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Unformatted text preview: gutierrez (ig3472) HW3 ditmire (58216) 1 This printout should have 24 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A target lies flat on the ground 9 m from the side of a building that is 10 m tall, as shown below. The acceleration of gravity is 10 m / s 2 . Air resistance is negligible. A student rolls a 9 kg ball off the horizontal roof of the building in the direction of the target. 9 m 10m v 10m The horizontal speed v with which the ball must leave the roof if it is to strike the target is most nearly 1. v = 9 m/s. 2. v = 6 m/s. 3. v = 9 2 m/s. 4. v = 9 5 m/s. 5. v = 3 3 m/s. 6. v = 9 3 m/s. 7. v = 9 2 2 m/s. correct 8. v = 2 9 m/s. 9. v = 3 9 m/s. 10. v = 5 9 m/s. Explanation: m = 9 kg , not required h = 10 m , x = 9 m , and g = 10 m / s 2 . Observe the motion in the vertical direction only and it is a purely 1dimension movement with a constant acceleration. So the time need for the ball to hit the ground is t = radicalBigg 2 h g and the horizontal speed should be v = x t for the ball to hit the target. Therefore v = x radicalbigg g 2 h = (9 m) radicalBigg 10 m / s 2 2 (10 m) = 9 2 m / s = 9 2 2 m / s . 002 10.0 points A particle moving at a velocity of 7 . 1 m / s in the positive x direction is given an accelera tion of 4 . 5 m / s 2 in the positive y direction for 1 . 8 s. What is the final speed of the particle? Correct answer: 10 . 7713 m / s. Explanation: Let : v xf = v xi = 7 . 1 m / s . a y = 4 . 5 m / s 2 , and t = 1 . 8 s . gutierrez (ig3472) HW3 ditmire (58216) 2 The vertical velocity undergoes constant ac celeration: v yf = v yi + a t = 0 + (4 . 5 m / s 2 )(1 . 8 s) = 8 . 1 m / s . Thus v f = radicalBig v 2 xf + v 2 yf = radicalBig (7 . 1 m / s) 2 + (8 . 1 m / s) 2 = 10 . 7713 m / s . 003 10.0 points A mass slides with negligible friction on a circular track of 1 m radius oriented vertically. Its speed at the position shown in the figure is 3.13 m/s. The acceleration of gravity is 9.8 m/s 2 . v At the position shown in the figure, which of the labeled arrows best represents the di rection of the acceleration of the mass? 1. 2. 3. 4. correct 5. The mass is traveling at a constant veloc ity, therefore it has no acceleration. 6. 7. 8. 9. Explanation: The magnitude of the centripetal is a r = v 2 r = (3 . 13 m / s) 2 1 m 9 . 8 m / s 2 . The centripetal acceleration is inward r and gravity is down k . a a r g 004 (part 1 of 7) 10.0 points A projectile of mass 0 . 687 kg is shot from a cannon. The end of the cannons barrel is at height 6 . 9 m, as shown in the figure. The initial velocity of the projectile is 11 m / s ....
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This note was uploaded on 04/13/2010 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics

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