What is the final velocity of the five-car train after a series of collisions?

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Homework Help Overview

The problem involves a series of collisions between train cars, specifically focusing on the final velocity of a five-car train formed after these collisions. The context is centered around the conservation of momentum in a system of identical train cars moving at different speeds.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the conservation of momentum and question the calculations related to momentum before and after collisions. Some express confusion about the implications of identical masses on the final velocity.

Discussion Status

There is an ongoing exploration of the problem, with participants questioning the original poster's calculations and assumptions. Some guidance has been offered regarding the relationship between initial and final momentum, but no consensus has been reached on the correct approach or final outcome.

Contextual Notes

Participants note that all train cars are identical, which raises questions about whether mass can be ignored in calculations. There is also a suggestion that the original poster may benefit from starting a new thread for clarity.

grewas8
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Homework Statement


Three identical train cars, coupled together, are rolling east at 3.97 m/s. A fourth car traveling east at 4.59 m/s catches up with the three and couples to make a four-car train. A moment later, the train cars hit a fifth car that was at rest on the tracks, and it couples to make a five-car train. What is the speed of the five-car train?

Homework Equations



p=mv
Momentum is conserved


The Attempt at a Solution



Given- V(1,2,3)= 3.97 m/s
v(initial of 4)= 4.59

Find- v(1,2,3,4,5)
p at the instant 4 hits 1,2,3
p at the instant 5 hits 1,2,3,4

Cart 4- P=mv
P= 4.59m (initially)

Cart 1,2,3
P=mv
P=3m(3.97)=11.91m

When 4 collides with 1,2,3
p(of 4)=p(of 1,2,3)
4.59m=11.91m
Change in momentum=11.91m-4.59m=7.32m

When 1,2,3,4 collides with 5
P(1,2,3,4)=7.32m
P(1,2,3,4,5)=7.32m-mv

Really stuck after this.:confused: ..is anything i did above incorrect?
any help will be appreciated greatly :!)
 
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grewas8 said:
When 4 collides with 1,2,3
p(of 4)=p(of 1,2,3)
4.59m=11.91m
Change in momentum=11.91m-4.59m=7.32m

Huh? How can 4.59m=11.91m? You are making this much too complicated. What is the total momentum of the five cars before any collisions? How should this relate to the total momentum of the 5 car train after all of the collisions? So write down an expression for the final momentum in terms of the unknown final velocity and solve for it.
 
hi i was trying to do this question and i was wondering since all the masses are the same for the carts do we just ignore the masses?? then wouldn't the final velocity of the 5cars be the same as the 4cars?? can you please explain how one would go about doing this question??
 
No, it won't be the same. Use conservation of momentum. The momentum of the 4 cars is the same as the momentum of the 5 cars, but the velocity isn't because the mass has changed. In the future, try to post a new question instead of adding onto an old thread. You'll probably get a much quicker response.
 

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