Let the initial direction of A be positive. [Q1] To just avoid a collision, the final velocity of A ($v_A$) must be equal to the final velocity of B with the box ($v_B$). Let this common velocity be $v_f$. The process can be analyzed in two steps, both conserving momentum.

1. A pushes the box:

$$(M+m)v_0 = Mv_A + mv_{box}$$

2. B catches the box:

$$mv_{box} + M(-v_0) = (M+m)v_B$$

Set the condition $v_A = v_B = v_f$ and solve the system for $v_{box}$:

$$(M+m)v_0 = Mv_f + mv_{box} \quad (1)$$ $$mv_{box} - Mv_0 = (M+m)v_f \quad (2)$$

From (2), $v_f = \frac{mv_{box} - Mv_0}{M+m}$. Substitute into (1):

$$(M+m)v_0 = M\left(\frac{mv_{box} - Mv_0}{M+m}\right) + mv_{box}$$ $$(M+m)^2 v_0 = Mmv_{box} - M^2 v_0 + m(M+m)v_{box}$$ $$(M^2+2Mm+m^2+M^2)v_0 = (Mm+Mm+m^2)v_{box}$$ $$(2M^2+2Mm+m^2)v_0 = (2Mm+m^2)v_{box}$$ $$v_{box} = \frac{2M^2+2Mm+m^2}{m(2M+m)}v_0$$

Substituting values: $M=30$ kg, $m=15$ kg, $v_0=2.0$ m/s.

$$v_{box} = \frac{2(30)^2+2(30)(15)+15^2}{15(2 \cdot 30+15)}(2.0) = \frac{1800+900+225}{15(75)}(2.0) = \frac{2925}{1125}(2.0) = 5.2 \text{ m/s}$$

[Q2] By the work-energy theorem, the work done by A on the box equals the change in the box's kinetic energy.

$$W_A = \Delta K_{box} = \frac{1}{2}m v_{box}^2 - \frac{1}{2}m v_0^2 = \frac{1}{2}m(v_{box}^2 - v_0^2)$$

Substituting values:

$$W_A = \frac{1}{2}(15)(5.2^2 - 2.0^2) = 7.5(27.04-4.0) = 7.5(23.04) = 172.8 \text{ J}$$