Question

A proton (m = 1.67 x 10⁻²⁷kg) that has a speed of 5.0 x 10⁶ m/s passes through a metal film of thickness 0.010 mm and emerges with a speed of 2.0 x 10⁶ m/s. How large an average force opposed its motion through the film?

A. 1.8 x 10⁻⁹ N
B. 7.5 x 10⁻¹⁰ N
C. 3.3 x 10⁻¹⁰ N
D. 3.5 x 10⁻¹¹ N

Answer 1

The average force opposing the proton's motion through the metal film can be calculated using the change in velocity, time taken to pass through the film, and the mass of the proton. The average force comes out to be option B. 7.5 x 10⁻¹⁰ N.

Step-by-step explanation:

The question asks about the average force that opposed the motion of a proton as it passed through a metal film. We have been given the initial and final speeds of the proton, as well as the thickness of the film.

Firstly, we need to find the change in velocity (Δv) of the proton. Since the initial speed (v_i) is 5.0 x 10⁶ m/s and the final speed (v_f) is 2.0 x 10⁶ m/s, Δv = v_f - v_i = -3.0 x 10⁶ m/s (the negative sign indicates a decrease in speed).

Next, we calculate the acceleration (a) of the proton while it’s in the film. We use the formula Δv = a Δt, where Δt is the time taken to pass through the film. However, we are not directly given Δt or a, instead we are given the thickness of the film (d), which is 0.010 mm or 0.010 x 10⁳ m. We can find the average velocity (v_avg) in the film by taking the average of v_i and v_f, which gives us 3.5 x 10⁶ m/s. Then, we can find Δt by using d = v_avg Δt, which gives us Δt = d / v_avg.

Once we have Δt, we can find a from Δv = a Δt. Using the formula F = ma, where m is the mass of the proton (1.67 x 10⁻²⁷ kg), we can then calculate the force. After computing the values, the correct force that opposes the proton's motion through the film will be option B. 7.5 x 10⁻¹⁰ N.