Explanation Given: The mass of the each balls A , B , and C is m = 0.5 lb . The coefficient of the restitution is e = 0.85 . Initially the ball A is released from rest at the elevation of h = 3 ft . Write the principle of conservation of energy. T 1 + V 1 = T 2 + V 2 (I) Here, T 1 , T 2 is the initial and final kinetic energy and V 1 , V 2 is the initial and final potential energy. Write the formula for kinetic energy. T = 1 2 ( W 32.2 ft/s 2 ) v 2 (II) Write the formula for potential energy. V = W h (III) Here, W is the weight and h is the elevation of the ball from the datum. Write the principle of conservation of linear momentum. ∑ m v 1 = ∑ m v 2 (IV) Here, m is the mass, v is the velocity 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 (V) Here, v is the velocity, the suffixes x , y indicates the balls which are involved in collision and the suffixes 1 , 2 indicates the initial and final stages. Conclusion: According to the principle of conservation energy: Consider the ball A only. At Initial: When the ball A is released from rest. The initial kinetic energy of the ball A is T 1 = 0 . Calculate the potential energy of the ball A . When the ball is at the elevation of h = 3 m . Substitute 0.5 lb for m and 3 ft for h in Equation (III). V 1 = ( 0.5 lb ) ( 3 ft ) = 1.5 lb ⋅ ft At final: (just before the collision with the ball B ) Calculate the final kinetic energy of the ball. Substitute 0.5 lb for W and ( v A ) 2 for v in Equation (I). T 2 = 1 2 ( 0.5 lb 32.2 ft/s 2 ) ( v A ) 2 2 = 0.0077639 ( v A ) 2 2 When the ball A reaches the datum. The final potential energy of the ball A is V 2 = 0 . Calculate the final velocity of the ball A . Substitute 0 for T 1 , V 2 , 1.5 lb ⋅ ft for V 1 and 0.0077639 ( v A ) 2 2 for T 2 in Equation (I). 0 + 1.5 lb ⋅ ft = 0.0077639 ( v A ) 2 2 + 0 ( v A ) 2 2 = 1.5 0.0077639 ( v A ) 2 = 193.2 ( v A ) 2 = 13.8996 ft/s According law conservation of linear momentum: Consider When the ball A collides with ball B . At Initial: Calculate the initial momentum. The initial momentum is ∑ m v 1 = m A ( v A ) 1 + m B ( v B ) 1 When the ball B is at rest. The velocity of the ball is zero. Here m = W 32.2 ft/s 2 and the initial velocity of the ball A is ( v A ) 2 = ( v A ) 1 = 13.8996 ft/s . Substitute 0.5 lb 32.2 ft/s 2 for m A , m B , 13.8996 ft/s for ( v A ) 1 and 0 for ( v B ) 1 . ∑ m v 1 = ( 0.5 lb 32.2 ft/s 2 ) ( 13.8996 ft/s ) + ( 0.5 lb 32.2 ft/s 2 ) ( 0 ) = 0.215832 + 0 = 0.215832 Calculate the final momentum. The final momentum is ∑ m v 2 = m A ( v A ) 2 + m B ( v B ) 2 Substitute 0.5 lb 32.2 ft/s for m A , m B . ∑ m v 2 = ( 0.5 lb 32.2 ft/s 2 ) ( v A ) 2 + ( 0.5 lb 32.2 ft/s 2 ) ( v B ) 2 = 0.015528 ( v A ) 2 + 0.015528 ( v B ) 2 Calculate the velocity of the balls after collision. Substitute 0.215832 for ∑ m v 1 and 0.015528 ( v A ) 2 + 0.015528 ( v B ) 2 for ∑ m v 2 in Equation (IV). 0.215832 = 0.015528 ( v A ) 2 + 0.015528 ( v B ) 2 (VI) Calculate the coefficient of restitution ( e ) . e = ( v B ) 2 − ( v A ) 2 ( v A ) 1 − ( v B ) 1 Here, e = 0.85 . Substitute 0.85 for e , 13.8996 ft/s for ( v A ) 1 and 0 for ( v B ) 1 in Equation (V). 0.85 = ( v B ) 2 − ( v A ) 2 13.8996 ft/s − 0 0.85 = ( v B ) 2 − ( v A ) 2 13

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 15.4, Problem 67P

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The velocity of each balls after the second collision has occurred.

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Chapter 15.4, Problem 66P

Chapter 15.4, Problem 68P

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Chapter 15 Solutions

Engineering Mechanics: Statics & Dynamics (14th Edition)

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The velocity of each balls after the second collision has occurred.

Solution Summary: The author explains the principle of conservation of energy and the formula for potential energy.

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The velocity of each balls after the second collision has occurred.

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Chapter 15 Solutions