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30 June, 06:41

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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  1. 30 June, 09:39
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    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
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