Here, m₁ = 1 kg, m₂ = 5 kg

u₁ = u, u₂ = 0

v₁ = $$-$$2 m s^$$-$$1, v₂ = v

By the law of conservation of linear momentum, we get

$${m_1}{u_1} + {m_2}{u_2} = {m_1}{u_1} + {m_2}{u_2}$$

$$1 \times u + 5 \times 0 = 1 \times ( - 2) + 5 \times v$$

$$u = 5v - 2$$ ...... (i)

By the definition of coefficient of restitution,

$$e = {{{v_2} - {v_1}} \over {{u_1} - {u_2}}}$$

For a perfectly elastic collision, e = 1

$$\therefore$$ $$1 = {{v + 2} \over u}$$ or $$u = v + 2$$ ..... (ii)

Solving equations (i) and (ii), we get

u = 3 m s^$$-$$1, v = 1 m s^$$-$$1

Before collision,

Total momentum of the system = 1 $$\times$$ 3 + 5 $$\times$$ 0 = 3 kg m s^$$-$$1

After collision,

Total momentum of the system = 1 $$\times$$ ($$-$$2) + 5 $$\times$$ (1) = 3 kg m s^$$-$$1

Hence, option (a) is correct.

Momentum of 5 kg mass after collision = 5 $$\times$$ 1 = 5 kg m s^$$-$$1.

Hence, option (b) is incorrect.

Velocity of centre mass is $${v_{cm}} = {{1 \times 3 + 5 \times 0} \over {1 + 5}} = {1 \over 2}$$ m s ^$$-$$1

Kinetic energy of the centre of mass

$$ = {1 \over 2}{m_{system}}v_{CM}^2 = {1 \over 2} \times 6 \times {\left( {{1 \over 2}} \right)^2} = 0.75$$ J

Hence, option (c) is correct.

Before collision,

Total kinetic energy of the system

$$ = {1 \over 2} \times 1 \times {3^2} + {1 \over 2} \times 5 \times {0^2} = 4.5$$ J

After collision,

Total kinetic energy of the system

$$ = {1 \over 2} \times 1 \times {( - 2)^2} + {1 \over 2} \times 5 \times {(1)^2} = 4.5$$ J

Hence, option (d) is incorrect.