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10 January, 17:08

A 9.0 g wad of sticky clay is hurled horizontally at a 90 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 (in m/s) immediately before impact?

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  1. 10 January, 17:23
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    107.58 m/s

    Explanation:

    from the question we are given the following:

    mass of the stick (M₁) = 9 g = 0.009 kg

    mass of the block (M₂) = 90 g = 0.09 kg

    total mass (M₁+M₂) = 99 g = 0.099 kg

    distance (s) = 7.5 m

    coefficient of friction (Uk) = 0.650

    acceleration due to gravity (g) = 9.8 m/s^{2}

    apply the equation below

    change in kinetic energy = net work done

    final kinetic energy (K₂) - final kinetic energy (K₁) = Uk x force (F) x distance (s)

    final kinetic energy is zero because the clay and the wood comes to a stop

    kinetic energy = 0.5 x m x v^{2} and

    force = m x g so the equation now becomes

    0.5 x m x v^{2} = Uk x m x g x s

    0.5 x 0.099 x v^{2} = 0.650 x 0.099 x 9.8 x 7.5

    V = 9.78 m/s

    from the conservation of momentum

    M₁ V₁ + M₂ V₂ = (M₁ + M₂) V

    V₂ is zero since the block is initially at rest, so our equation becomes

    M₁ V₁ = (M₁ + M₂) V

    0.009 x V₁ = 0.099 x 9.78

    V₁ = 107.58 m/s

    the speed of the clay before impact is 107.58 m/s
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