Explanation Given: The mass of the ball A is m A = 3 kg . The mass of the ball B is m B = 2 kg . The ball A traveling at the initial velocity of ( v A ) 1 = 8 m/s . The ball B traveling at the initial velocity of ( v B ) 1 = 3 m/s . The coefficient of the restitution is e = 0.7 . Write the principle of conservation of linear momentum. ∑ m v 1 = ∑ m v 2 m A ( v A ) 1 + m B ( v B ) 1 = m A ( v A ) 2 + m B ( v B ) 2 (I) Here, m is the mass, v is the velocity, the suffixes A , B indicates the ball A and ball B and the suffixes 1 , 2 indicates the initial and final stages. Write the formula for coefficient of restitution ( e ) . e = ( v B ) 2 − ( v A ) 2 ( v A ) 1 − ( v B ) 1 (II) Here, v is the velocity, the suffixes A , B indicates the ball A and ball B and the suffixes 1 , 2 indicates the initial and final stages. Conclusion: According law conservation of linear momentum: At Initial: Calculate the initial momentum. The initial momentum of the system is ∑ m v 1 = m A ( v A ) 1 + m B ( v B ) 1 The balls A and B moves in opposite direction facing each other and collide. Hence the velocity of ball A positive (moving right) and the velocity of ball B is negative (moving left). Substitute 3 kg for m A , 8 m/s for ( v A ) 1 , 2 kg for m B and − 4 m/s for ( v B ) 1 . ∑ m v 1 = 3 kg ( 8 m/s ) + ( 2 kg ) ( − 4 m/s ) = 24 − 8 = 16 N ⋅ s Calculate the final momentum. Write the final momentum of the system. ∑ m v 2 = m A ( v A ) 2 + m B ( v B ) 2 Substitute 3 kg for m A and 2 kg for m B . ∑ m v 2 = 3 kg ( v A ) 2 + 2 kg ( v B ) 2 = 3 ( v A ) 2 + 2 ( v B ) 2 Calculate the velocity of the balls after collision

15th Edition

ISBN: 9780137514724

Author: HIBBELER

Publisher: PEARSON

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Chapter 15, Problem 64P

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The velocity of each balls A and B just after collision

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Chapter 15, Problem 63P

Chapter 15, Problem 65P

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The velocity of each balls A and B just after collision

Solution Summary: The author explains the principle of conservation of linear momentum and the formula for coefficient of restitution.

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Chapter 15 Solutions