OLDHW22 - oldhomewk 22 PAPAGEORGE MATT Due Apr 1 2008 4:00...

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oldhomewk 22 – PAPAGEORGE, MATT – 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 head-on 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. If the surface were inclined, the report could be true. 2. If potential energy were released to the objects during the collision, the report could be true. correct 3. Momentum was not conserved; therefore the report is false. 4. If the objects had different masses, the report could be true. 5. If there were no friction between the ob- jects and the surface, 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) 7 kg object moving with a speed of 7 . 7 m / s collides with a(n) 19 kg object moving with a velocity of 8 m / s in a direction 24 from the initial direction of motion of the 7 kg object. 7 . 7 m / s 8 m / s 19 kg 7 kg 24 What is the speed of the two objects after the collision if they remain stuck together? Correct answer: 7 . 7858 m / s (tolerance ± 1 %). Explanation: Let : m 1 = 7 kg , m 2 = 19 kg , m f = m 1 + m 2 = 26 kg , v 1 = 7 . 7 m / s , v 2 = 8 m / s , p 1 = m 1 v 1 = 53 . 9 kg m / s , p 2 = m 2 v 2 = 152 kg m / s , p 1 p 2 = m 1 v 1 m 2 v 2 = 16385 . 6 kg 2 m 2 / s 2 , p x = p 1 + p 2 cos θ = 192 . 759 kg m / s , p y = p 2 sin θ = 61 . 824 kg m / s , θ = 24 , and π - θ = 156 . v 1 v 2 m 2 m 1 θ v f φ 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( π - θ ) = p 2 f .
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oldhomewk 22 – PAPAGEORGE, MATT – 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( π - θ ) m 1 + m 2 = bracketleftbigg 1 (7 kg) + (19 kg) bracketrightbigg × bracketleftBig (2905 . 21 kg 2 m 2 / s 2 ) 2 + (23104 kg 2 m 2 / s 2 ) 2 - (16385 . 6 kg 2 m 2 / s 2 ) cos(156 ) bracketrightBig 1 / 2 = 7 . 7858 m / s . Since the direction of the total momentum before and after the collision does not change, the initial and final momentum direction is φ = arctan parenleftbigg p y p x parenrightbigg = arctan parenleftbigg 61 . 824 kg m / s 192 . 759 kg m / s parenrightbigg = 17 . 7827 . Alternate Solution: Since p 2 f = p 2 x + p 2 y , we have v f = radicalBig p 2 x + p 2 y m 1 + m 2 = bracketleftbigg 1 (7 kg) + (19 kg) bracketrightbigg × bracketleftBig (37156 kg 2 m 2 / s 2 ) + (3822 . 2 kg 2 m 2 / s 2 ) bracketrightBig 1 / 2 = 7 . 7858 m / s .
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