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 2.90 times the speed of the proton initially at rest, find the following (a) the speed of each proton after the collision in terms of v initially moving protorn 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 Need Help? Read It

Explanation / Answer

before collsion

m1 (moving proton) = m

v1i = vi i

m2(rest proton) = m

v2i = 0

after collision

v1f = 2.9*v2i j = 0

v2f = ?

from momentum conservation

m1*v1i + m2*v2i = m1*v1f + m2*v2f

m1*vi i + 0 = m1*0 j + m2*v2f

m1 = m2

vi i = v2f

v2f = vi i

v2f = vi

part(a)

initially moving proton = 0 vi

initially at rest proton = 1 vi

part(b)

intially moving proton = 90

initiallly at rest proton = 0

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