# Five train cars and their final speed

• brunettegurl
In summary, the problem involves three identical train cars rolling east at 1.13 m/s, a fourth car traveling east at 6.43 m/s that catches up and couples to make a four-car train, and a fifth car at rest on the tracks that couples to make a five-car train. The speed of the five-car train can be found using the conservation of momentum equation, with the initial and final momentums being equal. Through this calculation, the speed of the five-car train is determined to be 7.56 m/s.

## Homework Statement

Three identical train cars, coupled together, are rolling east at 1.13 m/s. A fourth car traveling east at 6.43 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

pinitial = pfinal
m1v1+m2v2 = mtotalvfinal

## The Attempt at a Solution

so this is what i did

3mV+ mV= 4mV [the masses cancel]
(1.13)+(6.43)= V
7.56 m/s = v

4mV+ mv = 5mv
Vtotal+0 = vfinal
7.56 m/s = vfinal

can someone pls tell me what I'm doing wrong thanks Looks like this exact same problem was recently discussed here:

I haven't read it, but it might help quite a bit. I also recommend you label your velocities like v1, v2, etc.

Edit: Seems you noticed it already, I see your post in there now.

You do not add the velocities. You use the conservation of momentum equation. So for the first part you would have...

3m*(1.13 m/s) + m*(6.43 m/s) = 4m*v_final

Then you solve for v_final. You do the same for the last part of the problem.

thank u sooo much i feel so foolish that was really straighforward..thanx again