Question

Two protons are released from rest when they are 0.715 nm apart. Part A: What is the maximum speed they will reach?
Part B: what is the maximum acceleration they will achieve?

Answer 1

Answer:

(A). The maximum speed is 13891 m/s.

(B). The maximum acceleration is $26.98\times10^{16}\ m/s^2$

Explanation:

Given that,

Distance = 0.715 nm

(A). We need to calculate the maximum speed

Using formula of speed

$v=\sqrt{k\dfrac{e^2}{m}(\dfrac{1}{d}-\dfrac{1}{x})}$

Where, m = mass of proton

d = distance

$v=\sqrt{9\times10^{9}\times\dfrac{(1.6\times10^{-19})^2}{1.67\times10^{-27}}(\dfrac{1}{0.715\times10^{-9}}-\dfrac{1}{\infty})}$

$v=13891\ m/s$

(B). We need to calculate the maximum acceleration

Using formula of acceleration

$a=k\dfrac{e^2}{mr^2}$

$a=9\times10^{9}\dfrac{(1.6\times10^{-19})^2}{1.67\times10^{-27}\times(0.715\times10^{-9})^2}$

$a=26.98\times10^{16}\ m/s^2$

Hence, (A). The maximum speed is 13891 m/s.

(B). The maximum acceleration is $26.98\times10^{16}\ m/s^2$