# Possible mistake in the solutions? [closed]

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: And the solution given in the soln manual: ## closed as off-topic by Kyle Kanos, Jon Custer, John Rennie, Emilio Pisanty, stafusaJul 27 at 18:02

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