A point mass of $1\,\text{kg}$ collides elastically with a stationary point mass of $5\,\text{kg}$. After their collision, the $1\,\text{kg}$ mass reverses its direction and moves with a speed of $2\,\text{m s}^{-1}$. Which of the following statement(s) is (are) correct for the system of these two masses?

Step 1: Define the initial parameters.

Let $m_1$ and $m_2$ be the masses, $u$ be the initial velocity of $m_1$, and $v_1'$ and $v_2'$ be the velocities after collision.

$$m_1 = 1\,\text{kg}$$ $$m_2 = 5\,\text{kg}$$ $$v_1' = -2\,\text{m s}^{-1}$$

Step 2: Apply conservation of momentum.

The total momentum before collision equals the total momentum after collision.

$$m_1 u = m_1 v_1' + m_2 v_2'$$ $$u = -2 + 5v_2'$$

Step 3: Apply the elastic collision condition.

For an elastic collision, the relative velocity of separation equals the relative velocity of approach.

$$u = v_2' - v_1'$$ $$u = v_2' + 2$$

Step 4: Solve for the velocities.

Equate the expressions for $u$.

$$-2 + 5v_2' = v_2' + 2$$ $$v_2' = 1\,\text{m s}^{-1}$$ $$u = 3\,\text{m s}^{-1}$$

Step 5: Check the total momentum of the system.

Calculate the total momentum $p_{\text{total}}$.

$$p_{\text{total}} = m_1 u$$ $$p_{\text{total}} = 1 \cdot 3 = 3\,\text{kg m s}^{-1}$$

This confirms Statement (1) is correct.

Step 6: Check the momentum of the $5\,\text{kg}$ mass after collision.

Calculate the final momentum $p_2'$ of the second mass.

$$p_2' = m_2 v_2'$$ $$p_2' = 5 \cdot 1 = 5\,\text{kg m s}^{-1}$$

This shows Statement (2) is incorrect.

Step 7: Check the kinetic energy of the center of mass.

Find the velocity of the center of mass $V_{\text{cm}}$.

$$V_{\text{cm}} = \frac{p_{\text{total}}}{m_1 + m_2}$$ $$V_{\text{cm}} = \frac{3}{6} = 0.5\,\text{m s}^{-1}$$

Calculate the kinetic energy associated with the center of mass motion $K_{\text{cm}}$.

$$K_{\text{cm}} = \frac{1}{2} (m_1 + m_2) V_{\text{cm}}^2$$ $$K_{\text{cm}} = \frac{1}{2} (6) (0.5)^2 = 0.75\,\text{J}$$

This confirms Statement (3) is correct.

Step 8: Check the total kinetic energy of the system.

Since the collision is elastic, the total kinetic energy is conserved.

$$K_{\text{total}} = \frac{1}{2} m_1 u^2$$ $$K_{\text{total}} = \frac{1}{2} (1) (3)^2 = 4.5\,\text{J}$$

This shows Statement (4) is incorrect.

Answer: The correct statements are (1) and (3).