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

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

(B). The maximum acceleration is
`26.98*10^(16)\ m/s^2`

Step-by-step explanation:

Given that,

Distance = 0.715 nm

(A). We need to calculate the maximum speed

Using formula of speed

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

Where, m = mass of proton

d = distance

`v=\sqrt{9*10^(9)*((1.6*10^(-19))^2)/(1.67*10^(-27))((1)/(0.715*10^(-9))-(1)/(\infty))}`

`v=13891\ m/s`

(B). We need to calculate the maximum acceleration

Using formula of acceleration

`a=k(e^2)/(mr^2)`

`a=9*10^(9)((1.6*10^(-19))^2)/(1.67*10^(-27)*(0.715*10^(-9))^2)`

`a=26.98*10^(16)\ m/s^2`

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

(B). The maximum acceleration is
`26.98*10^(16)\ m/s^2`