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Unformatted text preview: homework 13 RAHMAN, TARIQUE Due: Mar 21 2008, 3:00 am 1 Question 1, chap 10, sect 1. part 1 of 2 10 points When the velocity of an object is doubled, by what factor is its momentum changed? 1. 4 2. 3 3. 2 correct 4. 1 Explanation: Momentum is proportional to speed. Question 2, chap 10, sect 1. part 2 of 2 10 points When the velocity of an object is doubled, by what factor is its kinetic energy changed? 1. 2 2. 3 3. 1 4. 4 correct Explanation: Kinetic Energy is proportional to speed squared. Question 3, chap 10, sect 1. part 1 of 1 10 points Bill (mass m ) plants both feet solidly on the ground and then jumps straight up with velocity v . The earth (mass M ) then has velocity 1. V Earth = + v man . 2. V Earth = + parenleftBig m M parenrightBig v man . 3. V Earth = parenleftbigg M m parenrightbigg v man . 4. V Earth = + radicalbigg m M v man . 5. V Earth = parenleftBig m M parenrightBig v man . correct 6. V Earth = radicalbigg m M v man . 7. V Earth = v man . 8. V Earth = + parenleftbigg M m parenrightbigg v man . Explanation: The momentum is conserved. We have m v man + M V Earth = 0 So V Earth = parenleftBig m M parenrightBig v man . Question 4, chap 10, sect 1. part 1 of 1 10 points An open train car moves with speed 14 . 7 m / s on a flat frictionless railroad track, with no engine pulling the car. It begins to rain. The rain falls straight down and begins to fill the train car. The speed of the car 1. stays the same. 2. decreases. correct 3. increases. Explanation: Using Newtons second law, we have dP dt = 0 , since no external forces act on the train in the horizontal direction. With no rain, the train will move at a constant velocity; however, when it starts to rain, and the rain starts to fill the car, the mass of the train changes. Thus, m dv dt = v dm dt . Since dm dt is positive; i.e. , the mass of the train is increasing with accumulating rain, homework 13 RAHMAN, TARIQUE Due: Mar 21 2008, 3:00 am 2 dv dt should be negative; i.e. , the speed of the train should decrease. Question 5, chap 10, sect 1. part 1 of 2 10 points A revolutionary war cannon, with a mass of 1980 kg, fires a 18 . 6 kg ball horizontally. The cannonball has a speed of 103 m / s after it has left the barrel. The cannon carriage is on a flat platform and is free to roll horizontally. What is the speed of the cannon immedi- ately after it was fired? Correct answer: 0 . 967576 m / s (tolerance 1 %). Explanation: Let : m = 18 . 6 kg , M = 1980 kg , and v = 103 m / s . The cannons velocity immediately after it was fired is found by using conservation of momentum along the horizontal direction: M V + mv = 0 V = m M v where M is the mass of the cannon, V is the velocity of the cannon, m is the mass of the cannon ball and v is the velocity of the cannon ball. Thus, the cannons speed is | V | = m M | v | = 18 . 6 kg 1980 kg (103 m / s) = . 967576 m / s ....
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