Question

In the figure shown, a plate of mass $60\text{gm}$ is at rest and in equilibrium. A particle of mass $m=30\text{gm}$ is released from height $\frac{4.5mg}{K}$ above the plate. The particle hits the plate and sticks to it. If the time interval between the collision of the particle and the time at which it comes to momentary rest for the first time is given by $\frac{\pi }{x}$, then find $x$.
Spring has force constant $1\text{N/m}$. Neglect the duration of collision

Answer 1

Velocity of the particle just before collision
$u=\sqrt{2g\times \frac{4.5mg}{K}}=3g\sqrt{\frac{m}{K}}$
Just after collision, let velocity of the system be $\left(V\right)$
Then, from momentum conservation,
$mu=3mV$
$\Rightarrow V=\frac{u}{3}=g\sqrt{\frac{m}{K}}$
Now the system performs SHM with time period
$T=2\pi \sqrt{\frac{3m}{K}}$
And mean position will be $\frac{mg}{K}$ distance below the point of collision, corresponding to balancing of weight of $m$.
Let the equation of motion be
$y=A\sin (\omega t+\varphi )$ ..... (i)
$\Rightarrow v=\frac{dy}{dt}=A\omega \cos (\omega t+\varphi )$ ......(ii)
At $t=0,y=\frac{mg}{K}$ and $v=g\sqrt{\frac{m}{K}}$
From Eqs. (i) and (ii), $\frac{mg}{K}=A\sin \varphi$ ...(iii)
and $\sqrt{3}\frac{mg}{K}=A\cos \varphi$ ... (iv)
[because $\omega =\sqrt{\frac{K}{3m}}$]
Squaring and adding (iii) and (iv),
$A=\frac{2mg}{K}$ ....(v)
From Eqs. (iii) and (v), $\tan \varphi =\frac{1}{\sqrt{3}}$ or $\varphi =\frac{5\pi }{6}$
Therefore, $y=\frac{2mg}{K}\sin (\sqrt{\frac{K}{3m}}t+\frac{5\pi }{6})$
The plate will be at rest again when $y=-A$ [maximum compression]
i.e $-A=-\frac{2mg}{K}=\frac{2mg}{K}\sin (\sqrt{\frac{K}{3m}}t+\frac{5\pi }{6})$
$\Rightarrow \sqrt{\frac{K}{3m}}t+\frac{5\pi }{6}=\frac{3\pi }{2}$
$\Rightarrow t=\frac{2\pi }{3}\sqrt{\frac{3m}{K}}$
Using values $K=1\text{N/m},m=30\text{gm}$
$\Rightarrow t=\frac{\pi }{5}s$
Thus, $x=5$ is the answer.