NAME: KEY
PHGN341: Thermal Physics
Quiz 1
January 18, 2013
1. During a hailstorm, hailstones with an average mass of m and a speed of v strike a car windshield of area, A, which is
at an angle, θ, with respect to the horizontal. The hailstones strike at an average rate of r and scatter elastically from
the windshield. Assume the car is at rest.
(a) In terms of the quantities given, what is the average pressure the hailstones exert on the windshield?
Solution: The pressure is the force per unit area. The force is given by the rate of momentum delivered by the hailstones
to the windshield. Assuming an elastic collision, the momentum change for one collision is ∆p = 2mv cos θ (Note that
θ is the angle the hail strikes with respect to the normal to the windshield.) Thus, the pressure is
P =
2mvr cos θ
.
A
(b) Evaluate the pressure for m = 3 g, v = 20 m/s, θ = 30◦ , A = 0.6 m2 , and r = 40 per second, and express your
answer as a fraction of atmospheric pressure (1.01 × 105 Pa).
Solution: Plugging in the above values gives: P = 6.93 Pa or P/PA = 6.86 × 10−5 .
(c) Now assume the car is traveling at 20 m/s. What is the pressure now?
√
√
Solution: Consider the frame where the car is at rest. The hail is now moving at a speed of 202 + 202 m/s = 2 × 20
m/s. Note
√ that the rate the hail strikes the windshield is proportional to the speed; so now the new rate of hail
strikes is 2 r. (This phenomenon is related to the concept of the incoming flux in scattering theory. The incoming
flux is defined as the number of projectiles per unit time per unit area. If the density of projectiles is ρ then the
flux is vρ. The bottom line is that the pressure is proportional to the speed squared.) Also note that the hail is
now striking the
at an angle of 15◦ with respect to the normal of the windshield. Thus, the pressure is
√ windshield
√
P(c) = P(a) × 2 × 2 × (cos 15◦ / cos 30◦ ) = 15.46 Pa.
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