Posted by Steve Simon on February 3, 2013, 11:15 am, in reply to "Q 2.6a"
To give a bit more detail. Consider the center of mass degree of freedom. X_{CM} = (x1 + x2)/2. So the center of mass velocity is V_{CM} = (v1 + v2)/2. However (if the two masses are the same) the mass associated with the center of mass is M_{CM} = m1 + m2 = 2m. So the center of mass momentum is V_{CM} M_{CM} = [(v1 + v2)/2 ] [2 m] = p1 + p2. 
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