A block of mass m = 1 kg slides with velocity $\upsilon =6m/s$  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:  $\left(take g=10m/s^2\right)$

Initial angular momentum of block about hinge before Collison = Final angular momentum of combine system after collision

  $\Rightarrow mvl=L ---\left(1\right)$
Now, using energy conservation

  $\begin{array}{l}Loss in rotational kinetic energy of combine system = Gain in potential energy of combine system\\ \text{}\Rightarrow \frac{L^2}{2I}=mgl\left(1-\cos \theta \right)+Mg\frac{l}{2}\left(1-\cos \theta \right) \left(\because \cos \theta =\frac{l-h}{l}\Rightarrow h=l\left(1-\cos \theta \right)\right)\end{array}$
 $\begin{array}{l}\Rightarrow \frac{{\left(mvl\right)}^2}{2\left(m{l}^2+\frac{M{l}^2}{3}\right)}=\left(m+\frac{M}{2}\right)gl\left(1-\cos \theta \right) \\ \left(\because I=I_{block}+I_{rod}=m{l}^2+\frac{M{l}^2}{3}\& from \left(1\right)\right)\text{}\\ \Rightarrow \frac{{\left(1\times 6\times 1\right)}^2}{2\left(1\times 1^2+\frac{2\times 1^2}{3}\right)}=\left(1+\frac{2}{2}\right)10\times 1\left(1-\cos \theta \right)\\ \Rightarrow \cos \theta =\frac{23}{50}\\ \left(\because m=1kg, l=1m and M=2kg\right)\text{}\\ \Rightarrow \theta ={63}^0\end{array}$