1. Given data

• Mass of first body: $m_1 = 2\,\text{kg}$
• Initial speed of first body: $u_1 = 4\,\text{m/s}$
• Second body is initially at rest: $u_2 = 0$
• After elastic collision, first body continues in the same direction with one-fourth of its initial speed: $v_1 = \frac{1}{4} \times 4 = 1\,\text{m/s}$

We need the speed of the centre of mass.

2. Use conservation of momentum

Let mass of second body be $m_2$ and its final speed be $v_2$.

Initial momentum: $p_i = m_1 u_1 + m_2 u_2 = 2\cdot 4 + m_2\cdot 0 = 8$

Final momentum: $p_f = m_1 v_1 + m_2 v_2 = 2\cdot 1 + m_2 v_2 = 2 + m_2 v_2$

So, $8 = 2 + m_2 v_2$ $m_2 v_2 = 6 \quad ...(1)$

3. Use conservation of kinetic energy

Since collision is elastic, $\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$

Substitute values: $\frac{1}{2}(2)(4^2) = \frac{1}{2}(2)(1^2) + \frac{1}{2}m_2v_2^2$ $16 = 1 + \frac{1}{2}m_2v_2^2$ $\frac{1}{2}m_2v_2^2 = 15$ $m_2v_2^2 = 30 \quad ...(2)$

4. Find $m_2$

From (1): $v_2 = \frac{6}{m_2}$

Substitute into (2): $m_2\left(\frac{6}{m_2}\right)^2 = 30$ $m_2\cdot \frac{36}{m_2^2} = 30$ $\frac{36}{m_2} = 30$ $m_2 = \frac{36}{30} = \frac{6}{5}\,\text{kg}$

5. Find centre of mass speed

Velocity of centre of mass remains constant: $V_{\text{cm}} = \frac{\text{total initial momentum}}{\text{total mass}}$

So, $V_{\text{cm}} = \frac{8}{2 + \frac{6}{5}}$ $V_{\text{cm}} = \frac{8}{\frac{16}{5}} = 8\cdot \frac{5}{16} = \frac{5}{2}\,\text{m/s}$

Given, $V_{\text{cm}} = \frac{x}{10}\,\text{m/s}$

Therefore, $\frac{x}{10} = \frac{5}{2}$ $x = 25$

6. Final answer

$\boxed{25}$