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

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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,
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern