Calculating the Speed of a Hopper Car After Dumping Coal

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The discussion centers on calculating the speed of a hopper car after it dumps coal. Participants debate whether the car's speed remains unchanged or increases after releasing the coal. Some argue that momentum is conserved, suggesting the car should maintain its speed, while others believe the car should pick up speed due to the coal's downward motion. The time factor of 4 seconds is questioned, with some suggesting it influences the forces at play. Ultimately, the consensus leans towards the idea that the car's speed does not change if the coal falls straight down without any horizontal force acting on it.
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A railroad hopper car has mass 50000 kg when empty and contains 50000 kg of coal. As it coasts along the track at 10 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?

i applied the mv(initial) = mv (final) equation...

i got something like

(100000 kg) (10m/s) = (50000 kg) V final

apparently it's incorrect, any hints?
 
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How is that incorrect? That should be exactly what you get. Now just solve for V.
 
Technically... shouldn't it travel the same speed? There wasnt a momentum transfer and there's no readily apparent force acting on the car.
 
Pengwuino said:
Technically... shouldn't it travel the same speed? There wasnt a momentum transfer and there's no readily apparent force acting on the car.
It should pick up speed. The momentum changed. Think of it as a reverse inelastic collision.
 
But there is no collision, no momentum has changed because there's no force.

The only force I coudl see is if there was one generated by the coal which means a force of 125000 kg*m/s^2 was generated for 4 seconds... which still arrives at 20m/s for some reason... Well I am lost.
 
momentum is conserved, i kept getting 20 m/s

does the time, 4 sec matter in answering this question?
 
huskydc said:
does the time, 4 sec matter in answering this question?

It has to... there's some sort of force acting here and I think this is where it occurs (at the 4 seconds). Did you put the whole question in the post?
 
Pengwuino said:
But there is no collision, no momentum has changed because there's no force.

The only force I coudl see is if there was one generated by the coal which means a force of 125000 kg*m/s^2 was generated for 4 seconds... which still arrives at 20m/s for some reason... Well I am lost.
There is no collision in an explosion (another reverse inelastic collsion), yet momentum changes.
 
An explosion?

When are the geniuses going to come to this thread and show us how this is actually done?
 
  • #10
may i ask what explosion? or collision? the only thing happening is that the cart is dropping off coal..
 
  • #11
Pengwuino said:
An explosion?

When are the geniuses going to come to this thread and show us how this is actually done?

...
Pengwuino said:
Technically... shouldn't it travel the same speed? There wasnt a momentum transfer and there's no readily apparent force acting on the car.

Because someone already had the right idea?

Unless the coal is getting launched backwards relative to the hopper car, the velocity does not change. If we assume that the downward movement of the coal is negligble, then the coal is moving at the same velocity as the car when it falls out of the car. A few moments later, it collides with the platform, but it is not longer connected to the train car at that point. The velocity of the car should be unchanged.
 
  • #12
See that's what I am saying! I was waiting for someoen to come in and tell one of us that we were right because we had 2 different opinions on what was going on.
 
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