# Puck momentum problem

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1. Nov 1, 2016

### ChrisBrandsborg

1. The problem statement, all variables and given/known data
A hockey puck of mass M hits two other, identical pucks of mass m. The two pucks fly off with the same speed vf at angles of ±θ relative to the direction the original puck was travelling (see figure). The original puck had initial speed vi, and the two other pucks were initially at rest. We will assume that the pucks are sliding frictionlessly on the ice.

a) If the collision is elastic and the first puck ends up at rest after the collision, what is the final speed vf of the two other pucks and the angle θ?

b) What relation must one require between m and M in order for the scenario of a) to be possible? What happens if this requirement on m and M is not met?

For each question, provide an algebraic expression in terms of (some or all of) M, m, vi.

2. Relevant equations
Ek: 1/2mvi2 = 1/2mvf2

3. The attempt at a solution
a) Ek conserved:

1/2Mvi2 = 2(1/2mvf2)

vf = √(Mvi)2/2m

How do I find the angle?

Px = 0 -> 0 = 2mvf⋅sinθ
Py = 0 -> Mvi = 2mvf⋅cosθ

I know the angle should be 45°, but can't figure out exactly how to get there.

b) 1/2Mvi = 2(1/2mvf)

Make an equation for M:

M = 2m (vf2/vi2)

Which means that if this requirement isn't met, than its not elastic and K is not conserved.

Last edited: Nov 1, 2016
2. Nov 1, 2016

### PeroK

This is correct. Almost: I didn't notice those brackets.

This last equation is not what they are looking for. This is simply energy conservation. Instead, you need to combine this with an equation you get from conservation of momentum.

3. Nov 1, 2016

### ChrisBrandsborg

Mvi = 2 (mvf) ? That is conservation of momentum?

4. Nov 1, 2016

### PeroK

No. Momentum is a vector. It has $x$ and $y$ components. As you had in your first post.

5. Nov 1, 2016

K.E. = ½mv²

6. Nov 1, 2016

### PeroK

Yes, I didn't look closely enough. I've edited my original response.

7. Nov 1, 2016

### ChrisBrandsborg

Why isn't the first one correct?

8. Nov 1, 2016

### PeroK

You seem to be struggling with latex a bit. Look at post #5. You've defined KE with $v$ rather than $v^2$. Then you get $(Mv)^2$ rather than $Mv^2$ and finally you seem to get the right answer for $M$, which all suggests you may just be mistyping.

9. Nov 1, 2016

### ChrisBrandsborg

Oh yeah, I see that I´ve been mistyping.. I meant v2

10. Nov 1, 2016

### PeroK

Okay, just to take stock. You have, from conservation of energy:

$v_f^2 = \frac{Mv_i^2}{2m}$

Now, conservation of momentum?

11. Nov 1, 2016

### ChrisBrandsborg

Yes, and from conservation of momentum:

Px: 0 = 2mvf⋅sinθ
Py: Mvi = 2mvf⋅cosθ

12. Nov 1, 2016

### PeroK

The first equation is clearly not correct. That implies $\theta = 0$. Remember that momentum is a vector so you must respect the direction the object is moving. The first equation is not needed for this question anyway. But, you need to understand why it's wrong.

What can you do with the second?

13. Nov 1, 2016

### ChrisBrandsborg

Do I put my equation for vf into my two equations for momentum?

14. Nov 1, 2016

### PeroK

The first equation is of little interest in this problem. Focus on the second.

15. Nov 1, 2016

### ChrisBrandsborg

Oh, I thought since they are moving only in y-direction, the x equation has to be = 0 (but I guess it´s just = 0), not 0 = 2mvfsinθ ?
Hmm.. what I can I do with the second one.. Hmm. Can I put $v_f = √\frac{Mv_i^2}{2m}$ in for vf ?

16. Nov 1, 2016

### ChrisBrandsborg

Oh okay.. hmm

17. Nov 1, 2016

### ChrisBrandsborg

Do I need something else? Or can I just divide by 2mvf to get cosθ alone?

18. Nov 1, 2016

### PeroK

Sounds like a plan!

19. Nov 1, 2016

### PeroK

Why not just do it and see?

20. Nov 1, 2016

### ChrisBrandsborg

I thought you told me that the first equation wasn't relevant for finding the angle?