Shooting a bullet at a system of blocks So, I made this question up myself.... and I'm curious about the answer.  It requires only secondary-school-level knowledge of physics:
You have a surface (ground) with a certain coefficient of friction.  On this surface is a rectangular block of mass $M_{1}$, which has length $L$.  On top of this block is another block (but square) of mass $M_{2}$ which has side-length of $L/4$ ... the very left edge of the square block is perfectly placed at $L/2$ of the rectangular block.  In other words, if the rectangular block has length $L$, than the left edge of the square block (placed on top of the rectangular block) is placed at $L/2$ of the rectangular block.  In addition, the top side of the rectangular block that is making contact with the square block has coefficient of friction = 0.
You decide to shoot a bullet (horizontally) of mass $m << M_{1}$ and $ m << M_{2}$ which travels with velocity $v_{i}$ right at the center of the rectangular block on the yz-plane.  So: if the blocks are sitting in 2D (xy-plane)... then the bullet is hitting the center of the rectangular block at its LEFT edge.  Let's say that the bullet penetrates the rectangular block and "becomes part of it."
Now, this is the question:
Find the expression for the bullet velocity, $v_{i}$, that will JUST make the square block to fall off the rectangular block; let's call this the threshold velocity.  Imagine that the ground goes off to infinity.
 A: First, using the law of conservation of momentum:
$P_{initial} = P_{final}$
$P_{initial} = mv_i$  (since only the bullet is moving)
$P_{final} = (m_1 + m)u$, where I have introduced $u$ as the velocity of the mass 1 and bullet system after the collision.  I haven't include mass 2 here because there is no friction between mass 1 and mass 2.
Equating gives:
$mv_i = (m_1 + m)u$, and solving for $u$ gives:
$u = \frac{mv_i}{(m_1 + m)}$
Now this is the velocity of the bottom block immediately after the bullet hits it.  It will then slow down due to friction, with an acceleration $a$.  We need the block to come to a halt, after travelling a distance of $L/2 + L/8$.  The reason for this, is because the left edge of the top block is $L/2$ from the left of the bottom block, and the center of gravity is an additional $L/8$ units from this left edge.  Therefore the bottom block must move a distance of $5L/8$ before coming to rest.
We use the following equation to set the distance the block travels before coming to rest:
$v^2 = u^2 + 2as$
Therefore,
$0 = \frac{m^2v_i^2}{(M_1 + m)^2} + 2a\frac{5L}{8}$
Therefore,
$\sqrt{-a\frac{5L(M_1 + m)^2}{4m^2}} = v_i$
We can write acceleration in terms of the coefficient of friction.  
$a = -\frac{\mu N}{M_1 + m} = -\frac{\mu g(M_1 + M_2 + m}{M_1 + m}$   (This time I have included M_2, since this contributes to the normal force).
Therefore,
$v_i = \sqrt{\frac{5L\mu g (M_1+M_2+m)(M_1 + m)^2}{4m^2(M_1 + m)}} = \sqrt{\frac{5L\mu g (M_1+M_2+m)(M_1 + m)}{4m^2}}$ 
which is the velocity of the bullet required to shift the bottom  mass a distance of $5L/8$, which will consequently cause the top block to fall off.
