chem5314_hwk4_soln - Chem 5314 homework#4 out of 25 marks...

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

View Full Document Right Arrow Icon
Chem 5314 homework #4 – out of 25 marks Problem 1 – 10 marks a – 3 marks) Consider two particles of masses m 1 and m 2 in one dimension, interacting through a potential that depends only upon their relative separation ( x 1 - x 2 ), so that V ( x 1 , x 2 ) = V ( x 1 - x 2 ). Given that the force acting upon the j th particle is f j = - ( ∂V/∂x j ), show that f 1 = - f 2 . What law is this? solution: Define x = x 1 - x 2 . Then we can write f 1 = - ∂V ∂x 1 = - ∂V ∂x ∂x ∂x 1 = - ∂V ∂x (1) and f 2 = - ∂V ∂x 2 = - ∂V ∂x ∂x ∂x 2 = ∂V ∂x (2) since ( ( x 1 - x 2 ) /∂x 1 ) = 1 and ( ( x 1 - x 2 ) /∂x 2 ) = - 1. solution: This is Newton’s third law: For every action, there is an equal and opposite reaction. b – 2 marks) Newton’s equations for the two particles are m 1 d 2 x 1 dt 2 = - ∂V ∂x 1 and m 2 d 2 x 2 dt 2 = - ∂V ∂x 2 (3) Now introduce center of mass and relative coordinates by X m 1 x 1 + m 2 x 2 M (4) x x 1 - x 2 (5) where M = m 1 + m 2 , and solve for x 1 and x 2 to obtain x 1 = X + m 2 M x and x 2 = X - m 1 M x (6) Show that Newton’s equations in these coordinates are m 1 d 2 X dt 2 + m 1 m 2 M d 2 x dt 2 = - ∂V ∂x (7) 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
and m 2 d 2 X dt 2 - m 1 m 2 M d 2 x dt 2 = + ∂V ∂x (8) solution: To solve for x 1 and x 2 , we multiply Eq. (4) by M and Eq. (5) by m 2 and add them to get x 1 , and then we multiply Eq. (4) by M and Eq. (5) by m 1 and substract them to get x 2 . The only tricky thing here is to show that ∂V ∂x = ∂V ∂x 1 (9) and ∂V ∂x = - ∂V ∂x 2 (10) but we actually did this in Eqs. (1) and (2).
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