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Unformatted text preview: Question 1, chap 4, sect 1. part 1 of 1 10 points A ball initially moves horizontally with ve locity vectorv i , as shown. It is then struck by a stick. After leaving the stick, the ball moves vertically with a velocity vectorv f , which has the magnitude as vectorv i . v i v f Which of the following vectors best repre sents the direction of the average force that the stick exerts on the ball? 1. 2. 3. 4. 5. None of these graphs is correct. 6. correct 7. 8. 9. Explanation: Basic Concept: Vector addition of velocities Solution: The vectorv f velocity is the resultant of vectorv i and vectorv s , so it is the diagonal of a parallelogram whose sides are vectorv i and vectorv s . v i v f v s Question 2, chap 4, sect 3. part 1 of 1 10 points Two 45 N forces and a 90 N force act on a hanging box as shown. 45 N 45 N 90 N Will the box experience acceleration? 1. Yes; downward. correct 2. Yes; upward. 3. Unable to determine; the angle is needed. 4. No; it is balanced. Explanation: The horizontal components of the two 45 N forces cancel, leaving an upward force that is less than 90 N. Thus, the net force on the box is down, causing it to accelerate downward. Question 3, chap 4, sect 3. part 1 of 3 10 points A 4.27 kg ball is dropped from the roof of a building 175.0 m high. While the ball is falling to Earth, a horizontal wind exerts a constant force of 13.9 N on the ball. The acceleration of gravity is 9 . 81 m / s 2 . a) How long does it take to hit the ground? Correct answer: 5 . 9731 s (tolerance 1 %). Explanation: Basic Concept: y = 1 2 g ( t ) 2 since v i,y = 0 m/s. Given: m = 4 . 27 kg y = 175 . 0 m g = 9 . 81 m / s 2 Solution: t = radicalBigg 2 y g = radicalBigg 2( 175 m) 9 . 81 m / s 2 = 5 . 9731 s Question 4, chap 4, sect 3. part 2 of 3 10 points b) How far from the building does the ball hit the ground? Correct answer: 58 . 0706 m (tolerance 1 %). Explanation: Basic Concepts: F x = ma x x = 1 2 a x ( t ) 2 since v i,x = 0 m/s. Given: F wind = 13 . 9 N Solution: a x = F wind m = 13 . 9 N 4 . 27 kg = 3 . 25527 m / s 2 x = 1 2 ( 3 . 25527 m / s 2 ) (5 . 9731 s) 2 = 58 . 0706 m Question 5, chap 4, sect 3. part 3 of 3 10 points c) What is its speed when it hits the ground? Correct answer: 61 . 7379 m / s (tolerance 1 %). Explanation: Basic Concepts: v x = a x t v y = g t v = radicalBig v 2 x + v 2 y Solution: v x = ( 3 . 25527 m / s 2 ) (5 . 9731 s) = 19 . 444 m / s v y = ( 9 . 81 m / s 2 ) (5 . 9731 s) = 58 . 5961 m / s Thus v = radicalBig (19 . 444 m / s) 2 + ( 58 . 5961 m / s) 2 = 61 . 7379 m / s Question 6, chap 4, sect 4. part 1 of 2 10 points The following 2 questions refer to the colli sions between a car and a truck whose weight is much heavier than the car ( M m ). For each description of a collision below, choose the one answer from the possibilities that best describes the size (or magnitude) of the forces between the car and the truck....
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This note was uploaded on 12/04/2008 for the course PHY 302K taught by Professor Kaplunovsky during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Kaplunovsky

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