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Test #1 -solutions

# Test #1 -solutions - Version 022/AABBC Test#1...

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Version 022/AABBC – 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 9.0points A rubber ball is dropped from rest onto the floor, and bounces back up to the same height from which it started. Ignore the force of fric- tion due to the air. Which of the following sets of plots most accurately depict this mo- tion? (The force plots depict the force on the ball by the environment.) 1. t y t v y t F y 2. t y t v y t F y 3. t y t v y t F y 4. t y t v y t F y correct 5. t y t v y t F y Explanation: Since there is a constant force, gravity, downward, the plot of the y-position should be parabolic with negative slope whose mag- nitude is increasing over time, until it hits the floor. At this point the slope should switch sign and point upward and decrease in mag- nitude over time. The velocity should be ini- tially 0 m/s since the ball started from rest, and decrease linearly until it hits the floor, at which point it is very quickly given a boost to a positive value, and again continues to de- crease linearly. The force should be always a negative constant value, except for when the ball hits the floor, at which point it should be a narrow positive peak to indicate the brief upward force exerted on the ball by the floor. 002(part1of3)3.0points A basketball player wishes to throw the ball to a teammate across the court. If he throws the ball at a 30 angle, with what initial speed must he throw the ball so that his teammate receives the ball in t flight = 0 . 69 s? Use g = 9 . 8 m / s 2 . 1. 14.994 2. 10.29 3. 6.762 4. 8.82 5. 9.408 6. 7.644 7. 14.7 8. 15.288 9. 15.582 10. 12.348 Correct answer: 6 . 762 m / s. Explanation: Using the momentum principle in the y direction, p f,y = p i,y + F net ,y Δ t 0 = p i,y m g parenleftbigg t flight 2 parenrightbigg v i,y = g parenleftbigg t flight 2 parenrightbigg and since v i,y = | vectorv i | sin θ , | vectorv i | = g t flight 2 sin θ = 6 . 762 m / s .

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Version 022/AABBC – Test #1 – Antoniewicz – (56445) 2 003(part2of3)3.0points What is the highest point along the trajec- tory, vectorr apex = ( r apex,x , r apex,y , 0 ) , relative to the thrower’s hand ? (Take the initial loca- tion to be the r i = ( 0 , 0 , 0 ) m.) First give r apex ,x . 1. 6.11068 2. 6.73702 3. 6.42003 4. 13.2945 5. 2.02034 6. 5.51488 7. 3.91083 8. 4.94965 9. 3.43725 10. 9.93367 Correct answer: 2 . 02034 m. Explanation: The highest point occurs at Δ t = t flight / 2. For the x coordinate, r apex ,x = v i,x t flight 2 = ( v i cos θ ) t flight 2 = 2 . 02034 m . 004(part3of3)3.0points Find r apex,y . 1. 1.6769 2. 0.927202 3. 1.0595 4. 0.583223 5. 2.03852 6. 3.45744 7. 3.8378 8. 2.8676 9. 0.74529 10. 0.63504 Correct answer: 0 . 583223 m. Explanation: Since the net force in the y direction is nonzero, we have to use the appropriate posi- tion update formula here: r apex ,y = v i,y t flight 2 1 2 g parenleftbigg t flight 2 parenrightbigg 2 = ( v i sin θ ) t flight 2 1 8 g t 2 flight = 0 . 583223 m . 005(part1of3)3.0points A soccer ball is kicked at an angle of 59 to the horizontal, with an initial speed of 18 m / s.
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