Calculating the Speed of a Hopper Car After Dumping Coal

  • Thread starter Thread starter huskydc
  • Start date Start date
  • Tags Tags
    Car Coal Speed
Click For Summary
SUMMARY

The discussion centers on the physics of a railroad hopper car with a mass of 100,000 kg (including 50,000 kg of coal) coasting at 10 m/s while releasing coal over 4 seconds. Participants debate the final speed of the car after coal dumping, with some asserting that the car should maintain its speed due to conservation of momentum, while others suggest it should accelerate. The consensus indicates that if the coal falls straight down without any horizontal velocity, the car's speed remains unchanged at 10 m/s, as there is no net external force acting on the system.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the principle of conservation of momentum
  • Basic knowledge of inelastic collisions
  • Concept of forces acting on a system
NEXT STEPS
  • Study Newton's laws of motion in detail
  • Explore conservation of momentum in various systems
  • Learn about inelastic collisions and their implications
  • Investigate the effects of external forces on moving objects
USEFUL FOR

Physics students, educators, and anyone interested in understanding the dynamics of moving systems and the principles of momentum conservation.

huskydc
Messages
78
Reaction score
0
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?
 
Physics news on Phys.org
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.
 

Similar threads

Replies
9
Views
6K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 13 ·
Replies
13
Views
7K
  • · Replies 27 ·
Replies
27
Views
4K
  • · Replies 12 ·
Replies
12
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K