A ball of mass 1 kg is attached to an inextensible string. The ball is released from the position shown in figure. The impulse imparted by the string to the ball immediately after the string becomes taut is $\sqrt{\alpha }{\mathrm{kgms}}^{-1}$ . Calculate $\alpha$. $\left[\mathrm{Take} \mathrm{g}=10{\mathrm{ms}}^{-2}\right]$ 

The string will taut when the particle will fall through a distance 2 m in downard direction. Applying impulse momentum equation, we get mv – J = 0

$\Rightarrow \mathrm{J}=\mathrm{mv}=\mathrm{m}\sqrt{2\mathrm{gh}}$

$\Rightarrow \mathrm{J}=1\sqrt{2\times 10\times 2}=2\sqrt{10}{\mathrm{kgms}}^{-1}$

$=\sqrt{40}{\mathrm{kgms}}^{-1}\Rightarrow \alpha =40$

At the bottom most point, just before collision,

${\mathrm{v}}_{\mathrm{m}}=\sqrt{4\mathrm{ga}}\text{ and }{\mathrm{v}}_{2\mathrm{m}}=0$

Also, we know that

${\mathrm{v}}_1=\left(\frac{{\mathrm{m}}_1-{\mathrm{em}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}\right){\mathrm{u}}_1\text{ and }{\mathrm{v}}_2=\left(\frac{(1+\mathrm{e}){\mathrm{m}}_1}{{\mathrm{m}}_1+{\mathrm{m}}_2}\right){\mathrm{u}}_1$

$\Rightarrow {\mathrm{v}}_{\mathrm{m}}^{\mathrm{\prime }}=\left(\frac{\mathrm{m}-\frac{1}{2}\times 2\mathrm{m}}{\mathrm{m}+2\mathrm{m}}\right)\sqrt{4\mathrm{ga}}=0$

$\Rightarrow {\mathrm{v}}_{2\mathrm{m}}^{\mathrm{\prime }}={\left(\frac{\left(1+\frac{1}{2}\right)\mathrm{m}}{\mathrm{m}+2\mathrm{m}}\right)}^{\mathrm{\prime }}\sqrt{4\mathrm{ga}}=\frac{1}{2}\sqrt{4\mathrm{ga}}$

After second impact

$\Rightarrow {\mathrm{v}}_{\mathrm{m}}^{\mathrm{\prime }}=\left(\frac{\left(1+\frac{1}{2}\right)2\mathrm{m}}{\mathrm{m}+2\mathrm{m}}\right)\left(\frac{-\sqrt{4\mathrm{ga}}}{2}\right)=-\frac{\sqrt{4g\mathrm{a}}}{2}$

$\Rightarrow {\mathrm{v}}_{2\mathrm{m}}=\left(\frac{2\mathrm{m}-\frac{1}{2}\times \mathrm{m}}{2\mathrm{m}+\mathrm{m}}\right)\left(\frac{-\sqrt{4\mathrm{ga}}}{2}\right)=-\frac{\sqrt{-4g\mathrm{a}}}{2}$

$\text{ So, }{\mathrm{h}}_{\mathrm{m}}=\frac{{\left({\mathrm{v}}_{\mathrm{m}}^{\mathrm{\prime }}\right)}^2}{2\mathrm{g}}=\frac{\mathrm{a}}{2}\text{ and }{\mathrm{h}}_{2\mathrm{m}}=\frac{{\left({\mathrm{v}}_{2\mathrm{m}}\right)}^2}{2\mathrm{g}}=\frac{\mathrm{a}}{8}$

$\Rightarrow x=2\text{ and }y=8$