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30 March, 15:50

A 12404.7 kg railroad car travels alone on a level frictionless track with a constant speed of 11.6 m/s. a 5579.1 kg load, initially at rest, is dropped onto the car. what will be the car's new speed

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  1. 30 March, 18:36
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    The momentum must be conserved during both interactions

    In the first case, the momentum was

    P = m*v

    where m is the mass

    v is the speed

    P1 = 12404.7 kg * 11.6 m/s = 143894.52 kg*m/s

    P2 = (12404.7 + 5579.1) kg * V2 = 143894.52 kg*m/s

    Now, we find V2 (Car's new speed)

    V2 = 143894.52 [kg*m/s]/17983.8[kg] = 8 m/s
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