# Momentum and relative velocity

Attached is P14.2, a simple conservation of momentum problem with relative velocities. I have solved the problem on the second page. My answer is wrong because the signs of the velocities are wrong. I have no idea why. I am very bad at assuming directions and being consistent but I cant figure out how I messed this one up...

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• Momentum.pdf
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Doc Al
Mentor
Looks to me like you took "to the left" as positive, yet you called the speed of the flatcar +0.34 m/s (it should be negative).

Yes I took left to be positive, why should the flatcar be negative? Its moving to the right? How do I know that? I assumed that car A, car B and the flatcar had a constant final velocities to the left, which made them all positive.

Doc Al
Mentor
Yes I took left to be positive, why should the flatcar be negative? Its moving to the right? How do I know that?
From conservation of momentum.

I assumed that car A, car B and the flatcar had a constant final velocities to the left, which made them all positive.
And that's why you get a negative mass for the flatcar, a nonsensical answer!

Usually, but not necessarily the sign convention is that velocities pointed left are NEGATIVE (viz. -2.55 m/s and -2.5 m/s). Velocities to the right are usually given a positive value as the problem does +.34 m/s. Using these values you will get the correct answer. Velocity is a vector quantity having magnitude and direction. In a 1-D problem, + is to the right and - is to the left, by convention. (You can reverse the convention, FOR ALL THE VELOCITIES IN THE PROBLEM, and you also get the right answer.)

Doc Al
Mentor
Usually, but not necessarily the sign convention is that velocities pointed left are NEGATIVE (viz. -2.55 m/s and -2.5 m/s). Velocities to the right are usually given a positive value as the problem does +.34 m/s. Using these values you will get the correct answer. Velocity is a vector quantity having magnitude and direction. In a 1-D problem, + is to the right and - is to the left, by convention. (You can reverse the convention, FOR ALL THE VELOCITIES IN THE PROBLEM, and you also get the right answer.)
The issue was not one of sign convention, but of assigning the correct direction to the flatcar's motion.

Doc Al, you said I know the flatcar is moving to the right from conservation of momentum...Initially I dont know any of the directions for the velocities. I assumed them and got an answer that was obviously wrong. So I know that my assumption is wrong. But how to I know from conservation of momentum that the cars are going in the oposite direction of the flatcar?

Doc Al
Mentor
Doc Al, you said I know the flatcar is moving to the right from conservation of momentum...Initially I dont know any of the directions for the velocities.
At some point, physical intuition will kick in and that will save you time. If the cars go left, the flatcar must go right. (The problem creater assumed you would know this.)
I assumed them and got an answer that was obviously wrong. So I know that my assumption is wrong. But how to I know from conservation of momentum that the cars are going in the oposite direction of the flatcar?
They start out from rest so total momentum equals zero. If everything moved in the same direction, then total momentum could not be zero!

At some point, physical intuition will kick in and that will save you time. If the cars go left, the flatcar must go right. (The problem creater assumed you would know this.)

They start out from rest so total momentum equals zero. If everything moved in the same direction, then total momentum could not be zero!

When the cars start to move, yes the momentum is zero, they are moving in no direction at all. Lets say at time t=1s, they can all start moving in the same direction, we would have: (mv)A+(mv)B+(mv)F=constant
That constant would just be a larger number than if I assumed that the flatcar was going in the other direction, in which case that same equation would look like this:
(mv)A+(mv)B-(mv)F=constant
I dont see why one is wrong and the other isnt.

What I am thinking is similiar to the following situation:
Lets say that I were on a train, sitting in my seat, at rest while the train too is at rest (lets say its at a stop). Then after a couple of minutes the train starts to move again, it moves east. I could get up from my seat and walk east too. I dont see why thats against physical intuition. The same thing for the cars, where car A and car B are me and the train is the flatcar.

Doc Al
Mentor
When the cars start to move, yes the momentum is zero, they are moving in no direction at all.
Before the cars move, everything is at rest. The momentum of flatcar + cars starts and remains zero since we presume no external force acts on the flatcar.
Lets say at time t=1s, they can all start moving in the same direction, we would have: (mv)A+(mv)B+(mv)F=constant
That constant would just be a larger number than if I assumed that the flatcar was going in the other direction, in which case that same equation would look like this:
(mv)A+(mv)B-(mv)F=constant
I dont see why one is wrong and the other isnt.
The only way everything can start moving in the same direction is if an external force acts on them. Presumably, there is no such force acting. (The tracks are frictionless.)

What I am thinking is similiar to the following situation:
Lets say that I were on a train, sitting in my seat, at rest while the train too is at rest (lets say its at a stop). Then after a couple of minutes the train starts to move again, it moves east.
The train starts to move because the tracks exert a force on it.
I could get up from my seat and walk east too. I dont see why thats against physical intuition. The same thing for the cars, where car A and car B are me and the train is the flatcar.
No, the cases are different, due to the presence of an external force. To see this more clearly, let's use a different example: You are on a rowboat floating on a calm lake. (Assume no resistive force from the water on the boat.) If you start to walk east, what happens to the boat?

Thats brilliant...thanks. I'm learning more here than I am in class.