physics question (i don't understand how to do this problem:( )

Question

A proton of mass M collides obliquely with another proton. The first proton is moving with a speed of 6.0 x 106 m/s before it hits the second, stationary proton. Assuming the collision is perfectly elastic (no kinetic energy is lost) and using the fact that the first proton is moved 30° from its initial path after the collision, figure out the speed and direction of each proton after the collision. Note that after the collision the two protons travel at 90˚ to each other.

(answer: Struck proton 3.0 x 106 m/s, 60° from the incident proton’s original direction; incident ball 5.2 x 106 m/s)

Answer 1

As this is an elastic collision, both momentum and kinetic energy are conserved.

The total momentum of the protons before collision = MV₁ + M₂ (0) = MV₁. The initial kinetic energy is

1/2MV₁². Assume it is traveling in the positive x direction,

After collision, the first proton is traveling at a velocity V_{1 post} in a direction 30° to the positive x direction (positive angle). The second proton would be in a direction 90° from the post-collision direction of proton #1 or at -60°.

Now calculate the velocity components of each proton is the x and y directions after collision.

V_{1post x} = V1post x cos30 and V_{1post y} = V1post x sin 30.

V_{2 post x} = V2post x cos-60 and V_{2post y} = V2post x sin -60. V1post and V2post are the magnitudes of the velocity vectors after collision.

Equate the x and y components before collision to the x and y components after collision.

V1x_pre = V1post x cos30 + V2post x cos-60

0 = V1post x sin 30 + V2post x sin -60

The last two equations each have two unknowns, V1post and V2post which can be solved simultaneously.

Multiply the first equation by sin30 and the second by -cos30 then add to the first equation after substituting the values for the trig functions.

The math is messy, but I did get V1post = 3 x 10⁶ m/s and V2post = 5.2 x 10⁶ m\s.