Q#21 (Past Exam Paper – March 2018 Paper 12 Q10)
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
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
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