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Unformatted text preview: homework 08 HETTIGER, CHRISTOF Due: Feb 5 2008, 4:00 am 1 Question 1, chap 4, sect 4. part 1 of 3 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 R bardbl < bardbl vectorv Q bardbl 2. bardbl vectorv P bardbl < bardbl vectorv Q bardbl < bardbl vectorv R bardbl 3. bardbl vectorv Q bardbl < bardbl vectorv R bardbl < bardbl vectorv P bardbl 4. bardbl vectorv R bardbl < bardbl vectorv Q bardbl < bardbl vectorv P bardbl 5. bardbl vectorv Q bardbl < bardbl vectorv P bardbl = bardbl vectorv R bardbl correct Explanation: The speed of the ball in the x-direction is constant. Because of gravitational accelera- tion, the speed in the y-direction 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 velocity). The answer is then v Q < v P = v R . Question 2, chap 4, sect 4. part 2 of 3 10 points Which of the following diagrams best indi- cates the direction of the acceleration, if any, on the ball at point R ? 1. 2. 3. The ball is in free-fall and there is no acceleration at any point on its path. 4. 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 3, chap 4, sect 4. part 3 of 3 10 points Which of the following diagrams best indi- cates the direction of the net force, if any, on the ball at point Q ? 1. 2. 3. 4. The ball is in free-fall and there is no homework 08 HETTIGER, CHRISTOF Due: Feb 5 2008, 4:00 am 2 acceleration at any point on its path. 5. 6. 7. correct 8. 9. Explanation: By Newtons second law, the force is in the direction of the acceleration (downward). Question 4, 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 6 m at your feet, then a horizontal shelf of 5 m , then another drop of 4 m to the bottom of the canyon, which has a horizontal floor. You kick a 0 . 44 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. 5 m x 10m v 6 m 4 m What initial horizontal velocity v will be required to barely clear the edge of the shelf below you? Correct answer: 4 . 51848 m / s (tolerance 1 %). Explanation: Let : y 1 = 6 m , y 2 = 4 m , and x = 5 m ....
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