Practice Homework 3 Solutions

Practice Homework 3 Solutions - practice 03 – ALIBHAI...

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Unformatted text preview: practice 03 – ALIBHAI, ZAHID – Due: Feb 4 2007, 4:00 am 1 Question 1 part 1 of 2 10 points A particle travels between two parallel ver- tical walls separated by 13 m. It moves to- ward the opposing wall at a constant rate of 7 . 2 m / s. It hits the opposite wall at the same height. The acceleration of gravity is 9 . 8 m / s 2 . 13 m 9 . 8m / s 2 7 . 2 m / s a) What will be its speed when it hits the opposing wall? Correct answer: 11 . 4067 m / s (tolerance ± 1 %). Explanation: Let : d = 13 m , v x = 7 . 2 m / s , g = 9 . 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 g 8 . 84722m / s 7 . 2 m / s 1 1 . 4 6 7 m / s 3 9 . 1 3 9 2 3 . 99354 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 = (13 m) (7 . 2 m / s) = 1 . 80556 s . is the time for the particle to reach the oppo- site wall. Vertically, the particle reaches its maxi- mum height at t h = t f 2 = d 2 v x , so the initial vertical velocity v y is v y = g t h = g d 2 v x = (9 . 8 m / s 2 ) (13 m) 2 (7 . 2 m / s) = 8 . 84722 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 (7 . 2 m / s) 2 + (8 . 84722 m / s) 2 = 11 . 4067 m / s . Question 2 part 2 of 2 10 points b) At what angle with the wall will the particle strike? Correct answer: 39 . 1392 ◦ (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 the angle, so tan θ = v x v y , so θ = arctan parenleftbigg v x v y parenrightbigg = arctan parenleftbigg 7 . 2 m / s 8 . 84722 m / s parenrightbigg = 39 . 1392 ◦ . practice 03 – ALIBHAI, ZAHID – Due: Feb 4 2007, 4:00 am 2 Note: The particle reaches a height h = 1 2 g t f 2 2 = 1 2 (9 . 8 m / s 2 ) (1 . 80556 s) 2 2 = 3 . 99354 m . Question 3 part 1 of 1 10 points A target lies flat on the ground 7 m from the side of a building that is 10 m tall, as shown below. The acceleration of gravity is 10 m / s 2 . Air resistance is negligible. A student rolls a 7 kg ball off the horizontal roof of the building in the direction of the target. 7 m 10m v 10m The horizontal speed v with which the ball must leave the roof if it is to strike the target is most nearly 1. v = √ 2 7 m/s. 2. v = 7 √ 3 m/s. 3. v = √ 5 7 m/s. 4. v = 7 √ 3 3 m/s. 5. v = 3 √ 7 m/s. 6. v = 2 √ 7 m/s. 7. v = 7 √ 5 m/s. 8. v = 7 √ 2 m/s. 9. v = 7 √ 2 2 m/s. correct 10. v = √ 3 7 m/s. Explanation: m = 7 kg , not required h = 10 m , x = 7 m , and g = 10 m / s 2 . Observe the motion in the vertical direction only and it is a purely 1-dimension movement with a constant acceleration. So the time need for the ball to hit the ground is t = radicalBigg 2 h g and the horizontal speed should be v = x t for the ball to hit the target. Therefore v = x radicalbigg g 2 h = (7 m) radicalBigg 10 m / s 2 2 (10 m) = 7 √ 2 m / s = 7 √ 2...
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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Practice Homework 3 Solutions - practice 03 – ALIBHAI...

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