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7 May, 17:15

A 45 g bullet strikes and becomes embedded in a 1.55 kg block of wood placed on a horizontal surface just in front of the gun. If the coefficient of kinetic friction between the block and the surface is 0.28, and the impact drives the block a distance of 13.0 m before it comes to rest, what was the muzzle speed of the bullet in meters/second?

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  1. 7 May, 17:17
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    299.51 m/s

    Explanation:

    m = mass of the bullet = 45 g = 0.045 kg

    M = mass of the block = 1.55 kg

    v = muzzle speed of the bullet

    V = speed of bullet-block combination after the collision

    μ = Coefficient of friction between the block and the surface = 0.28

    d = distance traveled by the block = 13 m

    V' = final speed of the bullet-block combination = 0 m/s

    acceleration of the bullet-block combination due to frictional force is given as

    a = - μg

    using the kinematics equation

    V'² = V² + 2 a d

    0² = V² + 2 ( - μg) d

    0 = V² - 2 (μg) d

    0 = V² - 2 (0.28) (9.8) (13)

    V = 8.45 m/s

    Using conservation of momentum for collision between bullet and block

    mv = (M + m) V

    (0.045) v = (1.55 + 0.045) (8.45)

    v = 299.51 m/s
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