Five train cars and their final speed

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SUMMARY

The discussion centers on a physics problem involving the conservation of momentum to determine the final speed of a five-car train. Initially, three train cars are moving at 1.13 m/s, and a fourth car at 6.43 m/s couples with them. The correct approach involves using the equation for momentum conservation: 3m*(1.13 m/s) + m*(6.43 m/s) = 4m*v_final, leading to a final speed of 7.56 m/s for the four-car train. The same principle applies when the fifth car, initially at rest, couples with the four-car train.

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brunettegurl
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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 :))
 
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Looks like this exact same problem was recently discussed here:

https://www.physicsforums.com/showthread.php?t=155891

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
 

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