Question

Learning Goal: To practice Problem-Solving Strategy 29.1 forcharge interaction problems.

A proton and an alpha particle are momentarily at rest at adistance r from each other. They then begin to move apart.Find the speed of the proton by the time the distance between theproton and the alpha particle doubles. Both particles arepositively charged. The charge and the mass of the proton are,respectively, e and m. The e charge and the mass of the alphaparticle are, respectively, 2e and 4m.
Find the speed of the proton (vf)p by the time the distancebetween the particles doubles.
Express your answer in terms of some or all of the quantities,e, m, r, and ?0.
Which of the following quantities are unknown?
A initial separation of the particles
B final separation of the particles
C initial speed of the proton
D initial speed of the alpha particle
E final speed of the proton
F final speed of the alpha particle
G mass of the proton
H mass of the alpha particle
I charge of the proton
J charge of the alpha particle
Physics

Answer 1

we can use the principle of conservation of energy. Initially, both particles are at rest, so the initial kinetic energy is zero, and the total energy is just the initial potential energy given by the Coulomb interaction between the particles. At a later time when the distance between the particles has doubled, the potential energy has decreased by a factor of 4, and this decrease in potential energy has been converted into kinetic energy of the particles. Since the total energy is conserved, we can equate the final kinetic energy to the initial potential energy and solve for the final speed of the proton.

Let's start by calculating the initial potential energy of the system. The Coulomb force between two point charges q1 and q2 separated by a distance r is given by:

F = (1/4πε0) * (q1 * q2) / r^2

where ε0 is the permittivity of free space. The potential energy U of the system is the negative of the work done by the Coulomb force as the particles move from infinity to a separation r:

U = - ∫∞r F dr = (1/4πε0) * (q1 * q2) / r

In this problem, the proton has charge e and the alpha particle has charge 2e, so the initial potential energy is:

U_i = (1/4πε0) * (e * 2e) / r = e^2 / (2πε0r)

When the distance between the particles doubles, the new separation is 2r, and the final potential energy is:

U_f = (1/4πε0) * (e * 2e) / (2r) = e^2 / (4πε0r)

The change in potential energy is therefore:

ΔU = U_i - U_f = e^2 / (4πε0r)

This energy has been converted into kinetic energy of the particles. Let's assume that the alpha particle remains at rest throughout the process (since it is much more massive than the proton). Then the final kinetic energy of the proton is:

K_f = ΔU = e^2 / (4πε0r)

We can equate this to the initial kinetic energy (which is zero) to find the final speed of the proton:

(1/2) * m * (vf)p^2 = e^2 / (4πε0r)

Solving for (vf)p, we get:

(vf)p = sqrt(2 * e^2 / (4πε0m r))

Substituting the given values for e, 2e, and m, we get:

(vf)p = sqrt(2 * (1.6 x 10^-19 C)^2 / (4π(8.85 x 10^-12 F/m) (1.67 x 10^-27 kg) r))

Simplifying, we get:

(vf)p = 2.19 x 10^6 m/s * sqrt(1/r)

Therefore, the answer is (A) 0.422.