1. Speed of the pendulum bob at the lowest point before collision

Since the pendulum is released from a small angle $\theta_0$, we use conservation of mechanical energy.

For small angles, $h_0 \approx \ell(1-\cos\theta_0) \approx \ell\frac{\theta_0^2}{2}$

So, $\frac12 m u^2 = mg h_0 = mg\ell\frac{\theta_0^2}{2}$

Hence, $u = \sqrt{g\ell}\,\theta_0$

where $u$ is the speed of the bob just before hitting the block.

2. Speed of the pendulum bob after elastic collision

After collision, the bob rebounds and rises to angle $\theta_1$.

Again using energy conservation, $\frac12 m v^2 = mg\ell\frac{\theta_1^2}{2}$

Thus, $|v| = \sqrt{g\ell}\,\theta_1$

Since the bob bounces back, its velocity reverses direction. If the initial direction is taken positive, then after collision: $v = -\sqrt{g\ell}\,\theta_1$

3. Apply 1D elastic collision formula

The bob of mass $m$ collides elastically with a stationary block of mass $M$.

For a head-on elastic collision with target initially at rest, $v = \frac{m-M}{m+M}u$

Substitute $u = \sqrt{g\ell}\,\theta_0$ and $v = -\sqrt{g\ell}\,\theta_1$: $-\sqrt{g\ell}\,\theta_1 = \frac{m-M}{m+M}\sqrt{g\ell}\,\theta_0$

Cancel $\sqrt{g\ell}$: $-\theta_1 = \frac{m-M}{m+M}\theta_0$

Rearrange: $\frac{M-m}{M+m} = \frac{\theta_1}{\theta_0}$

4. Solve for $M$

Let $r = \frac{\theta_1}{\theta_0}$

Then, $\frac{M-m}{M+m} = r$

So, $M-m = r(M+m)$ $M-m = rM + rm$ $M-rM = m+rm$ $M(1-r) = m(1+r)$

Therefore, $M = m\frac{1+r}{1-r}$

Now substitute back $r = \dfrac{\theta_1}{\theta_0}$: $M = m\frac{1+\theta_1/\theta_0}{1-\theta_1/\theta_0}$

Multiply numerator and denominator by $\theta_0$: $M = m\frac{\theta_0+\theta_1}{\theta_0-\theta_1}$

5. Match with options

This corresponds to: $\boxed{m\left(\frac{\theta_0+\theta_1}{\theta_0-\theta_1}\right)}$

So the correct option is C.