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A 1190-kg flatbed train car rolls along frictionless tracks at a speed of 0.132 m/s. A 62.2-kg man stands on the flatbed and, in order to stop the car, runs to the edge and jumps off.

0.132 m/s If the cars final velocity is zero, what is the final speed v of the man? m/s

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Answer #1

V = initial velocity of train-man combination = 0.132 m/s

M = mass of train = 1190 kg

m = 62.2 kg

Vf = final velocity of train = 0

vf = final velocity of man = v

using conservation of momentum

(M + m) V = M Vf + m vf

(1190 + 62.2) (0.132) = (1190) (0) + (62.2) v

v = 2.7 m/s

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