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**[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 θ

_{1}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 θ

_{1}?

## Homework Equations

v

_{1i}- v

_{1f}*cos(θ

_{1}) = v

_{2f}*cos(θ

_{2})

v

_{1f}*sin(θ

_{1}) = v

_{2f}*sin(θ

_{2})

v

_{1i}

^{2}= v

_{1f}

^{2}+ v

_{2f}

^{2}

KE

_{1f}= 0.5*m*v

_{1f}

^{2}

## 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

_{1i}

^{2}= v

_{2f}

^{2}*((sin

^{2}(θ

_{2}) / sin

^{2}(θ

_{1})) + 1)

tan(θ

_{2}) = v

_{1f}*sin(θ

_{1}) / (v

_{1i}- v

_{1f}*cos(θ

_{1}))

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

_{1i}

^{2}- 2*v

_{1i}*v

_{1f}*cos(θ

_{1}) + v

_{1f}

^{2}= v

_{2f}

^{2}

I substituted using the kinetic energy balance equation, did the algebra, and came up with

v

_{1i}

^{2}- 2*v

_{1i}*v

_{1f}*cos(θ

_{1}) + v

_{1f}

^{2}= v

_{1i}

^{2}- v

_{1f}

^{2}

Simplifying this gave:

v

_{1f}= v

_{1i}*cos(θ

_{1})

Therefore the kinetic energy of ball 1 after the collision is

KE

_{1f}= 0.5*m*(v

_{1i}*cos(θ

_{1}))

^{2}

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