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10 February, 23:46

A father racing his son has half the kinetic energy of the son, who has half the mass of the father. The father speeds up by 1.0 m/s and then has the same kinetic energy as the son. What are the original speeds of (a) the father and (b) the son

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  1. 11 February, 00:10
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    a. 2.41 m/s

    b. 4.82 m/s

    Explanation:

    Father's KE is:

    K. E. = ½mv²

    Sons k. e. is:

    k. e. = ½MV²

    Since M = ½m:

    k. e. = ¼mV²

    We are given that K. E. = ½k. e., therefore:

    ½mv² = ½ (¼mV²)

    ½mv² = ⅛mV²

    => v² = V²/4

    => v = V/2 and V = 2v

    When the father's speed is (v + 1) m/s, K. E. = k. e.:

    K. E. = ½m (v + 1) ²

    => ½m (v + 1) ² = ¼mV²

    (v + 1) ² = ½V²

    Since V = 2v

    => (v + 1) ² = ½ (2v) ²

    v² + 2v + 1 = ½ (4v²)

    v² + 2v + 1 = 2v²

    => v² - 2v - 1 = 0

    => v = 2.41m/s or - 0.41m/s

    Since speed cannot be negative:

    v = 2.41m/s

    Recall that V = 2v

    => V = 2 * 2.41 = 4.82m/s

    Hence, the father's speed is 2.41m/s and the son's speed is 4.82m/s.
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