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: oldhomewk 22 YOO, HEE Due: Apr 1 2008, 4:00 am 1 Question 1, chap 11, sect 2. part 1 of 1 10 points Objects having the same mass travel toward each other on a flat surface, each with a speed of 1.0 meter per second relative to the surface. The objects collide headon and are reported to rebound after the collision, each with a speed of 2.0 meters per second relative to the surface. Which of the following assessments of this report is most accurate? 1. Momentum was not conserved; therefore the report is false. 2. If there were no friction between the ob jects and the surface, the report could be true. 3. If the surface were inclined, the report could be true. 4. If potential energy were released to the objects during the collision, the report could be true. correct 5. If the objects had different masses, the report could be true. Explanation: P i = m 1 v 1 m 2 v 2 = 0 P f = m 2 v 2 m 1 v 1 = 0 The total momentum is conserved here. However, v 1 = v 2 = 2 v 1 = 2 v 2 , so energy must be added to the system to make the process physically realistic. Thus if potential energy were released to the objects during the collision, the report could be true. Question 2, chap 11, sect 4. part 1 of 1 10 points A(n) 6 kg object moving with a speed of 8 . 5 m / s collides with a(n) 17 kg object moving with a velocity of 7 . 2 m / s in a direction 19 from the initial direction of motion of the 6 kg object. 8 . 5 m / s 7 . 2 m / s 17 kg 6 kg 19 What is the speed of the two objects after the collision if they remain stuck together? Correct answer: 7 . 45337 m / s (tolerance 1 %). Explanation: Let : m 1 = 6 kg , m 2 = 17 kg , m f = m 1 + m 2 = 23 kg , v 1 = 8 . 5 m / s , v 2 = 7 . 2 m / s , p 1 = m 1 v 1 = 51 kg m / s , p 2 = m 2 v 2 = 122 . 4 kg m / s , p 1 p 2 = m 1 v 1 m 2 v 2 = 12484 . 8 kg 2 m 2 / s 2 , p x = p 1 + p 2 cos = 166 . 731 kg m / s , p y = p 2 sin = 39 . 8495 kg m / s , = 19 , and  = 161 . v 1 v 2 m 2 m 1 v f m f The final momentum is p f = ( m 1 + m 2 ) v f . (1) Momentum is conserved vectorp 1 + vectorp 2 = vectorp f . (2) Using the law of cosines, we have p 2 1 + p 2 2 2 p 1 p 2 cos(  ) = p 2 f . oldhomewk 22 YOO, HEE Due: Apr 1 2008, 4:00 am 2 Solving for v f , we have v f = radicalBig p 2 1 + p 2 2 2 p 1 p 2 cos(  ) m 1 + m 2 = bracketleftbigg 1 (6 kg) + (17 kg) bracketrightbigg bracketleftBig (2601 kg 2 m 2 / s 2 ) 2 + (14981 . 8 kg 2 m 2 / s 2 ) 2 (12484 . 8 kg 2 m 2 / s 2 ) cos(161 ) bracketrightBig 1 / 2 = 7 . 45337 m / s . Since the direction of the total momentum before and after the collision does not change, the initial and final momentum direction is = arctan parenleftbigg p y p x parenrightbigg = arctan parenleftbigg 39 . 8495 kg m / s 166 . 731 kg m / s parenrightbigg = 13 . 4418 ....
View
Full
Document
This note was uploaded on 05/04/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics, Mass

Click to edit the document details