# Perfectly elastic collisions (proof)

1. Dec 16, 2004

### Phyzix

An object collides elastically (perfectly) with another object (identical object) at rest. If it is not a head on collision how can i PROVE that the angle between them afterwards is 90 degrees?

I have NO IDEA on what to do...it's been puzzling me and my friends for a little while now...

help anyone?

2. Dec 16, 2004

### Evil_Kyo

In a elastic collision momentum and KE are conserved:

$$m_1\overrightarrow{v}_1_i=m_1\overrightarrow{v}_1_f+m_2\overrightarrow{v}_2_f$$

$$\frac{1}{2}m_1v^2_1_i=\frac{1}{2}m_1v^2_1_f+\frac{1}{2}m_2v^2_2_f$$

in this case the masses are the same so we can simplify:

$$\overrightarrow{v}_1_i=\overrightarrow{v}_1_f+\overrightarrow{v}_2_f$$

$$v^2_1_i=v^2_1_f+v^2_2_f$$

Consider now the first equation, look at the vectors $$\overrightarrow{v}_1_f$$ and $$\overrightarrow{v}_2_f$$, if they are added they are equal to $$\overrightarrow{v}_1_i$$, so you can picture the 3 of them as forming a triangle.

If you picture this you can see that $$v^2_1_i=v^2_1_f+v^2_2_f$$ is the Theorem of Pythagoras. $$\overrightarrow{v}_1_i$$ is the hypotenuse so the other two sides form 90º.

I hope the explanation is clear enough.