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Problem 1 (15pts) A round trip flight: An airplane flies between two cities separated by a distance D. Assume the wind blows directly from one city to the other at a speed Va and the speed of the airplane is V 0 relative to the air. (a) How long does it take for the airplane to make a round trip between the two cities? (b) To an observer on the ground, what is the average speed of the airplane for such a round trip? (c) To an observer on the ground, what is the average velocity for the round trip?
City A D City B
wind
Va
V 0
Solution: (a) Total time = (^) V 0 D+Va + (^) V 0 D−Va = (^) V^2 02 DV −V^0 a 2.
(b) Average speed is Total distance traveledTotal time. We find the average speed = (^2) DV 0 /^2 (VD 02 −V (^) a (^2) ) = V^02 V− 0 V^ a^2
(c) Average velocity is Net displacementTotal time = 0 since the net displacement is zero.
Problem 2 (15pts) Two walkers: Two persons start from the same location O and walk around a square in opposite directions with constant speeds. The square is 30m by 30m. A’s speed is 2m/s and B’s speed is 1m/s. (a) Find the coordinates of the point where A and B will meet for the first time. (b) Find the distance between the meeting place and the origin O. (c) Find the average velocity VA of A and the average velocity VB of B between the time when they first start and the time when they first meet. (Either give the components of VA and VB or their magnitudes and directions.)
O
y
x
B (^) A
30m 30m
Solution: (a) To meet A travels 80m and B travels 40m. They meet at (10m, 30 m). (b) The distance is √ 102 + 30^2 = 10√10m. (c) The average velocities of A and B are the same since they have the same net displacement during the same time. A and B traveled for (^140) m/sm = 40s. Thus VA = VB = (10 40 ,30)s m = (0. 25 m/s, 0. 75 m/s).
Problem 4 (15pts) A balloon and a block: A balloon is tied to a block. The mass of the block is 2kg. The tension of the string between the balloon and the block is 30N. Due to the wind, the string has an angle θ relative to the vertical direction. cos θ = 4/5 and sin θ = 3/5. Assume the acceleration of gravity is g = 10m/s^2. Also assume the block is small so the force on the block from the wind can be ignored. (a) Find the x-component and the y-component of the force F exerted on the block by the string. (b) Find the x-component and the y-component of the acceleration a of the block. (c) Assume the mass of the balloon is zero and the force of the wind on the balloon is in the x-direction. Find the magnitude of the force of the wind on the balloon.
2kg
θ Balloon
wind
T = 30N
y
x
Solution: (a) The magnitude of the force (from the string) is T = 30N. The x-component = T sin θ = 30 × 35 = 18N. The y-component = T cos θ = 30 × 45 = 24N. (b) The total force on the block is: the x-component = 18N. the y-component = 24 − mg = 24 − 20 = 4N. The x-component of the acceleration = 18N/ 2 kg = 9m/s^2. The y-component of the acceleration = 4N/ 2 kg = 2m/s^2. (c) Since the mass of the balloon is zero, the net force on the balloon must be zero. The x-component of the force on the balloon by the string is − 18 N. The force from the wind on the balloon must balance that force and thus must be 18N.
