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Test #1-solutions

# Test #1-solutions - Version 065/ABAAB Test#1...

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Version 065/ABAAB – Test #1 – Antoniewicz – (56445) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 9.0 points A planet with a mass of 1 × 10 23 kg travels around a star in a nearly circular obrbit in the xy plane as shown in the diagram. Its speed is constant at 15000 m / s. Star C A D B Which arrow best describes the direction of Δ P in going from B to C? a e c g h b d f 1. g . 2. a 3. c 4. e 5. h 6. d 7. b correct 8. f Explanation: We know that direction of Δ P is given by P B + Δ P = P C Δ P = P C - P B Thus, from the directions of P C and P B it follows that the correct answer is b 002 (part 1 of 3) 3.0 points A soccer ball is kicked at an angle of 60 to the horizontal, with an initial speed of 22 m / s. Assume for the moment that we can neglect air resistance. g = - 9 . 8 m / s 2 . How long is the ball in the air? 1. 4.0354 2. 3.34348 3. 3.88828 4. 3.80756 5. 3.76546 6. 3.11528 7. 3.14878 8. 3.21289 9. 3.30169 10. 3.38383 Correct answer: 3 . 88828 s. Explanation: To determine how long the ball is in the air, we can use the position update formula derived from the momentum principle: y f = y i + v i,y Δ t + 1 2 a y Δ t 2 . We just need to fill in the quantities that we know. Clearly, y f = y i = 0 m and a y = g = - 9 . 8 m / s 2 . The only other thing we need to solve for the time is the initial velocity in the y direction. We obtain this from trigonometry: v i,y = v sin 60 = (22 m / s) sin 60 = 19 . 0526 m / s . So, rearranging the above equation to solve for the time, we obtain

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Version 065/ABAAB – Test #1 – Antoniewicz – (56445) 2 0 = v i,y Δ t + 1 2 a y Δ t 2 v i,y Δ t = - 1 2 a y Δ t 2 Δ t = - 2 v i,y a y = 2 (19 . 0526 m / s) 9 . 8 m / s 2 = 3 . 88828 s . 003 (part 2 of 3) 3.0 points How far does the ball go (horizontal distance along the field)? 1. 26.7471 2. 46.4093 3. 33.8383 4. 35.348 5. 42.7711 6. 37.8442 7. 32.1637 8. 45.118 9. 26.0526 10. 30.6538 Correct answer: 42 . 7711 m. Explanation: There is no force in the x direction, so v x is constant. This makes this part simple: v x = Δ x Δ t Δ x = v x Δ t = ( v cos 60 )( Δ t ) = (22 m / s) cos 60 (3 . 88828 s) = 42 . 7711 m . 004 (part 3 of 3) 3.0 points How high does it go? 1. 18.5204 2. 19.2513 3. 16.5698 4. 11.8886 5. 18.8898 6. 11.3615 7. 13.6941 8. 18.1435 9. 14.3545 10. 13.1235 Correct answer: 18 . 5204 m. Explanation: To find the height, we can use the position update formula again, although this time we are solving for a di ff erent variable. Now we want to find y f , and we need to plug in the correct time that it takes to reach the peak. Since the trip up and the trip down take the same amount of time, we can simply divide the total flight time in half to find that Δ t peak = Δ t 2 = 1 . 94414 s . Now we just plug into the formula above: y f = y i + v i,y Δ t peak + 1 2 a y Δ t 2 peak = 0 + (19 . 0526 m / s)(1 . 94414 s) + 1 2 ( - 9 . 8 m / s 2 )(1 . 94414 s) 2 = 18 . 5204 m . 005 9.0 points A proton of mass m p is moving along the -ˆ z direction and undergoes a change in its speed from 0.994c to 0.998c. What is the magnitude and direction of the impulse acting on the proton?
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