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

Explanation / Answer

m1 = m2 = mp (mass of proton )

v1i = vi

v2i = 0

after collison

v1f = 2.3*v2f j   ( + y direction )

v2f = ?

momentum before collison

Pi = m1*v1i + m2*v2i

momentum after collision

Pf = m1*v1f + m2*v2f

from momentum conservation

momnetum before collision = momentum after collision

Pf = Pi

mp*vi + mp*0 = mp*2.3*v2f j + mp*v2fxi + mp*v2fy j

comparing i

vi = v2fx

v2fx = vi

comparing j

0 = 2.3*v2f + v2fy

0 = 2.3*vi + v2fy

v2fy = -2.3 i

(a)

speed of moving proton = 2.3 *v i

speed of rest proton = sqrt(V2fx^2+v2fy^2) = sqrt(vi + (2.3*vi)^2 = 2.51*vi

(b)

initial moving proton = 90

initially at rest proton = tan^-1(v2fy/v2fx) = +293.5

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