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17 June, 19:03

A (n) 9.9 g bullet is fired into a (n) 138 g block of wood at rest on a horizontal surface and stays inside. After impact, the block slides 13 m before coming to rest. The acceleration of gravity is 9.8 m/s^2.

If the coefficient of friction between the surface and the block is 0.6, find the speed of the bullet before impact. Answer in units of m/s.

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  1. 17 June, 21:19
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    185.25 m/s

    Explanation:

    consider the motion of the combination of bullet and block after the collision

    v₀ = initial speed just after the collision

    v' = final speed = 0 m/s

    μ = Coefficient of friction = 0.6

    g = acceleration due to gravity = 9.8 m/s²

    a = acceleration of the combination = - μ g = - (0.6) (9.8) = - 5.88 m/s²

    d = stopping distance = 13 m

    using the kinematics equation

    v'² = v₀² + 2 a d

    0² = v₀² + 2 ( - 5.88) (13)

    v₀ = 12.4 m/s

    m = mass of the bullet = 9.9 g = 0.0099 kg

    M = mass of the wood = 138 g = 0.138 kg

    v = speed of bullet before collision

    v₀ = speed of combination after the collision = 12.4 m/s

    Using conservation of momentum

    m v = (m + M) v₀

    (0.0099) v = (0.0099 + 0.138) (12.4)

    v = 185.25 m/s
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