A solid sphere of mass $$2$$ kg is making pure rolling on a horizontal surface with kinetic energy $$2240$$ J. The velocity of centre of mass of the sphere will be ______ m s$$^{-1}$$.

For a solid sphere of mass $$m$$ making pure rolling on a horizontal surface, the total kinetic energy is the sum of translational and rotational kinetic energy.

For a solid sphere, the moment of inertia is $$I = \frac{2}{5}mr^2$$, and for pure rolling, $$\omega = \frac{v}{r}$$.

Total KE = Translational KE + Rotational KE:

$$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right)\left(\frac{v}{r}\right)^2$$

$$KE = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$$

Given: $$m = 2$$ kg, $$KE = 2240$$ J.

$$2240 = \frac{7}{10} \times 2 \times v^2$$

$$2240 = \frac{14}{10} v^2 = 1.4\,v^2$$

$$v^2 = \frac{2240}{1.4} = 1600$$

$$v = 40 \text{ m s}^{-1}$$

The velocity of the centre of mass of the sphere is $$\mathbf{40}$$ m s$$^{-1}$$.