A 0.015-kg bullet is fired straight up at a falling wooden block that has a mass

Question

A 0.015-kg bullet is fired straight up at a falling wooden block that has a mass of 2.5 kg. the bullet has a speed of 1000m/s when it strikes the block. the block originally was dropped from rest from the top of the building and had been falling for a time t when the collision with the bullet occurred. as a result of the collision, the block (with the bullet in it) reverses direction, rises, and comes to a momentary half at the top of the building. find the time t.

Explanation / Answer

velocity of bullet before collision(u1) = 1000j (+j is for direction) mass of bullet (m1) = 0.015 kg Let velocity of block before collision(u2) = -wj (-j for direction) mass of block(m2) = 2.5 kg As the block halts again at top of building after collision, velocity of both after collision(v) should be wj (+j for direction) From conservation of momentum, m1u1 + m2u2 = (m1+m2)v => 15j - 2.5wj = 2.515*wj => 15 = w*(2.5+2.515) => w = 15/5.015 But v = u+at => w = 0+gt => t = w/g = 15/(5.015*9.81) = 0.3 sec(approx)

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