Two particles A and B of equal masses lie close together on a horizontal table and are connected by a light inextensible string of length $l$ . A is projected vertically upwards with a velocity $\sqrt{10\mathrm{g}l}$ . The velocity with which it reaches the table again is $\sqrt{xgl}$ . Find $x$.

String becomes taut when A moves upwards by a distance $l$ under influence of gravity. Let $v_1$be the velocity of A at this moment, then

${\mathrm{v}}_1^2=(\sqrt{10\mathrm{g}l}{)}^2-2\mathrm{g}l=8\mathrm{g}l\Rightarrow {\mathrm{v}}_1=\sqrt{8\mathrm{g}l}$

Let $v_2$ be the common velocities of both A and B, just after string becomes taut. Then from law of conservation of linear momentum, we have

$m=\sqrt{8gl}+m\left(0\right)=(m+m)v_2$

$\Rightarrow {\mathrm{v}}_2=\frac{{\mathrm{v}}_1}{2}=\frac{\sqrt{8\mathrm{g}l}}{2}$

Both particles return to their original height with the same speed v₂ in downward direction and the string becomes loose as soon as B strikes the ground and the speed v with which A strikes the ground is,

${\mathrm{v}}^2={\mathrm{v}}_2^2+2\mathrm{g}l=\frac{8\mathrm{g}l}{4}+2\mathrm{g}l\Rightarrow {\mathrm{v}}^2=4\mathrm{g}l$

$\Rightarrow \mathrm{v}=2\sqrt{\mathrm{g}l}=\sqrt{4\mathrm{g}l}\Rightarrow \mathrm{x}=4$