Momentum (I think)

1. Oct 26, 2008

Chrisleo13

1. The problem statement, all variables and given/known data
Two pucks collide on an air hockey table. Puck A has a mass of 21 g and is initially traveling in the +x direction at 6.2 m/s. Puck B has a mass of 84 g and is initially at rest. After the pucks collide, puck A moves away at an angle of 49 above the +x axis, while puck B travels at an angle of 42 below the +x axis.

a. Calculate puck A's final speed.
b. Calculate puck B's final speed.
c. What fraction of kinetic energy was lost in this collision?

I really have no clue how to start this question. Thanks in advance.

2. Oct 26, 2008

Sheneron

Well you know that momentum is going to be conserved. So the initial momentum is going to equal the final momentum of the 2 pucks. You will also have to break it down into the x and y directions.
So the initial momentum in the x direction will equal the final momentum in the x direction. Same for the y. You have the masses, and the initial speed, and I surmise that you will end up with 2 equations and 2 unknowns, which means its cake.

3. Oct 27, 2008

Chrisleo13

Thanks for the help, but I still have the equations wrong I guess. I understand what to do, but I guess I am setting them up wrong. If you could help me with that I would appreciate it.

Thanks

4. Oct 28, 2008

Sheneron

$$p = mv$$

$$p_{i} = p_{f}$$ In both x and y.

For x:
$$m_{1}v_{1ix} = m_{1}v_{1fx} + m_{2}v_{2fx}$$
$$m_{1}v_{1i}cos(\Theta) = m_{1}v_{1f}Cos(\Theta) + m_{2}v_{2f}Cos(\Theta)$$

For y:

$$m_{1}v_{1iy} = m_{1}v_{1fy} + m_{2}v_{2fy}$$
$$m_{1}v_{1i}sin(\Theta) = m_{1}v_{1f}Sin(\Theta) + m_{2}v_{2f}sin(\Theta)$$
$$0 = m_{1}v_{1f}Sin(\Theta) + m_{2}v_{2f}sin(\Theta)$$

0 because there is no initial velocity in the y direction.

So you have two equations there, and two unknowns. Solve one equation in terms of something, and plug that into the other equation in order to only have 1 unknown.

Make sure to keep track of your subscripts in order to not get confused. You don't want to add masses or velocities if they are for 2 different objects.

5. Oct 28, 2008

Chrisleo13

Hmm, I should solve for something in the the X direction because in the Y direction Vi = 0.

Are there two separate ones for each Puck? For the final speed of A and B?

6. Oct 29, 2008

Sheneron

It doesn't matter that Vi is 0 in the y direction equation. You are solving for Vf. The only two things you don't know are V1f and V2f. You have two equations though. Solve one of them for one of the Vf's and plug it into the other equation to get rid of the two unknowns in that equation.

7. Oct 29, 2008

Chrisleo13

Alright, I think I solved one equation right to plug into the other.

I solved the first one

m1vix = m1vfx1 + m2vf2

vix = m1vfx1 + m2vf2 / m1

Am I on the right track?

8. Oct 29, 2008

Sheneron

Yes and no. You did indeed solve for one of the knowns, but you did so incorrectly, and you dropped a couple of subscripts you need to keep track of, and where did the cosines go? Stay organized. Plus why would you solve your equation for vix when that is already one of your knowns? Solve it for one of your unknowns.

$$m_{1}v_{1i}cos(\Theta) = m_{1}v_{1f}Cos(\Theta) + m_{2}v_{2f}Cos(\Theta)$$

$$\frac{m_{1}v_{1i}cos(\Theta) - m_{2}v_{2f}Cos(\Theta)}{m_{1}Cos(\Theta)} = v_{1f}$$

Last edited: Oct 29, 2008
9. Oct 29, 2008

Chrisleo13

I plugged that equation into the first equation and when I canceled everything, I got m2v2cos(theta)s to cancel because 1 is negative and 1 is positive.

10. Oct 30, 2008

Sheneron

I have pretty much helped you the most possible extent so this will more than likely be my last post. It seems you are missing some fundamental concept and you may just need to review things before you try to solve this. I will explain this one more time:

You have two equations. One for the x direction and one for the y direction.
1)
$$m_{1}v_{1i}cos(\Theta) = m_{1}v_{1f}Cos(\Theta) + m_{2}v_{2f}Cos(\Theta)$$

2)
$$0 = m_{1}v_{1f}Sin(\Theta) + m_{2}v_{2f}sin(\Theta)$$

You also have two unknowns, V1f and V2f. You solve one equation down to one of your unknowns, and then plug what you just solved for into the unknown of the other equation. Then you can get an answer for that unknown.

11. Oct 30, 2008

Chrisleo13

I got it, thanks. I was plugging in the wrong mass. Thanks a lot though.

12. Oct 30, 2008