# Glancing Elastic Collison

1. Apr 18, 2014

### Fizics14

1. Two masses, m and M are involved in a glacing collision as seen below where θ and ø= pi/2.
If M = nm what must n be such that the collision is elastic?

Remember if θ+ø=pi/2 then cos(θ)=sin(ø) and cos(ø)=sin(θ)

2. I am suppose to find an number for n.

3. ∑KEo=∑KEf
1/2m$_{1}$v$^{2}_{0}$+0=1/2m$_{1}$v$^{2}_{f1}$+1/2m$_{2}$v$^{2}_{f2}$

substitute m$_{1}$n for m$_{2}$ and cancel the 1/2m$_{1}$

v$^{2}_{o1}$=v$^{2}_{f1}$+nv$^{2}_{f2}$
n=$\frac{v^{2}_{o1}-v^{2}_{f1}}{v^{2}_{f2}}$

Not sure what to do from here please help. I know I'm probably suppose to use the θ and ø, but I'm not sure how to incorporate it.

1. The problem statement, all variables and given/known data

Known: M=nm, θ+ø=pi/2
Unknown: n

2. Relevant equations

W$_{NC}$=ΔKE+ΔPE
KE=$\frac{1}{2}$mv$^{2}$
Momentum=∑p=mv$_{f}$-mv$_{o}$

3. The attempt at a solution

PS sorry I kinda messed this up it is my first post.

Last edited: Apr 18, 2014
2. Apr 18, 2014

### Fizics14

3. Apr 18, 2014

### Andrew Mason

From conservation of energy:

mv02 = nmv12 + mv22

which can be rewritten:

(1) $v_0^2 = nv_1^2 + v_2^2$

Now draw the vectors for the initial and final momenta of each body. What is the relationship between the initial total momentum and the final total momentum?

Since the final momentum vectors are at right angles to each other, what is the length of the sum of those two vectors (i.e. the magnitude of the final total momentum)? How is that length related to the length of the initial momentum vector?

Now compare that relationship with (1).

AM

4. Apr 19, 2014

### Fizics14

The momentum should be equal since it is elastic so I'm assuming the vector before the collision should be equal to the vectors after the collision right? So I solved for two different equations. Separated them by x and y components. I'm not sure what to solve for now. I really need help I have to present this to the class on Monday!

5. Apr 19, 2014

### dauto

while that method will work, it is easier to follow Andrew Mason's advice and draw the diagram representing the addition of the two vectors (the momenta of the two objects after collision). Every diagram representing the addition of two vectors is a triangle. For this problem that triangle turns out to be a square triangle which allows you to use Pythagoras theorem to relate the momenta of the particles before and after the collision.

6. Apr 19, 2014

### Fizics14

How does finding that vector help me solve for n?

7. Apr 19, 2014

### dauto

It will give you another equation. right now you have only one equation $v^{2}_{o1}=v^{2}_{f1}+nv^{2}_{f2},$ but that isn't enough because you don't know the speeds.

8. Apr 19, 2014

### Fizics14

Okay so is this correct? Can you square both sides of that equation?
Here is my attempt and now I'm stuck...

9. Apr 19, 2014

### dauto

You're almost there. Your triangle looks right, but you labeled it with speeds v. speed isn't a conserved quantity. you should've labeled the triangle with momenta p where p = mv. Momentum is a conserved quantity.

10. Apr 19, 2014

### Fizics14

When I change it to momentum I think it just solves for the equation we already have right?
By the way thank you so much for helping!!! I'm sorry I'm so bad at this :/

Last edited: Apr 19, 2014
11. Apr 19, 2014

### dauto

No, not right. You get a different equation. Do it.

12. Apr 19, 2014

### Fizics14

My attempt:

13. Apr 19, 2014

### dauto

Yes, that's the equation. Now, what can be canceled out of that equation at the bottom of the page?

14. Apr 19, 2014

### Andrew Mason

Ok. So what value of n satisfies that relationship:

$v_{f1}^2 + (nv_{f2})^2 = (v_{f1}^2 + nv_{f2}^2)$

AM

Last edited: Apr 19, 2014
15. Apr 19, 2014

### Fizics14

Is it just 1?

16. Apr 19, 2014

### dauto

ding ding ding!!!

17. Apr 19, 2014

### Fizics14

Sweet haha thanks you guys so much!

18. Apr 19, 2014

### Andrew Mason

Very good. Is that a guess? Work it out algebraically to show that n = 1 and then we will give you full marks.

AM

19. Apr 19, 2014

### dauto

There is another solution. Can you see it?

20. Apr 19, 2014

### Fizics14

I'm guessing zero so I should probably put both