eq. (ii) by Eq. (i), h h M M m ¢ = + i.e., h M M m h ¢ = + æ è ç ö ø ÷ Option (a) is correct. 9. Now, v e v 1 2 1 2 ¢ = + (entering plase) \ 10 1 2 16 = + × e i.e., e= 1 4 Option (b) is correct. 10. y l b l L b l L l b L b CM = × + × æ è ç ö ø ÷ + æ è ç ö ø ÷ × + × æ è ç ö ø ÷ ( ) ( ) 2 1 2 3 1 2 l bl bL l L b bl bL = + + + 2 2 2 2 6 2 y l CM according toquestion = é ë ê ê ê ù û ú ú ú or l l L l L l L + é ë ê ù û ú = + + 2 2 2 6 2 2 or l L 2 2 2 6 = or l L = 3 11. mv m v mv 1 1 2 2 1 2 = - â€¦(i) e =1 \ 2 2 1 v v = \ m v m v m v 1 2 2 2 1 2 2 ( ) = - \ 3 1 2 m m = or m m 1 2 1 3 = 12. As the resultant of the velocities of 1st and 2nd are just opposite to that of 3rd, the 4th particle will travel in the line in which 3rd is travelling. Let the velocity of 4th particle is u as shown in figure. \ u v v cos cos 45 45 ° + ° = i.e., u v = - ( ) 2 1 Total energy released = + + + - 1 2 1 2 1 2 1 2 2 1 2 2 2 2 mv mv mv m v [ ( )] = + - 1 2 3 2 1 2 2 mv [ ( ) ] = - mv 2 3 2 [ ] Option (a) is correct. 13. 192 | Mechanics-1 1st 2nd 45° 4th u 3rd v v v m m A B A B Rest â€“1 v = 16 ms 2 2 â€“1 v = 2gs = 10 ms 1 ' v 2 mu mv = 2 1 i.e., v u 1 2 = C will increase the tension in the string. mu mv = 2 2 i.e., v u 2 2 = C will also increase the weight of B when collision takes place. Thus, mu m mv = + ( ) 2 3 i.e., v u 3 3 = A m 1 v 1 B m 1 Þ A v 2 B v 2 A B C u 2 mg mg A B C u 2 mg mg A B C u mg mg mg mg mg mg 2mg 2mg 2mg 2mg \ v v v 1 2 3 : : = u u u 2 2 3 : : = 3 3 2 : : Option (b) is correct. 14. v v CM = ° cos 30 = v 3 2 Option (a) is correct. 15. r i ® = CM ^ r 1 ® (position vector of lighter piece) = + - 3 2 4 i j k ^ ^ ^ r r r ® ® ® = + + CM m m m m 1 1 2 2 1 2 r r r 2 1 2 1 1 2 ® ® ® = + - ( ) m m m m CM = - + - 2 2 3 3 2 4 4 3 i i j k ^ ^ ^ ^ ( ) = - + - 1 4 6 2 3 2 4 [ ( )] ^ ^ ^ ^ i i j k = - + 1 4 4 8 [ ] ^ ^ j k = - + j k ^ ^ 2 \ The heavier part will be at ( , , ) 0 1 2 - . Option (d) is correct. 16. Motion of A : h gt 2 1 2 2 = , v A (at time t) = = g t gh i.e., t h g = Motion of B : h v h g g h g 2 1 2 = - i.e., h v h g = \ v gh = v gh g h g B = - × =0 Collision of A and B at time t : m gh m = 3 w \ w = gh 3 Velocity of the combined mass when it reach ground v gh ¢ = + 2 2 2 w = + gh g gh 2 i.e., v gh ¢ = 19 3 Option (d) is correct. 17. u = velocity of man w.r.t. cart Let v = velocity of cart w.r.t. ground \ Velocity of man w.r.t. ground = + u v m u v mv m ( ) + + = × 2 3 0 \ v u = - 3 Work done = KE gained by man and cart = + + 1 2 1 2 2 2 2 m u v mv ( ) Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 193 30° v v cos 30° CM h/2 h/2 h Place of collision w 3 m m A time = t (say) v 2m B g A B A B m + 2m 2m m gh Rest w Page 5 Objective Questions (Level 2) Single Correct Option 1. m v m m g l m v 1 1 1 1 0 5 3 + ´ = + m v m g l 1 1 2 3 5 × = \ v m m g l 1 1 3 2 5 = Option (b) is correct. 2. mu mv = 2 Þ v u = 2 0 2 2 2 = + - v g H ( ) Þ H v g = 2 2 = ( / ) u g 2 2 2 = u g 2 8 or H gl g = 2 8 or l l ( cos ) 1 4 - = a or 1 1 4 - = cos a or a = - cos 1 3 4 Option (c) is correct. 3. Here | | | | v v 2 1 ® ® = = v say Net momentum of the two elements as shown in figure = = ® dp | | dp =2vdmsinq = æ è ç ö ø ÷ 2v M R Rd p q q ( )sin = 2Mv d p q q sin p Mv d = ò 2 0 2 p q q p / sin = - 2 0 2 Mv p q p [cos ] / = - - 2 0 1 Mv p [ ] = 2Mv p Option (b) is correct. 4. 1 2 2 1 2 2 0 2 ( ) m v kx = or v k m x = 2 0 or 2 2 0 mv mk x = or F t mkx av ´ = D 2 0 or F mk t x av = 2 0 D Option (b) is correct. 190 | Mechanics-1 l v 1 m 1 m Rest u = Ö5gl v /3 1 m H l m a 2m m v m m Rest u = Ö2gl R q q d q d q 2q dm v 2 v 1 dp dm v 1 ® ® ® ® dl kx 0 kx 0 5. As the collisions of the striker and the walls of the carrom are perfectly elastic the striker will follow the path OPQROP ... Change in KE = work done against friction 1 2 0 2 mv mgs - = m (m = mass of striker, s = displacement of the striker) Þ s v g = 2 2m = ´ ´ ( ) 2 2 10 2 0.2 =1 m PQ OP OA = = 2 = æ è ç ö ø ÷ = 1 2 2 2 1 2 m OP PQ + = + = 1 2 1 2 1 m \ Striker will stop at point Q where co-ordinates are 1 2 2 1 2 , æ è ç ö ø ÷ . Option (a) is correct. 6. As no force is acting on the system along horizontal, the CM of the system will not shift horizontally. 4 1 1 p p = - ( ) Þ p = 1 5 m Displacement ( ) x of bar when pendulum becomes vertical x p =sinq x p = sinq = ° 1 5 30 sin = 1 10 m = 0.1 m When the ball reaches the other extreme end the bar will further shift to the left by distance x and as such the net displacement of the bar will be 2x i.e., 0.2 m. Option (b) is correct. 7. Momentum imparted to the floor in 1st collision = - - = + p ep p e ( ) ( ) 1 2nd collision = - - = + ep e p ep e ( ) ( ) 2 1 3rd collision = - - = + e p e p e p e 2 3 2 1 ( ) ( ) As theoretically there will be infinite collision, total momentum imparted to floor = + + + + + + ¥ p e ep e e p e ( ) ( ) ( ) .... 1 1 1 2 = + + + + ¥ p e e e ( )[ ] 1 1 2 K = + - p e e ( ) 1 1 1 = + - æ è ç ö ø ÷ p e e 1 1 Option (d) is correct. 8. Let F be the frictional force applied by plate when bullet enters into it \ 1 2 2 mu Fh = â€¦(i) If plate was free to move mu M m u + = + ¢ 0 ( ) \ u m M m u ¢ = + New KE of bullet = - + ¢ 1 2 1 2 2 2 mu M m u ( ) (entering plate) Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 191 x q p 4 kg 4 kg x 1 kg CM 1 â€“ p x CM ep p During first collision Floor 2 ep ep 3 ep 2 ep Floor During Second collision During third collision 45° P R Q O A 45° = - + + æ è ç ö ø ÷ 1 2 1 2 2 2 mu M m m M m u ( ) = - + é ë ê ù û ú 1 2 1 2 mu m M m = + æ è ç ö ø ÷ 1 2 2 mu M M m = ¢ Fh â€¦(ii) Dividing Eq. (ii) by Eq. (i), h h M M m ¢ = + i.e., h M M m h ¢ = + æ è ç ö ø ÷ Option (a) is correct. 9. Now, v e v 1 2 1 2 ¢ = + (entering plase) \ 10 1 2 16 = + × e i.e., e= 1 4 Option (b) is correct. 10. y l b l L b l L l b L b CM = × + × æ è ç ö ø ÷ + æ è ç ö ø ÷ × + × æ è ç ö ø ÷ ( ) ( ) 2 1 2 3 1 2 l bl bL l L b bl bL = + + + 2 2 2 2 6 2 y l CM according toquestion = é ë ê ê ê ù û ú ú ú or l l L l L l L + é ë ê ù û ú = + + 2 2 2 6 2 2 or l L 2 2 2 6 = or l L = 3 11. mv m v mv 1 1 2 2 1 2 = - â€¦(i) e =1 \ 2 2 1 v v = \ m v m v m v 1 2 2 2 1 2 2 ( ) = - \ 3 1 2 m m = or m m 1 2 1 3 = 12. As the resultant of the velocities of 1st and 2nd are just opposite to that of 3rd, the 4th particle will travel in the line in which 3rd is travelling. Let the velocity of 4th particle is u as shown in figure. \ u v v cos cos 45 45 ° + ° = i.e., u v = - ( ) 2 1 Total energy released = + + + - 1 2 1 2 1 2 1 2 2 1 2 2 2 2 mv mv mv m v [ ( )] = + - 1 2 3 2 1 2 2 mv [ ( ) ] = - mv 2 3 2 [ ] Option (a) is correct. 13. 192 | Mechanics-1 1st 2nd 45° 4th u 3rd v v v m m A B A B Rest â€“1 v = 16 ms 2 2 â€“1 v = 2gs = 10 ms 1 ' v 2 mu mv = 2 1 i.e., v u 1 2 = C will increase the tension in the string. mu mv = 2 2 i.e., v u 2 2 = C will also increase the weight of B when collision takes place. Thus, mu m mv = + ( ) 2 3 i.e., v u 3 3 = A m 1 v 1 B m 1 Þ A v 2 B v 2 A B C u 2 mg mg A B C u 2 mg mg A B C u mg mg mg mg mg mg 2mg 2mg 2mg 2mg \ v v v 1 2 3 : : = u u u 2 2 3 : : = 3 3 2 : : Option (b) is correct. 14. v v CM = ° cos 30 = v 3 2 Option (a) is correct. 15. r i ® = CM ^ r 1 ® (position vector of lighter piece) = + - 3 2 4 i j k ^ ^ ^ r r r ® ® ® = + + CM m m m m 1 1 2 2 1 2 r r r 2 1 2 1 1 2 ® ® ® = + - ( ) m m m m CM = - + - 2 2 3 3 2 4 4 3 i i j k ^ ^ ^ ^ ( ) = - + - 1 4 6 2 3 2 4 [ ( )] ^ ^ ^ ^ i i j k = - + 1 4 4 8 [ ] ^ ^ j k = - + j k ^ ^ 2 \ The heavier part will be at ( , , ) 0 1 2 - . Option (d) is correct. 16. Motion of A : h gt 2 1 2 2 = , v A (at time t) = = g t gh i.e., t h g = Motion of B : h v h g g h g 2 1 2 = - i.e., h v h g = \ v gh = v gh g h g B = - × =0 Collision of A and B at time t : m gh m = 3 w \ w = gh 3 Velocity of the combined mass when it reach ground v gh ¢ = + 2 2 2 w = + gh g gh 2 i.e., v gh ¢ = 19 3 Option (d) is correct. 17. u = velocity of man w.r.t. cart Let v = velocity of cart w.r.t. ground \ Velocity of man w.r.t. ground = + u v m u v mv m ( ) + + = × 2 3 0 \ v u = - 3 Work done = KE gained by man and cart = + + 1 2 1 2 2 2 2 m u v mv ( ) Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 193 30° v v cos 30° CM h/2 h/2 h Place of collision w 3 m m A time = t (say) v 2m B g A B A B m + 2m 2m m gh Rest w = - æ è ç ö ø ÷ + - æ è ç ö ø ÷ 1 2 3 1 2 2 3 2 2 m u u m u = × + 1 2 4 3 1 2 2 9 2 2 m u m u = + 2 3 9 2 2 mu mu = 7 9 2 mu Option (d) is correct. 18. v m m m m CM = + + 30 50 = 40 m/s upwards. If the velocity of CM becomes zero at displacements 0 40 2 10 2 2 = + - ( )s Þ s =80 m \ Maximum height attained by CM = + 20 80 m m =100 m Option (c) is correct. 19. As the masses are equal and the collision is elastic, the particles will exchange their velocities as shown in figure. Gain in KE of 1st particle = - - 1 2 2 1 2 2 2 m v mu ( ) = - 2 1 2 2 2 mv mv = 3 2 2 mv = 3 2 2 p m Option (c) is correct. 20. According to question ( ) ( ) ( ) ( ) 4 4 2 4 4 4 4 m x m a m m m x m a m m + + = + + or x a x a 2 4 4 5 + = + i.e., x a = 6 Option (b) is correct. 21. x CM of Fig. 1 will as that of Fig. 2. x a a CM = × + × + 40 2 50 3 2 40 50 = + 20 75 90 a a = = 95 90 19 18 a a 194 | Mechanics-1 m 40 m 20 m m 50 m/s Initial position of CM 30 m/s v 2v B A In terms of velocity v 2v B A After collision Before collision 4m P CM 4m x a CM 4m x a R m O 20 10 40 20 a a a a a x y a O 50 40 a a x y a Fig. 1 Fig. 2 p â€“ 2 p B A In terms of momentum Before collisionRead More

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