# test02 - Version 044/AACDA TEST2 Tsoi (58020) 1 This...

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Unformatted text preview: Version 044/AACDA TEST2 Tsoi (58020) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Starting from rest at a height equal to the radius of the circular track, a block of mass 15 kg slides down a quarter circular track under the influence of gravity with friction present (of coefficient ). The radius of the track is 49 m. The acceleration of gravity is 9 . 8 m / s 2 . 49 m 15 kg If the kinetic energy of the block at the bottom of the track is 2700 J, what is the work done against friction? 1. 4085.8 2. 2810.0 3. 1704.4 4. 3662.4 5. 7394.0 6. 3773.4 7. 2930.8 8. 4302.0 9. 7283.8 10. 4503.0 Correct answer: 4503 J. Explanation: W = W f + K W f = mg R- K = (15 kg) (9 . 8 m / s 2 ) (49 m)- (2700 J) = 4503 J . 002 10.0 points A spring is compressed between two cars on a frictionless airtrack. Car A has four times the mass of car B, M A = 4 M B , while the springs mass is negligible. Both cars are initially at rest. When the spring is released, it pushes them away from each other. Which of the following statements correctly describes the velocities, the momenta, and the kinetic energies of the two cars after the spring is released? Note : Velocities and momenta are given below as vectors. 1. vectorv A =- 1 3 vectorv B vectorp A =- 2 3 vectorp B K A = 4 3 K B 2. vectorv A =- 1 4 vectorv B vectorp A =- vectorp B K A = 1 4 K B correct 3. vectorv A =- 1 2 vectorv B vectorp A =- 2 vectorp B K A = K B 4. vectorv A = + 1 5 vectorv B vectorp A = + 4 5 vectorp B K A = 4 25 K B 5. vectorv A =- vectorv B vectorp A =- 4 vectorp B K A = 16 K B 6. vectorv A = + 1 4 vectorv B vectorp A = + vectorp B K A = 4 K B 7. vectorv A =- 4 vectorv B vectorp A =- 16 vectorp B K A = 64 K B 8. vectorv A =- 2 vectorv B vectorp A =- 8 vectorp B K A = 16 K B Version 044/AACDA TEST2 Tsoi (58020) 2 9. vectorv A =- 1 4 vectorv B vectorp A =- vectorp B K A = 4 K B 10. vectorv A =- vectorv B vectorp A =- vectorp B K A = K B Explanation: Let : M A = 4 M B . There are no external forces acting on the cars, so their net momentum vectorp A + vectorp B is con- served. The initial net momentum is obvi- ously zero, hence after the spring is released, vectorp A + vectorp B = ; i.e., vectorp A =- vectorp B . The velocities and the kinetic energies fol- low from the momenta: vectorv = vectorp M . So given M A = 4 M B , it follows that vectorv A =- 1 4 vectorv B . Likewise, K = 1 2 M vectorv 2 = vectorp 2 2 M . Since vectorp A =- vectorp B , K A = 1 4 K B . 003 (part 1 of 2) 10.0 points A football is thrown hard horizontally and it hits and sticks in a cart that is on a track at position A in the diagram. The cart with the football in it then moves along the hilly frictionless track to position B, where the cart stops....
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## This note was uploaded on 04/15/2009 for the course PHY 58020 taught by Professor Tsoi during the Spring '09 term at University of Texas at Austin.

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test02 - Version 044/AACDA TEST2 Tsoi (58020) 1 This...

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