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Unformatted text preview: test 02 ALLEN, HUNTER Due: Mar 20 2008, 6:00 pm 1 Question 1, chap 9, sect 11. part 1 of 1 10 points An open cart on a level surface rolls without frictional loss through a downpour of rain. The rain falls vertically downward as shown below. As the cart rolls, an appreciable amount of rain water accumulates in the cart. v rain rain water cart The speed of the cart will 1. decrease because of conservation of mo mentum. correct 2. remain the same because the raindrops are falling perpendicular to the direction of carts motion. 3. increase because of conservation of mo mentum. 4. increase because of conservation of me chanic energy. 5. decrease because of conservation of me chanic energy. Explanation: This is an inelastic collision in the direction along which the cart is rolling. Only mo mentum vectorp along that direction is conserved. Because the raindrops fall vertically, they do not carry momentum horizontally. Assume m of rain water accumulates on the cart: p i = p f mv = ( m + m ) v . Therefore v = m m + m v v < v . The speed of the cart will decrease because of conservation of momentum. Question 2, chap 8, sect 4. part 1 of 1 10 points Tim, with mass 74 kg, climbs a gymnasium rope a distance of 5 . 1 m. The acceleration of gravity is 9 . 8 m / s 2 . How much potential energy does Tim gain? 1. 2782 . 42 J 2. 2870 . 22 J 3. 2960 . 78 J 4. 3057 . 6 J 5. 3156 . 38 J 6. 3256 . 34 J 7. 3368 . 46 J 8. 3476 . 65 J 9. 3584 . 84 J 10. 3698 . 52 J correct Explanation: Let : m = 74 kg , h = 5 . 1 m , and g = 9 . 8 m / s 2 . Potential energy is U = mg h = (74 kg) (9 . 8 m / s 2 ) (5 . 1 m) = 3698 . 52 J . Question 3, chap 10, sect 4. part 1 of 1 10 points A potters wheel moves from rest to an angular speed of 0.42 rev/s in 35.9 s. Assuming constant angular acceleration, what is its angular acceleration in rad/s 2 ? 1. . 0626669 rad / s 2 2. . 0647751 rad / s 2 3. . 066798 rad / s 2 4. . 0688854 rad / s 2 5. . 0712526 rad / s 2 6. . 073508 rad / s 2 correct test 02 ALLEN, HUNTER Due: Mar 20 2008, 6:00 pm 2 7. . 0757972 rad / s 2 8. . 0781779 rad / s 2 9. . 0807015 rad / s 2 10. . 083202 rad / s 2 Explanation: Basic Concept: f = i + t = t since i = 0 rad / s. Given: f = 0 . 42 rev / s t = 35 . 9 s Solution: = f t = . 42 rev / s 35 . 9 s 2 rad 1 rev = 0 . 073508 rad / s 2 Question 4, chap 10, sect 10. part 1 of 1 10 points A 26 kg mass and a 15 kg mass are sus pended by a pulley that has a radius of 8 . 3 cm and a mass of 7 kg. The cord has a negligible mass and causes the pulley to rotate without slipping. The pulley rotates without friction. The masses start from rest 4 . 1 m apart. Treat the pulley as a uniform disk....
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 Spring '07
 KOPP
 Physics, Friction

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