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 2.00 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 .8944 (Correct) vi initially at rest proton .4472 (Correct )

(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 63.58 (Incorrect: Your answer is incorrect.)° relative to the +x direction initially at rest proton ° relative to the +x direction

Explanation / Answer

b) momentum in y direction has to be zero

0.8944 sin theta = 0.4472 sin alpha...........equation 1

momentum in x direction should be mvi or take it as 1 unit

1 = 0.8944 cos theta + 0.4472 cos alpha....equation 2

solving eq 1,

sin alpha = 2 sin theta

cos alpha = sqrt [ 1 - 4 sin^2 theta] = sqrt [1-4 (1-cos^2 theta)] = sqrt[1-4+4 cos^2 theta] = sqrt[4cos^2 theta -3]

putting this value in equation 2,

1 - 0.8944 cos theta =0.4472 sqrt[4cos^2 theta -3]

let cos theta = x

1- 0.8944x=0.4472 sqrt[4x^2 -3]

(1- 0.8944x)^2=0.4472^2[4x^2 -3]

x = 0.89443

cos theta = 0.89443

theta = 26.56 degree

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.