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?
4/5 R
1/4 R
1/2 R
1/16 R

Answer 1

Answer:

$R_f=\dfrac{4}{5}R$

Explanation:

Here, the kinetic and potential energy is conserved

$K_f-K_i=U_f-U_i\\\Rightarrow \dfrac{1}{2}mv_i^2-\dfrac{1}{2}mv_f^2=\dfrac{1}{4\pi \epsilon}\dfrac{q_1q_2}{r}\\\Rightarrow \dfrac{1}{2}m(\dfrac{1}{2}v_f)^2-\dfrac{1}{2}mv_f^2=\dfrac{1}{4\pi \epsilon}\dfrac{q_1q_2}{R}$

For the second case

$K_f-K_i=U_f-U_i\\\Rightarrow \dfrac{1}{2}mv_i^2-\dfrac{1}{2}mv_f^2=\dfrac{1}{4\pi \epsilon}\dfrac{q_1q_2}{r}\\\Rightarrow \dfrac{1}{2}m(\dfrac{1}{4}v_f)^2-\dfrac{1}{2}mv_f^2=\dfrac{1}{4\pi \epsilon}\dfrac{q_1q_2}{R_f}$

Subtract the equations we get

$\dfrac{1}{2}m(\dfrac{1}{2}v_f)^2-\dfrac{1}{2}mv_f^2=\dfrac{1}{4\pi \epsilon}\dfrac{q_1q_2}{R}\\ \dfrac{1}{2}m(\dfrac{1}{4}v_f)^2-\dfrac{1}{2}mv_f^2=\dfrac{1}{4\pi \epsilon}\dfrac{q_1q_2}{R_f}$

$\\\Rightarrow \dfrac{1}{4}mv_f^2-mv_f^2=\dfrac{1}{2\pi \epsilon}\dfrac{q_1q_2}{R}\\ \dfrac{1}{16}mv_f^2-mv_f^2=\dfrac{1}{2\pi \epsilon}\dfrac{q_1q_2}{R_f}\\\Rightarrow -\dfrac{3}{4}mv_f^2=\dfrac{1}{2\pi \epsilon}\dfrac{q_1q_2}{R}\\ -\dfrac{15}{16}}mv_f^2=\dfrac{1}{2\pi \epsilon}\dfrac{q_1q_2}{R_f}$

Divide the equations

$\dfrac{\dfrac{3}{4}}{\dfrac{15}{16}}=\dfrac{R_f}{R}\\\Rightarrow R_f=\dfrac{4}{5}R$

$R_f=\dfrac{4}{5}R$