Steel pellets, each with a mass of 0.60 g, fall vertically onto a horizontal plate at a rate of 100 pellets per minute.

Question 21

Steel pellets, each with a mass of 0.60 g, fall vertically onto a horizontal plate at a rate of 100 pellets per minute. They strike the plate with a velocity of 5.0 m s^-1 and rebound with a velocity of 4.0 m s^-1.

What is the average force exerted on the plate by the pellets?

A 0.0010 N                  B 0.0054 N                  C 0.0090 N                  D 0.54 N

Reference: Past Exam Paper – March 2018 Paper 12 Q10

Solution:

Answer: C.

Force is defined as the rate of change of momentum.

Force F = Δp / Δt

As the pellets strike the plate, their momentum changes. The velocity of the pellets changes direction as they rebound.

Consider one steel pellet,

Initial velocity = 5.0 m s^-1

Final velocity = – 4.0 m s^-1      (the direction changes)

Change in momentum Δp = mΔv = 0.6×10^-3 × (–4.0 – 5.0) = (–) 5.4×10^-3 Ns

This is the change in momentum of ONE steel pellet.

Rate of fall of pellets on the plate = 100 pellets per min

1 min - - > 100 pellets

60 s - - > 100 pellets

1 s -- > 100 / 60 = 1.667 pellets

In 1 second, an average of 1.667 pellets fall on the plate.

Force F = Δp / Δt        (force is the change in momentum per second)

Average force = rate of change of momentum of all pellets

Average force = 1.667 × 5.4×10^-3 = 0.0090 N