# Railway/Cannon - Conservation of Momentum HW Problem

Hello forum.

I have a HW question that I don't fully grasp just yet. It was multiple choice and somehow I guessed the right answer based on the work I did complete, but I want to know how to get to the solution and which steps I'm leaving out. I'll follow the format to write out the question.

1. Homework Statement

A cannon is mounted on a railway flatcar; both are moving to the right at speed, Vo, relative to the tracks. When the cannon fires a cannonball, the cannon and flatcar recoil so they move to the left at speed, Vo. If M is the total mass of the cannon and railway car without the cannonball, and m is the mass of the cannonball, what is the speed of the cannonball relative to the tracks after it has been fired?

## Homework Equations

Momentum:
p = mv
P (initial) = P(final)

## The Attempt at a Solution

My attempt at a solution started with knowing that momentum would be conserved which meant that p(initial) could be rewritten as M(initial) * V(initial) , and p(final) would be rewritten as M(final) + Vo + m(cannonball) * v(cannonball)

So overall line by line I have:

M(i) * Vo (x^ - unit vector) = M(f)*Vo(-x - unit vector) + m(cannonball) * v(cannonball)

After solving this algebraically, I got the answer for Vo to equal (2*M(i) * Vo / m(cannonball) ), and this answer wasn't listed among the answer choices and the correct answer that I guessed on is not quite this.

I can't seem to find the step that I'm missing.

I also would love if someone could give me a more thorough understanding of what the unit vector notation is, because I keep wanting to think of that x as a variable just from dealing with so much math. But apparently in this instance, it only is an indicator of direction? And it doesn't stand for any value to be solved for?

This is giving me a headache... I appreciate anyone who can help clarify this for me.

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TSny
Homework Helper
Gold Member
Does the cannonball have any momentum before the cannon is fired?

Does the cannonball have any momentum before the cannon is fired?
The question is typed out exactly as I wrote it there. So I'm not really sure what they want me to assume regarding the cannonball BEFORE it was fired. I'm guessing the cannonball itself is at rest, but if it is in the cannon and traveling on the cart, then it would be part of the total mass of the system. so thus would have its own momentum, but I'm not totally sure.

TSny
Homework Helper
Gold Member
but if it is in the cannon and traveling on the cart, then it would be part of the total mass of the system. so thus would have its own momentum
Yes. It is natural to assume that the ball would be at rest with respect to the cannon just before it is fired. So, the ball, cannon, and cart would all be moving together relative to the ground just before firing.

goodness, i feel like an idiot smh.

This was bothering me all night and into today, and now as I read it I see where I was confused. When they said the TOTAL MASS WITHOUT THE CANNONBALL represented by M, I assumed that meant the total mass of the entire system. But now as I read it, it indicates the M is the total mass of just the cannon + flatcar. m itself is just the mass of the cannonball. So both masses TOGETHER (M + m) would be part of the system both before and after the firing. And I think I forgot to include that as part of the p(initial) calculation.

So really the P(initial) should be = M (cannon+flatcar) * Vo + m(cannonball) * Vo

I didn't include the m(cannonball) * Vo part. I think that may have been the mistake.

By the way, when it comes to the x^ symbol to represent the unit vector, was I right in saying that this isn't like an x variable that we're solving for in normal mathematics equations? I get confused by the notation sometimes. Can it only be +x^ or -x^ to just indicate the direction of the velocity vector (east or west)?

TSny
Homework Helper
Gold Member
So really the P(initial) should be = M (cannon+flatcar) * Vo + m(cannonball) * Vo

I didn't include the m(cannonball) * Vo part. I think that may have been the mistake.
Yes. Good.

TSny
Homework Helper
Gold Member
By the way, when it comes to the x^ symbol to represent the unit vector, was I right in saying that this isn't like an x variable that we're solving for in normal mathematics equations?
Yes, that's right.
I get confused by the notation sometimes. Can it only be +x^ or -x^ to just indicate the direction of the velocity vector (east or west)?
Yes.

Thanks so much. I appreciate your help!

TSny
Homework Helper
Gold Member
You're welcome.