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Unformatted text preview: schroeder (cs39963) – Quiz 4: Two Dimensional Motion  Projectile and Circular Moton – balasubraman This printout should have 31 questions. Multiplechoice 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 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 = v cos θ = (166 m / s) cos23 ◦ = 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 = schroeder (cs39963) – Quiz 4: Two Dimensional Motion  Projectile and Circular Moton – balasubraman (4 m / s 2 )ˆ . 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 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) + vectorat , so vectorv (0) = vectorv ( t ) − vectorat = (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 = (ˆ ı − ˆ ) (4 m)....
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This note was uploaded on 12/02/2010 for the course PHYS 1401 taught by Professor Staff during the Fall '08 term at Texas A&M University, Corpus Christi.
 Fall '08
 STAFF
 Physics

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