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Your answer is partially correct. Try again. To view an interactive solution to a problem that is similar to this one, select Interactive Solution 7.24. A 0.00746-kg bullet is fired straight up at a falling wooden block that has a mass of 3.76 kg. The bullet has a speed of 799 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. Unit the tolerance is +/-296

Explanation / Answer

|v| = g t = 9.8 t

Applying momentum conservation,

0.00746 x 799 - 3.76 x 9.8 x t = (0.00746 + 3.76)(9.8 t)

5.96 - 36.85t = 36.93t

t = 0.081 sec

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