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Unformatted text preview: Answer, Key – Homework 10 – David McIntyre – 45123 – May 10, 2004 1 This printout should have 26 questions. Multiplechoice questions may continue on the next column or page – find all choices before making your selection. The due time is Central time. Chapter 9 problems. 001 (part 1 of 1) 0 points An object of mass m is moving with speed v to the right on a horizontal frictionless surface, as shown, when it explodes into two pieces. Subsequently, one piece of mass 2 5 m moves with a speed v 2 / 5 = v 2 to the left. v m The speed k ~v 3 / 5 k of of the other piece the object is 1. k ~v 3 / 5 k = v 2 . 2. k ~v 3 / 5 k = v 3 . 3. k ~v 3 / 5 k = 7 v 5 . 4. k ~v 3 / 5 k = 3 v 2 . 5. k ~v 3 / 5 k = 2 v . correct Explanation: The horizontal component of the momen tum is conserved, so 0 + mv = 2 5 mv 2 / 5 + 3 5 mv 3 / 5 0 + mv = 2 5 m ‡ v 2 · + 3 5 mv 3 / 5 mv = 1 5 mv + 3 5 mv 3 / 5 3 5 v 3 / 5 = 6 5 v k ~v 3 / 5 k = 2 v . 002 (part 1 of 1) 0 points An open train car moves with speed 19 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 Newton’s 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, dv dt should be negative; i.e. , the speed of the train should decrease. 003 (part 1 of 2) 5 points Consider the set up of a ballistic pendu lum where m 1 = 11 . 4 g, m 2 = 3 . 97 kg, h = 0 . 243 m. The acceleration of gravity is 9 . 8 m / s 2 . m 1 v 1 v f 2 m + m 1 2 m h Find the final velocity of the system ( m 1 + m 2 ) immediately after the collision and before the pendulum starts to swing upwards. Correct answer: 2 . 18238 m / s. Explanation: From the setup, the final velocity of the col lision is the initial velocity of the subsequent Answer, Key – Homework 10 – David McIntyre – 45123 – May 10, 2004 2 swing. During the swinging process the total energy is conserved E i = E f 1 2 ( m 1 + m 2 ) v 2 f = ( m 1 + m 2 ) g h. Therefore v f = p 2 g h = q 2(9 . 8 m / s 2 )(0 . 243 m) = 2 . 18238 m / s . 004 (part 2 of 2) 5 points Find v 1 , the initial speed of m 1 . Correct answer: 762 . 188 m / s. Explanation: The linear momentum is conserved in a collision p i = p f m 1 v 1 = ( m 1 + m 2 ) v f Therefore v 1 = m 1 + m 2 m 1 v f = (0 . 0114 kg) + (3 . 97 kg) (0 . 0114 kg) (2 . 18238 m / s) = 762 . 188 m / s ....
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 Spring '08
 RIJSSENBEEK
 Energy, Kinetic Energy, Mass, Momentum, Work, Correct Answer

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