# Elastic collision in 2 dimensions

[SOLVED] Elastic collision in 2 dimensions

## Homework Statement

A ball with mass m and initial velocity v1i 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, v1i, and θ1?

## Homework Equations

v1i - v1f*cos(θ1) = v2f*cos(θ2)
v1f*sin(θ1) = v2f*sin(θ2)
v1i2 = v1f2 + v2f2
KE1f = 0.5*m*v1f2

## 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:
v1i2 = v2f2*((sin22) / sin21)) + 1)
tan(θ2) = v1f*sin(θ1) / (v1i - v1f*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
v1i2 - 2*v1i*v1f*cos(θ1) + v1f2 = v2f2

I substituted using the kinetic energy balance equation, did the algebra, and came up with
v1i2 - 2*v1i*v1f*cos(θ1) + v1f2 = v1i2 - v1f2

Simplifying this gave:
v1f = v1i*cos(θ1)

Therefore the kinetic energy of ball 1 after the collision is
KE1f = 0.5*m*(v1i*cos(θ1))2

Last edited:

tiny-tim
Homework Helper
Welcome to PF!

Hi Necropolitan! Welcome to PF! EDIT: I squared the first and second equations and added them together, so now I have
v1i2 - 2*v1i*v1f*cos(θ1) + v1f2 = v2f2
Now I'll try to see if this gets me anywhere.

Yup, that should do it! Thanks tiny-tim!