# Momentum of five-car train

## Homework Statement

Three identical train cars, coupled together, are rolling east at 3.97 m/s. A fourth car travelling 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. ..is anything i did above incorrect?:grumpy:
any help will be appreciated greatly :!!)

## The Attempt at a Solution

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Dick
Homework Helper
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 wouldnt the final velocity of the 5cars be the same as the 4cars?? can you please explain how one would go about doing this question??

Dick