A block moving horizontally on a smooth surface with a speed of 40 m s$$^{-1}$$ splits into two parts with masses in the ratio of 1 : 2. If the smaller part moves at 60 m s$$^{-1}$$ in the same direction, then the fractional change in kinetic energy is:

Let the original mass of the single block be $$M$$ and its initial speed be $$u = 40\ \text{m s}^{-1}$$. The initial kinetic energy is given by the formula $$K = \dfrac12 m v^{2}$$, so we have

$$K_i = \dfrac12\,M\,u^{2} = \dfrac12\,M\,(40)^{2} = \dfrac12\,M\,(1600) = 800\,M.$$

The block breaks into two pieces whose masses are in the ratio $$1:2$$. We denote the smaller mass by $$m$$ and the larger mass by $$2m$$. Because the total mass must remain the same,

$$m + 2m = 3m = M \;\;\Longrightarrow\;\; m = \dfrac{M}{3}, \qquad 2m = \dfrac{2M}{3}.$$

The problem states that the smaller piece moves forward with speed $$v_1 = 60\ \text{m s}^{-1}$$. We still have to find the speed $$v_2$$ of the larger piece.

The surface is smooth, so no external horizontal force acts on the system; therefore linear momentum is conserved. The law of conservation of momentum says

$$\text{(total momentum before)} = \text{(total momentum after)}.$$

This gives

$$M\,u = m\,v_1 + (2m)\,v_2.$$

Substituting $$u = 40\ \text{m s}^{-1}$$, $$m = \dfrac{M}{3}$$ and $$v_1 = 60\ \text{m s}^{-1}$$, we obtain

$$M(40) = \dfrac{M}{3}(60) + \dfrac{2M}{3}\,v_2.$$

Dividing every term by $$M$$ to cancel the common factor,

$$40 = \dfrac{60}{3} + \dfrac{2}{3}\,v_2.$$

Since $$\dfrac{60}{3} = 20$$, we have

$$40 = 20 + \dfrac{2}{3}\,v_2 \;\;\Longrightarrow\;\; 40 - 20 = \dfrac{2}{3}\,v_2 \;\;\Longrightarrow\;\; 20 = \dfrac{2}{3}\,v_2.$$

Multiplying both sides by $$\dfrac{3}{2}$$ to isolate $$v_2$$, we find

$$v_2 = 20\left(\dfrac{3}{2}\right) = 30\ \text{m s}^{-1}.$$

Now we calculate the kinetic energy after the split. Using $$K = \dfrac12 m v^{2}$$ for each part,

$$\begin{aligned} K_f &= \dfrac12\,m\,v_1^{2} + \dfrac12\,(2m)\,v_2^{2} \\ &= \dfrac12\left(\dfrac{M}{3}\right)(60)^{2} + \dfrac12\left(\dfrac{2M}{3}\right)(30)^{2}. \end{aligned}$$

Simplifying each term one at a time:

First term:

$$\dfrac12\left(\dfrac{M}{3}\right)(60)^{2} = \dfrac{M}{6}\,(3600) = 600\,M.$$

Second term:

$$\dfrac12\left(\dfrac{2M}{3}\right)(30)^{2} = \dfrac{M}{3}\,(900) = 300\,M.$$

Adding these contributions gives

$$K_f = 600\,M + 300\,M = 900\,M.$$

The change in kinetic energy is

$$\Delta K = K_f - K_i = 900\,M - 800\,M = 100\,M.$$

The fractional change in kinetic energy is the ratio $$\dfrac{\Delta K}{K_i}$$, so

$$\dfrac{\Delta K}{K_i} = \dfrac{100\,M}{800\,M} = \dfrac{1}{8}.$$

Hence, the correct answer is Option D.