
Force due to beam assuming complete reflection F=C2P=C2dtdE;P is power
So change in momentum of mirror.
m( V−0)=∫Fdt=C2∫dE=C2E
Now using work energy theorem ...(1)
$\begin{aligned}
& \mathrm{W}_{\mathrm{g}}=\Delta \mathrm{k} \
& -\mathrm{mg} \ell(1-\cos \theta)=0-\frac{1}{2} \mathrm{mv}^2 \
& \mathrm{~g} \ell\left(2 \sin ^2 \frac{\theta}{2}\right)=\frac{\mathrm{v}^2}{2}
\end{aligned}$
as θ is small
gℓ2(2θ)2=21 m2c24E2 gℓθ2= m2c24E2θ=mcgℓ2E