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

`R_f=(4)/(5)R`

Step-by-step explanation:

Here, the kinetic and potential energy is conserved

`K_f-K_i=U_f-U_i\\\Rightarrow (1)/(2)mv_i^2-(1)/(2)mv_f^2=(1)/(4\pi \epsilon)(q_1q_2)/(r)\\\Rightarrow (1)/(2)m((1)/(2)v_f)^2-(1)/(2)mv_f^2=(1)/(4\pi \epsilon)(q_1q_2)/(R)`

For the second case

`K_f-K_i=U_f-U_i\\\Rightarrow (1)/(2)mv_i^2-(1)/(2)mv_f^2=(1)/(4\pi \epsilon)(q_1q_2)/(r)\\\Rightarrow (1)/(2)m((1)/(4)v_f)^2-(1)/(2)mv_f^2=(1)/(4\pi \epsilon)(q_1q_2)/(R_f)`

Subtract the equations we get

`(1)/(2)m((1)/(2)v_f)^2-(1)/(2)mv_f^2=(1)/(4\pi \epsilon)(q_1q_2)/(R)\\ (1)/(2)m((1)/(4)v_f)^2-(1)/(2)mv_f^2=(1)/(4\pi \epsilon)(q_1q_2)/(R_f)`

`\\\Rightarrow (1)/(4)mv_f^2-mv_f^2=(1)/(2\pi \epsilon)(q_1q_2)/(R)\\ (1)/(16)mv_f^2-mv_f^2=(1)/(2\pi \epsilon)(q_1q_2)/(R_f)\\\Rightarrow -(3)/(4)mv_f^2=(1)/(2\pi \epsilon)(q_1q_2)/(R)\\ -(15)/(16)}mv_f^2=(1)/(2\pi \epsilon)(q_1q_2)/(R_f)`

Divide the equations

`((3)/(4))/((15)/(16))=(R_f)/(R)\\\Rightarrow R_f=(4)/(5)R`

`R_f=(4)/(5)R`