Answer three questions from this part. All questions carry equal marks.

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

$$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$$

Substitute the given values:

$$(1.0 \text{ kg})(4.0 \text{ m/s}) + (2.0 \text{ kg})(0 \text{ m/s}) = (1.0 \text{ kg})v_1 + (2.0 \text{ kg})v_2$$ $$4.0 = v_1 + 2v_2 \quad (*)$$

For a perfectly elastic collision, the relative speed of approach equals the relative speed of separation:

$$u_1 - u_2 = v_2 - v_1$$

Substitute the given values:

$$4.0 \text{ m/s} - 0 \text{ m/s} = v_2 - v_1$$ $$4.0 = v_2 - v_1 \quad (**)$$

From equation $(**)$, express $v_2$ in terms of $v_1$:

$$v_2 = v_1 + 4.0$$

Substitute this expression for $v_2$ into equation $(*)$:

$$4.0 = v_1 + 2(v_1 + 4.0)$$ $$4.0 = v_1 + 2v_1 + 8.0$$ $$4.0 = 3v_1 + 8.0$$ $$3v_1 = 4.0 - 8.0$$ $$3v_1 = -4.0$$ $$v_1 = -\frac{4.0}{3} \text{ m/s}$$ $$v_1 \approx -1.33 \text{ m/s}$$

$$v_2 = v_1 + 4.0$$ $$v_2 = -\frac{4.0}{3} + 4.0$$ $$v_2 = -\frac{4.0}{3} + \frac{12.0}{3}$$ $$v_2 = \frac{8.0}{3} \text{ m/s}$$ $$v_2 \approx 2.67 \text{ m/s}$$

The velocities of the balls after collision are:
Velocity of the first ball: $\boxed{-1.33 \text{ m/s}}$
Velocity of the second ball: $\boxed{2.67 \text{ m/s}}$