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7 May, 19:30

A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.550 m/s. The total mass of the sled, man, and rock is 96.5 kg. The mass of the rock is 0.300 kg and the man can throw it with a speed of 17.5 m/s. Both speeds are relative to the ground. Determine the speed of the sled if the man throws the rock forward (i. e. in the direction the sled is moving).

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  1. 7 May, 21:37
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    Answer: 0.5 m/s

    Explanation:

    Given

    Speed of the sled, v = 0.55 m/s

    Total mass, m = 96.5 kg

    Mass of the rock, m1 = 0.3 kg

    Speed of the rock, v1 = 17.5 m/s

    To solve this, we would use the law of conservation of momentum

    Momentum before throwing the rock: m*V = 96.5 kg * 0.550 m/s = 53.08 Ns

    When the man throws the rock forward

    rock:

    m1 = 0.300 kg

    V1 = 17.5 m/s, in the same direction of the sled with the man

    m2 = 96.5 kg - 0.300 kg = 96.2 kg

    v2 = ?

    Law of conservation of momentum states that the momentum is equal before and after the throw.

    momentum before throw = momentum after throw

    53.08 = 0.300 * 17.5 + 96.2 * v2

    53.08 = 5.25 + 96.2 * v2

    v2 = [53.08 - 5.25 ] / 96.2

    v2 = 47.83 / 96.2

    v2 = 0.497 ~ = 0.50 m/s
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