A proton, moving with a velocity of viI, collides elastically with another proto

Question

A proton, moving with a velocity of viI, collides elastically with another proton that is initially at rest. Assuming that after the collision the speed of the initially moving proton is 2.70 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, initially moving proton x vi initially at rest proton x 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 degree relative to the +x direction initially at rest proton degree relative to the +x direction

Explanation / Answer

If the proton collides eleastically , then kinetic energy is conerved. If  the mass of the protons be m and the speed of the target proton after the collision be v. Then

Given that

A proton, moving with a velocity of Vi (i vector), collides elastically with another proton that is initially at rest(vi2) =0

Then (1/2)mvi2 =(1/2)m(0.6v)2+(1/2)mv2

vi2 =(0.36+1)v2 =1.36v2

Now vi =1.16v=====> v =(1.16)vi =0.857vi

Therefore the speed of the rest proton is given by

v =0.60vi =0.60(1.16v) =0.696vi

b)

The impulses between the bodies have a sent the bodies off at angles theta1 and theta2 to x-axis, from the conservation of momentum along the x-axis is given by

m1vi =m1v1fcos(x)+m2v2fcosx.....(1)'

Mass cancel because the masses are equal

vi =v1fcos(x)+v2fcosx

Now substituting the values final and intial velocites we get

cosx =1.16/1.553 =0.746

then for the rest proton x =41.75 degrees with +x-axis

assuming the initially proton scatters toward the positive y direction then 90-41.75=48.25 degrees with +x-axis

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