A proton, moving with a velocity of v i î , collides elastically with another pr

Question

A proton, moving with a velocity of viî, 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 following

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

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

Explanation / Answer

a] Let initially moving proton be A angle with x axis, and rest proton be B angle below x axis and their speed be 2.4v and v respectively.

by energy conservation, 0.5 m vi^2 = 0.5 mv^2 + 0.5 m (2.4v)^2

vi^2 = 6.76v^2

vi= 2.6v

speed of initially moving proton = vi*2.4/2.6 = 0.923 vi

speed of proton at rest = vi/2.6 = 0.385 vi

b] By momentum conservation in both axes,

m*vi = m 0.923 vi cos A + m*0.385 vi cos B and. 0 = m 0.923 vi sin A - m*0.385 vi sin B

1 = 0.923 cos A + 0.385 cos B and 0 = 0.923 sin A + 0.385 sin B

A = 22.62 degree

B = -67.4 degree

= 22.62   ° relative to the +x direction

initially at rest proton = - 67.4 degree relative to the +x direction

= 22.62   ° relative to the +x direction

initially at rest proton = - 67.4 degree relative to the +x direction

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