# Using the Center of mass; string-balls system

One ball is projected upwards, vertically, and one's supposed to find the maximum height to which the center of mass of the system rises. Why can't you just find the center-of-mass- velocity at the initial instant and use $$h = {u_{\rm COM}^2 \over 2g}~?$$ The book solves it differently, and the answers are different.

• Because the height of each ball is different from the height of the center of mass. – John Alexiou Mar 26 at 13:51
• Yes, but isn't it the COM height you're supposed to find? – Harry Holmes Mar 26 at 13:59
• Sorry, yes, I misread the question. So if you calculate the initial $u_{\rm COM}$ correctly then the answer should be valid. But you need to show what you have tried, Please edit the question to show your work. – John Alexiou Mar 26 at 14:04

The problem with your analysis is that is assumes the that the "collision" mediated by the string is elastic.

If initially, one balls is given momentum $$p_1=p$$, the total momentum of the system is:

$$P = p_1 + p_2 = p + 0 = p$$

with total kinetic energy:

$$T = \frac{p_1^2}{2m} + \frac{p_2^2}{2m} =\frac{p^2}{2m}$$

Then the string pulls the other ball along for the ride, and in this step, momentum is conserved. With $$p_1'= p_2'\equiv p'$$:

$$P'= p_1'+p_2' = 2p' = P$$

so that:

$$p' = \frac p 2$$

The kinetic energy is:

$$T' = \frac{(p/2)^2}{2m} + \frac{(p/2)^2}{2m}$$

$$T' = \frac {p^2}{4m} = \frac 1 2 T$$

Energy was lost.

• Sorry, by u(COM), I meant the velocity after the string had pulled the other one up. So by your analysis, that velocity would be p/2(2m), or p/4m. – Harry Holmes Mar 26 at 15:26