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23 October, 15:48

A 26-g steel-jacketed bullet is fired with a velocity of 630 m/s toward a steel plate and ricochets along path CD with a velocity 500 m/s. Knowing that the bullet leaves a 50-mm scratch on the surface of the plate and assuming that it has an average speed of 600 m/s while in contact with the plate, determine the magnitude and direction of the impulsive force exerted by the plate on the bullet.

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  1. 23 October, 18:41
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    F = - 3.53 10⁵ N

    Explanation:

    This problem must be solved using the relationship between momentum and the amount of movement.

    I = F t = Δp

    To find the time we use that the average speed in the contact is constant (v = 600m / s), let's use the uniform movement ratio

    v = d / t

    t = d / v

    Reduce SI system

    m = 26 g (1 kg/1000g) = 26 10⁻³ kg

    d = 50 mm (1m / 1000 mm) = 50 10⁻³ m

    Let's calculate

    t = 50 10⁻³ / 600

    t = 8.33 10⁻⁵ s

    With this value we use the momentum and momentum relationship

    F t = m v - m v₀

    As the bullet bounces the speed sign after the crash is negative

    F = m (v-vo) / t

    F = 26 10⁻³ (-500 - 630) / 8.33 10⁻⁵

    F = - 3.53 10⁵ N

    The negative sign indicates that the force is exerted against the bullet
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