30 kg block on spring scale, k = 30 kN/m

Data: $m_A=30$ kg, $m_B=10$ kg, $h=2$ m, $k=30$ kN/m, perfectly plastic impact ($e=0$), $g=9.81\ \text{m/s}^2$.

Step 1: block just before impact

$$v_A=\sqrt{2gh}=\sqrt{2(9.81)(2)}=6.264\ \text{m/s}$$

Step 2: plastic impact (pan initially at rest)

Momentum is conserved and both move together:

$$30(6.264)=(30+10)v'\ \Rightarrow\ v'=4.698\ \text{m/s}$$

Step 3: initial compression of the spring due to the pan

$$x_1=\frac{m_Bg}{k}=\frac{10(9.81)}{30000}=0.00327\ \text{m}$$

Step 4: work-energy from just after impact to the lowest point

Let $x_2$ be the maximum total compression and $M=40$ kg. Kinetic energy just after impact: $\tfrac12Mv'^2=441.45$ J.

$$\tfrac12Mv'^{2}+\tfrac12kx_1^{2}+Mg(x_2-x_1)=\tfrac12kx_2^{2}$$ $$15000x_2^{2}-392.40x_2-\left(441.45+0.1604-1.2831\right)=0$$ $$x_2=0.1849\ \text{m}$$

The pan moves down $x_2-x_1=0.1816$ m from its position at impact.

Answer: maximum deflection of the pan $\approx0.182$ m below its position at impact (total spring compression $\approx0.185$ m).