ch09-p109 - d 1 we obtain d 1 = 0.2999 m 30 cm. (b) A very...

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109. Using Eq. 9-67 and Eq. 9-68, we have after the collision v 1 = m 1 m 2 m 1 + m 2 v 1 i = 0.6 m 1 1.4 m 1 v 1 i = 3 7 (4 m/s) v 2 = 2 m 1 m 1 + m 2 v 1 i = 2 m 1 1.4 m 1 v 1 i = 1 7 (4 m/s) . (a) During the (subsequent) sliding, the kinetic energy of block 1 K 1 f = 1 2 m 1 v 1 2 is converted into thermal form ( Δ E th = μ k m 1 gd 1 ). Solving for the sliding distance
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Unformatted text preview: d 1 we obtain d 1 = 0.2999 m 30 cm. (b) A very similar computation (but with subscript 2 replacing subscript 1) leads to block 2s sliding distance d 2 = 3.332 m 3.3 m....
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