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# solution_5pdf - armington(kma786 – hw0208 – fiete...

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Unformatted text preview: armington (kma786) – hw0208 – fiete – (57165) 1 This print-out should have 12 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 particle travels horizontally between two parallel walls separated by 18 . 4 m. It moves toward the opposing wall at a constant rate of 5 . 5 m / s. Also, it has an acceleration in the direction parallel to the walls of 1 . 5 m / s 2 . 18 . 4 m 1 . 5m / s 2 5 . 5 m / s What will be its speed when it hits the opposing wall? Correct answer: 7 . 44528 m / s. Explanation: Let : d = 18 . 4 m , v x = 5 . 5 m / s , a = 1 . 5 m / s 2 , Basic Concepts Kinematics equations v = v o + g t s = s o + v o t + 1 2 g t 2 d a 5 . 01818m / s 5 . 5 m / s 7 . 4 4 5 2 8 m / s 4 7 . 6 2 2 8 ◦ 8 . 39405 m The horizontal motion will carry the parti- cle to the opposite wall, so d = v x t f and t f = d v x = (18 . 4 m) (5 . 5 m / s) = 3 . 34545 s . is the time for the particle to reach the oppo- site wall. Horizontally, the particle reaches the maxi- mum parallel distance when it hits the oppo- site wall at the time of t = d v x , so the final parallel velocity v y is v y = a t = a d v x = (1 . 5 m / s 2 ) (18 . 4 m) (5 . 5 m / s) = 5 . 01818 m / s . The velocities act at right angles to each other, so the resultant velocity is v f = radicalBig v 2 x + v 2 y = radicalBig (5 . 5 m / s) 2 + (5 . 01818 m / s) 2 = 7 . 44528 m / s . 002 (part 2 of 2) 10.0 points armington (kma786) – hw0208 – fiete – (57165) 2 At what angle with the wall will the particle strike? Correct answer: 47 . 6228 ◦ . Explanation: When the particle strikes the wall, the ver- tical component is the side adjacent and the horizontal component is the side opposite the angle, so tan θ = v x v y , so θ = arctan parenleftbigg v x v y parenrightbigg = arctan parenleftbigg 5 . 5 m / s 5 . 01818 m / s parenrightbigg = 47 . 6228 ◦ . Note: The distance traveled parallel to the walls is y = 1 2 a t 2 = 1 2 (1 . 5 m / s 2 ) (3 . 34545 s) 2 = 8 . 39405 m . 003 10.0 points Initially (at time t = 0) a particle is mov- ing vertically at 7 . 9 m / s and horizontally at 0 m / s. Its horizontal acceleration is 1 . 8 m / s 2 . At what time will the particle be traveling at 39 ◦ with respect to the horizontal? The acceleration due to gravity is 9 . 8 m / s 2 ....
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solution_5pdf - armington(kma786 – hw0208 – fiete...

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