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Unformatted text preview: arellano (aa39398) – Practice Homework 3 Solutions – Weathers – (17101) 1 This printout should have 57 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 particle travels horizontally between two parallel walls separated by 18 . 4 m. It moves toward the opposing wall at a constant rate of 8 . 4 m / s. Also, it has an acceleration in the direction parallel to the walls of 4 . 8 m / s 2 . 18 . 4 m 4 . 8m / s 2 8 . 4 m / s What will be its speed when it hits the opposing wall? Correct answer: 13 . 4577 m / s. Explanation: Let : d = 18 . 4 m , v x = 8 . 4 m / s , a = 4 . 8 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 10 . 5143m / s 8 . 4 m / s 1 3 . 4 5 7 7 m / s 3 8 . 6 2 1 8 ◦ 11 . 5156 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) (8 . 4 m / s) = 2 . 19048 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 = (4 . 8 m / s 2 ) (18 . 4 m) (8 . 4 m / s) = 10 . 5143 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 (8 . 4 m / s) 2 + (10 . 5143 m / s) 2 = 13 . 4577 m / s . 002 (part 2 of 2) 10.0 points arellano (aa39398) – Practice Homework 3 Solutions – Weathers – (17101) 2 At what angle with the wall will the particle strike? Correct answer: 38 . 6218 ◦ . 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 8 . 4 m / s 10 . 5143 m / s parenrightbigg = 38 . 6218 ◦ . Note: The distance traveled parallel to the walls is y = 1 2 a t 2 = 1 2 (4 . 8 m / s 2 ) (2 . 19048 s) 2 = 11 . 5156 m . 003 10.0 points Initially (at time t = 0) a particle is moving vertically at 5 . 2 m / s and and horizontally at 0 m / s. The particle accelerates horizontally at 1 . 9 m / s 2 . The acceleration of gravity is 9 . 8 m / s 2 . At what time will the particle be traveling at 54 ◦ with respect to the horizontal? Correct answer: 0 . 418844 s. Explanation: Basic Concept Kinematic Equation v = v o + a t . Solution The vertical velocity is v y t = v y g t . The horizontal velocity is v x t = v y + a t = a t . v x t v y t v t 54 ◦ The vertical component is the opposite side and the horizontal component is the adjacent side to the angle, so tan θ = v y t v x t = v y g t a t a t tan θ = v y g t a t tan θ + g t = v y t = v y a tan θ + g = (5 . 2 m / s) (1 . 9 m / s 2 ) tan(54 ◦ ) + (9 . 8 m / s 2 ) = . 418844 s ....
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This note was uploaded on 10/31/2010 for the course PHYSICS 1710 taught by Professor Weathers during the Fall '10 term at North Texas.
 Fall '10
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