This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: markowitz (am45362) – HW06 – distler – (56295) 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 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 spring’s 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 = − 2 vectorv B vectorp A = − 8 vectorp B K A = 16 K B 2. vectorv A = + 1 5 vectorv B vectorp A = + 4 5 vectorp B K A = 4 25 K B 3. vectorv A = − 1 3 vectorv B vectorp A = − 2 3 vectorp B K A = 4 3 K B 4. vectorv A = − vectorv B vectorp A = − 4 vectorp B K A = 16 K B 5. vectorv A = − vectorv B vectorp A = − vectorp B K A = K B 6. vectorv A = − 4 vectorv B vectorp A = − 16 vectorp B K A = 64 K B 7. vectorv A = + 1 4 vectorv B vectorp A = + vectorp B K A = 4 K B 8. vectorv A = − 1 2 vectorv B vectorp A = − 2 vectorp B K A = K B 9. vectorv A = − 1 4 vectorv B vectorp A = − vectorp B K A = 1 4 K B correct 10. vectorv A = − 1 4 vectorv B vectorp A = − vectorp B K A = 4 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 . 002 10.0 points Two objects of mass 9 g and 6 g , respectively, move parallel to the xaxis, as shown. The 9 g object overtakes and collides with the 6 g object. Immediately after the collision, markowitz (am45362) – HW06 – distler – (56295) 2 the ycomponent of the velocity of the 9 g object is 6 m / s upward. v i 9 g v i 6 g What is the ycomponent of the velocity vectorv of the 6 g object immediately after the collision? 1. vectorv y = 8 m / s downward 2. vectorv y = 3 . 5 m / s downward 3. vectorv y = 6 . 66667 m / s downward 4. vectorv y = 14 m / s downward 5. vectorv y = 9 m / s downward correct 6. vectorv y = 7 m/s upward 7. vectorv y = 5 m/s upward 8. vectorv y = 3 m / s downward Explanation: Let : m 1 = 9 g , m 2 = 6 g , and v 1 y = 6 m / s ....
View
Full
Document
This note was uploaded on 11/29/2011 for the course PHY 302K taught by Professor Kaplunovsky during the Spring '08 term at University of Texas.
 Spring '08
 Kaplunovsky
 Physics, Friction

Click to edit the document details