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Unformatted text preview: homework 05 – RAHMAN, TARIQUE – Due: Feb 11 2008, 3:00 am 1 Question 1, chap 4, sect 4. part 1 of 1 10 points A ball is thrown and follows the parabolic path shown. Air friction is negligible. Point Q is the highest point on the path. Points P and R are the same height above the ground. Q R P Which of the following diagrams best indi cates the direction of the acceleration, if any, on the ball at point R ? 1. 2. 3. 4. The ball is in freefall and there is no acceleration at any point on its path. 5. correct 6. 7. 8. 9. Explanation: Since air friction is negligible, the only ac celeration on the ball after being thrown is that due to gravity, which acts straight down. Question 2, chap 4, sect 4. part 1 of 1 10 points A ball is thrown and follows the parabolic path shown. Air friction is negligible. Point Q is the highest point on the path. Points P and R are the same height above the ground. Q R P How do the speeds of the ball at the three points compare? 1. bardbl vectorv P bardbl < bardbl vectorv Q bardbl < bardbl vectorv R bardbl 2. bardbl vectorv P bardbl = bardbl vectorv R bardbl = bardbl vectorv Q bardbl 3. bardbl vectorv Q bardbl < bardbl vectorv P bardbl = bardbl vectorv R bardbl correct 4. bardbl vectorv P bardbl = bardbl vectorv R bardbl < bardbl vectorv Q bardbl 5. bardbl vectorv R bardbl < bardbl vectorv Q bardbl < bardbl vectorv P bardbl 6. bardbl vectorv Q bardbl < bardbl vectorv R bardbl < bardbl vectorv P bardbl Explanation: The speed of the ball in the xdirection is constant. Because of gravitational accelera tion, the speed in the ydirection is zero at point Q . Since points P and R are located at the same point above ground, by symmetry we see that they have the same vertical speed component (though they do not have the same homework 05 – RAHMAN, TARIQUE – Due: Feb 11 2008, 3:00 am 2 velocity). The answer is then “ v Q < v P = v R ”. Question 3, chap 4, sect 4. part 1 of 3 10 points A projectile is fired with an initial speed v at t = 0. The angle between the initial velocity v and the horizontal plane is α . x y v α y max R A B The time t max it takes for the projectile to reach its maximum height is 1. t max = v cos α g 2. t max = v cos α 2 g 3. t max = v g 4. t max = g v 5. t max = v sin α g correct 6. t max = v g 7. t max = v 8. t max = v sin α 2 g Explanation: Basic Concepts: For two dimensional projectile motion in a gravitational field the acceleration is due to gravity only and acts exclusively on the y component of velocity....
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This note was uploaded on 02/03/2010 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner
 Physics, Friction, Work

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