Coal Car and horizontal impulse

  • Thread starter turandorf
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  • #1
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Homework Statement


A railroad hopper car has a mass of 30000 kg when empty and contains 50000 kg of coal. As it coasts along the track at 5 m/s the hopper opens and steadily releases all the coal onto a platform below the rails over a period of 4 s. What is the horizontal impulse delivered to coal by the platform?


Homework Equations


Pi=Pf
J=change in p/change in t


The Attempt at a Solution


I tried 40,000 (which is total mass*velovity)=15,000(which is car*velocity)x which i thought was the change in momentum so I divided that by 4 sec. This was wrong. Any ideas?
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
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Welcome to PF.

Well they want the horizontal impulse. So how much is dropped over the 4 seconds?
Presumably all of the 50,000kg.

That amount of coal was moving at 5 m/s.
So that's your initial momentum.
The final momentum is seemingly 0.
And 4 seconds is the interval ...
 
  • #3
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Thanks for the help! I really appreciate it! (btw the answer was 250000 kg*m/s)
 
  • #4
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I seem to be stuck on this part now
e) 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?
I think the equation I need is:
.5(m1Vi)^2-.5(m2Vf)^2=miVi-m2Vf
where m1 is the mass of coal and the car and m2 is just the empty car. Vi=5 and Vf is what I am looking for. So I have the change in KE equals the change in momentum. I tried solving it and got nowhere. Please let me know if this is the right approach. Thanks in advance for the help.
 
  • #5
LowlyPion
Homework Helper
3,097
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Maybe think of it in terms of conservation of momentum?

That won't change now will it?

Since the velocity of the coal, through the efforts of the eager to learn physics students, will have all of its momentum stripped away and retained by the car and the shoveling classmates, you have the whole load of coal acquiring a momentum of 0, by the time it's all shoveled off.

For momentum before to be the same as momentum after all the hard work ...

Or you could look at it as a center of mass problem. What speed does the car need to go for the center of mass to remain in motion at 5 m/s, if the load goes to 0 velocity.
 
  • #6
Doc Al
Mentor
45,204
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I think the equation I need is:
.5(m1Vi)^2-.5(m2Vf)^2=miVi-m2Vf
where m1 is the mass of coal and the car and m2 is just the empty car.
:yuck: That equation isn't dimensionally consistent, so it can't be right. (Compare the units of KE to the units of momentum.)
Vi=5 and Vf is what I am looking for. So I have the change in KE equals the change in momentum.
KE and momentum are different beasts with different units and cannot be set equal to each other.

Hint: Forget about kinetic energy--it's not conserved. (The person threw the coal, adding energy.) Just apply conservation of momentum. What's the final speed of the coal?
 

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