AP Physics 2 Exam Question

A particle accelerator is used to increase the kinetic energy of two particles, particle A and particle B. Particle A has a mass of 2.0 x 10^-27 kg and a charge of +1.6 x 10^-19 C, while particle B has a mass of 1.0 x 10^-27 kg and a charge of -1.6 x 10^-19 C. The particles are accelerated through a potential difference of 100 V.

a) Calculate the change in potential energy for both particle A and particle B as they pass through the potential difference. b) Assuming the particles start from rest, calculate the final kinetic energy of particle A and particle B. c) Determine the final velocity of particle A and particle B, given that the speed of light in a vacuum is 3.0 x 10^8 m/s.

Answer with Step-by-Step Explanation

a) The change in potential energy for a charged particle accelerated through a potential difference can be calculated using the equation:

ΔPE = q * ΔV

where ΔPE is the change in potential energy, q is the charge, and ΔV is the potential difference.

For particle A: ΔPE_A = (+1.6 x 10^-19 C) * (100 V) = +1.6 x 10^-17 J

For particle B: ΔPE_B = (-1.6 x 10^-19 C) * (100 V) = -1.6 x 10^-17 J

b) The final kinetic energy of a particle can be obtained by converting the change in potential energy into kinetic energy.

KE_final = ΔPE

For particle A, KE_Afinal = ΔPE_A = +1.6 x 10^-17 J

For particle B, KE_Bfinal = ΔPE_B = -1.6 x 10^-17 J

c) The final kinetic energy can be equated to the kinetic energy formula:

KE_final = (1/2) * m * v^2

Rearranging the equation, we get:

v = √((2 * KE_final) / m)

For particle A: v_A = √((2 * 1.6 x 10^-17 J) / (2.0 x 10^-27 kg)) = 1.264 x 10^10 m/s

For particle B: v_B = √((2 * -1.6 x 10^-17 J) / (1.0 x 10^-27 kg)) = -2.529 x 10^10 m/s (Note: The negative sign indicates the direction opposite to the initial acceleration)

Therefore, the final velocity of particle A is 1.264 x 10^10 m/s, and the final velocity of particle B is -2.529 x 10^10 m/s.

(Note: Since the final velocity of particle B is negative, it means the particle is moving in the opposite direction to its initial acceleration.)