# Momentum Question

1. Oct 30, 2005

### amcavoy

Here is a link to the image: http://img500.imageshack.us/img500/2681/171812nv.gif

I found the initial momentum to be $\left(50\cos{30^o},-50\sin{30^o}\right)$ and the final momentum to be $\left(50\cos{25^o},50\sin{25^o}\right)$. To find the change, I just subtracted these, but the answer is incorrect. Note: The method I used to find the initial and final vectors was to multiply the magnitude of the velocity by sine and cosine respectively, and then by 2 (the mass).

Can someone give me a hint as to where I went wrong?

Thank you.

2. Oct 30, 2005

### Chi Meson

What values did you get when you subtracted these momenta?

(Did you subtract the negative)

3. Oct 30, 2005

### amcavoy

Yes, I did. This is what I came up with (rounded off):

$$\Delta\vec{p}=\left<2.01,46.1\right>\text{N}-\text{s}$$

...I hope it's just an arithmetic error. Please let me know what you think if you get a chance.

Thanks again!

4. Oct 30, 2005

### Erwin Schrodinger

∆px = px-px = 50cos25[E]-50cos30[E] ≈ 2.014[E]

∆py = py-py = 50sin25[N]-50sin30 = 50sin25+50sin30[N] ≈ 46.131 [N]

Now you know the horizontal and verticial components of the impulse, but you need to add the vectors together to find the total impulse. Drawing these 2 vectors will form a right triangle in which you can use pythagorean theorem to find the total impulse.

Last edited: Oct 30, 2005
5. Oct 30, 2005

### amcavoy

I need to find the vector change though, not the magnitude.

6. Oct 30, 2005

### Erwin Schrodinger

You mean the horizontal and vertical changes in momentum? Your answer looks correct then. What answer does your book give?

7. Oct 30, 2005

### amcavoy

I don't have an answer for it, but it does give me a hint. It says that the impulse delivered to the mass by the plate only has a y-component. Does this mean I can get rid of my x component and use the y? For this problem, does that hint mean I would use (0,46.1) instead of (2.01,46.1)?

Thanks again.

8. Oct 30, 2005

### Erwin Schrodinger

Hmm, that's strange. If ∆px = 0, then px` = px but that clearly isn't the case here since the angle of incidence and the angle of deflection are not equal. I don't see how they can make that claim. I'm just as stumped as you are now. :yuck:

9. Oct 30, 2005

### amcavoy

Hmm... I found the answer, which is the one I suggested above (0,46.1). I find that strange as well, but I guess I'll just have to work with some similar problems.

Thanks a lot for your help.

10. Oct 30, 2005

### Chi Meson

Dang! I was just coming back to this to point out that the frictionless steel surface could not have changed the x-momentum of the ball.

Therefore, the final speed could not be 50 m/s because 50 cos30 has to equal 2*v'*cos 25. From this you find v'.

11. Oct 30, 2005

### Erwin Schrodinger

Oh okay, I see the problem we both made here. We assumed that the final velocity of the ball was equal to the inital velocity but the question didn't state that. Tricky...