A body of mass 2 kg moving with a speed of 4 m s$$^{-1}$$ makes an elastic collision with another body at rest and continues to move in the original direction but with one fourth of its initial speed. The speed of the two body centre of mass is $$\frac{x}{10}$$. Find the value of x.

We have two bodies. The first body has mass $$m_1 = 2\ \text{kg}$$ and is moving with initial speed $$u_1 = 4\ \text{m s}^{-1}$$. The second body has mass $$m_2$$ (unknown for now) and is initially at rest, so $$u_2 = 0$$.

Because the collision is perfectly elastic, two basic conservation laws can be applied:

1. Conservation of linear momentum $$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2.$$

2. Conservation of kinetic energy $$\dfrac12 m_1 u_1^{2} + \dfrac12 m_2 u_2^{2} = \dfrac12 m_1 v_1^{2} + \dfrac12 m_2 v_2^{2}.$$

After the collision, the first body continues in the same direction but with one-fourth of its initial speed. Hence

$$v_1 = \dfrac{u_1}{4} = \dfrac{4}{4} = 1\ \text{m s}^{-1}.$$

Let the speed of the second body after collision be $$v_2$$ (to be found). We now substitute the known numbers into the two conservation equations.

Step 1: Momentum conservation $$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$$ $$2 \times 4 + m_2 \times 0 = 2 \times 1 + m_2 v_2$$ $$8 = 2 + m_2 v_2.$$

Rearranging gives $$m_2 v_2 = 8 - 2 = 6. \quad -(1)$$

Step 2: Kinetic-energy conservation $$\dfrac12 m_1 u_1^{2} + 0 = \dfrac12 m_1 v_1^{2} + \dfrac12 m_2 v_2^{2}$$ $$\dfrac12 \times 2 \times 4^{2} = \dfrac12 \times 2 \times 1^{2} + \dfrac12 m_2 v_2^{2}$$ $$1 \times 16 = 1 \times 1 + \dfrac12 m_2 v_2^{2}$$ $$16 = 1 + \dfrac12 m_2 v_2^{2}.$$

Subtracting 1 from both sides, we obtain $$\dfrac12 m_2 v_2^{2} = 15$$ $$m_2 v_2^{2} = 30. \quad -(2)$$

Step 3: Solving for $$v_2$$ We divide equation (2) by equation (1): $$\dfrac{m_2 v_2^{2}}{m_2 v_2} = \dfrac{30}{6}$$ $$v_2 = 5\ \text{m s}^{-1}.$$

Step 4: Determining $$m_2$$ (only for completeness) Using equation (1): $$m_2 = \dfrac{6}{v_2} = \dfrac{6}{5} = 1.2\ \text{kg}.$$

Step 5: Velocity of the centre of mass The centre of mass (COM) velocity is constant throughout the motion. It can be found from the initial momenta: $$v_{\text{cm}} = \dfrac{m_1 u_1 + m_2 u_2}{m_1 + m_2}.$$

Substituting the values, we get $$v_{\text{cm}} = \dfrac{2 \times 4 + 1.2 \times 0}{2 + 1.2} = \dfrac{8}{3.2}\ \text{m s}^{-1}.$$

Now, $$\dfrac{8}{3.2} = 2.5\ \text{m s}^{-1}.$$

The problem states that the COM speed can be expressed as $$\dfrac{x}{10}\ \text{m s}^{-1}$$. So we set

$$\dfrac{x}{10} = 2.5 \quad\Longrightarrow\quad x = 2.5 \times 10 = 25.$$

So, the answer is $$25$$.