A proton is projected toward a fixed nucleus of charge +Ze with velocity vo. Ini

Question

A proton is projected toward a fixed nucleus of charge +Ze with velocity vo. Initially the two particles are very far apart. When the proton is a distance R from the nucleus its velocity has decreased to 1/2vo. How far from the nucleus will the proton be when its velocity has dropped to 1/4vo? A proton is projected toward a fixed nucleus of charge +Ze with velocity vo. Initially the two particles are very far apart. When the proton is a distance R from the nucleus its velocity has decreased to 1/2vo. How far from the nucleus will the proton be when its velocity has dropped to 1/4vo?

4/5 R

1/4 R

1/2 R

1/16 R

None of these.

Explanation / Answer

answer is 4/5 R

1) KEf - KEi= PEf - PEi
[1/2m(1/2vo)^2] - [1/2m(vo)^2] = [1/(4??0)q1q2/R] - 0

(2) KEf-KEi=PEf-PEi
[1/2m(1/4vo)^2] - [1/2m(vo)^2] = [1/(4??0)q1q2/Rf] - 0

You combine like terms to reduce the equations to:
(1)-3/8m(vo)^2 = 1/(4??0)q1q2/R
(2)-15/32m(vo)^2 = 1/(4??0)q1q2/Rf

You divide equation (1) by (2) or (2) by (1) [it doesn't really matter], and you solve for Rf and get the final result of:

Rf=4/5R

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