# Angles in Two-Dimensional Elastic Collision

1. Nov 13, 2009

### axcess

1. The problem statement, all variables and given/known data
The setup is there are two pucks (m1, m2) on an air table and they undergo elastic collision while m2 is stationary. I am asked to find the angle between the two velocities but I can't figure out how without being giving at least one angle.

Variables:
m1 = 28.0 g = 0.28 kg
m2 = 102 g = 0.102 kg
v1 = 0.785 m/s
v2 = ?

2. Relevant equations
Ki = (1/2)m1v1i2
Kf = Ki
Kf = (1/2)m1v1f2 + (1/2)m2v2f2 = Ki

3. The attempt at a solution
First I went about finding what v2 was by using Kf = Ki

(1/2)(0.028)(0.785)2 + (1/2)(0.102)v2f2 = (1/2)(0.028)(1)2
(0.008) + (1/2)(0.102)v2f2 = (0.014)
v2f2 = (0.117)
v2f = 0.343 m/s

Now this is the part I'm stuck, since they want me to find the angle between v1 and v2. If they had given me at least an angle for m1 I could use [0 = m1v1fsin$$\theta$$ - m2v2fsin$$\theta$$] but I have no angles to work with, and I'm not sure how I should arrange the equations to find the angle. If I were to guess I would say that they angle would be 90o since m2 was stationary, but I'm not sure how to prove that and I don't think it applies to this case since I have two different masses.

If anyone could help me understand this problem, that would be great!

2. Nov 13, 2009

### ApexOfDE

Denote the angles of 2 v after collision along x-axis is alpha and beta. Then you write down conservation of momentum in x and y-axis. You will have 2 equations that contains cos, sin. Do some stuff and you can find 2 angles :D

3. Nov 13, 2009

### ideasrule

You can define your coordinate axes so that the moving puck is initially travelling along the x-axis. You can then assume its angle (to the x-axis) becomes theta1 after the collision while the other puck goes off at theta2. Now there are three unknowns, v2f, theta1, and theta2, and three equations: 2 for momentum and 1 for energy.