A ball of mass 10 kg moving with a velocity $$10\sqrt{3}$$ m s$$^{-1}$$ along the $$x$$-axis, hits another ball of mass 20 kg which is at rest. After the collision, the first ball comes to rest while the second ball disintegrates into two equal pieces. One piece starts moving along $$y$$-axis with a speed of 10 m s$$^{-1}$$. The second piece starts moving at an angle of 30° with respect to the $$x$$-axis. The velocity of the ball moving at 30° with $$x$$-axis is $$x$$ m s$$^{-1}$$.The configuration of pieces after the collision is shown in the figure below. The value of $$x$$ to the nearest integer is ________.

A ball of mass $$10 \text{ kg}$$ moving at $$10\sqrt{3} \text{ m/s}$$ along the $$x$$-axis hits a $$20 \text{ kg}$$ ball at rest. After the collision, the first ball stops and the second ball breaks into two equal pieces of $$10 \text{ kg}$$ each.

By conservation of momentum, the total momentum before collision is $$10 \times 10\sqrt{3} = 100\sqrt{3} \text{ kg m/s}$$ along the $$x$$-axis, and zero along the $$y$$-axis.

One piece moves along the $$y$$-axis at $$10 \text{ m/s}$$, so its momentum is $$10 \times 10 = 100 \text{ kg m/s}$$ along $$y$$ and $$0$$ along $$x$$. Let the second piece move at speed $$v$$ at $$30°$$ with the $$x$$-axis. Its momentum components are $$10v\cos 30°$$ along $$x$$ and $$10v\sin 30°$$ along $$y$$.

Applying conservation of momentum along the $$x$$-axis: $$10v\cos 30° = 100\sqrt{3}$$, so $$10v \times \dfrac{\sqrt{3}}{2} = 100\sqrt{3}$$, giving $$v = 20 \text{ m/s}$$.

We can verify with the $$y$$-component: the first piece has $$+100 \text{ kg m/s}$$ along $$y$$, and the second piece must have $$-100 \text{ kg m/s}$$ along $$y$$ (since total $$y$$-momentum is zero). Indeed, $$10 \times 20 \times \sin 30° = 100$$, and this piece moves at $$30°$$ below the $$x$$-axis, confirming the balance.

The answer is $$\boxed{20}$$.