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29 May, 02:59

In a ballistics test, a 24 g bullet traveling horizontally at 1300 m/s goes through a 25-cm-thick 360 kg stationary target and emerges with a speed of 930 m/s. The target is free to slide on a smooth horizontal surface. How long is the bullet in the target? What average force does it exert on the target? What is the target's speed just after the bullet emerges?

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  1. 29 May, 05:40
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    A) 0.00022s b) 40363.6N c) 0.025m/s

    Explanation:

    Mass = 24g = 0.024kg, distance though the target = thickness of the target = 25cm = 0.25m

    Initial speed of the bullet = 1300m/s, final speed = 930m/s

    Using equation of motion

    Distance = 1/2 (vf+vi) * t (time in seconds)

    t = 0.25*2 / (1300+930) = 0.00022s

    B) force exerted on the body

    F = ma = m * (vf-vi) / t = 0.024 * (930-1300) / 0.00022

    F = - 40363N, it is negative because the body decelerated during this motion

    C) using law of conservation of momentum,

    M1*U1 + M2*U2 (M2and U1 are the mass and initial speed of the body) = M1V1 + M2V2

    The target was at rest so initial speed U2 = 0

    0.024*1300 + 360*0 = 0.024*930 + 360*V2

    31.2 = 22.32+360*V2

    31.2-22.33 = 360*V2

    V2 = 8.88/360 = 0.025m/s
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