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To view an interactive solution to a problem that is similar to this one, select Interactive Solution 7.24. A 0.0118-kg bullet is fired straight up at a falling wooden block that has a mass of 1.05 kg. The bullet has a speed of 606 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

initially block having velocity v downward after collision it willhave speed v but upward.

Applying momentum conservation,

(0.0118 x 606) + ( - 1.05 v) = (1.05 + 0.0118) v

7.1508 = 1.05 v + 1.0618 v

v = 3.386 m/s

Now applying vf = vi + a t

3.386 = 0 + 9.8 t

t = 0.346 sec ....Ans

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