A man of mass M1 = 114 kg is standing on a M2 = 5700 kg railroad flatcar that ro

Question

A man of mass M1 = 114 kg is standing on a M2 = 5700 kg railroad flatcar that rolls without friction to the right with speed 3.4 m/s. Calculate the change in velocity (in m/s) of the flatcar, if the man runs to the left so that his speed relative to the original velocity of the car is 4.9 m/s.

Hint:
Momentum has to be conserved. The man and flatcar have to be moving with the same momentum before and after he starts walking. Before both flatcar and man are going to the right with the same velocity. Afterwards the car is moving to the right but the man is moving to the left with the difference of the two velocities.
[M(car)+M(man)]v(initial) = M(man)[v(initial)-v(man rel to car)] + M(car)v(final)
Note that it asks for the change in velocity

Explanation / Answer

vmc = Vm - Vc = 4.9

Vm = 4.9 + Vc

using momentum conservation.

5700 x 3.4 = 5700 x Vc - 114 x Vm

19380 = 5700Vc - 114(4.9 + Vc)

19938.6 = 5586Vc

Vc = 3.57 m/s3

change in velocity = 3.57 - 3.4 = 0.169 m/s

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