The total momentum and energy of the system (A + B + spring) are conserved.

[Q1] At maximum compression $x_{max}$, both blocks move with a common velocity $v_c$. By conservation of momentum:

$$m_A v_A + m_B v_B = (m_A + m_B)v_c$$ $$v_c = \frac{m_A v_A + m_B v_B}{m_A + m_B} = \frac{(2.0 \text{ kg})(10 \text{ m/s}) + (5.0 \text{ kg})(3 \text{ m/s})}{2.0 \text{ kg} + 5.0 \text{ kg}} = 5.0 \text{ m/s}$$

By conservation of energy, the initial kinetic energy equals the final kinetic energy plus the stored potential energy:

$$\frac{1}{2}m_A v_A^2 + \frac{1}{2}m_B v_B^2 = \frac{1}{2}(m_A + m_B)v_c^2 + \frac{1}{2}kx_{max}^2$$ $$x_{max} = \sqrt{\frac{m_A v_A^2 + m_B v_B^2 - (m_A+m_B)v_c^2}{k}}$$ $$x_{max} = \sqrt{\frac{2.0(10)^2 + 5.0(3)^2 - (7.0)(5.0)^2}{1120}} = \sqrt{\frac{200 + 45 - 175}{1120}} = \sqrt{\frac{70}{1120}} = 0.25 \text{ m}$$

[Q2] The overall interaction is an elastic collision. Let the final velocities be $v_A'$ and $v_B'$. Conservation of momentum: $m_A v_A + m_B v_B = m_A v_A' + m_B v_B'$

$$35 = 2.0 v_A' + 5.0 v_B'$$

For elastic collisions, the relative velocity is conserved: $v_A - v_B = v_B' - v_A'$

$$10 - 3 = 7 = v_B' - v_A' \implies v_B' = 7 + v_A'$$

Substituting $v_B'$ into the momentum equation:

$$35 = 2.0 v_A' + 5.0(7 + v_A') = 2.0 v_A' + 35 + 5.0 v_A' = 7.0 v_A' + 35$$

This gives $v_A' = 0$ m/s, and thus $v_B' = 7 + 0 = 7$ m/s.