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 vi initially moving0.928 proton 0.372 initially Your response is within 10% of the correct value. This may be due to at rest roundoff error, or you could have a mistake in your calculation. Carry proton out all intermediate results to at least four-digit accuracy to minimize roundoff error. × v, (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 68.2 initially at rest proton 21.80 1 ° relative to the +x direction relative to the +x direction Need Help?Read It

Explanation / Answer

(A) Applying momentum conservation,

m vi + 0 = m (v1fxi + v1fy j) + m (v2fx i + v2fy j )

so, v1fx + v2fx = vi

and v1fy + v2fy = 0

given that v1f = 2.30 v2f

v1fx^2 + v1fy^2 = 5.29 v2x^2 + 5.29 v2fy^2

and applying energy conservation,

m vi^2 /2 + 0 = m (v1fx^2 + v1fy^2)/2 + m( v2x^2 + v2fy^2 ) /2

vi^2 = v1fx^2 + v1fy^2 + v2x^2 + v2fy^2

solving,

v1fx = 0.841 v , v1fy = 0.366 vi , v2fx = 0.159 vi, v2fy = - 0.366 vi

(A) v1f = sqrt(v1fx^2 + v2fy^2) vi = 0.917 vi

Ans: 0.917

v2f = sqrt(v2fx^2 + v2fy^2) vi = 0.399 vi

Ans: 0.399 Or 0.400

(b) theta1 = 23.5 deg

theta2 = - 66.5 deg

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