Solve the following :

Solution:

(a) For initial velocities,

Use C.O.E.L

$\frac{1}{2} m V_{1}^{2}=m g l$

$V_{1}=\sqrt{2 g l}$

$\frac{1}{2}(2 m) V_{2}^{2}=(2 m) g l$

Use C.O.L.M

$m \cdot V_{1}+2 m V_{2}=m V_{1}^{\prime}+(2 m) V_{2}^{\prime}$

$m(\sqrt{2 g l})+2 m(-\sqrt{2 g l})=m V_{1}^{\prime}+(2 m) V_{2}^{\prime}$

$V_{1}^{\prime}+2 V_{2}^{\prime}=-\sqrt{2 g l}$

$\therefore V_{1}-V_{2}=V_{2}^{2}-V_{1}^{2}$

$V_{1}^{\circ}-V_{2}^{\sigma}=-V_{1}+V_{2}$

$=-(\sqrt{2 g l}-(-\sqrt{2 g l}))$

$=-2 \sqrt{2 g l}$

Solve (1) and (2)

$-3 V_{2}^{\prime}=-\sqrt{2 g l} \Longrightarrow V_{2}^{\prime}=\frac{\sqrt{2 g l}}{3}$

Similarly,

$V_{1}^{\prime}=\frac{\sqrt{50 g l}}{3}$

(b) Use C.O.E.L

$\frac{1}{2} 2 m(0)^{2}-\frac{1}{2} 2 m\left(V_{2}\right)^{2}=-2 m g h$

$$

\Rightarrow h=\frac{l}{9}

$$

For mass m, use C.O.E.L,

$\frac{1}{2} m(0)^{2}-\frac{1}{2} m\left(V_{1}^{\prime}\right)^{2}=m g h^{\circ}$

$g h^{\prime \prime}=\frac{1}{2} \frac{50 g l}{9}$

$h^{\prime}=\frac{25 l}{9}$

$2 \mathrm{~m}$ will rise $u n$ to $h=\frac{l}{9}$

$\mathrm{m}$ will rise up to

$h^{\prime}=\frac{25 l}{9}$