Elastic collision in 2 dimensions

[Necropolitan]

[SOLVED] Elastic collision in 2 dimensions

Homework Statement

A ball with mass m and initial velocity v_1i collides elastically with a second ball of the same mass that is initially at rest. After the collision, the first ball moves away at an angle of θ₁ with respect to the initial velocity. What is the kinetic energy of the first ball after the collision, with respect to the variables m, v_1i, and θ₁?

Homework Equations

v_1i - v_1f*cos(θ₁) = v_2f*cos(θ₂)
v_1f*sin(θ₁) = v_2f*sin(θ₂)
v_1i² = v_1f² + v_2f²
KE_1f = 0.5*m*v_1f²

The Attempt at a Solution

I know I have enough information to solve for the final kinetic energy of the first ball, but I'm stumped on how to isolate the unknown variables from the system of equations.
So far I have:
v_1i² = v_2f²*((sin²(θ₂) / sin²(θ₁)) + 1)
tan(θ₂) = v_1f*sin(θ₁) / (v_1i - v_1f*cos(θ₁))
But I don't know where to go from here.

EDIT: I squared the first and second equations and added them together, so now I have
v_1i² - 2*v_1i*v_1f*cos(θ₁) + v_1f² = v_2f²

I substituted using the kinetic energy balance equation, did the algebra, and came up with
v_1i² - 2*v_1i*v_1f*cos(θ₁) + v_1f² = v_1i² - v_1f²

Simplifying this gave:
v_1f = v_1i*cos(θ₁)

Therefore the kinetic energy of ball 1 after the collision is
KE_1f = 0.5*m*(v_1i*cos(θ₁))²

 

[tiny-tim]

Welcome to PF!

Hi Necropolitan! Welcome to PF!

EDIT: I squared the first and second equations and added them together, so now I have
v_1i² - 2*v_1i*v_1f*cos(θ₁) + v_1f² = v_2f²
Now I'll try to see if this gets me anywhere.

Yup, that should do it!

 

[Necropolitan]

Thanks tiny-tim!