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Phys 1401: General Physics I Spring 2016

Practice Exam 1

1. (0 Points) What course is this? Survey Questions – no points a. PHYS 1401 b. PHYS 1402 c. PHYS 2425 d. PHYS 2426 2. (0 Points) Which exam is this? a. Exam 1 b. Exam 2 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 m/s 2 m ⋅ s2 1 squared? = = 2 2 2 (m/s) m m ⋅s a. m b. 1 c. s/m d. s e. 1/m

5. Express the following quantity using an SI prefix: 23 × 10−7 m 23 × 10−7 = 2.3 × 10−6 = 2.3 𝜇 a. 230 nm b. 2.3 μm c. 23 μm d. 230 μm e. 23 mm 6. In Newtonian Mechanics (which is what we study), under what conditions is the equation 𝑥 = 𝑥0 + 𝑣avg Δ𝑡 valid to use? This is an extension of the a. Stationary object only. definition of average velocity. b. Constant velocity only. c. Constant acceleration only. d. Only when the acceleration is varying. e. It is generally true. 7. 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. 17.5 m/s Say their trip to work has a length of 30𝑥. The total distance is 60𝑥. b. 20 m/s The time to work is 30𝑥/15 = 2𝑥, and the time home is 𝑥. c. 22.5 m/s The total time is then 3𝑥. d. 25 m/s 60𝑥 So the average speed is 3𝑥 = 20 m/s. e. 27.5 m/s 8. An object moves according to the equation Δ𝑥 = −𝑡2 . What is the magnitude of its acceleration? 1 a. 0 m/s² 1 Compare to Δ𝑥 = 𝑣𝑖 Δ𝑡 + 𝑎(Δ𝑡)2 . Since 𝑎 = −1, 𝑎 = −2, and |𝑎| = 2. 2 2 b. 1 m/s² c. 2 m/s² d. 3 m/s² e. 6 m/s² 9. An object moves according to the equation Δ𝑥 = −𝑡 2 . What its speed at time 𝑡 = 4? 1 a. 2 m/s Compare to Δ𝑥 = 𝑣𝑖 Δ𝑡 + 𝑎(Δ𝑡)2 . Note 𝑣𝑖 = 0. 2 b. 4 m/s Rebuild 𝑣𝑓 = 𝑣𝑖 + Δ𝑣 = 𝑣𝑖 + 𝑎Δ𝑡 = 0 + (−2)(4) = −8. c. 7 m/s Answer is |𝑣| = 8 m/s. d. 8 m/s e. 13 m/s 10. An object moves according to the equation Δ𝑥 = −𝑡 2 . What is it doing at time 𝑡 = 0? a. Temporarily not moving. Since 𝑣𝑖 = 0, it’s not moving at 𝑡 = 0. b. Moving at constant velocity. There is an acceleration, it will soon move. c. Moving and speeding up. d. Moving and slowing down. e. Moving and changing direction.

Phys 1401: General Physics I Spring 2016

Practice Exam 1

11. 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? Coming to a stop, the final speed is 𝑣𝑓 = 0. a. 0 m/s There is only one direction of motion, so average speed and average velocity b. 5 m/s are basically the same thing. c. 15 m/s 1 1 The average speed is 𝑣avg = 2 (𝑣𝑖 + 𝑣𝑓 ) = (30) = 15 m/s. d. 30 m/s 2 e. 60 m/s 12. Which diagram could represent the position of a car accelerating from rest? 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. 13. If you drop a ball from a height of 0.5 m, how much time does it take for it to hit the ground? 1 a. 0.05 s Start with Δ𝑦 = 𝑣𝑖 Δ𝑡 + 𝑎Δ𝑡2 , use Δ𝑦 = −ℎ, 𝑣𝑖 = 0, and 𝑎 = −𝑔 to get: 2 b. 0.10 s 2ℎ 1 ℎ = 2 𝑔𝑡2 , and solve for 𝑡 = √ 𝑔 = 0.32 s. c. 0.23 s d. 0.32 s e. 6.6 s 14. A plane is flying in a direction described by an azimuth of 100°. The definition of azimuth is the angle “clockwise from north”. What mathematical polar coordinate angle corresponds to this azimuth? This angle is 10° south of east. This is the a. −80° negative direction from our reference point. b. −10° c. 10° d. 80° e. 110° 15. If a 6 ft. tall vertical pole casts a shadow that is 2 ft. long, what is the angle of the sun above the horizon? a. 18° The opposite is 6 ft, and the adjacent is 2 ft. b. 19° 6 𝜃 = atan ( ) = 72° c. 30° 2 d. 60° e. 72° 16. 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 𝑣2 𝑣2 Landing is same height as take-off, so Range = 0 sin(2𝜃 ) = 0 𝑔 𝑔 b. 80 m c. 160 m d. 327 m e. 653 m 17. A ball is tossed straight up into the air. When is its acceleration pointing opposite to its velocity? a. On the way up. Accel is down for the whole motion. b. At the top. Vel is opposite (up) only while ball is going up. c. On the way down. d. Both (a) and (b). e. All of the above.

Phys 1401: General Physics I Spring 2016

Practice Exam 1

A ball is fired horizontally from a launcher with an initial speed of 12 m/s. It is fired from 11 m above the ground. 18. The amount of time the ball in the air is: 1 1 a. 0.5 s Δ𝑦 = − 2 𝑔𝑡 2 which becomes ℎ = 2 𝑔𝑡 2 b. 1 s c. 1.5 s 2ℎ 2 ⋅ 11 𝑡=√ =√ d. 2 s 𝑔 9.8 e. 3 s 19. At the point of landing, the 𝑦-component of the ball’s velocity is: a. 0 Knowing the time, b. −5 m/s 𝑣𝑦 = 𝑣0𝑦 − 𝑔𝑡 c. −9.8 m/s = (0) − (9.8)(1.5) d. −15 m/s = −14.7 e. −30 m/s 20. The range (horizontal distance) of the ball’s flight is: Knowing the time, a. 3 m 𝑥 = 𝑣𝑥 𝑡 b. 6 m = (12)(1.5) c. 11 m = 18.0 d. 15 m Closest choice is 15 m. e. 22 m 21. 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 2 = 𝑣02 + 2𝑎Δ𝑥 and let 𝑣 = 0. b. 6.0 m 𝑣02 = 2|𝑎|Δ𝑥 c. 9.0 m So Δ𝑥 is proportional to 𝑣02 . d. 18 m If 𝑣0 is 3× smaller, Δ𝑥 should be 9× smaller. e. 31 m 22. For a horizontally-launched projectile, which is the correct formula for the distance fallen ( ℎ) depending on the horizontal range (𝑥)? 𝑥 = 𝑣𝑥 𝑡, so 𝑡 = 𝑥/𝑣𝑥 a. ℎ = 12𝑔𝑥/𝑣 b. ℎ = 12𝑔𝑥 2 /𝑣 2 1 𝑥 2 1 1 ℎ = 𝑔𝑡2 = 𝑔 ( ) = 𝑔𝑥 2 /𝑣𝑥2 c. ℎ = 12𝑔𝑥/𝑣 2 2 2 𝑣𝑥 2 d. ℎ = 12𝑔𝑣 2 /𝑥 2 23. Two forces are known to be 𝐹1 = 6.0 N in the 𝑥-direction, and 𝐹2 = 8.0 N at an angle of 130°. What is the direction of a the resultant 𝐹1 + 𝐹2 ? a. 48° b. 75° c. 82° d. 113° e. 120°

𝐹𝑥 = 𝐹1𝑥 + 𝐹2𝑥 = (6.0) + (8.0) cos(130°) = 0.858 𝐹𝑦 = 𝐹1𝑦 + 𝐹2𝑦 = (0) + (8.0) sin(130°) = 6.13 𝜃 = atan(𝐹𝑦 /𝐹𝑥 ) = atan(6.13/0.858) = 82°

Phys 1401: General Physics I Spring 2016

Practice Exam 1

24. 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 1 (5)𝑡 2 𝑥 = 30𝑡 and 𝑥 = c. 360 m 2 1 2 ⋅ 30 d. 720 m 30𝑡 = (5)𝑡 2 , so 𝑡 = = 12 2 5 e. It never catches up. 𝑥 = (30)(12) = 360

25. If an object is thrown straight downward with a speed 𝑣0 from the top of a building of height ℎ, which is the most correct equation for the position as a function of time? Take 𝑦 = 0 to be ground level. 1

a. 𝑦 = ℎ − 𝑣0 𝑡 − 𝑔𝑡 2 2 1

b. 𝑦 = ℎ + 𝑣0 𝑡 + 𝑔𝑡2 2

1

d. 𝑦 = −ℎ − 𝑣0 𝑡 + 𝑔𝑡 2 2 1

e. 𝑦 = −ℎ + 𝑣0 𝑡 − 𝑔𝑡 2 2

c. 𝑦 = 𝑣0 − 𝑔𝑡 26. A velocity vector has a 𝑦-component 𝑣𝑦 = 3.0 m/s and it is at an angle of 37° CCW from the 𝑥-axis. What is its magnitude? 𝑣𝑦 = 𝑣 sin(𝜃) a. 1.8 m/s 𝑣𝑦 3.0 b. 2.4 m/s = 𝑣= sin(𝜃) sin(37°) c. 3.7 m/s d. 5 m/s e. 6 m/s 27. One student throws a heavy red ball horizontally from a balcony of a tall building with an initial speed 𝑣𝑖 . 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. Both have 𝑣𝑦 = 0, like #13 and #18. b. The balls reach the ground at the same instant. 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. 28. With what speed do you need to throw a ball straight up to have it just barely reach a height of 3 m? a. 3 m/s 2 − 2𝑔Δ𝑦 𝑣𝑦2 = 𝑣0𝑦 b. 8 m/s 2 0 = 𝑣0𝑦 − 2(9.8)(3.0) c. 10 m/s d. 30 m/s 𝑣0𝑦 = √2(9.8)(3.0) = 7.7 e. 60 m/s 29. To drive to the store from home, somebody first travels 5 km west, then they travel 5 km northwest. What is the straight line distance between their home and the store? a. 4 km Δ𝑥 = (−5) + (5) cos(135°) = −8.54 b. 5 km Δ𝑦 = (0) + (5)sin(135°) = 3.54 c. 7 km Δ𝑟 = √8.542 + 3.542 = 9.24 d. 9 km e. 10 km...