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Unformatted text preview: homework 03 – KIM, JI – Due: Sep 19 2007, 4:00 am 1 Question 1, chap 4, sect 3. part 1 of 2 10 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 6 . 7 m / s. Also, it has an acceleration in the direction parallel to the walls of 2 . 9 m / s 2 . It hits the opposite wall at a higher point. 18 . 4 m 2 . 9m / s 2 6 . 7 m / s a) What will be its speed when it hits the opposing wall? Correct answer: 10 . 4076 m / s (tolerance ± 1 %). Explanation: Let : d = 18 . 4 m , v x = 6 . 7 m / s , a = 2 . 9 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 7 . 96418m / s 6 . 7 m / s 1 . 4 7 6 m / s 4 . 7 2 8 ◦ 10 . 9359 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) (6 . 7 m / s) = 2 . 74627 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 = (2 . 9 m / s 2 ) (18 . 4 m) (6 . 7 m / s) = 7 . 96418 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 (6 . 7 m / s) 2 + (7 . 96418 m / s) 2 = 10 . 4076 m / s . Question 2, chap 4, sect 3. part 2 of 2 10 points homework 03 – KIM, JI – Due: Sep 19 2007, 4:00 am 2 b) At what angle with the wall will the particle strike? (Here the angle is between 0 to 90 degrees.) Correct answer: 40 . 0728 ◦ (tolerance ± 1 %). Explanation: When the particle strikes the wall, the ver tical component is the side adjacent and the horizontal component is the side opposite to the angle, so tan θ = v x v y , so θ = arctan parenleftbigg v x v y parenrightbigg = arctan parenleftbigg 6 . 7 m / s 7 . 96418 m / s parenrightbigg = 40 . 0728 ◦ . Note: The distance traveled parallel to the walls is y = 1 2 a t 2 = 1 2 (2 . 9 m / s 2 ) (2 . 74627 s) 2 = 10 . 9359 m . Question 3, chap 4, sect 4. part 1 of 2 10 points You are standing at the top of a cliff that has a stairstep configuration. There is a verti cal drop of 7 m at your feet, then a horizontal shelf of 9 m , then another drop of 3 m to the bottom of the canyon, which has a horizontal floor. You kick a 0 . 27 kg rock, giving it an initial horizontal velocity that barely clears the shelf below. The acceleration of gravity is 9 . 8 m / s 2 . Consider air friction to be negligi ble. 9 m Δ x 10m v 7 m 3 m What initial horizontal velocity v will be required to barely clear the edge of the shelf below you? Correct answer: 7 . 52994 m / s (tolerance ± 1 %)....
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This note was uploaded on 05/05/2008 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner
 Physics, Work

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