Question

In the figure shown, a plate of mass 60 grams is at rest and in equilibrium. A particle of mass m = 30 grams is released from height $\frac{4.5mg}{k}$ from the plate. The particle sticks to the plate. Neglecting the duration of collision, the time from the collision of the particle and the plate to the moment when the spring has maximum compression is in the form of Xπ ms (millisecond) where X is (Spring has force constant of 100 N/m)

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 of system of plate + particle
$mu=3mV\Rightarrow V=\frac{u}{3}=g\sqrt{\frac{m}{k}}$
Now system perform's SHM with time period $T=2\pi \sqrt{\frac{3m}{k}}$ and mean position as $\frac{mg}{k}$ distance below the point of collision.
Let the equation of motion $y=\sin (\omega t+\varphi )$
for $t=0$, $y=\frac{mg}{k}$
$\frac{mg}{k}=A\sin \varphi \left(1\right)NowforamplitudeV=\omega \sqrt{A^2-y^2}g\sqrt{\frac{m}{k}}=\sqrt{\frac{k}{3m}}\sqrt{A^2-\frac{m^2{g}^2}{k^2}}\Rightarrow A=\frac{2mg}{k}\left(2\right)$By (1) and (2)
$y=\frac{A}{2}toy=0\Rightarrow t=\frac{T}{12}y=0toy=A\Rightarrow t=\frac{T}{4}$
Total time $t=\frac{T}{12}+\frac{T}{4}=\frac{2\pi }{3}\sqrt{\frac{3m}{k}}$

Using values, $t=20\pi ms$