January 18, 2008
PHY2053 Discussion
Quiz 1 (Chapter 1 & 2.1-2.4)
Name:
UFID:
*1. (5 pts) Newton’s law of universal gravitation is represented by F = GMm/r², where F
is the gravitational force, M and m are masses, and r is a length. What is the dimension
of the proportionality constant G ? (Hint: Force is given by mass times acceleration.)
Since the dimension of both sides should be the same, we get
[F] = [G][Mm]/[r²].
Plug [F] = [M][a] = MLTˉ², [Mm] = M² and [r²] = L² into the equation, and solve it for [G]:
MLTˉ² = [G]M²/L² ⇒ [G] = Mˉ¹L³Tˉ²
***2. (5 pts) Two boats start together and race across 50 km-wide lake and back. The
boat that completes the round trip first wins. Boat A goes across at 40 km/h and returns
at 40km/h. Boat B goes across at 30km/h at first, but it doubles its speed when it passes
boat A. Which boat wins and by how much distance?
We calculate the time when two boats pass each other. Since when they pass each other,
the sum of the distance traveled by the two boats equal to 100 km, we get
40t+30t = 100 ⇒ t = 100/70 = 1.43 h
The distance traveled by the boat B at this moment is given by
x = vt = 30×1.43 = 42.9 km
Now we calculate how long it takes for the boats to complete a round trip:
A: t’ = 100/40 = 2.5 h
B: t’’ = 42.9/30 + (100-42.9)/60 = 2.38 h
Thus boat B wins. When B reaches the goal, the distance covered by A is:
x’ = 40×2.38 = 95.2 km.
Thus B wins by
Δx = 100 – 95.2 = 4.8 km.
**3. (5 pts) A tortoise runs with a speed of 0.25 m/s and a hare runs 10 times as fast. In a
race they start together, but the hare stops to rest for 5.0 minute. The hare wins by 15 m.
What is the length of the race?
Assuming the hare runs for t seconds, then the tortoise runs for t+300 (5min = 300 sec).
At the end of the race, the hare wins by 15 m, therefore we get
15 = Xh-Xt = 2.5t-0.25(t+300) ⇒ 2.25t = 90 ⇒ t = 40 s.
Thus the length of the race is given by
X = vt = 2.5×40 = 100m
*4. (5 pts) A certain car slows down at a rate of 4.00 m/s² when the driver slams on the
brakes. How long does it take for this car to stop if it initially travels at 30 mi/h? (Hint: 1
mi = 1609 m)
Converting mi/h to m/s, we get
30 mi/h = 30(mi/h)×1609/1(m/mi)×1/3600(h/s) = 13.4 m/s
Using the definition of the acceleration, we get
-4 = (0-13.4)/Δt ⇒ Δt = 13.4/4 = 3.35 s