HW06-solutions

# HW06-solutions - markowitz(am45362 HW06 distler(56295 This...

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markowitz (am45362) – HW06 – distler – (56295) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points 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 = 0 ; 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 10.0points Two objects of mass 9 g and 6 g , respectively, move parallel to the x -axis, as shown. The 9 g object overtakes and collides with the 6 g object. Immediately after the collision,

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markowitz (am45362) – HW06 – distler – (56295) 2 the y -component of the velocity of the 9 g object is 6 m / s upward. v i 9 g v i 6 g What is the y -component 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 . Momentum vectorp is conserved for collision m 1 vectorv 1 = m 2 vectorv 2 . In particular, the y -component of the mo- mentum must be the same before or after the collision. Since the two objects move horizon- tally along the x -axis, the y -component of the momentum before the collision is zero.
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