A block of mass \(m =1\, kg\) slides with velocity \(v=6\, m / s\) on a frictionless horizontal surface and collides with a uniform vertical rod and sticks to it as shown. The rod is pivoted about \(O\) and swings as a result of the collision making angle \(\theta\) before momentarily coming to rest. If the rod has mass \(M =2 \,kg \) and length \(l=1\, m\) the value of \(\theta\) is approximately (Take \(\left.g=10 \,m / s ^{2}\right)\)

\(mv l=\frac{ Ml ^{2}}{3} \omega+ m l^{2} \omega\)

\(\Rightarrow \omega=\frac{1 \times 6 \times 1}{\frac{2}{3}+1}=\frac{18}{5}\)

Now using energy consevation

\(\frac{1}{2}\left( M \frac{l^{2}}{3}\right) \omega^{2}+\frac{1}{2}\left( m l^{2}\right) \omega^{2}\)

\(=( m + M ) r _{ cm }(1-\cos \theta)\)

\(=( m + M )\left(\frac{ m l+\frac{ M l}{2}}{ m + M }\right) g (1-\cos \theta)\)

\(\frac{5}{6} \times\left(\frac{18}{5}\right)^{2}=20(1-\cos \theta)\)

\(\Rightarrow 1-\cos \theta=\frac{18}{5} \times \frac{3}{20}\)

\(\cos \theta=1-\frac{27}{50}\)

\(\cos \theta=\frac{23}{50} \Rightarrow \theta \simeq 63^{\circ}\)