# Collisions in Two Dimensions (Perfectly Elastic)

1. Nov 12, 2008

### calvert11

1. The problem statement, all variables and given/known data
Assume an elastic collision (ignoring friction
and rotational motion).
A queue ball initially moving at 4.3 m/s
strikes a stationary eight ball of the same size
and mass. After the collision, the queue ball’s
final speed is 2.3 m/s .

Find the queue ball’s angle with respect
to its original line of motion.

v1i = 4.3
v1f = 2.3

2. Relevant equations
p1i + p2i = p1f + p2f
KEi = KEf

3. The attempt at a solution

sqrt(4.3^2 - 2.3^2) = v2f = 3.6 m/s

4.3 = 2.3*cos θ + 3.6*cos φ (for x direction)

0 = 2.3*sin θ - 3.6*sin φ (for y direction)

2.3*sin θ = 3.6*sin φ

Not quite sure how to proceed from here.

2. Nov 12, 2008

### asleight

$$m_1u_1+m_2u_2=m_1_v_1+m_2v_2\rightarrow m(u_1+0)=m(v_1+v_2)$$.

3. Nov 12, 2008

### calvert11

I'm not sure how that helps . The masses are equal so they drop from the equations and the velocities are known.

4. Nov 12, 2008

### asleight

4.3 = 2.3 + v...

5. Nov 12, 2008

### calvert11

Still don't know what you're getting at :).

But I managed to solve it using pythagoras: (4.3-2.3 cos θ)^2 + (2.3 sin θ)^2 = 3.6^2

With that I get around 57 degrees which is correct.

Just out of curiosity, would there have been a simpler way of solving this? My solution required using power-reducing formulas for sin/cos...which seems too complicated considering the level of the course.