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PS1solutions - CHEM 120A Problem Set 1 Solutions David...

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CHEM 120A Problem Set 1 Solutions David Hoffman , Tara Yacovitch , Doran Bennett 1/27/2010 These solutions are intended to be a guide, or roadmap, to help you answer the homework questions, they are not intended to be step-by-step instructions. For this reason, the majority of the mathematical manipulations have been intentionally left out so as to force you to work through the questions yourselves. If you find that this guide is not enough then please e-mail us or come to our office hours or the discussion sections where we will be more than happy to help you. 1. The solution can be found in the McQuarrie Solutions manual. 2. The solution can be found in the McQuarrie Solutions manual. 3. (a) We have the following equation for the total energy E = N X i =1 p 2 i / 2 m i + V ( x 1 , x 2 , . . . , x N ) (1a) We can differentiate both sides with respect to time d E d t = 0 = N X i =1 v i ˙ p i + ∂V ∂x i (1b) In order for the right hand side to be equal to zero the term in parentheses must equal zero. This because the v i ’s can be chosen arbitrarily for an arbitrary system, yet the equality must hold. Thus, ˙ p i = - ∂V ∂x i , i (1c) (b) Note that p = ˆ ı p x + ˆ p y + ˆ k p z . Thus ˙p = -∇ V follows directly. 1 4. (a) We start with the expression for total energy in normal coordinates and then transform to center of mass ( COM ) coordinates. 1 We will use bolded items for vector quantities and non-bold items for either the magnitude of the vector or a scalar. Ex: p is the momentum vector whereas p = | p | is the magnitude of the momentum, which is a scalar . Also, the dot above a variable indicates its time derivative, i.e. ˙p = d d t p 1
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The total energy is, E tot = 1 2 m 1 v 2 1 + 1 2 m 2 v 2 2 + V ( r ) (2a) Now we can begin to perform some algebra. 2 After multiplying by M/M and completing the square you’ll find that indeed, E tot = M 2 v 2 cm + μ 2 v 2 + V ( r ) (2b) QED 3 (b) Now we need to prove that the COM momentum is conserved. There are two ways of solving this problem: i. First we could realize that the Coulomb potential is proportional to 1
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