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16 September, 14:16

A 8.3-g wad of sticky clay is hurled horizontally at a 82-g wooden block initially at rest on a horizontal surface. The clay sticks to the block. After impact, the block slides 7.50 m before coming to rest. If the coefficient of friction between block and surface is 0.650, what was the speed of the clay immediately before impact?

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  1. 16 September, 14:28
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    The speed of the clay before the impact was 106.35 m/s.

    Explanation:

    the only force doing work on the system is the frictional force, f, the work done by f is given by:

    Wf = ΔK = Kf - Ki

    The clay and the block will come to rest after sliding Δx = 7.50 m, if their intial speed is v and the combined mass is m and μ is the coefficient of friction and g is gravity, then:

    f*Δx = Ki

    m*g*Δx*μ = 1/2*m*v^2

    v^2 = 2*g*Δx*μ

    = 2 * (9.8) * (7.50) * (0.650)

    = 95.55

    v = 9.78 m/s

    This is the veloty of clay and block after the clay hit the block.

    if the velocity the clay and block attains after the impact is v and the initial speed of the clay is v1 and the mass is m and the speed of the block initially is V = 0 m/s and the mass is M, then according to the conservation of linear momentum:

    m*v1 + M*V = v (m + M)

    m*v1 = v (m + M)

    v1 = v (m + M) / m

    v1 = (9.78) (8.3*10^-3 + 82*10^-3) / (8.3*10^-3)

    v1 = 106.35 m/s

    Therefore, the speed of the clay before the impact was 106.35 m/s.
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