A fixed smooth wedge has two faces, each inclined at an angle θ to...

Let the speed of each block at time $t$ be $v$. Since they slide down smooth inclines of angle $\theta$, their acceleration is $a = g\sin\theta$.

The speed at time $t$ is $v = (g\sin\theta)t$.

The velocity vectors of the two blocks are directed downwards along the inclines. Their horizontal components are equal and opposite, thus cancelling each other out. Their vertical components are both directed downwards with magnitude $v\sin\theta$.

The magnitude of the total linear momentum of the system is $P = 2m(v\sin\theta) = 2m(g\sin\theta t)\sin\theta = 2mgt\sin^2\theta$.

The momentum of a single block of mass $m$ falling freely for time $t$ is $p = mgt$.

Equating the two momenta: $2mgt\sin^2\theta = mgt$

$\sin^2\theta = \dfrac{1}{2}$

$\sin\theta = \dfrac{1}{\sqrt{2}}$

$\theta = 45^\circ$.

If one incorrectly assumes the total momentum is simply $2mv$ without resolving into components, they would get $2mgt\sin\theta = mgt \Rightarrow \theta = 30^\circ$.

Answer: $45^\circ$