Question

In the figure shown a plate a plate of mass $60gm$ is at rest and in equilibrium, A particle of mass $m=30gm$ is released from height $\frac{4.5mg}{k}$ from the plate.The particle sticks to the plate. Neglecting the duration of collision find time from the collision of the particle and the plate to the moment when the spring has maximum compression. Spring has force constant $1N/m$. Calculate value of time in the form $\pi /x$ and find the value of $x.$

Answer 1

Velocity of the particle just before collision.
$u=\sqrt{2g\times \frac{4.5mg}{K}}\Rightarrow u=3g\sqrt{\frac{m}{K}}$

Now it collides with the plate.

Now just after collision velocity $\left(V\right)$ of the system of plate + particle.

$mu=3mV\Rightarrow V=\frac{u}{3}=g\sqrt{\frac{m}{K}}$

Now the system performs $SHM$ with time period

$T=2\pi \sqrt{3m/K}(\omega =\sqrt{\frac{K}{3m}})$ and mean position as $mg/K$ distance below the point of collision.

Let the equation of motion is

$y=A\sin (\omega t+\varphi )$

$v=\frac{dy}{dt}=A\omega \cos (\omega t+\varphi )$

At $t=0,y=\frac{mg}{K}$ and $v=g\sqrt{\frac{m}{K}}$
from Eqs (i) and (ii) $\frac{mg}{K}=A\sin \varphi$

$A=\frac{2mg}{K}$
from Eqs (iii) and (iv) $\Rightarrow \varphi =\frac{5\pi }{6}$

$y=\frac{2mg}{K}\sin (\sqrt{\frac{K}{3m}}t+\frac{5\pi }{6})$

Hence equation of $SHM$ should be

$y=-A=-\frac{2mg}{K}=\frac{2mg}{K}\sin (\sqrt{\frac{K}{3m}}t+\frac{5\pi }{6})$
The plate will be at rest again when

$y=-A=-\frac{2mg}{K}=\frac{2mg}{K}\sin (\sqrt{\frac{K}{3m}}t+\frac{5\pi }{6})$

$\Rightarrow \sin (\sqrt{\frac{K}{3m}}t+\frac{5\pi }{6})=-1=\sin \frac{3\pi }{2}$

$\Rightarrow \sqrt{\frac{K}{3m}}t+\frac{5\pi }{6}=\frac{3\pi }{2}\Rightarrow t=\frac{2\pi }{3}\sqrt{\frac{3m}{K}}$

Using value $t=\pi /5s$.