Please Help!! In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The

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Please Help!! In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The collision is cushioned by a spring (k = 1200 N/m). a) Determine the velocity of each cart after the collision. b) Determine the maximum compression of the spring.

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Tryphena 2 months 2021-08-22T11:40:50+00:00 1 Answers 8 views 0

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    2021-08-22T11:41:55+00:00

    Answer:

    initial momentum 0.6kg * -5m/s + 0.8kg * 2m/s = -1.4 kg·m/s

    final momentum = -1.4 kg·m/s = 0.6kg * u + 0.8kg * v

    For an elastic head-on collision, we know that the

    relative velocity of approach = relative velocity of separation, or

    7m/s = u – v, or

    u = v + 7

    Plug this into momentum eqn:

    -1.4 = 0.6(v+7) + 0.8v = 0.6v + 4.2 + 0.8v

    -5.6 = 1.4v

    v = -4 m/s ← 0.8kg cart, now moving West

    u = v + 7 = 3 m/s ← 0.6kg cart, now moving East

    initial Ek = ½ * 0.6kg * (5m/s)² + ½ * 0.8kg * (2m/s)² = 9.1 J

    This gets converted into spring energy,

    U = 9.1 J = ½kx² = ½ * 1200N/m * x²

    x = 0.123 m = 12.3 cm

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )