June 6, 2008
PHY2053 Discussion
Quiz 3 (Chapter 3)
Name:
UFID:
**1. (5pts) City 1 and city 2 are 50.0 km apart and city 2 is east from city 1. A couple
arranges to drive from city 1 and meet a couple driving form city 2 at a lake, which is
northeast from city 1. The couple from city 1 travels at 80 km/h, and it takes them 2.5 h
to get the lake. The other couple from city 2 travels at 60 km/h. How long does it take
them to get the lake?
The distance between city 1 and the lake is
x1L = vt = 80×2.5 = 200 km
We express the position of the lake and city 2 relative to city 1 in components:
x1L = (200cos45˚, 200sin45˚) = (141 km, 141 km) , x12 = (50 km, 0 km)
The position of the lake relative to city 1 is
x2L = x1L – x12 = (141, 141) - (50, 0) = (91 km, 141 km)
The magnitude of this vector is the distance between city 1 and the lake.
x2L = √(912+1412) = 168 km
Thus the driving time of the second couple is
t’ = x2L/v’ = 168/60 = 2.80 h
*2. (5pts) A boat moves through the water of a river at 8.50 m/s relative to the water,
regardless of the boat’s direction. If the water in the river is flowing at 1.5 m/s, how long
does it take the boat to make a round trip consisting of a 500-m displacement
downstream followed by a 500-m displacement upstream?
When the boat is traveling downstream/upstream, the velocities of the boat relative to
the shore are respectively
vdown = vBR +vRE = 8.5+1.5 = 10.0 m/s
vup = vBR - vRE = 8.5-1.5 = 7.00 m/s
Thus the total time is
ttot = tdown+tup = x/vdown+x/vup = 500/10+500/7 = 121 s
**3. (5pts) A car is parked on a cliff overlooking the ocean that makes an angle of 20.0˚
below the horizontal. The car rolls from rest down the incline with a constant
acceleration of 3.00 m/s2 over a distance of 40.0 m to the edge of a cliff, which is 50.0 m
above the ocean. What is the car’s position relative to the base of the cliff when the car
lands in the ocean?
The initial velocity of the projectile motion is the velocity of the car at the edge of the
cliff. Thus we have
v02 = 2aL ⇒ v0 = √(2aL) = 15.5 m/s, -20˚ to the horizontal
The x- and y- components of the initial velocity are
(v0x, v0y) = (15.5cos(-20˚), 15.5sin(-20˚)) = (14.6 m/s, -5.30 m/s)
We can calculate the time of flight using the formula for the vertical displacement.
Δy = v0yt-(1/2)gt² ⇒ 4.9t²+5.3t-50 = 0 ⇒ t = (-5.3+√(5.3²+4×4.9×50))/9.8 = 2.70 s
Thus the horizontal distance is
Δx = v0xt = 14.6×2.7 = 39.4 m
***4. (5pts) We derived that when you throw a ball on a flat ground, you can maximize
the horizontal range by throwing the ball at an angle of 45˚. Now suppose you throw a
ball on a 15˚ incline. At what angle do you need to throw the ball to get the maximum
range? Hint: Express the incline as Δy = Δxtanα, where α = 15˚. Find the time of
flight in terms of initial velocity and projection angle, then maximize the horizontal
displacement. Useful formula: sinAcosB±sinBcosA = sin(A±B).
For convenience, we measure the projection angle φ from the surface of the incline.
When the ball hits the incline, the horizontal and vertical displacements satisfies Δy =
Δx×tanα, thus we have,
v₀sin(α+φ)t-(1/2)gt² = v₀cos(α+φ)t×tanα
⇒ t = (2/g) v₀ [sin(α+φ)-cos(α+φ)tanα] = 2 v₀sinφ/(gcosα)
The horizontal displacement is
Δx = v₀cos(α+φ)t = v₀cos(α+φ)×2 v₀sinφ/(gcosα) = (v₀²/gcosα)[sin(2φ+α)-sinα]
The sine function takes the maximum value when
2φ+α = 90˚ ⇒ φ = (90˚-α)/2 = (90-15)/2 = 37.5˚ above the incline