# Elastic Collision in two dimensions question

## Homework Statement

[/B]A 2.0 kg ball moving with a speed of 3.0 m/s hits, elastically, an identical stationary ball as shown (The first ball is travelling to the right). If the first ball moves away with angle 30° to the original path, determine
1. the speed of the first ball after the collision.
2. the speed and direction of the second ball after the collision.

## Homework Equations

m1v1 + m2v2 = m1v1' + m2v2'
1/2 m1v1^2 + 1/2m2v2^2 = 1/2 m1v1'^2 + 1/2 m2v2'^2

## The Attempt at a Solution

[/B]
So far I understand that the conservation of momentum and energy can be used, however with the amount of different ways these formulas can be used I don't know where to go with them. Does anyone have any advice to help break this down algebraically?

haruspex
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## Homework Statement

[/B]A 2.0 kg ball moving with a speed of 3.0 m/s hits, elastically, an identical stationary ball as shown (The first ball is travelling to the right). If the first ball moves away with angle 30° to the original path, determine
1. the speed of the first ball after the collision.
2. the speed and direction of the second ball after the collision.

## Homework Equations

m1v1 + m2v2 = m1v1' + m2v2'
1/2 m1v1^2 + 1/2m2v2^2 = 1/2 m1v1'^2 + 1/2 m2v2'^2

## The Attempt at a Solution

[/B]
So far I understand that the conservation of momentum and energy can be used, however with the amount of different ways these formulas can be used I don't know where to go with them. Does anyone have any advice to help break this down algebraically?
Remember that momentum is vector, so it is conserved in all directions. That should give you two equations, plus the energy equation.
Pick an xy system of coordinates. Either create separate variables for velocities in different directions (x and y) or introduce variables to represent angles. Then rewrite the relevant equations using these.

The formulas I have come up with are

- v1'sin theta = v2'sin theta
v1 - v1'cos theta = v2'cos theta

and
v1^2 - V1'^2 = V2'^2

I am unsure of how I join these equations though, I know I need to square the momentum equation and sub them into the kinetic equation but I'm not quite sure how, do I need to divide over the cos' and sin's to get them on the same side or what? (v2' = -v1'sin theta/sin theta)

- v1'sin theta = v2'sin theta
v1 - v1'cos theta = v2'cos theta
You are using the same angle, θ, for V1' and V2'. It was given in the problem statement that the angle of V1' is 30°.

Edit: Welcome to Physics Forums. Except for using the same angle θ for both V1' and V2', I think your equations are correct.

Ok, I am still unsure of how to continue on this question though, I don't really even know what I'm trying to do next (get rid of the angles?) but I don't know how to join the equations properly and it seems no matter what algebra I do I just end up going in a circle.

I don't really even know what I'm trying to do next (get rid of the angles?)
There is one unknown angle, θ, which is the angle of V2'. The V1' angle is 30°.

I remember solving this type of problem before, and yes, it is kind of a tricky solution. I am heading out the door in about 5 minutes so don't have time right now. But it definitely involves one or two trig. identities. The first one may involve sin2 + cos2 = 1, but I don't remember for sure.

haruspex