Question

A proton, moving with a velocity of vi, collides elastically with another proton that is initially at rest. Assuming that after the collision the speed of the initially moving proton is 2.90 times the speed of the proton initially at rest, find the following.

(a) the speed of each proton after the collision in terms of vi

I already figured out part a which the answer is:
initially moving proton: 0.94537 ✕ vi
initially at rest proton: 0.32599 ✕ vi

(b) the direction of the velocity vectors after the collision (assume that the initially moving proton scatters toward the positive y direction)

initially moving proton____° relative to the +x direction
initially at rest proton_____° relative to the +x direction

Part B is what I need help with.

Answer 1

Assuming your part a is correct, now consider momenetum conservation.
If vi is called the x direction, then momentum is conserved
mvi=m(.94537)vi cosTheta11+m(.32599)vi*cosTheta2
and momentum in the perpendiuclar y directioni is conserved
0=m(.94537)vi*sinTheta1+m*.32599Vi*sinTheta2

solve for theta1, theta2,angles with respect to the x direction.

Answer 2

I do not know how to solve this. I tried.

Answer 3

well ur stupid

Answer 4

well too bad for u