# Homework Help: 2D elastic collision

1. Aug 5, 2016

### amanda.ka

1. The problem statement, all variables and given/known data
A 2D elastic collision:
Two pucks (masses m1 = 0.5 kg and m2 = 0.3 kg) collide on a frictionless air-hockey table. Puck 1 has an initial velocity of 4 m/s in the positive x direction and a final velocity of 2 m/s in an unknown direction, θ. Puck 2 is initially at rest. Find the final speed of puck 2 and the angles θ and φ.

I get stuck at the end of this problem when I have to use the two equations to solve for 2 unknown angles. If someone could show me how to do that last step that would be great. Thanks in advance!
2. Relevant equations

3. The attempt at a solution
Since the collision is elastic I found the final kinetic energy using Ei = Ef and it equals 4.47 m/s

Then conservation of the x and y components of total momentum:

X DIRECTION: Pi = Pf
m1v1 + m2v2i = m1v1 + m2v2f
(0.5)(4) = (0.5)(2cosθ) + (0.3)(4.47cosφ)
2 = cosθ + 1.341cosφ

Y DIRECTION: Pi = Pf
m1v1 + m2v2i = m1v1 + m2v2f
0 = (O.5)(2sinθ) -(0.3)(4.47sinφ)
0 = sinθ - 1.341sinφ

2. Aug 5, 2016

### TSny

Hint: cos2θ + sin2θ = 1

(Your work looks good so far.)

3. Aug 5, 2016

### amanda.ka

So I solve the first equation for cosφ and the second equation for sinφ?
1.341cosφ = 2-cosθ
and
-1.341sinφ = sinθ

then square each term:
1.3412cos2φ = 22-cos2θ
and
-1.3412sin2φ = sin2θ

then I added them but I'm still a bit lost/don't know the next step :/

1.3412cos2φ + (-1.341)2sin2φ = 22-cos2θ + sin2θ

4. Aug 5, 2016

### TomHart

Take another look at (2-cosθ)2. It doesn't look right. Also, I thought it might have been a little simpler to solve for sinθ and cosθ, then square, then add the equations and then implement TSny's hint.

5. Aug 5, 2016

### TSny

Note that (2 - cosθ)2 ≠ 22-cos2θ. In general (a + b)2 ≠ a2 + b2.

In order to use the trig identity cos2θ + sin2θ = 1, you could solve the first equation for cosθ. You already have an expression for sinθ, except I believe you have a sign error in -1.341sinφ = sinθ.