Question

A proton and an alpha particle are momentarily at rest at a distance r from each other. They then begin to move apart. Find the speed of the proton by the time the distance between the proton and the alpha particle doubles. Both particles are positively charged. The charge and the mass of the proton are, respectively, e and m. The charge and the mass of the alpha particle are, respectively, 2e and 4m.

Answer 1

Answer:

v = 0.525 /√r

Explanation:

For this problem we must use Newton's second law where force is electric given by Coulomb's law.

             F = m a

             k q₁ q₂ / x² = m dv / dt

Where index 1 we will use it for the proton and index 2 for the alpha particle

             k / m q₁q₂ / x₂ = dv/dx   dx/dt = dv/dx   v

            (k / m q₁q₂) dx / x² = v dv

We integrate

           (k / m q₁q₂) (-1 / x) = v² / 2

We evaluate between the lower limit x = r and v = 0 to the upper limit x = 2r

           (k / mq₁q₂) (-1 / 2r + 1 / r) = ½ (v²-0)

           v² = 2k/m  q₁q₂ (1 / 2r)

           q₁ = e

           q₂ = 2e

           v² = k 2e² /m  1 /r

           v² = 2k e²/m  1 /r

Let's replace the values

          v² = 2 8.99 10⁹ (1.6 10⁻¹⁹)² /1.67 10⁻²⁷   1 / r

          v = √( 27.562 10⁻² 1 / r)

          v = 5.25 10⁻¹  1 / √r

          v = 0.525 /√r