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

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Unformatted text preview: 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 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...
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This note was uploaded on 02/08/2012 for the course CHEM 5314 taught by Professor Nielsen during the Fall '11 term at University of Texas at Dallas, Richardson.

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

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