Explanation Given: The mass of the balls A , B , C , D is m . Initially the ball A and B each moves with the velocity of v . The collision elastic and the coefficient of the restitution is e = 1 . Here the collision takes place in following order. Ball B collides with ball C 1) . Ball C collides with ball D 2) . Ball A collides with ball B 3) . Again Ball B collides with ball C 4) . Here, mass of the all balls are equal. Write the principle of conservation of linear momentum. ∑ m v 1 = ∑ m v 2 m [ ( v x ) 1 + ( v y ) 1 ] = m [ ( v x ) 2 + ( v y ) 2 ] (I) Here, m is the mass, v is the velocity, the suffixes x , y indicates the balls which involves in collision and the suffixes 1 , 2 indicates the initial and final stages. Write the formula for coefficient of restitution ( e ) . e = ( v y ) 2 − ( v x ) 2 ( v x ) 1 − ( v y ) 1 (II) Here, v is the velocity, the suffixes x , y indicates the balls which involves in collision and the suffixes 1 , 2 indicates the initial and final stages. Conclusion: According law conservation of linear momentum: 1) Ball B collides with ball C : At Initial: (Before collision) Calculate the initial momentum. The initial momentum of the system is ∑ m v 1 = m [ ( v B ) 1 + ( v C ) 1 ] The ball C is at rest until the ball B collides with it. Hence the initial velocity of the ball C is zero. ( v C ) 1 = 0 Substitute v for ( v B ) 1 and 0 for ( v C ) 1 . ∑ m v 1 = m ( v + 0 ) = m v At Final: (After collision) Calculate the final momentum. After collision the balls moves in opposite direction to each other. The ball B moves to left (negative) and ball C moves to right (positive). The final momentum of the system is ∑ m v 2 = m [ − ( v B ) 2 + ( v C ) 2 ] Calculate the final velocity of the balls after collision. Substitute m v for ∑ m v 1 and m [ − ( v B ) 2 + ( v C ) 2 ] for ∑ m v 2 in Equation (I). m v = m [ − ( v B ) 2 + ( v C ) 2 ] v = ( v c ) 2 − ( v B ) 2 (III) Calculate the coefficient of restitution ( e ) . Refer Equation (II). The coefficient of restitution is e = ( v C ) 2 − ( v B ) 2 ( v B ) 1 − ( v C ) 1 Here, e = 1 . Substitute 1 for e , v for ( v B ) 1 and 0 for ( v C ) 1 . 1 = ( v C ) 2 − [ − ( v B ) 2 ] v − 0 1 = ( v C ) 2 + ( v B ) 2 v v = ( v C ) 2 + ( v B ) 2 (IV) Calculate the final velocity of the balls B and C . Substitute ( v C ) 2 − ( v B ) 2 for v in Equation (IV). ( v C ) 2 − ( v B ) 2 = ( v C ) 2 + ( v B ) 2 ( v C ) 2 − ( v C ) 2 = ( v B ) 2 + ( v B ) 2 2 ( v B ) 2 = 0 ( v B ) 2 = 0 Substitute 0 for ( v B ) 2 in Equation (IV). v = ( v C ) 2 + 0 ( v C ) 2 = v The final velocity of the balls B and C is ( v B ) 2 = 0 ( v C ) 2 = v 2) Ball C collides with ball D : At Initial: (Before collision) Here, the initial velocity of the ball C is ( v C ) 1 = v . Calculate the initial momentum. The initial momentum of the system is ∑ m v 1 = m [ ( v C ) 1 + ( v D ) 1 ] The ball D is at rest until the ball C collides with it. Hence the initial velocity of the ball D is zero. ( v D ) 1 = 0 Substitute v for ( v C ) 1 and 0 for ( v D ) 1 . ∑ m v 1 = m ( v + 0 ) = m v At Final: (After collision) Calculate the final momentum. After collision the balls moves in opposite direction to each other. The ball C moves to left (negative) and ball D moves to right (positive). The final momentum of the system is ∑ m v 2 = m [ − ( v C ) 2 + ( v D ) 2 ] Calculate the final velocity of the balls after collision. Substitute m v for ∑ m v 1 and m [ − ( v C ) 2 + ( v D ) 2 ] for ∑ m v 2 in Equation (I). m v = m [ − ( v C ) 2 + ( v D ) 2 ] v = ( v D ) 2 − ( v C ) 2 (V) Calculate the coefficient of restitution ( e ) . Refer Equation (II). The coefficient of restitution is e = ( v D ) 2 − ( v C ) 2 ( v C ) 1 − ( v D ) 1 Here, e = 1 . Substitute 1 for e , v for ( v C ) 1 and 0 for ( v D ) 1 . 1 = ( v D ) 2 − [ − ( v C ) 2 ] v − 0 1 = ( v D ) 2 + ( v C ) 2 v v = ( v D ) 2 + ( v C ) 2 (VI) Calculate the final velocity of the balls C and D . Substitute ( v D ) 2 − ( v C ) 2 for v in Equation (VI). ( v D ) 2 − ( v C ) 2 = ( v D ) 2 + ( v C ) 2 ( v D ) 2 − ( v D ) 2 = ( v C ) 2 + ( v C ) 2 2 ( v C ) 2 = 0 ( v C ) 2 = 0 Substitute 0 for ( v C ) 2 in Equation (VI). v = ( v D ) 2 + 0 ( v D ) 2 = v The final velocity of the balls C and D is ( v C ) 2 = 0 ( v D ) 2 = v 3) Ball A collides with ball B : At Initial: (Before collision) Here, the initial velocity of the ball A is ( v A ) 1 = v

15th Edition

ISBN: 9780137514724

Author: HIBBELER

Publisher: PEARSON

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Chapter 15, Problem 62P

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Why after collision the balls C and D each move off with velocity of v and why the ball D doesn’t move off with the velocity of 2 v .

Solution Summary: The author explains that the balls Candd move off with the same velocity of v and the ball D doesn't.

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To explain:

Why after collision the balls C and D

each move off with velocity of v

and why the ball D

doesn’t move off with the velocity of 2v .

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Chapter 15 Solutions

Why after collision the balls C and D each move off with velocity of v and why the ball D doesn’t move off with the velocity of 2 v .

Solution Summary: The author explains that the balls Candd move off with the same velocity of v and the ball D doesn't.

Concept explainers

Videos

To explain: Why after collision the balls C and D each move off with velocity of v and why the ball D doesn’t move off with the velocity of 2v .

Want to see the full answer?

Chapter 15 Solutions

To explain:

Why after collision the balls C and D

each move off with velocity of v

and why the ball D

doesn’t move off with the velocity of 2v .