solution6_pdf - arellano (aa39398) – Practice Homework 3...

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Unformatted text preview: arellano (aa39398) – Practice Homework 3 Solutions – Weathers – (17101) 1 This print-out should have 57 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 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.

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solution6_pdf - arellano (aa39398) – Practice Homework 3...

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