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Question

To view an interactive solution to a problem that is similar to this one, select Interactive Solution 7.24. A 0.0120-kg bullet is fired straight up at a falling wooden block that has a mass of 2.66 kg. The bullet has a speed of 674 m/s when it strikes the block. The block originally was dropped from rest from the top of a building and had been falling for a time t when the collision with the bullet occurs. As a result of the collision, the block (with the bullet in it) reverses direction, rises, and comes to a momentary halt at the top of the building. Find the time t.

Explanation / Answer

kinetic energy of the bullet = 0.5 * 0.00120* 674^2 N

potential energy of the block lost = m g h = 2 g (0.5 g t2) =g2 t2

0.5 * 0.00120 * 674^2  = g2 t2 + (0.5 g t2) (0.00120) (g)

272.56= g2 t2 + 0.0006 g2 t2

272.56 = 1.0006 g2 t2

t2 = 272.56 / 1.0006 g2

t = 2.84 s

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