# Homework Help: Problem Involving Momentum & Collision

1. Jun 10, 2008

### Pat2666

Here's the problem :

A ball of mass m = 2.2 kg moving with a speed of v0 = 20.1 m/s strikes an identical ball which was initially at rest. After the collision, the incoming ball (ball 1) goes off at q1 = 24° relative to its original direction and the struck ball (ball 2) moves off at q2 = 34° as shown in the diagram.

I can't post the diagram, but ball 1 shoots off 24 degrees above the x-axis and ball 2 shoots 34 degress below the x-axis.

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a) What is the magnitude of the initial momentum of the two ball system?
pi = 44.22 kg m/s

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b) What is the ratio of the balls' final speeds?
|v1|/|v2| = 1.37

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c) What is the final speed of the incoming ball?
|v1| = m/s

Okay, I have another problem that I'm stuck on! I was able to figure out part A & B (answers above), but I can't seem to get part C.

I tried setting the horizontal momentum of the two objects after the collision equal to the initial horizontal momentum (44.22) and then solving for one variable.

44.22 = m1v1 + m2v2 or 44.22 = m (v1+v2)

Then I thought I would have to use the ratio of v1/v2 in order to solve for a variable. So I did the following:

v1/v2 = 1.37 ... v2 = v1/1.37

I then plugged that into the momentum equation above and solved for the horiztonal velocity and got 11.619m/s for v1.

Then I took that number and divided it by COS(24) to get the Vf since the ball shoots off at an angle. I got 12.72m/s, but it isn't right. Where have I gone wrong? :(

2. Jun 10, 2008

### G01

OK. I think you may have made a calculation mistake while finding v_1. I am getting v_1=8.375m/s.

Check your calculations, again.

3. Jun 10, 2008

### Pat2666

I'm still getting the same answer as I have before.

And I tried plugging your value in to solve for V1final and it was still wrong. :-\

4. Jun 10, 2008

### tiny-tim

Hi Pat!

(A very well laid-out answer, by the way … I wish other people were as clear! )

Hint: You must take horizontal components of momentum!

5. Jun 10, 2008

### Pat2666

Isn't that what I was doing though? I thought v1 and v2 that I was solving for in the equation you quoted were the two horizontal component?

6. Jun 10, 2008

### G01

TT I think he means that v_1=horizontal component of momentum of the first ball.

Then hes saying:

$$v_1=v_{f1}\cos24^o$$

$$v_{f1}=\frac{v_1}{\cos24^o}$$

where v_f1 = final velocity magnitude.

Am I correct, Pat?

7. Jun 10, 2008

### Pat2666

Yeah, I think so but I thought that's what I've been doing? I mentioned that I found v1 and then divided it by Cos24, but I must be getting the wrong value for v1.

8. Jun 10, 2008

### tiny-tim

v = v1cos24º + v2cos34º

9. Jun 10, 2008

### Pat2666

Now I'm a bit more confused sorry! lol

I don't know V do I? I thought you had to add the two horizontal components of momentum and set them equal to the initial momentum of the first ball? I wouldn't know V in your equation would I?

10. Jun 10, 2008

### tiny-tim

v = 20.1​

11. Jun 10, 2008

### Pat2666

I finally got it! Thanks!

I have to ask though...do the horizontal components always equal the sum of the initial horizontal compents? I just want to understand what I did :)

And thanks again!

12. Jun 10, 2008

### tiny-tim

Yup!

Good ol' Newton's second law, and conservation of momentum, are vector equations, so they're valid on all the components in any one direction.

13. Jun 10, 2008

### Pat2666

Oh okay I understand now! Thank you so much :)