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.

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Physics

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24 September, 06:34

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.

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Silas · Answer 1

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