A solid sphere of mass 1 kg rolls without slipping on a plane surface. Its kinetic energy is $$7 \times 10^{-3}$$ J. The speed of the centre of mass of the sphere is ______ cm s$$^{-1}$$.

Total kinetic energy of a rigid body that rolls without slipping is the sum of
(i) translational kinetic energy of the centre of mass and
(ii) rotational kinetic energy about the centre of mass.

Let the linear speed of the centre of mass be $$v$$. For a solid sphere of mass $$m$$ and radius $$R$$:

Moment of inertia about its centre is
$$I_{\text{CM}} = \frac{2}{5}\,mR^{2}$$.

Because the sphere rolls without slipping, the rolling condition is
$$v = \omega R$$, where $$\omega$$ is the angular speed.

Write each part of the kinetic energy:

Translational part: $$K_{\text{trans}} = \frac{1}{2}\,m v^{2}$$.

Rotational part: $$K_{\text{rot}} = \frac{1}{2}\,I_{\text{CM}}\omega^{2} = \frac{1}{2}\left(\frac{2}{5}mR^{2}\right)\left(\frac{v}{R}\right)^{2} = \frac{1}{5}\,m v^{2}$$.

Total kinetic energy therefore is
$$K_{\text{total}} = \frac{1}{2}m v^{2} + \frac{1}{5}m v^{2} = \frac{7}{10}\,m v^{2} \quad -(1)$$.

The problem gives $$K_{\text{total}} = 7 \times 10^{-3}\,\text{J}$$ and $$m = 1\,\text{kg}$$. Substitute these into $$(1)$$:

$$7 \times 10^{-3} = \frac{7}{10}\,(1)\,v^{2}$$

Divide both sides by $$\tfrac{7}{10}$$:

$$v^{2} = 10^{-2}$$

Taking the positive square root (speed is positive):

$$v = 0.1\,\text{m s}^{-1}$$.

Convert metres per second to centimetres per second:

$$0.1\,\text{m s}^{-1} = 0.1 \times 100 = 10\,\text{cm s}^{-1}$$.

Hence, the speed of the centre of mass of the sphere is 10 cm s$$^{-1}$$.