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This is from Kleppner's introduction to mechanics and I've been having trouble making sense of the solution in the solution manual for quite a while and I've concluded that its wrong but I need some sort of confirmation. I have trouble grasping why $\dot x_1$ and $\dot x_2$ are assumed to be the same $wl/2$ instead of using conservation of momentum to find the new combined velocity. I mean, where did the extra energy even come from? The question: enter image description here

And the solution given in the soln manual:enter image description here

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I believe you are correct. According to the solution, $m_1$ instantaneously goes from being motionless to moving with velocity $\frac{1}{2}\omega l$, requiring an instantaneous non-zero impulse applied on $m_1$, which this system cannot provide. As you indicated, conservation of momentum can be used after the moment $x_2 = l$ to describe the motion of $m_1$ and $m_2$.

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They seem to assume that the spring cannot be extended to a length beyond $l$ so after it is fully extended the two bodies indeed move with the same speed.

Without that assumption we can proceed as follows: after $t = \frac{\pi}{2\omega}$ the equations of motion are $$m_1\ddot{x_1} = k(x_2-x_1-l)$$ $$m_2\ddot{x_2} = k(l-x_2+x_1)$$ so adding the two equations gives $m_1\ddot{x_1} + m_2\ddot{x_2} = 0$, or $m_1\dot{x_1} + m_2\dot{x_2} = \text{const}$.

Therefore $$\dot{X} =\frac{m_1\dot{x_1} + m_2\dot{x_2}}{m_1+m_2} = \frac{m_1\dot{x_1}\left(\frac{\pi}{2\omega}\right) + m_2\dot{x_2}\left(\frac{\pi}{2\omega}\right)}{m_1+m_2} = \frac{m_2\omega l}{2(m_1+m_2)}$$

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