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To view an interactive solution to a problem that is similar to this one, select Interactive Solution 7.24. A 0.0132-kg bullet is fired straight up at a falling wooden block that has a mass of 2.53 kg. The bullet has a speed of 520 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. Number Units

Explanation / Answer

after colliosn bullet + block raises to h

(1/2)*(Mbullet + Mblock)*Vafter = (Mblock+Mbullet)*g*h

vafter = sqrt(2*g*h)

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for freely falling block

speed of the block

Vblock = -sqrt(2*g*h)

for bullet

Vbullet = 520 m/s

momentum before collison = momentum after collison

Mbullet*Vbullet - Mblock*Vblock = (Mblock + Mbullet)*Vafter

(0.0132*520) - (2.53*9.8*sqrt(2*9.8*h)) = (0.0132+2.53)*sqrt(2*9.8*h)

the ball has fallen a height h = 0.0032 m

for freely falling

h = (1/2)*g*t^2

0.0032 = (1/2)*9.8*t^2

t = 0.026 s <<<<<=======ANSWER

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