Quiz 4 Solutions

# Quiz 4 Solutions - schroeder(cs39963 Quiz 4 Two Dimensional...

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schroeder (cs39963) – Quiz 4: Two Dimensional Motion - Projectile and Circular Moton – balasubraman This print-out should have 31 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A projectile is fired straight upward at 166 m / s. How fast is it moving at the instant it reaches the top of its trajectory? Correct answer: 0 m / s. Explanation: At the top of the trajectory the vertical component of the speed of the projectile is zero, so the speed is simply the horizontal component of the initial velocity. v = v x 0 = 0 m / s 002 (part 2 of 2) 10.0 points How fast is it moving at the instant it reaches the top of its trajectory if the projectile is fired upward at 23 from the horizontal? Correct answer: 152 . 804 m / s. Explanation: v = v x 0 = v cos θ = (166 m / s) cos 23 = 152 . 804 m / s . 003 (part 1 of 3) 10.0 points A boy throws a ball upward. Compare the magnitudes of the gravitational accelerations at three points along the path of the ball. Point A is before the ball reaches the top. a) Point A on the way up. b) Point B is at the top. c) Point C is after it has passed the top and on the way down. B A C The magnitudes of the acceleration are re- lated as 1. a A < a B . 2. a A = g . correct Explanation: The gravitational acceleration near the sur- face of the earth is a constant, so the correct choice is a A = g . 004 (part 2 of 3) 10.0 points The magnitudes of the acceleration are re- lated as 1. a B > a C . 2. a B = a A . correct 3. a B = 0 . Explanation: The gravitational acceleration near the sur- face of the earth is a constant, so the correct choice is a B = a A . 005 (part 3 of 3) 10.0 points The magnitudes of the acceleration are re- lated as 1. a C = a B = a A . correct 2. a C < a B and a A < a B . Explanation: The gravitational acceleration near the sur- face of the earth is a constant, so the correct choice is a C = a B = a A . 006 (part 1 of 2) 10.0 points A particle has a constant acceleration vectora =

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schroeder (cs39963) – Quiz 4: Two Dimensional Motion - Projectile and Circular Moton – balasubraman (4 m / s 2 . At t = 2 sec, it has velocity vectorv = (ˆ ı + ˆ ) (2 m / s). What is its velocity at t = 0? 1. ˆ ı (4 m / s) + ˆ (4 m / s) 2. vector 0 3. ˆ ı (2 m / s) + ˆ (6 m / s) 4. ˆ ı (2 m / s) ˆ (6 m / s) correct 5. ˆ ı (4 m / s) ˆ (4 m / s) Explanation: vectorv ( t ) = vectorv (0) + vectora t , so vectorv (0) = vectorv ( t ) vectora t = (ˆ ı + ˆ ) (2 m / s) ˆ (4 m / s 2 )(2 s) = ˆ ı (2 m / s) + ˆ (2 m / s 8 m / s) = ˆ ı (2 m / s) ˆ (6 m / s) . 007 (part 2 of 2) 10.0 points After 2 sec, its position is found to be vectorr = ı ˆ ) (4 m). Where was it at t = 0? 1. ˆ ı (2 m) + ˆ (6 m) 2. ˆ ı (4 m) ˆ (4 m) 3. ˆ ı (2 m) ˆ (6 m) 4. vector 0 correct 5. ˆ ı (4 m) + ˆ (4 m) Explanation: vectorr ( t ) = vectorr (0) + vectorv (0) t + 1 2 vectora t 2 , so vectorr (0) = vectorr ( t ) vectorv (0) t 1 2 vectora t 2 = (ˆ ı ˆ ) (4 m) ı (2 m / s) ˆ (6 m / s)](2 s) 1 2 (4 m / s 2 (2 s) 2 = ˆ ı (4 m 4 m) ( 4 m + 12 m 8 m) = vector 0 .
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