A question

@Kushal Patankar @Akhil Bansal

@Kishore S Shenoy @Abhishek Sharma

WARNING : I don't believe this solution to be flawless...

At the time when the ball just sticks to the plank, given an inelastic collision, $v_0 = \sqrt{2gh}\\F = mg$

Assuming $\chi$ to be the stiffness of the spring.

Writing equation of position of the mass-plank system w.r.t initial position, let $\omega^2 = \dfrac \chi{M+m}$ $(M+m)\ddot y = mg - \chi y\\\ddot y = \dfrac m{M+m}g - \omega^2y$

Solving this differential equation, $y=\dfrac{mg}\chi - \dfrac{mg}\chi \cos\omega t+\dfrac {v_0} \omega \sin \omega t$

Time period, $T = \dfrac{2\pi}\omega = 2\pi\sqrt{\dfrac{M+m}\chi}$

Now, either using conservation of energy or taking maximum value of $y$ in the above equations, $h = \frac{b^2 \chi-2 b g m}{2 g m}$ OR $h = \frac{a^2 \chi^2+2 a g \chi m+2 a g\chi M-g^2 M^2}{2 g \chi m}$

I hope the final values, of $h$ are correct and that I did not miss out any terms in my equation... $\ddot\smile$

Hope this helps!

At the time when the body sticks to the pan of the spring, $\ddot y = \dfrac m{m+M}g - \dfrac\chi {m+M}y$ Here, downwards is taken to be positive.

At the equilibrium position, $\ddot y=0$

So, $y_{eq} = \dfrac{mg}\chi,\text{ downwards}$

Also, $\dfrac{a+b}2 = A,\text{ the amplitude}$

So $y_{eq} + b = A = \dfrac{a+b }2\\\Rightarrow \dfrac{mg}\chi = \dfrac{b-a}2\\\chi = \dfrac{2mg}{b-a}$

As we know, $T = 2\pi \sqrt{\dfrac{m+M}\chi}=2\pi\sqrt{\dfrac{(b-a)(m+M)}{2mg}}$

Now, to find $h$,$Mg = ky_0$$\begin{aligned}mg(h+b) + Mgb &= \dfrac k2(y_0+b)^2-\dfrac k2 y_0^2\\ &=\dfrac k 2 (b^2+2by_0)\end{aligned}\\ \boxed{mg(h+b) = \dfrac k 2 b^2}$ $\begin{aligned}mg(h-a)-Mga &= \dfrac k 2 (a-y_0)^2-\dfrac k 2 y_0^2\\&=\dfrac k 2 (a^2-2ay_0)\end{aligned} \\ \boxed{mg(h-a)=\dfrac k 2 a^2}$ Dividing, $\dfrac{h+b}{h-a} = \dfrac{b^2}{a^2} \\h+b = \dfrac{b^2}{b-a}\\h = \dfrac{ab}{b-a}$