As shown in the figure, a particle of mass 10 kg is placed at a point $$A$$. When the particle is slightly displaced to its right, it starts moving and reaches the point $$B$$. The speed of the particle at $$B$$ is $$x$$ m s$$^{-1}$$. (Take $$g = 10$$ m s$$^{-2}$$) The value of $$x$$ to the nearest integer is ________.

Let the vertical height of point $$A$$ above point $$B$$ be $$h$$. On releasing the particle from rest at $$A$$ and neglecting friction, mechanical energy is conserved.

Potential energy at $$A$$: $$U_A = m g h$$

Kinetic energy at $$A$$: the particle starts from rest, so $$K_A = 0$$

Potential energy at $$B$$: we choose point $$B$$ as the reference level, so $$U_B = 0$$

Kinetic energy at $$B$$: $$K_B = \tfrac12 m x^{2}$$, where $$x$$ is the speed to be found.

Conservation of mechanical energy gives $$U_A + K_A = U_B + K_B$$ $$\Rightarrow m g h = \tfrac12 m x^{2}$$ $$-(1)$$

The figure (given in the question) shows that the height drop from $$A$$ to $$B$$ is $$h = 5 \text{ m}$$.

Substituting $$m = 10 \text{ kg}$$, $$g = 10 \text{ m s}^{-2}$$ and $$h = 5 \text{ m}$$ into $$(1)$$:

$$10 \times 10 \times 5 = \tfrac12 \times 10 \times x^{2}$$

$$500 = 5 x^{2}$$

$$x^{2} = 100$$

$$x = 10 \text{ m s}^{-1}$$ (taking the positive root because speed is positive).

Therefore, the speed of the particle at point $$B$$, to the nearest integer, is $$\mathbf{10}$$.