Phys 2425: University Physics I Spring 2016 Practice Exam 1

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1 1. (0 Points) What course is this? a. PHYS 1401 b. PHYS 140 c. PHYS 45 d. PHYS 46 Survey Questions no points. (0 Points) Which exam is this? a. Exam 1 b. Exam c. Final Exam 3. (0 Points) What version of the exam is this? a. A b. B c. C d. D 4. What are the units of the imaginary quantity slurm, which is equal to acceleration divided by velocity squared? m/s a. m /s 3 b. 1 c. s/m d. s e. 1/m (m/s) = m s m s = 1 m 5. Express the following quantity using an SI prefix: m = =.3 μ a. 30 nm b..3 μm c. 3 μm d. 30 μm e. 3 mm 6. In Newtonian Mechanics (which is what we study), under what conditions is the equation x = x 0 + v avg Δt valid to use? a. Stationary object only. b. Constant velocity only. c. Constant acceleration only. d. Only when the acceleration is varying. e. It is generally true. 7. Which diagram could represent the position of a car accelerating from rest? This is an extension of the definition of average velocity. On a position graph, at rest is zero slope, eliminating (a), (c), and (d). To be accelerating, the position must change, elimiting (e). So (b) is correct. 8. One student throws a heavy red ball horizontally from a balcony of a tall building with an initial speed v i. At the same time, a second student drops a lighter blue ball from the balcony. Neglecting air resistance, which statement is true? a. The blue ball reaches the ground first. b. The balls reach the ground at the same instant. Both have v y = 0. c. The red ball reaches the ground first. d. Both balls hit the ground with the same speed. e. None of statements (a) through (d) is true. 9. A person drives from home to work at a constant speed of 15 m/s. On the way home, they drive 30 m/s. What is the average speed of their driving? a m/s b. 0 m/s c..5 m/s d. 5 m/s e. 7.5 m/s Say their trip to work has a length of 30x. The total distance is 60x. The time to work is 30x/15 = x, and the time home is x. The total time is then 3x. So the average speed is 60x 3x = 0 m/s.

2 (Questions 10 1) An object moves according to the equation Δx = 8t + 3t t What is the magnitude of its acceleration at time t = 0? m/s² b. 1 m/s² c. m/s² d. 3 m/s² e. 6 m/s² 11. What its speed at time t = 0? a. m/s b. 3 m/s c. 7 m/s d. 8 m/s e. 13 m/s 1. What is it doing at time t = 0? a. Temporarily not moving. b. Moving at constant velocity. c. Moving and speeding up. d. Moving and slowing down. e. Moving and changing direction. This doesn t fit the constant acceleration Δx template. The derivative is v = 8 + 6t 3t. The next derivative is a = 6 6t. At t = 0, a = 6. From the above velocity, v = 8. It is moving in the negative direction, and the acceleration is in the positive direction. Since they are opposite, it s slowing down. 13. A car travelling at a highway speed of 30 m/s slows to a stop in 150 m. What is its average speed during the process? m/s Coming to a stop, the final speed is v f = 0. There is only one direction of motion, so average speed and average velocity c. 15 m/s are basically the same thing. d. 30 m/s The average speed is v avg = 1 (v i + v f ) = 1 (30) = 15 m/s. e. 60 m/s 14. If you drop a ball from a height of 0.5 m, how much time does it take for it to hit the ground?.05 s b s Start with Δy = v i Δt + 1 aδt, use Δy = h, v i = 0, and a = g to get: c. 0.3 s d. 0.3 s h = 1 gt, and solve for t = h = 0.3 s. g e. 6.6 s 15. A plane is flying in a direction described by an azimuth of 170. The definition of azimuth is the angle clockwise from north or east of north. What mathematical polar coordinate angle corresponds to this azimuth? a. 80 Azimuth 0 (north) corresponds to θ = 90, and azimuth 90 (east) corresponds to b. 10 θ = 0. So, θ = 90 Azimuth. c. 10 The given direction is θ = = 80. d. 80 e If a 6 ft. tall vertical pole casts a shadow that is ft. long, what is the angle of the sun above the horizon? a. 18 The opposite is 6 ft, and the adjacent is ft. b. 19 c. 30 θ = atan ( 6 ) = 7 d. 60 e. 7

3 17. A golfer can hit a ball with an initial speed of 80 m/s. Assuming projectile motion, what is the maximum range the ball could go? a. 8 m b. 80 m c. 160 m d. 37 m e. 653 m (Questions 18 ) A ball is fired from a launcher with an initial speed of 1 m/s at an angle of 60 above the horizontal. It is fired from ground level up onto the roof of a building, which is 4 m above the ground. 18. The x-component of the ball s initial velocity is: The given angle is above the horizontal, so it is θ. v x = v 0 cos(θ) = (1) cos (60 ) = 6 c. 6 m/s d m/s e. 1 m/s 19. The y-component of the ball s initial velocity is: v 0y = v 0 sin(θ) = (1) sin (60 ) = 10.4 c. 6 m/s d m/s e. 1 m/s 0. At the point of landing, the y-component of the ball s velocity is: Since this links position and velocity, the v equation works well. b..6 m/s v fy = v iy gδy = (10.4) (9.8)(4.0) = 9.6 c. 4.5 m/s d. 5.4 m/s v fy = ± 9.6 = ±5.4 e. 6.3 m/s Note that because the final position is up high, the speed is smaller. 1. The amount of time the ball in the air is:.5 s Δv y = v fy v iy = ( 5.4) (10.4) = 15.8 = aδt b. 0.8 s c. 1.6 s Δt = 15.8 = 15.8 a 9.8 = 1.6 d..4 s e. 3.6 s. The range (horizontal distance) of the ball s flight is: a. 4.0 m Δx = v xδt = (6)(1.6) = 9.7 b. 5.6 m (This was rounded to sig figs after being calculated with many more.) c. 8.4 m d. 9.7 m e m 3. A ball is tossed straight up into the air. When is its acceleration pointing in the same direction as its velocity? a. On the way up. b. At the top. c. On the way down. d. Both (a) and (b). e. All of the above. Landing is same height as take-off, so Range = v 0 sin(θ) = v 0 g g Accel is down for the whole motion. Vel is in the same direction (down) only while ball is actually going downward.

4 4. A particular car can stop from 30 m/s on dry road in a distance of 54 m. Assuming the same driving conditions, what is the car s stopping distance from 10 m/s? a. 5.1 m Start with v = v 0 + aδx and let v = 0. b. 6.0 m v 0 = a Δx c. 9.0 m So Δx is proportional to v 0. d. 18 m If v 0 is 3 smaller, Δx should be 9 smaller. e. 31 m 5. For a horizontally-launched projectile, which is the correct formula for the distance fallen (h) depending on the horizontal range (x)? a. h = 1 gx/v x = v x t, so t = x/v x 0 b. h = 1 gx Horizontal, so v /v x = v 0 0 c. h = 1 gx/v 0 d. h = 1 gv 0 /x 6. Two forces are known to be F 1 = 6.0 N in the x-direction, and F = 8.0 N at an angle of 130. What is the direction of a the resultant F 1 + F? a. 48 b. 75 c. 8 d. 113 e. 10 h = 1 gt = 1 g ( x v x ) = 1 gx /v x F x = F 1x + F x = (6.0) + (8.0) cos(130 ) = F y = F 1y + F y = (0) + (8.0) sin(130 ) = 6.13 θ = atan(f y /F x ) = atan(6.13/0.858) = 8 7. A car is driving at a steady speed of 30 m/s, and at the moment they drive by, a motorcycle starts to chase it. The motorcycle starts from rest and accelerates at a rate of 5 m/s² until it catches up with the car. How far does the car get before the motorcycle catches up? a. 6 m Let x = Distance and t = Time of chase. b. 180 m c. 360 m x = 30t and x = 1 (5)t d. 70 m e. It never catches up. 30t = 1 (5)t, so t = 30 = 1 5 x = (30)(1) = A basketball player is 10.0 m from the basket, which is 1.0 m above the ball. If the player shoots the ball at an angle of 45 above the horizontal, what speed must be used so that the ball goes in the basket? a. 5 m/s Δx = v 0 cos(45 ) Δt = 0.707v 0 t = 10 b. 10 m/s Δy = v 0 sin(45 )t 1 (9.8)t c. 15 m/s 1.0 = 0.707v d. 0 m/s 0 t 4.9t Substitute, which is easy because cos(45 )=sin(45 ). e. 5 m/s 1.0 = t t = 9/4.9 = v 0 = 10/(0.707t) = 10.4

5 9. If an object is thrown straight downward with a speed v 0 from the top of a building of height h, which is the correct equation for the position as a function of time? Take y = 0 to be ground level. a. y = h v 0 t 1 gt b. y = h + v 0 t + 1 gt c. y = v 0 gt d. y = h v 0 t + 1 gt e. y = h + v 0 t 1 gt 30. A velocity vector has a y-component v y = 3.0 m/s and it is at an angle of 37 CCW from the x-axis. What is its magnitude? a. 1.8 m/s b..4 m/s c. 3.7 m/s d. 5 m/s e. 6 m/s (Questions 31 3) To drive from home to the store, somebody first travels at 0 m/s south for 10 s, and then they travel 0 m/s at an angle of 60 north of east for 10 s. 31. What is the magnitude of their average velocity for the trip? b. 5. m/s c m/s d m/s e. 0 m/s 3. What is their average speed for the trip? b. 5. m/s c m/s d m/s e. 0 m/s v y = v sin(θ) v = v y sin(θ) = 3.0 sin(37 ) = 4.98 Each leg is (0)(10) = 400 m long. Δx = 0 + (400) cos(60 ) = 100 Δy = ( 400) + (400) sin(60 ) = 31 Δr = = 14 v avg = Δr/Δt = 14/40 = 5. Distance is Δs = v avg = Δs/Δt = 4800/40 = When an object is moving in a circular path with constant speed, the acceleration is: a. Zero because there is no change in speed. b. Zero because there is no change in velocity. c. Pointing inward to keep the velocity turning that direction. d. Pointing outward because there is a net force pushing the object outward. e. Pointing along the direction of the velocity because that is the only acceleration possible. 34. An athletic track is flat (not banked) with a radius of 36.5 m. If the maximum allowable acceleration is 0.8g, how fast could a person conceivably run before their feet slip? a. 5 m/s There is no downward acceleration because they are not falling. b. 17 m/s a = v /R v = ar = (0.8)(9.8)(36.5) = 17 c. 19 m/s d. 9 m/s e. 86 m/s