# Conservation of Momentum question

## Homework Statement

Three identical train cars, coupled together, are rolling east at 3.71 m/s. A fourth car travelling east at 6.74 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. All 5 cars are identical. What is the speed of the five-car train?

## Homework Equations

m1v1+m2v2+... = m1v1'+m2v2'+...

Delphi51
Homework Helper
Welcome to PF, Janfor.
I can't help much until you attempt the problem yourself (bad for your education and boring for me).

You have two collisions in this problem. You can deal with them one at a time. For a collision, you write
MOMENTUM BEFORE = MOMENTUM AFTER
Put in "mv" for each moving object, before and after.
Then put in the numbers for each m and v.
Hopefully there will only be one unknown so you can solve for it!
Bon chance.

you have the right equation, so just plug in the variables:
the momentum of the train initially is the total of all three cars plus the total of the fourth cart:
(3m)*(3.71)+(6.74m) = (4m)(V)
the reason that this is the equation is because momentum before the collision equals the momentum after the collision. Since the cars are identical, mass can be identified as a constant m. also, when the fourth train impacts the three trains, the velocity will change because the momentum is moving faster, meaning we need a new variabe, V
When solving algebraically for v, you'll find that the m's cancel and you get v=4.47
now given the new velocity of the train, do the same thing:
(4m)*(4.47) = (5m)(v)
there is no addition because there is only one thing moving before and after the collision, since car 5 is initially at rest and the cars combine afterwards
solve for v again and get v = 3.57m/s