A small particle of mass m moving horizontally collides with a vertical hinged rod inelastically (e=0) as shown. The kinetic energy of rod and particle system is $\frac{m^{2} {v_{o}}^{2}}{x (m + \frac{M}{y})}$ . The value of $x y$ is

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Detailed Solution

Angular momentum before collision of system

$L_{i} = m v_{o} l$

After collision

$\begin{array}{l} L_{f} = (I_{1} + I_{2}) ω \\ = (m l^{2} + \frac{M l^{2}}{3}) ω \end{array}$

Using angular momentum conservation,

$\begin{array}{l} L_{f} = L_{i} \\ (m l^{2} + \frac{M l^{2}}{3}) ω = m V_{0} l \\ K E = \frac{1}{2} (m l^{2} + \frac{M l^{2}}{3}) ω^{2} \\ = \frac{1}{2} (m l^{2} + \frac{M l^{2}}{3}) \frac{{(m v_{0} l)}^{2}}{\begin{array}{l} {(m l^{2} + \frac{M l^{2}}{3})}^{2} \end{array}} \\ = \frac{m^{2} {v_{0}}^{2} l^{2}}{2 (m l^{2} + \frac{M l^{2}}{3})} = \frac{m^{2} {v_{0}}^{2}}{2 (m + \frac{M}{3})} \end{array}$