Answer
The smallest value of $v$ which will cause the pendulum to swing over the top of the arc is $v = \frac{m+M}{m}\sqrt{4gl}$
Work Step by Step
Let $m$ be the mass of the projectile and let $M$ be the mass of the block on the pendulum. Let $v$ be the initial speed of a projectile.
We can use conservation of momentum to find the speed $v'$ just after the collision.
$m~v = (m+M)~v'$
We can use conservation of energy to find the height $h$ reached by the pendulum as it swings up.
$\frac{1}{2}(m+M)~(v')^2 = (m+M)~gh$
$v' = \sqrt{2gh}$
We can replace $v'$ in the first equation above.
$m~v = (m+M)\sqrt{2gh}$
$v = \frac{m+M}{m}\sqrt{2gh}$
Since we want the pendulum to swing all the way over the top, we can let $h = 2l$, where $l$ is the length of the pendulum.
$v = \frac{m+M}{m}\sqrt{2g(2l)} = \frac{m+M}{m}\sqrt{4gl}$
The smallest value of $v$ which will cause the pendulum to swing over the top of the arc is:
$v = \frac{m+M}{m}\sqrt{4gl}$