HW4-soln - Physics 105 Mechanics Fall 2009 Homework 4...

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

View Full Document Right Arrow Icon
Physics 105: Mechanics Fall 2009 Homework 4 Solutions 2.P.39 SYSTEM = two rocks SURROUNDING = nothing significant All the speeds are in the range where γ 1. Δ ~ p tot = ~ F ext net Δ t = 0 . Initial: ~ p 1 i = m 1 i ~v 1 i = (9kg) h 4100 , - 2600 , 2800 i m / s = h 36900 , - 23400 , 25200 i kgm / s ~ p 2 i = m 2 i ~v 1 i = (6kg) h- 450 , 1800 , 3500 i m / s = h- 2700 , 10800 , 21000 i kgm / s ~ p tot,i = ~ p 1 i + ~ p 2 i = h 34200 , - 12600 , 46200 i kgm / s Final: ~ p 1 f = m 1 f ~v 1 f = (7kg) h 1300 , 200 , 1800 i m / s = h 9100 , 1400 , 12600 i kgm / s Total momentum of system is conserved: ~ p 1 f + ~ p 2 f = ~ p 1 i + ~ p 2 i ~ p 2 f = ~ p 1 i + ~ p 2 i - ~ p 1 f = h 25100 , - 14000 , 33600 i kgm / s ~v 2 f = ~ p 2 f /m 2 f = h 3137 , - 1750 , 4200 i m / s 2.P.45 System = junk + satellite Surroundings = no signficant forces exerted by other objects Assume γ = 1.
Image of page 1

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

View Full Document Right Arrow Icon
Let ~v 4 be the final velocity of the center of mass of the satellite. Initial momentum of system (just before collision): ~ p sys,i = m~v 2 + M~v 1 . x-component: p sys,i,x = - m | ~v 2 | + M | ~v 1 | . y-component: p sys,i,y = 0 . Final momentum of system (just after collision): ~ p sys,f = m~v 3 + M~v 4 . x-component: p sys,f,x = - m | ~v 3 | cos θ + Mv 4 ,x . y-component: p sys,f,y = m | ~v 3 | sin θ + Mv 4 ,y . Since there are no significant external forces acting on the system during the collision, ~ p sys,i = ~ p sys,f . Thus v 4 ,x = - m | ~v 2 | /M + | ~v 1 | + m | ~v 3 | cos θ/M. v 4 ,y = - m | ~v 3 | sin θ/M. 3.RQ.27 The magnitude of the gravitational due to the Earth on a mass m located a distance r from the center of the Earth (where r > R E ) is | ~ F g | = GM E m/r 2 . If the object is at the surface of the Earth, r = R E . Then the force becomes | ~ F g | = m ( GM E /R 2 E ). So the quantity in the parentheses is the magnitude of the gravitational field near the surface of the Earth, g = GM E /R 2 E .
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