Question

A 0.0121-kg bullet is fired straight up at a falling wooden block that has a mass of 4.99 kg. The bullet has a speed of 898 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.

Answer 1

Answer:

t = 0.11 s

Explanation:

The motion is complete inelastic so the motion can be model using

mb * vb + mB * vB = ( mb + mB ) vf

mb = 0.0121 kg ,  mB = 4.99 kg ,  vb = 898 m / s , vB = 0

Replacing to find vf

0.0121 * 898 m / s + 4.99 kg * 0 = ( 0.0121 + 4.99) * vf

vf = 2.18 m / s

Now to find the time take the acceleration as a= g = 9.8 m /s²

t = mb * vb / a * (mb + 2 * mB)

t = [0.0121 kg * 898 m /s] / 9.8 * ( 0.0121 kg + 2*4.99)

t = 0.11 s