Question

A disc of mass $3m$ and a disc of mass $m$ are connected by a massless spring of stiffness $k$. The heavier disc is placed on the ground with the spring vertical and lighter disc on top. From its equilibrium position, the upper disc is pushed down by a distance $\delta$ and released. Then

• A. if $\delta >3$mg/k the lower disc will bounce up
• B. if $\delta =2$mg/k, maximum normal reaction from ground on lower disc$=6$ mg.
• C. if $\delta =2$mg/k, maximum normal reaction from ground on lower disc $=4$ mg.
• D. if $\delta >4$ mg/k the lower disc will bounce up

Answer 1

Answer: (B), (D)

if $\delta =2$mg/k, maximum normal reaction from ground on lower disc$=6$ mg.
if $\delta >4$ mg/k the lower disc will bounce up

$x_{eqm}=\frac{mg}{k}$
Maximum normal force will be exerted when the smaller ball is at it's lower extreme
Thus $N=3mg+k(\frac{mg}{k}+\frac{2mg}{k})$
therefore option (B) is correct.
also for the heavier disc to stay on the ground
$k(\delta -\frac{mg}{k})\le 3mg$
thus for the disc to bounce up
$\delta >\frac{4mg}{k}$
Hence (D) is correct.