Angles in Two-Dimensional Elastic Collision

[axcess]

Homework Statement

The setup is there are two pucks (m_1, 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 = ?

Homework Equations

K_i = (1/2)m₁v_1i²
K_f = K_i
K_f = (1/2)m₁v_1f² + (1/2)m₂v_2f² = K_i

The Attempt at a Solution

First I went about finding what v₂ was by using K_f = K_i

(1/2)(0.028)(0.785)² + (1/2)(0.102)v_2f² = (1/2)(0.028)(1)²
(0.008) + (1/2)(0.102)v_2f² = (0.014)
v_2f² = (0.117)
v_2f = 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 = m₁v_1fsin\theta - m₂v_2fsin\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 90^o 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!

 

[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

 

[ideasrule]

You can define your coordinate axes so that the moving puck is initially traveling 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.