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2 June, 13:12

A 3.5-kg rifle fires a 15-g bullet at 840 m/s. With what speed does the rifle recoil (move backward toward the shooter's shoulder) ?

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  1. 2 June, 14:14
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    Momentum before the action = Momentum after

    So:

    Where m1 and v1 are for the gun and m2 and v2 are for the bullet

    (initial side) m1v1 + m2v2 = m1v1 + m2v2 (final side)

    Initially there is no momentum so we are left with

    0 = m1v1 + m2v2 (final)

    Subtract over m1v1

    -m1v1 = m2v2

    We are looking for v1 which is the final velocity of the gun

    so:

    v1 = m2v2 / - m1

    v1 = (15 x 10^-3 kg) (840) / - (3.5)

    v1 = - 3.6 m/s

    The negative indicates that it is travelling in the negative direction towards the shooter
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