Impact

m1, m2, m = 400.0, 100.0, 500.0
k  = 100E3
h  = 0.10
xM = 9.83E-3

Velocity at impact

The potential energy of body no.2 before release is $V = m_2 g h$ and its kinetic energy is $0$, the potential energy at impact point is $0$ and the kinetic energy is $T=m_2 v_2^2/2$. During the free fall no energy is dissipated, hence $$\begin{align} m_2 g h &= \frac12\,m_2v_2^2\rightarrow\ \rightarrow v_2 &= \sqrt{2 g h} \end{align}$$

Initial velocity of glued body

Using the conservation of momentum, the velocity $v_0$ at time $0^+$ can be derived as follows $$\begin{align} m_2 v_2 &= (m_1+m_2) v_0\rightarrow\ \rightarrow v_0 &= v_2\frac{m_2}{m_1+m_2} \end{align}$$

Response

The characteristics of the glued system are:

• stiffness $k$,
• mass $m = m_1+m_2$,
• natural frequency of vibration $\omega_n=\sqrt{k/m}$

We have the superposition of two different responses,

1. the response to the weight of body no.2, with $\Delta = m_2 g/k$ it is

   $$x_\Delta = \Delta\,(1-\cos\omega_nt).$$

2. the response to the initial velocity,

   $$x_v = \frac{v_0}{\omega_n}\,\sin\omega_nt$$

Summing up,

$$x(t) = \Delta + \left( \frac{v_0}{\omega_n}\sin\omega_nt-\Delta\cos\omega_nt\right)$$

and the maximum displacement $x_M$ is

$$x_M = \Delta + \sqrt{\Delta^2+\frac{v_0^2}{\omega_n^2}}.$$

Rearranging and squaring both members, it is

$$(x_M - \Delta)^2 = x_M^2-2x_M\Delta+\Delta^2 = \Delta^2 +\frac{v_0^2}{\omega_n^2}$$

and eventually we can inject the dependencies on $g$ to have

$$x_M\left(x_M-2\frac{m_2 g}k\right) = \frac{2 g h m_2^2}{m^2}\frac{m}k.$$

Collecting $g$ on one side of the equation and dividing both members by $x_M$ we have

$$2m_2g\,\left(1+\frac{h}{x_M}\frac{m_2}{m}\right) = k\,x_M$$

and it is possible to write the code fragment that gives us the value of the surface acceleration of gravity.

g = k*xM/2/m2/(1+h/xM*m2/m)
print('g = %.3f m/s^2'%g)

g = 1.620 m/s^2

A summary inspection of Wikipedia reveals that the surface gravity of the Moon is

$$g_\text{Moon} = 1.62\,\text{m}/\text{s}^2.$$