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Momentum - Throwing coal out of railroad car

  • Thread starter Bryon
  • Start date
  • #1
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Homework Statement



A railroad hopper car has a mass of 50000 kg when empty and contains 30000 kg of coal. As it coasts along the track at 8 m/s the hopper opens and steadily releases all the coal onto a platform below the rails over a period of 4 s.

How fast does the car travel after all the coal is dumped? 8m/s

What is the horizontal impulse delivered to coal by the platform? 240000

If the coal comes to rest 1 second after the dumping is finished, what was the average horizontal force on the platform? 48000

Let's say that instead of the coal falling out of the bottom of the coal car, some very strong physics student threw it backwards from the car, so that the velocity of each piece of coal was zero just after it left her shovel. How fast does the car travel after all the coal is thrown instead of dumped? ??? Clueless.

Homework Equations



0.5(Total weight)v^2 = 0.5*(80000)*8^2 = 2560000
0.5(Weight of coal)v^2 = ??
0.5(Weight of car)v^2 = ??

The Attempt at a Solution



m(coal)v(coal) + m(car empty)v(initial) = m(car + coal)v(initial)
 

Answers and Replies

  • #2
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m(coal)v(coal) + m(car empty)v(initial) = m(car + coal)v(initial)
This should be:

m(coal)v(coal final) + m(car empty)v(car final) = m(car + coal)v(initial)

You know everything except the final speed of the car.
 
  • #3
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Would the final speed of the coal be found using this:

m*g = 0.5*m*v^2?
 
  • #4
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Would the final speed of the coal be found using this:

m*g = 0.5*m*v^2?
you can do everything with conservation of momentum and you certainly don't need g. You only need the equation you already nearly posted correctly/
 
  • #5
99
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I got it! Thanks. What threw me off was that the velocity of the coal was zero.
 

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