Midterm 02-solutions

# Midterm 02-solutions - Version 048 Midterm 02 florin(56930...

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Version 048 – Midterm 02 – forin – (56930) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices be±ore answering. 001 10.0 points Three balls with masses: m , 2 m and 1 2 m , re- spectively are thrown ±rom the top o± a build- ing, all with the same initial speed v i . The Frst ball is thrown horizontally, the second at some angle θ 1 above the horizontal, and the third with some angle θ 2 < θ 1 below the hor- izontal. Neglecting air resistance, rank the speeds o± the balls as they reach the ground, ±rom the slowest to the ±astest: 1. 1, 2, 3 2. 2, 1, 3 3. 1, 3, 2 4. 3, 2, 1 5. 3, 1, 2 6. All three balls strike the ground with the same speed. correct 7. 2, 3, 1 Explanation: Consider each ball and the earth as three separate systems. There is no work done on these systems by the surroundings so Δ E sys = W surr = 0. Next, consider the system that consists o± the ball with mass m and the earth. Ini- tially, it has a total energy (ignoring rest mass energy) o± E sys,i = K i + U i = 1 2 mv 2 i + mg y i and a Fnal energy o± E sys,f = K f + U f = 1 2 2 f + f so Δ E sys = 1 2 m ( v 2 f v 2 i ) + mg ( y f y i ) = 0 Solving this equation ±or v f gives v f = r g ( y f y i ) + v 2 i The result does not depend on the mass o± the ball so the same expression can be derived ±or each one. All three balls begin with the same speed v i and experience the same change in height ( y f y i ) so the Fnal speed ±or all three is the same . 002 10.0 points A bucket containing a rock o± mass m is ro- tated in a vertical circle o± radius 1 . 469 m. What must be the minimum speed o± the pail at the top o± the circle so that the rock won’t ±all out? 1. 3.0043 2. 3.05123 3. 3.20475 4. 3.79423 5. 2.49656 6. 2.54515 7. 2.87085 8. 2.68019 9. 3.74745 10. 3.6695 Correct answer: 3 . 79423 m / s. Explanation: The system is the rock. In order ±or the rock to move in a circle o± radius 1 . 469 m, it ex- periences a net ±orce equal to the centripetal ±orce necessary to maintain the circular mo- tion. A ±ree body diagram o± the system includes only the ±orce o± gravity and the nor- mal ±orce exerted by the bottom o± the bucket, so F net = F N + F g . When moving with the minimum velocity, the rock is just about to ±all out, so the normal ±orce is zero and the centripetal acceleration is supplied by gravity alone.

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Version 048 – Midterm 02 – forin – (56930) 2 F N + F g = mg = = ma centripetal = m v 2 r . Then, mg = m v 2 r and so, v = g r = r (9 . 8 m / s 2 ) (1 . 469 m) = 3 . 79423 m / s . 003 10.0 points The escape speed From a very small asteroid is only 28 m / s. IF you throw a rock away From the asteroid at a speed oF 48 m / s, what will be its ±nal speed? G = 6 . 7 × 10 11 N · m 2 kg 2 . 1. 40.8901 2. 31.749 3. 37.081 4. 38.9872 5. 42.5323 6. 28.2666 7. 45.5961 8. 33.9411 9. 22.6495 10. 22.2711 Correct answer: 38 . 9872 m / s. Explanation: v esc = R 2 GM R v 2 esc = 2 GM R Use the Energy Principle. E i = E f U i + K i = U f + K f GMm r i + 1 2 mv 2 i = 0 + 1 2 mv 2 f Where U f is zero.
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Midterm 02-solutions - Version 048 Midterm 02 florin(56930...

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