Momentum conservation question.

  • Thread starter cragar
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  • #1
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Homework Statement


Two hobos each of mass [itex] m_h [/itex], are standing at one of a stationary railroad flatcar with frictionless wheels and mass [itex] m_c [/itex]. Either hobo can run to the other end of the flatcar and jump off with the same speed u relative to the car.
a) use conservation of momentum to find the speed of the recoiling car if the two men run and jump simultaneously.
b) What is it if the second man starts running only after the first has already jumped?
Which procedure gives the greater speed to the car?
Hint: the speed u is the speed of either hobo, realtive to the car just after he has jumped; it has the same value for either man and is the same in parts a) and b),

The Attempt at a Solution


a) [itex] m_c v_c=2m_h (u) [/itex]
[itex] v_c= \frac{2m_h u}{m_c} [/itex]

b) First I do when one hobo jumps off then I do it again in the rest frame and then add the speeds at the end.
[itex] (m_c+m_h)v=m_h(u) [/itex]
[itex] v= \frac{m_h u}{m_c+m_h} [/itex]
[itex] m_c v' =m_h u [/itex]
[itex] v'= \frac{m_h u }{m_c} [/itex]
[itex] v_{total}=v+v'=\frac{m_h u}{m_c+m_h} +\frac{m_h u }{m_c} [/itex]
If I assume [itex] m_h=m_c [/itex] just to make it easier to see which one is larger then
for part a) [itex] v=2u [/itex]
b) [itex] v=\frac{3u}{2} [/itex]
so a gives a faster cart speed.
 

Answers and Replies

  • #2
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Note that ##u## is relative to the car. In that frame, ##v_c = 0##, so your equations are meaningless. Use the ground frame. In that frame, the velocity of a hobo is ## v_g = v_c + u ##. Mind the signs, too.
 

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