A proton, moving with a velocity of v_ii^, collides elastically with another pro

Question

A proton, moving with a velocity of v_ii^, collides elastically with another proton that is initially at rest. Assuming that after the collision the speed of the initially moving proton is 1.30 times the speed of the proton initially at rest, find the following. the speed of each proton after the collision in terms of v_i initially moving proton initially at rest proton the direction of the velocity vectors after the collision (assume that the initially moving proton scatters toward the positive y direction) initially moving proton relative to the +x direction initially at rest proton relative to the +x direction

Explanation / Answer

given that

initial velocity of moving proton V1i = Vi i^

mass of both the protons = m

initial velocity of another proton is V2i = 0

Let velocity of resting proton after collision V2f.

so velocity of moving proton after collision V1f = 1.30 *Vf

by law of conservation of momentum

m1i*V1i +m2i*V21 = m1f*V1f + m2f*V2f

mVi + 0 = m*(1.30 Vf) + m*Vf

Vi = 2.30 Vf

Vf = Vi /2.30

velocity of resting proton after collision V2f = Vi/2.30

so velocity of moving proton after collision V1f = 1.30 *Vf = Vi*1.30 /2.30

V1f = 0.56* Vi

part (b)

the direction of velocity vectors after the collision

initially moving proton 90 degree relative to the +x direction .because it is given that moving partical scatters toward positive Y direction.

initially at rest partical 0 degree relative to the +x diretcion .because this partical moves in the the same direction in which the moving partical is move before collision .

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