Pivoted at its upper end, a metre stick weighing 240 g is able to...

Let the mass of the stick be $M = 240 \text{ g}$ and its length be $L = 1 \text{ m}$.
Let the mass of the particle be $m = 100 \text{ g}$.

The particle falls through a height $L$, so its velocity just before the collision is $v = \sqrt{2gL}$.
The angular momentum of the particle about the pivot just before the collision is:
$L_i = m v L = m L \sqrt{2gL}$

After the collision, the particle sticks to the lower end of the stick. The moment of inertia of the system (stick + particle) about the pivot is:
$I = I_{\text{stick}} + I_{\text{particle}} = \dfrac{M L^2}{3} + m L^2 = \left(\dfrac{M}{3} + m\right) L^2$

By conservation of angular momentum about the pivot during the collision:
$L_i = I \omega \Rightarrow \omega = \dfrac{m L \sqrt{2gL}}{\left(\dfrac{M}{3} + m\right) L^2}$

The kinetic energy of the system just after the collision is:
$K = \dfrac{1}{2} I \omega^2 = \dfrac{L_i^2}{2I} = \dfrac{m^2 L^2 (2gL)}{2 \left(\dfrac{M}{3} + m\right) L^2} = \dfrac{m^2 g L}{\dfrac{M}{3} + m}$

When the system swings to a maximum angle $\theta$, the center of mass of the stick rises by $\dfrac{L}{2}(1 - \cos\theta)$ and the particle rises by $L(1 - \cos\theta)$. The increase in potential energy is:
$U = M g \dfrac{L}{2} (1 - \cos\theta) + m g L (1 - \cos\theta) = \left(\dfrac{M}{2} + m\right) g L (1 - \cos\theta)$

By conservation of mechanical energy after the collision, $K = U$:
$\dfrac{m^2 g L}{\dfrac{M}{3} + m} = \left(\dfrac{M}{2} + m\right) g L (1 - \cos\theta)$

Substituting $M = 240 \text{ g}$ and $m = 100 \text{ g}$:
$\dfrac{100^2}{\dfrac{240}{3} + 100} = \left(\dfrac{240}{2} + 100\right) (1 - \cos\theta)$

$\dfrac{10000}{80 + 100} = (120 + 100) (1 - \cos\theta)$

$\dfrac{10000}{180} = 220 (1 - \cos\theta)$

$1 - \cos\theta = \dfrac{10000}{180 \times 220} = \dfrac{100}{396} = \dfrac{25}{99}$

$\cos\theta = 1 - \dfrac{25}{99} = \dfrac{74}{99} \approx 0.747$

Since $\cos 41^\circ \approx 0.754$ and $\cos 42^\circ \approx 0.743$, the angle $\theta$ is approximately $41^\circ$.

Answer: $\approx 41^\circ$