Page 1 Introductory Exercise 8.1 1. If a body is placed in a uniform gravitational field, the CG of the body coincides with the CM of the body. r r ® = ® = = CM S S i n i i i n i m m 1 1 while r r ® = ® = = CM S S i n i i i i n i i m g m g 1 1 If a body is placed in a uniformly increasing gravitational field ( ) g ® in the upward direction the CG of the body will be higher level than the CM. And, if the body is placed in a uniformly decreasing gravitational field in the upward direction the CG will be at a lower level the CM. CG shifts from CM according to the magnitude and direction of the gravitational field (by some other agency eg, earth) in which the body is placed. In zero gravitational field CG has no meaning while CM still exists, as usual. 2. The cen tre of mass of a rigid body may lie in side, on the sur face and even out side the body. The CM of a solid uni form sphere is at its cen tre. The CM of a solid ring is at the cen tre of the ring which lies out side the mass of the body thus, the state ment is false. (For fur ther de tails see an swer to 1 As ser tion and Rea son JEE corner). 3. Cen tre of mass al ways lies on the axis of sym me try of the body, if it ex ists. The state ment is thus true. 4. State ment is true. 5. As more mass is towards base. Distance < r 4 . 6. If two equal masses are kept at co-ordinates (R, 0) and ( , ) -R 0 , then their centre of mass will lie at origin. 7. X m X m X m m CM = + + 1 1 2 2 1 2 Y m y m y m m CM = + + 1 1 2 2 1 2 r X Y = + CM CM 2 2 = Distance of centre of mass from A g increasing CG CM CG CM g decreasing Centre of Mass, Conservation of Linear Momentum, Impulse and Collision 8 3 kg (1/2, 3/2) 2 kg (1, 0) 1 kg (0, 0) Page 2 Introductory Exercise 8.1 1. If a body is placed in a uniform gravitational field, the CG of the body coincides with the CM of the body. r r ® = ® = = CM S S i n i i i n i m m 1 1 while r r ® = ® = = CM S S i n i i i i n i i m g m g 1 1 If a body is placed in a uniformly increasing gravitational field ( ) g ® in the upward direction the CG of the body will be higher level than the CM. And, if the body is placed in a uniformly decreasing gravitational field in the upward direction the CG will be at a lower level the CM. CG shifts from CM according to the magnitude and direction of the gravitational field (by some other agency eg, earth) in which the body is placed. In zero gravitational field CG has no meaning while CM still exists, as usual. 2. The cen tre of mass of a rigid body may lie in side, on the sur face and even out side the body. The CM of a solid uni form sphere is at its cen tre. The CM of a solid ring is at the cen tre of the ring which lies out side the mass of the body thus, the state ment is false. (For fur ther de tails see an swer to 1 As ser tion and Rea son JEE corner). 3. Cen tre of mass al ways lies on the axis of sym me try of the body, if it ex ists. The state ment is thus true. 4. State ment is true. 5. As more mass is towards base. Distance < r 4 . 6. If two equal masses are kept at co-ordinates (R, 0) and ( , ) -R 0 , then their centre of mass will lie at origin. 7. X m X m X m m CM = + + 1 1 2 2 1 2 Y m y m y m m CM = + + 1 1 2 2 1 2 r X Y = + CM CM 2 2 = Distance of centre of mass from A g increasing CG CM CG CM g decreasing Centre of Mass, Conservation of Linear Momentum, Impulse and Collision 8 3 kg (1/2, 3/2) 2 kg (1, 0) 1 kg (0, 0) 8. Y A y A y A A CM = - - 1 1 2 2 1 2 9. A a x a y a 1 2 1 1 4 = = = , , A a x a y a 2 2 2 2 3 2 3 2 = = = , , In tro duc tory Ex er cise 8.2 1. Method 1 1 2 7 y y = - ( ) Þ y = 14 3 cm Displacement of CM = Position of CM ( ) f - Position of CM ( ) i = + - ( ) ( ) y x 2 = + æ è ç ö ø ÷ - æ è ç ö ø ÷ 14 3 2 20 3 =0 Method 2 Mv mv m v CM = + 1 1 2 2 = + - - - 1 2 2 1 1 1 ( ) ( ) ms ms =0 As velocity of CM is zero, there will not be any change in the position of the CM. 2. mg T ma sin 60° - = â€¦(i) T mg ma - ° = sin 30 â€¦(ii) Adding above equations, mg ma (sin sin ) 60 30 2 ° - ° = Þ a g = - æ è ç ö ø ÷ 3 1 4 a a 1 2 ® ® and are at right angle \ a a a CM ® ® ® = + = + m m m a a 1 2 1 2 2 1 2 ( ) or | | ( ) a CM ® = = - a g 2 3 1 4 2 162 | Mechanics-1 B 30° 60° a T mg sin 30° mg sin 60° A a + ve T (10 â€“ x) x 2 m A 1 kg At t = 0s t = 0 s A 1 kg 2 kg t = 0 s CM (i) 1x = 2 (10 â€“ x) Þ x = cm 20 3 y 7 â€“ y 1 m CM (f) 2 kg t = 1 s t = 1 s O y x Page 3 Introductory Exercise 8.1 1. If a body is placed in a uniform gravitational field, the CG of the body coincides with the CM of the body. r r ® = ® = = CM S S i n i i i n i m m 1 1 while r r ® = ® = = CM S S i n i i i i n i i m g m g 1 1 If a body is placed in a uniformly increasing gravitational field ( ) g ® in the upward direction the CG of the body will be higher level than the CM. And, if the body is placed in a uniformly decreasing gravitational field in the upward direction the CG will be at a lower level the CM. CG shifts from CM according to the magnitude and direction of the gravitational field (by some other agency eg, earth) in which the body is placed. In zero gravitational field CG has no meaning while CM still exists, as usual. 2. The cen tre of mass of a rigid body may lie in side, on the sur face and even out side the body. The CM of a solid uni form sphere is at its cen tre. The CM of a solid ring is at the cen tre of the ring which lies out side the mass of the body thus, the state ment is false. (For fur ther de tails see an swer to 1 As ser tion and Rea son JEE corner). 3. Cen tre of mass al ways lies on the axis of sym me try of the body, if it ex ists. The state ment is thus true. 4. State ment is true. 5. As more mass is towards base. Distance < r 4 . 6. If two equal masses are kept at co-ordinates (R, 0) and ( , ) -R 0 , then their centre of mass will lie at origin. 7. X m X m X m m CM = + + 1 1 2 2 1 2 Y m y m y m m CM = + + 1 1 2 2 1 2 r X Y = + CM CM 2 2 = Distance of centre of mass from A g increasing CG CM CG CM g decreasing Centre of Mass, Conservation of Linear Momentum, Impulse and Collision 8 3 kg (1/2, 3/2) 2 kg (1, 0) 1 kg (0, 0) 8. Y A y A y A A CM = - - 1 1 2 2 1 2 9. A a x a y a 1 2 1 1 4 = = = , , A a x a y a 2 2 2 2 3 2 3 2 = = = , , In tro duc tory Ex er cise 8.2 1. Method 1 1 2 7 y y = - ( ) Þ y = 14 3 cm Displacement of CM = Position of CM ( ) f - Position of CM ( ) i = + - ( ) ( ) y x 2 = + æ è ç ö ø ÷ - æ è ç ö ø ÷ 14 3 2 20 3 =0 Method 2 Mv mv m v CM = + 1 1 2 2 = + - - - 1 2 2 1 1 1 ( ) ( ) ms ms =0 As velocity of CM is zero, there will not be any change in the position of the CM. 2. mg T ma sin 60° - = â€¦(i) T mg ma - ° = sin 30 â€¦(ii) Adding above equations, mg ma (sin sin ) 60 30 2 ° - ° = Þ a g = - æ è ç ö ø ÷ 3 1 4 a a 1 2 ® ® and are at right angle \ a a a CM ® ® ® = + = + m m m a a 1 2 1 2 2 1 2 ( ) or | | ( ) a CM ® = = - a g 2 3 1 4 2 162 | Mechanics-1 B 30° 60° a T mg sin 30° mg sin 60° A a + ve T (10 â€“ x) x 2 m A 1 kg At t = 0s t = 0 s A 1 kg 2 kg t = 0 s CM (i) 1x = 2 (10 â€“ x) Þ x = cm 20 3 y 7 â€“ y 1 m CM (f) 2 kg t = 1 s t = 1 s O y x In tro duc tory Ex er cise 8.3 1. v v CM = + + + = 20 60 3 20 6 0 0 ( ) [As there is no force along horizontal direction]. Þ Velocity of trolley ( ) v = - 9 ms -1 Total energy produced by man = KE of man + KE of trolley = ´ ´ + ´ ´ - 1 2 60 3 1 2 20 9 2 2 ( ) = 1.08 kJ 2. On streching and then releasing the spring the restoring force on each block at instant will be same (according to Newtonâ€™s 3rd law of motion). Now, as force is same momentum p of each block will also be same (Dt being same) [As according to Newtonâ€™s second law of motion rate of change of momentum of a body is directly proportional to the net force applied on the body.] Now, as KE = p m 2 2 , KE of blocks at any instant will be inversely proportional to their respective masses. 3. As no external force acts on the system of particles, the velocity of CM shall not change. Thus, v CM = ´ + ´ + ´ + + 20 20 30 20 40 20 20 30 40 $ $ $ i i k v CM = ´ + + + + + ® 20 0 30 10 20 40 20 30 40 ( ) ^ ^ i k v 300 600 40 400 600 800 i j k i j k ^ ^ ^ ^ ^ ^ + + = + + Þ 40 100 600 200 v i j k ® = + + ^ ^ ^ i.e., v i j k ® = + + ( ) ^ ^ ^ 2.5 15 5 cms -1 4. Velocity of projectile at the highest point before explosion = 10 2 i ^ ms -1 As no extra external force would be acting during explosion, the velocity of CM will not change m m m 2 2 10 2 × + = ® ® 0 v i ^ Þ v i ® =20 2 ^ Range of rest half part = ´ R 2 2 (as the velocity of the projectile has doubled at the highest point) = R = u g 2 (as a = ° 45 ) = = ( ) 20 10 40 2 m Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 163 â€“1 (10i + 20k) cms 30 g 20 g z y x k j i Rest v ® R/2 20 cos 45° â€“1 = 10Ö2 ms 45° 20 x v = 10 Ö2 i ^ ® â€“1 20 cms â€“1 20 cms 30 g 20 g â€“1 20 cms z y x k j i 20 kg + ve v 20 kg 3 m/s 60 kg 60 kg Page 4 Introductory Exercise 8.1 1. If a body is placed in a uniform gravitational field, the CG of the body coincides with the CM of the body. r r ® = ® = = CM S S i n i i i n i m m 1 1 while r r ® = ® = = CM S S i n i i i i n i i m g m g 1 1 If a body is placed in a uniformly increasing gravitational field ( ) g ® in the upward direction the CG of the body will be higher level than the CM. And, if the body is placed in a uniformly decreasing gravitational field in the upward direction the CG will be at a lower level the CM. CG shifts from CM according to the magnitude and direction of the gravitational field (by some other agency eg, earth) in which the body is placed. In zero gravitational field CG has no meaning while CM still exists, as usual. 2. The cen tre of mass of a rigid body may lie in side, on the sur face and even out side the body. The CM of a solid uni form sphere is at its cen tre. The CM of a solid ring is at the cen tre of the ring which lies out side the mass of the body thus, the state ment is false. (For fur ther de tails see an swer to 1 As ser tion and Rea son JEE corner). 3. Cen tre of mass al ways lies on the axis of sym me try of the body, if it ex ists. The state ment is thus true. 4. State ment is true. 5. As more mass is towards base. Distance < r 4 . 6. If two equal masses are kept at co-ordinates (R, 0) and ( , ) -R 0 , then their centre of mass will lie at origin. 7. X m X m X m m CM = + + 1 1 2 2 1 2 Y m y m y m m CM = + + 1 1 2 2 1 2 r X Y = + CM CM 2 2 = Distance of centre of mass from A g increasing CG CM CG CM g decreasing Centre of Mass, Conservation of Linear Momentum, Impulse and Collision 8 3 kg (1/2, 3/2) 2 kg (1, 0) 1 kg (0, 0) 8. Y A y A y A A CM = - - 1 1 2 2 1 2 9. A a x a y a 1 2 1 1 4 = = = , , A a x a y a 2 2 2 2 3 2 3 2 = = = , , In tro duc tory Ex er cise 8.2 1. Method 1 1 2 7 y y = - ( ) Þ y = 14 3 cm Displacement of CM = Position of CM ( ) f - Position of CM ( ) i = + - ( ) ( ) y x 2 = + æ è ç ö ø ÷ - æ è ç ö ø ÷ 14 3 2 20 3 =0 Method 2 Mv mv m v CM = + 1 1 2 2 = + - - - 1 2 2 1 1 1 ( ) ( ) ms ms =0 As velocity of CM is zero, there will not be any change in the position of the CM. 2. mg T ma sin 60° - = â€¦(i) T mg ma - ° = sin 30 â€¦(ii) Adding above equations, mg ma (sin sin ) 60 30 2 ° - ° = Þ a g = - æ è ç ö ø ÷ 3 1 4 a a 1 2 ® ® and are at right angle \ a a a CM ® ® ® = + = + m m m a a 1 2 1 2 2 1 2 ( ) or | | ( ) a CM ® = = - a g 2 3 1 4 2 162 | Mechanics-1 B 30° 60° a T mg sin 30° mg sin 60° A a + ve T (10 â€“ x) x 2 m A 1 kg At t = 0s t = 0 s A 1 kg 2 kg t = 0 s CM (i) 1x = 2 (10 â€“ x) Þ x = cm 20 3 y 7 â€“ y 1 m CM (f) 2 kg t = 1 s t = 1 s O y x In tro duc tory Ex er cise 8.3 1. v v CM = + + + = 20 60 3 20 6 0 0 ( ) [As there is no force along horizontal direction]. Þ Velocity of trolley ( ) v = - 9 ms -1 Total energy produced by man = KE of man + KE of trolley = ´ ´ + ´ ´ - 1 2 60 3 1 2 20 9 2 2 ( ) = 1.08 kJ 2. On streching and then releasing the spring the restoring force on each block at instant will be same (according to Newtonâ€™s 3rd law of motion). Now, as force is same momentum p of each block will also be same (Dt being same) [As according to Newtonâ€™s second law of motion rate of change of momentum of a body is directly proportional to the net force applied on the body.] Now, as KE = p m 2 2 , KE of blocks at any instant will be inversely proportional to their respective masses. 3. As no external force acts on the system of particles, the velocity of CM shall not change. Thus, v CM = ´ + ´ + ´ + + 20 20 30 20 40 20 20 30 40 $ $ $ i i k v CM = ´ + + + + + ® 20 0 30 10 20 40 20 30 40 ( ) ^ ^ i k v 300 600 40 400 600 800 i j k i j k ^ ^ ^ ^ ^ ^ + + = + + Þ 40 100 600 200 v i j k ® = + + ^ ^ ^ i.e., v i j k ® = + + ( ) ^ ^ ^ 2.5 15 5 cms -1 4. Velocity of projectile at the highest point before explosion = 10 2 i ^ ms -1 As no extra external force would be acting during explosion, the velocity of CM will not change m m m 2 2 10 2 × + = ® ® 0 v i ^ Þ v i ® =20 2 ^ Range of rest half part = ´ R 2 2 (as the velocity of the projectile has doubled at the highest point) = R = u g 2 (as a = ° 45 ) = = ( ) 20 10 40 2 m Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 163 â€“1 (10i + 20k) cms 30 g 20 g z y x k j i Rest v ® R/2 20 cos 45° â€“1 = 10Ö2 ms 45° 20 x v = 10 Ö2 i ^ ® â€“1 20 cms â€“1 20 cms 30 g 20 g â€“1 20 cms z y x k j i 20 kg + ve v 20 kg 3 m/s 60 kg 60 kg Therefore, the rest half part will land at a distance of 40 40 2 + æ è ç ö ø ÷ m i.e., 60 m from the point of projection. 5. At point P Horizontal velocity = Initial horizontal velocity = ° - 20 2 45 1 cos ms =20 ms -1 Vertical velocity = ° - × 20 2 45 1 sin g (as t =1s) =70 ms -1 \ v ® (velocity of projectile at point P i.e., at t = 15 just before explosion) = + 20 10 i j ^ ^ Now, as the projectile breaks up into two equal parts and one part comes to rest, the velocity of other half part after explosion will be u v ® ® =2 = + 40 20 i j ^ ^ ms -1 Angle of projection ( ) a of 2nd half part after explosion at point P. a = é ë ê ù û ú - tan 1 20 40 = é ë ê ù û ú - tan 1 1 2 sin a = 1 5 Maximum height attained by second half part = Height of point P + Extra maximum height attained by second half part = ´ - ´ é ë ê ù û ú + 20 1 1 2 1 2000 2 2 2 2 g g ( ) sin a = + ´ ´ 15 2000 2 10 1 5 = + 15 20 =35 m 6. Momentum of platform + boy + stone along x-axis after throwing stone = before throwing stone \ ( ) 60 40 1 10 1 2 0 + + æ è ç ö ø ÷ = v Þ v = - 5 2 100 ms -1 Time of flight of stone 2 10 45 2 ´ ° = sin g s \ Horizontal displacement of platform (+ boy) =vT = - ´ 5 2 100 2 = - 1 10 m = - 10 cm 7. Thrust due to the upward component of the velocity of the bullet will rotate the movable end of the barrel and thus the bullet leaving the barrel will travelling at an angle greater than 45° when it comes out of the barrel. 164 | Mechanics-1 10 cm X T = Ö2 40 kg X 45° 1 kg â€“1 10 ms 60 kg s 1 kg â€“1 10 ms 45° 15 m â€“1 20Ö2 ms P â€“1 20 ms t = 15 a u 2 2 = Ö(40) + (20) â€“1 = Ö2000 ms t = 0 s a Ö5 1 2 R u Ö2 20 i q = 45° Bullet q' > q Bullet Page 5 Introductory Exercise 8.1 1. If a body is placed in a uniform gravitational field, the CG of the body coincides with the CM of the body. r r ® = ® = = CM S S i n i i i n i m m 1 1 while r r ® = ® = = CM S S i n i i i i n i i m g m g 1 1 If a body is placed in a uniformly increasing gravitational field ( ) g ® in the upward direction the CG of the body will be higher level than the CM. And, if the body is placed in a uniformly decreasing gravitational field in the upward direction the CG will be at a lower level the CM. CG shifts from CM according to the magnitude and direction of the gravitational field (by some other agency eg, earth) in which the body is placed. In zero gravitational field CG has no meaning while CM still exists, as usual. 2. The cen tre of mass of a rigid body may lie in side, on the sur face and even out side the body. The CM of a solid uni form sphere is at its cen tre. The CM of a solid ring is at the cen tre of the ring which lies out side the mass of the body thus, the state ment is false. (For fur ther de tails see an swer to 1 As ser tion and Rea son JEE corner). 3. Cen tre of mass al ways lies on the axis of sym me try of the body, if it ex ists. The state ment is thus true. 4. State ment is true. 5. As more mass is towards base. Distance < r 4 . 6. If two equal masses are kept at co-ordinates (R, 0) and ( , ) -R 0 , then their centre of mass will lie at origin. 7. X m X m X m m CM = + + 1 1 2 2 1 2 Y m y m y m m CM = + + 1 1 2 2 1 2 r X Y = + CM CM 2 2 = Distance of centre of mass from A g increasing CG CM CG CM g decreasing Centre of Mass, Conservation of Linear Momentum, Impulse and Collision 8 3 kg (1/2, 3/2) 2 kg (1, 0) 1 kg (0, 0) 8. Y A y A y A A CM = - - 1 1 2 2 1 2 9. A a x a y a 1 2 1 1 4 = = = , , A a x a y a 2 2 2 2 3 2 3 2 = = = , , In tro duc tory Ex er cise 8.2 1. Method 1 1 2 7 y y = - ( ) Þ y = 14 3 cm Displacement of CM = Position of CM ( ) f - Position of CM ( ) i = + - ( ) ( ) y x 2 = + æ è ç ö ø ÷ - æ è ç ö ø ÷ 14 3 2 20 3 =0 Method 2 Mv mv m v CM = + 1 1 2 2 = + - - - 1 2 2 1 1 1 ( ) ( ) ms ms =0 As velocity of CM is zero, there will not be any change in the position of the CM. 2. mg T ma sin 60° - = â€¦(i) T mg ma - ° = sin 30 â€¦(ii) Adding above equations, mg ma (sin sin ) 60 30 2 ° - ° = Þ a g = - æ è ç ö ø ÷ 3 1 4 a a 1 2 ® ® and are at right angle \ a a a CM ® ® ® = + = + m m m a a 1 2 1 2 2 1 2 ( ) or | | ( ) a CM ® = = - a g 2 3 1 4 2 162 | Mechanics-1 B 30° 60° a T mg sin 30° mg sin 60° A a + ve T (10 â€“ x) x 2 m A 1 kg At t = 0s t = 0 s A 1 kg 2 kg t = 0 s CM (i) 1x = 2 (10 â€“ x) Þ x = cm 20 3 y 7 â€“ y 1 m CM (f) 2 kg t = 1 s t = 1 s O y x In tro duc tory Ex er cise 8.3 1. v v CM = + + + = 20 60 3 20 6 0 0 ( ) [As there is no force along horizontal direction]. Þ Velocity of trolley ( ) v = - 9 ms -1 Total energy produced by man = KE of man + KE of trolley = ´ ´ + ´ ´ - 1 2 60 3 1 2 20 9 2 2 ( ) = 1.08 kJ 2. On streching and then releasing the spring the restoring force on each block at instant will be same (according to Newtonâ€™s 3rd law of motion). Now, as force is same momentum p of each block will also be same (Dt being same) [As according to Newtonâ€™s second law of motion rate of change of momentum of a body is directly proportional to the net force applied on the body.] Now, as KE = p m 2 2 , KE of blocks at any instant will be inversely proportional to their respective masses. 3. As no external force acts on the system of particles, the velocity of CM shall not change. Thus, v CM = ´ + ´ + ´ + + 20 20 30 20 40 20 20 30 40 $ $ $ i i k v CM = ´ + + + + + ® 20 0 30 10 20 40 20 30 40 ( ) ^ ^ i k v 300 600 40 400 600 800 i j k i j k ^ ^ ^ ^ ^ ^ + + = + + Þ 40 100 600 200 v i j k ® = + + ^ ^ ^ i.e., v i j k ® = + + ( ) ^ ^ ^ 2.5 15 5 cms -1 4. Velocity of projectile at the highest point before explosion = 10 2 i ^ ms -1 As no extra external force would be acting during explosion, the velocity of CM will not change m m m 2 2 10 2 × + = ® ® 0 v i ^ Þ v i ® =20 2 ^ Range of rest half part = ´ R 2 2 (as the velocity of the projectile has doubled at the highest point) = R = u g 2 (as a = ° 45 ) = = ( ) 20 10 40 2 m Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 163 â€“1 (10i + 20k) cms 30 g 20 g z y x k j i Rest v ® R/2 20 cos 45° â€“1 = 10Ö2 ms 45° 20 x v = 10 Ö2 i ^ ® â€“1 20 cms â€“1 20 cms 30 g 20 g â€“1 20 cms z y x k j i 20 kg + ve v 20 kg 3 m/s 60 kg 60 kg Therefore, the rest half part will land at a distance of 40 40 2 + æ è ç ö ø ÷ m i.e., 60 m from the point of projection. 5. At point P Horizontal velocity = Initial horizontal velocity = ° - 20 2 45 1 cos ms =20 ms -1 Vertical velocity = ° - × 20 2 45 1 sin g (as t =1s) =70 ms -1 \ v ® (velocity of projectile at point P i.e., at t = 15 just before explosion) = + 20 10 i j ^ ^ Now, as the projectile breaks up into two equal parts and one part comes to rest, the velocity of other half part after explosion will be u v ® ® =2 = + 40 20 i j ^ ^ ms -1 Angle of projection ( ) a of 2nd half part after explosion at point P. a = é ë ê ù û ú - tan 1 20 40 = é ë ê ù û ú - tan 1 1 2 sin a = 1 5 Maximum height attained by second half part = Height of point P + Extra maximum height attained by second half part = ´ - ´ é ë ê ù û ú + 20 1 1 2 1 2000 2 2 2 2 g g ( ) sin a = + ´ ´ 15 2000 2 10 1 5 = + 15 20 =35 m 6. Momentum of platform + boy + stone along x-axis after throwing stone = before throwing stone \ ( ) 60 40 1 10 1 2 0 + + æ è ç ö ø ÷ = v Þ v = - 5 2 100 ms -1 Time of flight of stone 2 10 45 2 ´ ° = sin g s \ Horizontal displacement of platform (+ boy) =vT = - ´ 5 2 100 2 = - 1 10 m = - 10 cm 7. Thrust due to the upward component of the velocity of the bullet will rotate the movable end of the barrel and thus the bullet leaving the barrel will travelling at an angle greater than 45° when it comes out of the barrel. 164 | Mechanics-1 10 cm X T = Ö2 40 kg X 45° 1 kg â€“1 10 ms 60 kg s 1 kg â€“1 10 ms 45° 15 m â€“1 20Ö2 ms P â€“1 20 ms t = 15 a u 2 2 = Ö(40) + (20) â€“1 = Ö2000 ms t = 0 s a Ö5 1 2 R u Ö2 20 i q = 45° Bullet q' > q Bullet In tro duc tory Ex er cise 8.4 1. To just lift rocket off the launching pad Thrust force = Weight or v dm dt mg - æ è ç ö ø ÷ = or - æ è ç ö ø ÷ = dm dt mg v = + ´ ´ ( ) 20 180 10 3 9.8 1.6 =1.225 kgs -1 (i) Rate of consumption of fuel = - 2 1 kgs \ Time required for the consumption of fuel t = = - 180 2 90 1 kg kgs s Ultimate speed gained by rocket v u gt v m m e = - + æ è ç ö ø ÷ log 0 â€¦(i) Substituting u = 0, v = 1.6 kms -1 , m 0 20 180 = + ( ) kg and m = 20 kg in Eq. (i). v = - ´ + ´ æ è ç ö ø ÷ 9.8 1.6 90 10 200 20 3 ln =2.8 kms -1 (iii) Rate of consumption of fuel = - 20 1 kgs \ t = = - 180 20 9 1 kg kgs s i.e., v = - ´ + ´ 9.8 1.6 9 10 200 20 3 ln = - 3.6kms 1 2. Mass at time t, m m t = - 0 m \ dm dt = -m ma = thrust force - mg or ma v dm dt mg = - æ è ç ö ø ÷ - or ma v mg = - m or ( ) ( ) m t d x dt u m t g 0 2 2 0 - = - - m m m (Q v u = ) 3. v u gt v m m = - + æ è ç ö ø ÷ ln 0 = - + - æ è ç ö ø ÷ 0 1 3 0 0 gt u m m t ln = - + g u 1 3 2 ln (at t = 1s) = - u g ln 3 2 Introductory Exercise 8.5 1. u ® (at t = 0 s) = + - ( ) ^ ^ 10 3 10 1 i j ms At t = 1 s Horizontal velocity = - 10 3 1 i ^ ms Vertical velocity = - × - ( ) ^ 10 1 1 g j ms = - 0 1 j ^ ms \ v i ® - = 10 3 1 ^ ms Change in velocity, D v v u ® ® ® = - = - + ( ) ( ) ^ ^ ^ 10 3 10 3 10 i i j = - - 10 1 j ^ ms i.e., Dv ® - =10 1 ms , downwards. 2. Impulse ( ) J ® imparted = Change in momentum in the time interval t=0 s to t = 2 s = m [(Velocity at t = 2 s) - (Velocity at t = 0 s)] = + - - - 2 4 4 4 1 1 kg ms ms [( ) ( ) ] ^ ^ ^ i j j = 8 i ^ Ns ( Q 1 1 1 kg ms Ns - = ) 3. Spring will become taut when the ball would go down by 2 m. v u g 2 2 2 2 = + ´ ´ or v =2 10 ms -1 D D p m v = = - 12 10 0 ( ) kgms -1 = 2 10 kg ms -1 \ Impulse imparted = 2 10 Ns. Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 165 60° 20 m/s 2 mRead More

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