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

Options

A. Total momentum of the system is `3 kg ms^(-1)`
B. Momentum of `5 kg` mass after collision is `4 kg ms^(-1)`
C. Kinetic energy of the centre of mass is `0.75 J`
D. Total kinetic energy of the system is `4 J `

Explanation

To solve the problem, we need to analyze the elastic collision between the two point masses: a 1 kg mass and a 5 kg mass. We will use the principles of conservation of momentum and conservation of kinetic energy, as both are conserved in elastic collisions.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- Mass \( m_1 = 1 \, \text{kg} \) (moving with an initial velocity \( u_1 \))
- Mass \( m_2 = 5 \, \text{kg} \) (initially at rest, so \( u_2 = 0 \))
- After the collision, the 1 kg mass reverses direction and moves with a speed of \( v_1 = -2 \, \text{m/s} \) (negative indicates the direction is reversed).
- Let the final velocity of the 5 kg mass be \( v_2 \).

2. **Apply Conservation of Momentum**:
\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]
Substituting the known values:
\[ 1 \cdot u_1 + 5 \cdot 0 = 1 \cdot (-2) + 5 \cdot v_2 \]
Simplifying gives:
\[ u_1 = -2 + 5v_2 \quad \text{(Equation 1)} \]

3. **Apply Conservation of Kinetic Energy**:
\[ \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]
Substituting the known values:
\[ \frac{1}{2} \cdot 1 \cdot u_1^2 + \frac{1}{2} \cdot 5 \cdot 0^2 = \frac{1}{2} \cdot 1 \cdot (-2)^2 + \frac{1}{2} \cdot 5 \cdot v_2^2 \]
Simplifying gives:
\[ \frac{1}{2} u_1^2 = 2 + \frac{5}{2} v_2^2 \quad \text{(Equation 2)} \]

4. **Solve the Equations**:
- From Equation 1, express \( u_1 \):
\[ u_1 = 5v_2 - 2 \]
- Substitute \( u_1 \) into Equation 2:
\[ \frac{1}{2} (5v_2 - 2)^2 = 2 + \frac{5}{2} v_2^2 \]
- Expanding and simplifying:
\[ \frac{1}{2} (25v_2^2 - 20v_2 + 4) = 2 + \frac{5}{2} v_2^2 \]
\[ 25v_2^2 - 20v_2 + 4 = 4 + 5v_2^2 \]
\[ 20v_2^2 - 20v_2 = 0 \]
Factor out:
\[ 20v_2(v_2 - 1) = 0 \]
Thus, \( v_2 = 0 \) or \( v_2 = 1 \).

5. **Conclusion**:
- The final velocities are:
- \( v_1 = -2 \, \text{m/s} \) (1 kg mass)
- \( v_2 = 1 \, \text{m/s} \) (5 kg mass)

### Final Statements:
- The system conserves both momentum and kinetic energy.
- The final velocities are consistent with the principles of elastic collisions.