The Conservation of Momentum
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N men, each with mass, m, stand on a railway flatcar of mass, M. They jump off one end of the flatcar with velocity u relative to the car. The car rolls in the opposite direction without friction.
a) What is the final velocity of the flatcar if all the men jump at the same time?
b) What is the final velocity of the flatcar if all the men jump off one at a time? (Can be left in form of a sum of terms)
Does case a or case b yield the largest final velocity of the flat car? Can you give a simple physical explanation for your answer?
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In the first case, from the conservation of momentum, we can write m1v1= m2v2.
Thus, when all the men jump: M.Vcar=N.m.u
Thus Vcar= N.m.u / M
The second case is different. We must first show what happens when the first man jumps. When he jumps, there will be N-1 men left on the car. Thus the mass of the car will be (M + (N-1)m ). Thus when we write the equation for conservation of momentum: (M + ...
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