EwS9zw6j0I9O_1210015966_jwk572

# EwS9zw6j0I9O_1210015966_jwk572 - homework 03 – KIM JI –...

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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.

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EwS9zw6j0I9O_1210015966_jwk572 - homework 03 – KIM JI –...

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