Homework Help: Linear momentum

1. Apr 6, 2013

Litcyb

1. The problem statement, all variables and given/known data
Im not sure if I did this problem correctly, can someone please reassure me?!

2. Relevant equations

Conservation of Linear momentum

pf=pi

m1v1=m2v2

3. The attempt at a solution

M(initial)=100,000 kg
M(final)= 100,000kg-15,000kg= 85,000kg

V(initial)=20.0m/s
V(final)=?

P(initial)=P(final)

m1v1=m2v2

(100,000)(20.0)=(85,000)(vf)

vf=23.53 m/s

did i do this correctly?

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2. Apr 6, 2013

rude man

It would be nice if one could speed up a train that way!

Time for a couple of thought experiments:
1. what if all the dropped sand were to fall on a frictionless plane embedded between the ties?
2. if instead of a frictionless plane the sand were stopped from moving by the ties, would the train know the difference?

3. Apr 6, 2013

SammyS

Staff Emeritus

See, it's not all that hard to include an image of the problem in your post so it's easy to read.

Who made up that problem ???

4. Apr 7, 2013

Litcyb

It is an example question from old quizzes. why? the problem doesnt make sense? can someone help me?

so would it be (100,000)(20)= (85,000)(vf) +(15,000)(20)?
because the sand that is falling out, has the same speed of the cart throughout the 30 mins

Last edited: Apr 7, 2013
5. Apr 7, 2013

kickingpaper

The 15000*20 term is unneeded since your accounting for the loss in mass in the 85000*vf term. The question makes sense to me btw. Your first answer is right and it's easy too check if you think about the conservation of momentum. For momentum to be conserved means that the momentum after some event is equal to the momentum before the event. If the train is losing mass it makes sense for the train to speed up to conserve momentum. Likewise if the train had gained mass, it would slow down.

6. Apr 7, 2013

Staff: Mentor

A change in momentum requires external forces to act on the object in the direction of that change. What force acts on the train in the direction of its motion if sand simply falls out the bottom?

Last edited: Apr 7, 2013
7. Apr 7, 2013

Much better!

8. Apr 7, 2013

Litcyb

@Rude man, im confused. so in general, M(total) *V(initial)= M(train)*V(final train) + m(of sand)* velocity( throughout the 30 mins)
which gives me vf=20m/s. which is sort of weird, since is the same velocity we started with.

Is it okay to think of it this way; that in order to have conservation of momentum, we need to still consider the mass that has fall out of the cart. thus, is the reason why we added (15,000)*(20).

however, is what kickingpaper said still valid?

The 15000*20 term is unneeded since your accounting for the loss in mass in the 85000*vf term. The question makes sense to me btw. Your first answer is right and it's easy too check if you think about the conservation of momentum. For momentum to be conserved means that the momentum after some event is equal to the momentum before the event. If the train is losing mass it makes sense for the train to speed up to conserve momentum. Likewise if the train had gained mass, it would slow down.

9. Apr 7, 2013

rude man

Yes. The sand has the same velocity and mass as before it was tossed, so there is no change in momentum of the system.

Note what gneill said: in order to change the momentum of a system you have to exert an external force F on it: ∫F dt = Δp or F = dp/dt where p = momentum.

No.

10. Apr 7, 2013

Litcyb

And since there are no external forces acting on it, then momentum is conserved. so, we have to finish with what we started.

if there was a external forces, such as in a rocket thrust problem, then we will have to involve impulse, in order to calculate the the Vf of the rocket, right?

11. Apr 7, 2013

SammyS

Staff Emeritus

Let's look at this as two objects:
1. The 15,000 of sand that leaks out of the railroad car. Is momentum conserved for this sand?

2. The railroad car along with the sand that hasn't leaked out. Is momentum conserved for this 85,000 kg compound object?

12. Apr 7, 2013

rude man

If the train car were a rocket there would be momentum associated with the expelled gases, depending of course on their mass and velocity. But the total momentum including the rocket "remains" and the expelled gases would still be conserved since the rocket is not an external force.

A corollary of this is that the center of mass of the rocket and all the expelled gases does not move!

EDIT: remember that this is conditional on no friction with the rails. If there were friction on the rails then momentum is still conserved but now you'd have to include the entire Earth in the picture!

13. Apr 7, 2013

rude man

No. That momentum is transferred to the Earth eventually.​

Last edited: Apr 7, 2013
14. Apr 7, 2013

SammyS

Staff Emeritus
I meant that question for OP. I should have made that clear.

The title of this thread includes momentum, but the problem statement doesn't . Why the big concern over conservation of momentum for this problem? (Rhetorical question)

15. Apr 7, 2013

rude man

Yeah, because I got the impression you were directing your remarks at more than one individual. My mistake ...

16. Apr 7, 2013

MostlyHarmless

As it was a bonus question, my guess is the problem was intended send you down the wrong track.

Edit: No pun intended. :)

17. Apr 7, 2013

Litcyb

Well guys! thank you either way!

18. Apr 7, 2013

rude man

The rationale for the answer, which i think most of us know, is still arguable via momentum conservation. See my post #2.