A proton, moving with a velocity of vi, collides elastically with another proton

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 2.40 times the speed of the proton initially at rest, find the (a) the speed of each proton after the collision in terms of v initially moving proton initially at rest proton (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 initially at rest proton o relative to the +x direction o relative to the +x direction

Explanation / Answer

a] By energy conservation,

0.5 m vi ^2 = 0.5m (2.4vf)^2 + 0.5m vf^2

vi^2 = 5.76 vf^2 + vf^2 = 6.76 vf^2

vi = 2.6 vf

vf = 0.3846 vi

initially moving proton's speed = 2.4*0.3846 vi

= 0.923 Vi

initially at rest proton's speed = 0.3846 vi

b] by momentum conservation, mvi = m 0.923vi cos theta1 + m 0.3846 vi cos theta2

1 = 0.923 cos theta1 + 0.3846 cos theta2

and also 0 = 0.923 sin theta1 + 0.3846 sin theta2

for moving proton theta1 = 22.62 degree,

for proton at rest, theta2 = -67.37 degree

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