RBD cYU 35

The bead $m_0$ falls from height $h_0$, striking the plank at distance $l_0$ from the pivot. The block $m$ is located at distance $l$ on the other side. Velocity of bead just before impact: $$ v_0 = \sqrt{2gh_0} $$

Let $J$ be the impulse between the bead and the plank. Let $J’$ be the impulse between the plank and the block $m$. Since the plank is inertia-less, the net torque of impulses about the pivot O must be zero: $$ J l_0 = J’ l \implies J’ = J \frac{l_0}{l} $$

For the bead ($m_0$): It rebounds upwards with velocity $v’_0$. $$ J = m_0 (v’_0 – (-v_0)) = m_0 (v’_0 + v_0) $$ For the block ($m$): It is launched upwards with velocity $v_m$. $$ J’ = m v_m $$

Let $\omega$ be the angular velocity of the plank immediately after impact. Velocity of point A (downwards) = $\omega l_0$. Velocity of point B (upwards) = $\omega l$. The block $m$ acquires the velocity of point B: $v_m = \omega l$.

Elastic Collision Condition ($e=1$): Relative velocity of separation = Relative velocity of approach. $$ v_{sep} = v’_0 + v_{plank\_A} = v’_0 + \omega l_0 $$ $$ v_{app} = v_0 $$ $$ v’_0 + \omega l_0 = v_0 \implies v’_0 = v_0 – \omega l_0 $$

Substitute $J’$ and $J$ into the torque balance equation ($J’ = J \frac{l_0}{l}$): $$ m v_m = m_0 (v’_0 + v_0) \frac{l_0}{l} $$ Substitute kinematic terms $v_m = \omega l$ and $v’_0 = v_0 – \omega l_0$: $$ m (\omega l) = \frac{m_0 l_0}{l} ( (v_0 – \omega l_0) + v_0 ) $$ $$ m \omega l^2 = m_0 l_0 (2 v_0 – \omega l_0) $$ $$ m \omega l^2 = 2 m_0 l_0 v_0 – m_0 l_0^2 \omega $$ $$ \omega (m l^2 + m_0 l_0^2) = 2 m_0 l_0 v_0 $$ $$ \omega = \frac{2 m_0 l_0 v_0}{m l^2 + m_0 l_0^2} $$

The launch velocity of block $m$ is $v_m = \omega l$. $$ v_m = \frac{2 m_0 l_0 l v_0}{m l^2 + m_0 l_0^2} $$ The height $h$ to which the block jumps is given by $h = \frac{v_m^2}{2g}$. $$ h = \frac{1}{2g} \left[ \frac{2 m_0 l_0 l \sqrt{2gh_0}}{m l^2 + m_0 l_0^2} \right]^2 $$ $$ h = \frac{1}{2g} \frac{4 m_0^2 l_0^2 l^2 (2gh_0)}{(m l^2 + m_0 l_0^2)^2} $$ Simplifying: $$ h = \frac{4 m_0^2 l_0^2 l^2 h_0}{(m l^2 + m_0 l_0^2)^2} $$