A uniform rod of mass m and length 2L on a smooth horizontal surface. A particle of mass m is connected to a string of length L whose other end is connected to the end ‘A’ of the rod. Initially the string is held taut perpendicular to the rod and the particle is given a velocity $v_0$ parallel to the initial position of the rod
 The acceleration of the centre of the rod immediately after the particle is projected is a .
The particle strikes the centre of the rod and sticks to it and the angular speed of the rod after this is $\omega .$ 

 Let tension in string immediately after the projection of the particle be T 
$\alpha$ = acceleration of COM of the rod 
$\alpha$ = angular acceleration of the rod
Acceleration of end A is 

${\mathrm{a}}_{\mathrm{A}}=\mathrm{a}+\mathrm{L\alpha }$
 

For the rod 

T = ma   …(i)
$\mathrm{TL}=\frac{1}{12}\mathrm{m}(2\mathrm{L}{)}^2\cdot \mathrm{\alpha }\Rightarrow \mathrm{T}=\frac{1}{3}\mathrm{m}\cdot \mathrm{L\alpha }\dots \dots \dots \text{ (ii) }$
$\text{ Motion of particle in the reference frame attached to point }\mathrm{A}$
 

$$\begin{gathered} T+{\mathrm{ma}}_{\mathrm{A}}\equiv \frac{{\mathrm{mv}}_0^2}{\mathrm{L}} \\ \mathrm{T}+\mathrm{ma}+\mathrm{mL\alpha }=\frac{{\mathrm{mv}}_0^2}{\mathrm{L}} \\ \mathrm{T}+\mathrm{T}+3\mathrm{T}=\frac{{\mathrm{mv}}_0^2}{\mathrm{L}}\therefore \mathrm{T}=\frac{{\mathrm{mv}}_0^2}{5\mathrm{L}} \\ \therefore \mathrm{a}=\frac{\mathrm{T}}{\mathrm{m}}=\frac{{\mathrm{v}}_0^2}{5\mathrm{L}} \end{gathered}$$
 ${\mathrm{v}}_{\mathrm{cm}}=\frac{{\mathrm{v}}_0}{2}$
 

In reference frame attached to the COM of the system the particle is initially moving to right with velocity $\frac{{\mathrm{v}}_0}{2}$
and the centre of the rod is moving to left with velocity $\frac{{\mathrm{v}}_0}{2}$
Let the angular speed after the particle sticks be $\omega$Note that the centre of mass continues to move with velocity $\frac{v_0}{2}$
$\text{ Angular momentum conservation about }\mathrm{COM}$
 

${\mathrm{L}}_{\mathrm{i}}={\mathrm{L}}_{\mathrm{f}}$
$$\begin{gathered} \mathrm{m}\frac{{\mathrm{v}}_0}{2}\frac{\mathrm{L}}{2}+\mathrm{m}\frac{{\mathrm{v}}_0}{2}\frac{\mathrm{L}}{2}=\frac{1}{12}\mathrm{m}(2\mathrm{L}{)}^2\cdot \mathrm{\omega } \\ \therefore {\mathrm{mv}}_0\mathrm{I}=\frac{2}{3}{\mathrm{mI}}^2\mathrm{\omega }\therefore \mathrm{\omega }=\frac{3}{2}\frac{{\mathrm{v}}_0}{\mathrm{L}} \end{gathered}$$