Two dimensional momentum problem

In summary: But this is wrong it doesn't work.In summary, a train with two joined carriages (m1=13563kg and m2=30394kg) separates with the initial velocity of 5.00m/s (u1). The final velocity of m2 is 0.96m/s (v2) and using the principles of conservation of momentum, the final velocity of m1 (v
  • #1
WherE mE weeD
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Homework Statement


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

Homework 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.

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.
 
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  • #2
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
The scenario is unclear, and your second post made it worse, not better :confused:

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
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
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
WherE mE weeD said:
Momentum of M2 after the separation = m2 x v2 = 30394 x 0.54 = 16412.76kg m/s
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
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
TomHart said:
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
Thats a my bad v2 is 0.96m/s
 
  • #9
WherE mE weeD said:
Momentum = (m1 x v1) (m2 x v2)
Can you check this formula?
 
  • #10
TomHart said:
Can you check this formula?
Momentum = (m1 x v1) + (m2 x v2)
 
  • #11
WherE mE weeD said:
Momentum = (m1 x v1) + (m2 x v2)
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?
 
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  • #12
Yes that's correct.
 
  • #13
WherE mE weeD said:
Momentum = (m1 x v1) (m2 x v2) to give the answer for v1.
But here you were multiplying the two parentheses together.
WherE mE weeD said:
219785 = (13563 x v1) 151970 / divide by 151970
And here you used the initial momentum of m2, not the final momentum.
 
  • #14
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
Thats an incorrect statement I made cheers Tom I am definitely getting the grips of momentum with your help.
 
  • #16
WherE mE weeD said:
219785 - 16412.76 = 13563 x v1
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.
 
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  • #17
Haha cheers well spotted.
 
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  • #18
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
WherE mE weeD said:
219785 = (13563 x v1) + 29178.24 / -29178.24
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
 
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  • #20
TomHart said:
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
It was a bit misleading I haven't quite got the hang of math lingo but I am 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?
 
  • #21
WherE mE weeD said:
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?
I don't know if I ever remember seeing any conservation of momentum problem that occurred during acceleration. That's not to say that you couldn't have one. But to answer your question I would say yes, you would use the final velocity of 5 m/s - or whatever the velocity happened to be when the event (collision or separation) occurred.
 
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Related to Two dimensional momentum problem

1. What is the definition of two dimensional momentum?

Two dimensional momentum is a measure of the quantity of motion an object has in both the horizontal and vertical directions. It is calculated by multiplying an object's mass by its velocity in both directions.

2. How is momentum conserved in a two dimensional system?

In a two dimensional system, momentum is conserved if the total momentum before an event is equal to the total momentum after the event. This means that the total momentum in both the horizontal and vertical directions remains the same.

3. How is the momentum of a system calculated?

The momentum of a system is calculated by adding the individual momentums of all the objects in the system. This can be done by multiplying an object's mass by its velocity in both the horizontal and vertical directions and then adding all of these values together.

4. What is the difference between elastic and inelastic collisions in terms of momentum?

In an elastic collision, the total momentum of the system remains the same before and after the collision. In an inelastic collision, the total momentum of the system changes due to some energy being lost to heat or sound.

5. How does the conservation of momentum apply to real-life situations?

The conservation of momentum applies to real-life situations such as car accidents or collisions between objects. In these situations, the total momentum of the system remains the same, but the individual momentums of the objects may change due to forces acting on them.

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