1. Initial Motion

Block A (2 kg) accelerates via friction ($a = \mu g = 2 \text{ m/s}^2$). Initial speed 1 m/s, distance 16 cm (0.16 m).

Velocity just before collision ($v_1$):

$$v_1^2 = 1^2 – 2(2)(0.16) = 1 – 0.64 = 0.36 \implies v_1 = 0.6 \text{ m/s}$$

2. Elastic Collision

$m_1 = 2, m_2 = 4$. $v_1 = 0.6, v_2 = 0$.

$$v_{1f} = \frac{m_1-m_2}{m_1+m_2}v_1 = \frac{-2}{6}(0.6) = -0.2 \text{ m/s}$$ $$v_{2f} = \frac{2m_1}{m_1+m_2}v_1 = \frac{4}{6}(0.6) = 0.4 \text{ m/s}$$

3. Post-Collision Distances

Both blocks stop due to friction ($a = 2$).

$S_1 = \frac{v_{1f}^2}{2a} = \frac{0.04}{4} = 0.01 \text{ m} = 1 \text{ cm}$.

$S_2 = \frac{v_{2f}^2}{2a} = \frac{0.16}{4} = 0.04 \text{ m} = 4 \text{ cm}$.

Total separation = $1 + 4 = 5 \text{ cm}$.