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20 September, 17:12

A 1.11 kg skateboard is coasting along the pavement at a speed of 6.82 m/s when a 0.880 kg cat drops from a tree vertically downward onto the skateboard. What is the speed of the skateboard-cat combination? Answer in units of m/s.

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  1. 20 September, 18:58
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    M1=1.11kg

    m2=0.880kg

    u1 = 6.82m/s

    u2 = 0m/s

    v1 = v2 = ?

    law of conservation of momentum: m1u1+m2u2 = m1v1+m2v2

    puttin in values: 1.11 (6.82) = 1.11v1+0.880v1

    7.57 = 1.99v1

    final speed v1 = 7.57/1.99 = 3.8 m/s
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