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17 July, 12:04

A 0.0140 kg bullet traveling at 205 m/s east hits a motionless 1.80 kg block and bounces off it, retracing its original path with a velocity of 103 m/s west. What is the final velocity of the block? Assume the block rests on a perfectly frictionless horizontal surface.

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  1. 17 July, 13:38
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    Final velocity of the block = 2.40 m/s east.

    Explanation:

    Here momentum is conserved.

    Initial momentum = Final momentum

    Mass of bullet = 0.0140 kg

    Consider east as positive.

    Initial velocity of bullet = 205 m/s

    Mass of Block = 1.8 kg

    Initial velocity of block = 0 m/s

    Initial momentum = 0.014 x 205 + 1.8 x 0 = 2.87 kg m/s

    Final velocity of bullet = - 103 m/s

    We need to find final velocity of the block (u)

    Final momentum = 0.014 x - 103 + 1.8 x u = - 1.442 + 1.8 u

    We have

    2.87 = - 1.442 + 1.8 u

    u = 2.40 m/s

    Final velocity of the block = 2.40 m/s east.
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