Explanation The fraction of more massive block’s kinetic energy transferred to lighter block is, f = 1 2 M V f 2 1 2 m v i 2 (I) Here, M is the mass of the heavier block, V f is the final velocity of the heavier block, m is the mass of the lighter block, and v i is the initial velocity of the lighter block. Write the expression for conservation of momentum on the head on collision. m v i = m v f + M V f (II) Here, v f is the final velocity of the lighter block. Rewrite the above relation. m ( v i − v f ) = M V f v i − v f = M m V f (III) Write the expression for conservation of kinetic energy. 1 2 m v i 2 = 1 2 m v f 2 + 1 2 M V f 2 Rewrite the above relation. m v i 2 = m v f 2 + M V f 2 m ( v i 2 − v f 2 ) = M V f 2 v i 2 − v f 2 = M m V f 2 (IV) Conclusion: Substitute 685 g for M and 232 g for m in the equation (III). v i − v f = 685 g 232 g V f = 2.95 V f Substitute 685 g for M and 232 g for m in the equation (IV)

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Chapter 9, Problem 39E

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The percentage of the more massive block’s kinetic energy transferred to lighter block.

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Chapter 9, Problem 38E

Chapter 9, Problem 40E

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