1. Let the initial mass be $M$

   The block moves with speed $40\,\text{m/s}$ on a smooth surface.

   Initial momentum: $p_i = M \cdot 40$

   Initial kinetic energy: $K_i = \frac{1}{2}M(40)^2 = 800M$

2. Masses after splitting

   The masses are in the ratio $1:2$.

   So, smaller part: $m_1 = \frac{M}{3}$

   Larger part: $m_2 = \frac{2M}{3}$

3. Use conservation of momentum

   The smaller part moves in the same direction with speed $60\,\text{m/s}$. Let the speed of the larger part be $v$.

   Since no external horizontal force acts, $M \cdot 40 = \frac{M}{3}\cdot 60 + \frac{2M}{3}v$

   Divide by $M$: $40 = 20 + \frac{2}{3}v$

   $20 = \frac{2}{3}v$

   $v = 30\,\text{m/s}$

4. Final kinetic energy

   $K_f = \frac{1}{2}\cdot \frac{M}{3}\cdot (60)^2 + \frac{1}{2}\cdot \frac{2M}{3}\cdot (30)^2$

   First term: $\frac{1}{2}\cdot \frac{M}{3}\cdot 3600 = 600M$

   Second term: $\frac{1}{2}\cdot \frac{2M}{3}\cdot 900 = 300M$

   Therefore, $K_f = 600M + 300M = 900M$

5. Change in kinetic energy

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

   Fractional change in kinetic energy: $\frac{\Delta K}{K_i} = \frac{100M}{800M} = \frac{1}{8}$

6. Option check

   • A: $\frac{1}{3}$ — incorrect
   • B: $\frac{2}{3}$ — incorrect
   • C: $\frac{1}{8}$ — correct
   • D: $\frac{1}{4}$ — incorrect

Therefore, the correct answer is: $\boxed{\frac{1}{8}}$