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**1. The problem statement, all variables and given/known data**

Consider a massless rod of length $L$ with a small mass $m1$ attached on one end, and $m2$ attached on the other end. The rod is initially in the vertical position at rest on a frictionless surface, with $m1$ on bottom and $m2$ on top. A small impulse is applied to the top of the rod in the horizontal direction, and it begins to fall.

Find the speeds of $m1$ and $m2$ when $m2$ is about to hit the ground.

**2. Relevant equations**

I.Conservation of energy: $m_2gL=1/2m_1v_1^2+1/2m_2v_2^2$

II.Conservation of momentum in the horizontal direction: $m_1v_1=m_2v_{2_x}$

III.Center of mass is located at $d_{cm}=(m_1*0+m_2*L)/(m_1+m_2)$ along the rod (with $m1$ at the origin).

**3. The attempt at a solution**

Since the only force acting on the system is gravity and the normal force on $m1$, which are both in the vertical direction, linear momentum is conserved in the horizontal direction, and therefore the center of mass of the rod does not change its position in the horizontal direction.

The net force on the system is $(m_1+m_2)g-N$ downwards, but this isn't very helpful since the normal force $N$ changes with time.

I had an idea to consider angular momentum about the mass $m1$ or maybe about the center of mass of the rod, but I'm not really sure how to proceed.

The problem is solved if we can find the horizontal component $v_{2_x}$ in terms of $v_2$, because then we can use equations I and II above to solve for $v1$ and $v2$. Equivalently, the problem is solved if we find $v_{2_y}$. But $v_{2_y}$ is related to the vertical speed of the center of mass of the system, $v_{cm}$ by a constant factor (using similar triangles). Therefore it suffices to find $v_{cm}$ just before $m_2$ hits the ground.

I am tempted to say, by conservation of energy, $(m_1+m_2)gd_{cm}=1/2(m_1+m_2)v_{cm}^2$, or $\sqrt{2gd_{cm}}=v_{cm}$. For some reason I don't think I can apply conservation of energy to the center of mass, but let's go on anyway.

By similar triangles, the height $h_cm$ of the center of mass above the ground is related to the height $h_{m_2}$ of $m_2$ above the ground, by: $h_{m_2}=h_{cm}*L/d_{cm}$. Thus $v_{2_y }=v_{cm}*L/d_{cm}$. I continued like this and I got an immediate contradiction. So I know I did something wrong, not with the algebra, but with the assumption that you can apply conservation of energy to the center of mass.

So now I'm stuck.