This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 = v x v y , so = 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 1dimension 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...
View Full
Document
 Spring '08
 Turner
 Work

Click to edit the document details