Explanation Given: The weight of the each stones A and B is m = 47 lb . The initial speed of the stone A is ( v A ) 1 = 8 ft/s . Initially the stone B is at rest. The coefficient of restitution between the stones is e = 0.8 . Draw the free body diagram of the stones 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 speed and the suffixes 1 , 2 indicates the initial and final stages. Refer Figure (1). Write the formula for coefficient of restitution ( e ) . e = ( v B ) x 2 − ( v A ) x 2 ( v A ) x 1 − ( v B ) x 1 (II) Here, v is the speed, the suffixes A , B indicates the stones A and B , 1 and 2 indicates the initial and final stages 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 stone A and stone B collides. At Initial: From Figure (1). ( v A ) x 1 = 8 ft/s ( cos 30 ° ) = 6.9282 ft/s When the stone B is at rest the speed of the stone is ( v B ) 1 = ( v B ) x 1 = 0 Calculate the initial momentum. ∑ m v 1 = m B ( v B ) x 1 + m A ( v A ) x 1 Substitute 47 lb for m A , m B , 6.9282 ft/s for ( v A ) x 1 and 0 for ( v B ) x 1 . ∑ m v 1 = ( 47 lb × 1 32.2 ft/s 2 ) × ( 6.9282 ft/s ) + ( 47 lb × 1 32.2 ft/s 2 ) ( 0 ) = 10.1126 + 0 = 10.1126 lb ⋅ s At Final: Calculate the final momentum. ∑ m v 2 = m B ( v B ) x 2 + m A ( v A ) x 2 Substitute 47 lb for m A , m B . ∑ m v 2 = ( 47 lb × 1 32.2 ft/s 2 ) ( v B ) x 2 + ( 47 lb × 1 32.2 ft/s 2 ) ( v A ) x 2 = 1.4596 ( v B ) x 2 + 1.4596 ( v A ) x 2 Calculate the speed of the stones after collision. Substitute 10.1126 lb ⋅ s for ∑ m v 1 and 1.4596 ( v B ) x 2 + 1.4596 ( v A ) x 2 for ∑ m v 2 in Equation (I). 10.1126 lb ⋅ s = 1.4596 ( v B ) x 2 + 1.4596 ( v A ) x 2 1.4596 ( v B ) x 2 + 1.4596 ( v A ) x 2 = 10.1126 (III) According to equation of the coefficient of restitution ( e ) between the stones. Here, e = 0.8 . Substitute 0.85 for e , − 1.1491 m/s for ( v A ) x 1 and 0 for ( v B ) x 1 in Equation (II). 0.8 = ( v B ) x 2 − ( v A ) x 2 6.9282 ft/s − 0 0.85 = ( v B ) x 2 − ( v A ) x 2 6.9282 ( v B ) x 2 − ( v A ) x 2 = 0

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 15.4, Problem 87P

To determine

The speeds of the stones A and B

just after collision:

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Chapter 15.4, Problem 86P

Chapter 15.4, Problem 88P

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The speeds of the stones 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 stones as shown in Figure (1).