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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 speed of 603 m/s when it strikes the block. The block originally was dropped from rest from the t 697-kg bullet is fired straight up at a falling wooden block that has a mass of 4.50 kg. The bullet has a dropped from rest from the top of a building and had been falling for a time t when the collision with the bullet oothe bullet occurs. As ? 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 the tolerance is +/-3%

Explanation / Answer

from the relation

v block = v0, block+a*t

a = -9.8 m/s^2

initially v0 = 0

v block = -9.8*t

during the collision form conservation of momentum

(mbullet+m block)*vf = m bullet*v bullet+m block*v block

vf = -v block substitute

t = -m bullet*v bullet/a(m bullet+2*m block)

t = -0.00697*603/-9.8(0.00697+2*4.5)

t = 0.0476 sec

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