chem5314_hwk4_soln

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

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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

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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).
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