Let the initial speed of the $1\,\text{kg}$ mass be $u$.
The $5\,\text{kg}$ mass is initially at rest.

After collision:

• $1\,\text{kg}$ mass has velocity $v_1=-2\,\text{m s}^{-1}$ (reversed direction)
• $5\,\text{kg}$ mass has velocity $v_2$

Use elastic collision condition.
For a 1D elastic collision, relative speed of separation = relative speed of approach: $v_2 - v_1 = u - 0$ $v_2 - (-2)=u$ $u=v_2+2$

Apply conservation of momentum. $1\cdot u + 5\cdot 0 = 1\cdot (-2) + 5v_2$ $u=-2+5v_2$

Substitute $u=v_2+2$: $v_2+2=-2+5v_2$ $4=4v_2$ $v_2=1\,\text{m s}^{-1}$

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

Now evaluate the options.

Option A: Total momentum of the system is $3\,\text{kg m s}^{-1}$

Total momentum is conserved, so $p=1\cdot 3=3\,\text{kg m s}^{-1}$ Also after collision: $p=1\cdot (-2)+5\cdot 1=-2+5=3$ Hence, A is correct.

Option B: Momentum of $5\,\text{kg}$ mass after collision is $4\,\text{kg m s}^{-1}$

$p_2=5\times 1=5\,\text{kg m s}^{-1}$ Hence, B is incorrect.

Option C: Kinetic energy of the centre of mass is $0.75\,\text{J}$

Centre of mass speed: $V_{\text{cm}}=\frac{\text{total momentum}}{\text{total mass}}=\frac{3}{1+5}=\frac{3}{6}=0.5\,\text{m s}^{-1}$

Kinetic energy of centre of mass motion: $K_{\text{cm}}=\frac{1}{2}(m_1+m_2)V_{\text{cm}}^2$ $=\frac{1}{2}(6)(0.5)^2=3\times 0.25=0.75\,\text{J}$ Hence, C is correct.

Option D: Total kinetic energy of the system is $4\,\text{J}$

Since collision is elastic, final KE = initial KE.

Final KE: $K=\frac{1}{2}(1)(2^2)+\frac{1}{2}(5)(1^2)$ $=2+2.5=4.5\,\text{J}$ Hence, D is incorrect.

Final answer
Correct options are: $\boxed{A,\ C}$

Comparison with stored correct answer
Stored correct answer: $A, C$
Derived answer: $A, C$
They match.