# Two dimensional momentum problem

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1. Mar 25, 2017

### WherE mE weeD

1. The problem statement, all variables and given/known data
So basically I have a train with two joined carriages which become separated (m1) Mass one =13563kg and (m2) Mass two =30394kg. the initial velocity of the joined carriages is 5.00m/s (u1).

When the carriages separate I have the final velocity of (m2) which is 0.96m/s (v2)

Using the conservation of momentum principles I want to find the final velocity of (m1) which will be (v1)
m1=13563kg
m2=30394kg
u1=5.00 m/s
v2=0.96 m/s
2. Relevant equations
I know Momentum Initial = Momentum Final
and Change in momentum: = m Δv = m (v – u)
and Velocity = Momentum/Mass
I was thinking finding the energy values using the KE formula KE=mv^2/2 but i'm honestly a bit lost.
3. The attempt at a solution
The masses together = 13563+30394 = 43957kg
Total Momentum = Total Mass x Velocity; M=((m1+m2)x5.00)=219785kg m/s
Momentum of m1 = m1 x Velocity = (m1x5.00)=67815kg m/s
Momentum of m2 = m2 x Velocity = (m2x5.00)=151970kg m/s

If velocity = Momentum/Mass
= Momentum of m2/m2
= 151970/30394 = 5m/s which is true
Momentum of M2 after the separation = m2 x v2 = 30394 x 0.96 = 29178.24kg m/s
Any help much appreciated.

Last edited: Mar 25, 2017
2. Mar 25, 2017

### WherE mE weeD

From both the carriages (m1) is the engine train carriage and (m2) a transport wagon with no means of acceleration. It is stated that m1 is pulling m2 at a constant velocity of 5.00 m/s (u1) this could mean m2 would continue at the constant velocity as it is the means of applied force.

The question asks for m1 velocity after separation which in my mind would increase with the loss of m2.

3. Mar 25, 2017

### Staff: Mentor

The scenario is unclear, and your second post made it worse, not better

It is not clear how or why the train carriages separated. Was it an explosive separation? Or maybe they are separated by a spring mechanism that thrusts them apart? What's the significance of one car being an engine? Does it mean there's a force due to the engine being applied to the track? Is the track smooth and level?

Can you post the original problem statement as it was given to you?

4. Mar 25, 2017

### TomHart

It is not clear to me what is happening in this problem. You have two train cars - an engine and another non-powered car. They are both initially connected together and moving at 5 m/s. Is the engine car powering the motion? If so, then is the system under acceleration, or is there another counter-force that balances the force from the engine car?. After separation, the non-powered car (m2) has a final velocity of 0.96 m/s. What has caused m2's velocity to decrease? Did the separation of the 2 cars cause the decrease? Is this really a conservation of momentum problem?

5. Mar 25, 2017

### WherE mE weeD

I will post up the question now thanks for prompt replys.

A diesel locomotive is putting a chemical transport wagon at a constant velocity of u (m/s). During the transport the engine and chemical wagon decouple to form two separate masses, m1 (kg) and m2 (kg). The two parts continue moving in the same direction with the chemical wagon moving at v2(m/s).

making use of the principle of momentum, determine the following:
i,) The velocity v1 of the locomotive after separation.
ii) The change in the total kinetic energy of the system.

6. Mar 25, 2017

### TomHart

Where did you get v2 = 0.54? I thought it was 0.96 m/s.

So it sounds like it's a fairly straightforward problem. You calculated the initial momentum correctly. For conservation of momentum, the final momentum must equal the initial momentum. ρinitial = ρfinal

7. Mar 25, 2017

### WherE mE weeD

I could transpose this formula;
Momentum = (m1 x v1) (m2 x v2) to give the answer for v1.

219785 = (13563 x v1) 151970 / divide by 151970
219785/151970 = 13563 x v1
1.446 = 13563 x v1 / divide by 13563
1.446/13563 = v1
= 0.000106614

8. Mar 25, 2017

### WherE mE weeD

Thats a my bad v2 is 0.96m/s

9. Mar 25, 2017

### TomHart

Can you check this formula?

10. Mar 25, 2017

### WherE mE weeD

Momentum = (m1 x v1) + (m2 x v2)

11. Mar 25, 2017

### TomHart

That looks better. So the final momentum is the sum of the individual final momentums and that sum has to equal the initial momentum, right?

12. Mar 25, 2017

### WherE mE weeD

Yes thats correct.

13. Mar 25, 2017

### TomHart

But here you were multiplying the two parentheses together.
And here you used the initial momentum of m2, not the final momentum.

14. Mar 25, 2017

### WherE mE weeD

So the no matter if its after the separation or before if you add the final or initial momentums of either m1 or m2 at any time they will equal the original total momentum?

Momentum = (m1 x v1) (m2 x v2)

219785 = (13563 x v1) + 16412.76 / -16412.76
219785 - 16412.76 = 13563 x v1
203372.24 = 13563 x v1 / divide by 13563
203372.24/13563 = v1
= 14.99m/s
Seems like v1 will be very fast.

15. Mar 25, 2017

### WherE mE weeD

Thats an incorrect statement I made cheers Tom Im definitely getting the grips of momentum with your help.

16. Mar 25, 2017

### TomHart

This would be correct, except for the fact that you calculated m2's final momentum using the incorrect velocity of 0.54. It should have been 0.96 m/s. That should result in a slightly lower final velocity for m1.
Edit: Oops. Changed m2 to m1.

17. Mar 25, 2017

### WherE mE weeD

Haha cheers well spotted.

18. Mar 25, 2017

### WherE mE weeD

219785 = (13563 x v1) + 29178.24 / -29178.24
219785 - 29178.24 = 13563 x v1
190606.76 = 13563 x v1 / divide by 13563
190606.76/13563 = v1
= 14.05m/s

19. Mar 25, 2017

### TomHart

I was kind of confused on this equation. You have a "/" followed by a "-" sign. That doesn't make sense to me.

Just to be clear, here is how I would write it out, where ρ stands for momentum.
ρinitial = (m1 + m2)vinitial
ρfinal = ρ1final + ρ2final = m1v1final + m2v2final
But we know that ρfinal = ρinitial (because of momentum conservation), so
ρinitial = m1v1final + m2v2final
Solving for v1final gives:
v1final = (ρinitial - m2v2final)/m1

20. Mar 26, 2017

### WherE mE weeD

It was a bit misleading I haven't quite got the hang of math lingo but im getting there.

I was just wondering in the case that the train began its journey from a rest initial velocity 0m/s and accelerated to a final velocity of 5m/s would I then use the final velocity?