Explanation Given: The mass of the each balls A and B is m = 200 g . The initial velocity of the ball A is ( v A ) 1 = 1.5 m/s . Initially the ball B is at rest. The coefficient of restitution between the balls is e = 0.85 . Draw the free body diagram of the balls as shown in Figure (1). Write the principle of conservation of linear momentum. ∑ m v 1 = ∑ m v 2 (I) Here, m is the mass, v is the velocity and the suffixes 1 , 2 indicates the initial and final stages. Refer Figure (1). Write the formula for coefficient of restitution ( e ) . e = ( v A ) x 2 − ( v B ) x 2 ( v B ) x 1 − ( v A ) x 1 (II) Here, v is the velocity, the suffixes A , B indicates the balls A and B and the suffixes x indicates along x − a x i s . Conclusion: According law conservation of linear momentum: (Along x − a x i s ) Consider when the ball A and ball B collides. At Initial: From Figure (1). ( v A ) x 1 = − 1.5 m/s ( cos 40 ° ) = − 1.1491 m/s Here, the negative sign indicates that the velocity acting in opposite direction (leftward). When the ball B is at rest the velocity of the ball is ( v B ) 1 = ( v B ) x 1 = 0 Calculate the initial momentum. ∑ m v 1 = m A ( v A ) x 1 + m B ( v B ) x 1 Substitute 200 g for m A , m B , − 1.1491 m/s for ( v A ) x 1 and 0 for ( v B ) x 1 . ∑ m v 1 = ( 200 g × 1 kg 1000 g ) × ( − 1.1491 m/s ) + ( 200 g × 1 kg 1000 g ) ( 0 ) = − 0.2298 + 0 = − 0.2298 N ⋅ s At Final: Calculate the final momentum. ∑ m v 2 = m A ( v A ) x 2 + m B ( v B ) x 2 Substitute 200 g for m A , m B . ∑ m v 2 = ( 200 g × 1 kg 1000 g ) ( v A ) x 2 + ( 200 g × 1 kg 1000 g ) ( v B ) x 2 = 0.2 ( v A ) x 2 + 0.2 ( v B ) x 2 Calculate the velocity of the balls after collision. Substitute − 0.2298 N ⋅ s for ∑ m v 1 and 0.2 ( v A ) x 2 + 0.2 ( v B ) x 2 for ∑ m v 2 in Equation (I). − 0.2298 N ⋅ s = 0.2 ( v A ) x 2 + 0.2 ( v B ) x 2 0.2 ( v A ) x 2 + 0.2 ( v B ) x 2 = − 0.2298 (III) According to equation of the coefficient of restitution ( e ) between the balls. Here, e = 0.85 . Substitute 0.85 for e , − 1.1491 m/s for ( v A ) x 1 and 0 for ( v B ) x 1 in Equation (II). 0.85 = ( v A ) x 2 − ( v B ) x 2 0 − ( − 1.1491 m/s ) 0.85 = ( v A ) x 2 − ( v B ) x 2 1.1491 ( v A ) x 2 − ( v B ) x 2 = 0.85 ( 1.1491 ) ( v A ) x 2 = 0.9767 + ( v B ) x 2 (IV) Calculate the final velocity of the balls A and B . Substitute 0.9767 + ( v B ) x 2 for ( v A ) x 2 in Equation (III)

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 15.4, Problem 86P

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The velocities of the balls A and B just after collision:

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Chapter 15.4, Problem 85P

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Engineering Mechanics: Statics & Dynamics (14th Edition)

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The velocities of the balls A and B just after collision:

Solution Summary: The author illustrates the principle of conservation of linear momentum by drawing the free body diagram of the balls.

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