$$ (y_{đầu} - H)^2 + \left(\frac{R}{2}\right)^2 = R^2 $$

$$ \implies y_{đầu} = H + \frac{\sqrt{3}}{2}R $$

 $W_{đầu} = Mgy_{đầu} = Mg(H + \frac{\sqrt{3}}{2}R)$

 $W_{cuối} = MgR + \frac{1}{2}Mv^2 + 2(\frac{1}{2}MV^2)$

$$ Mg(H + \frac{\sqrt{3}}{2}R) = MgR + \frac{1}{2}Mv^2 + MV^2 $$

$$ \implies g\left(H + R(\frac{\sqrt{3}}{2} - 1)\right) = \frac{1}{2}v^2 + V^2 \quad (*)$$

$$ V = v\sqrt{3} \quad (**) $$

Thay $(**)$ vào $(*)$:

$$ g\left(H + R(\frac{\sqrt{3}}{2} - 1)\right) = \frac{1}{2}v^2 + (v\sqrt{3})^2 $$

$$ g\left(H + R(\frac{\sqrt{3}}{2} - 1)\right) = \frac{1}{2}v^2 + 3v^2 $$

$$ g\left(H + R(\frac{\sqrt{3}}{2} - 1)\right) = \frac{7}{2}v^2 $$

$$ v^2 = \frac{2g}{7}\left(H + R(\frac{\sqrt{3}}{2} - 1)\right) $$

$$ v = \sqrt{\frac{2g}{7}\left(H + \frac{\sqrt{3}}{2}R - R\right)} $$