HW 03 solution

# HW 03 solution - gutierrez(ig3472 HW3 ditmire(58216 This...

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gutierrez (ig3472) – HW3 – ditmire – (58216) 1 This print-out should have 24 questions. Multiple-choice 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 10 m v 10 m 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 1-dimension 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 .

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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 - ˆ r and gravity is down - ˆ 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 cannon’s barrel is at height 6 . 9 m, as shown in the figure. The initial velocity of the projectile is 11 m / s . The projectile rises to a maximum height of Δ y above the end of the cannon’s barrel and strikes the ground a horizontal distance Δ x past the end of the cannon’s barrel. Δ x 11 m / s 56 Δ y 6 . 9 m Determine the vertical component of the initial velocity at the end of the cannon’s bar- rel, where the projectile began its trajectory.
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