Question

Two plates each of mass $m$ are connected by a massless spring as shown below.

A weight $w$ is put on the upper plate which compresses the spring further. When $w$ is removed, the entire assembly jumps up. The minimum weight $w$ needed for the assembly to jump up when the weight is removed is just more than

• A. m
• B. 2m
• C. 3m
• D. 4m

Answer 1

Answer: (B)

For block $m_2$ to lift off,
Spring force = Weight of block
$\Rightarrow$ $kx=mg$ or $x=\frac{mg}{k}$
Now, from energy conservation (position (I and II marked in the diagram),
Initial stored energy (due to $h$ compression) = Energy of extended spring + Potential energy of mass of upper block
$\Rightarrow \frac{1}{2}k{h}^2=mg\left(h+x\right)+\frac{1}{2}k{x}^3$
$\Rightarrow \frac{1}{2}k{h}^2=mg\left(h+\frac{mg}{k}\right)+\frac{1}{2}k{\left(\frac{mg}{k}\right)}^2$
$\Rightarrow \frac{1}{2}k{h}^2=mgh+\frac{m^2{g}^2}{k}+\frac{m^2{g}^2}{2k}$
$\Rightarrow \frac{k{h}^2}{2}=mgh+\frac{3m^2{g}^2}{2k}$
$\Rightarrow k^2{h}^2-(2mgk)h-3m^2{g}^2=0$
$h=\frac{2mgk\pm \sqrt{4m^2{g}^2{k}^2-4(k^2)(-3m^2{g}^2)}}{2k^2}$
$=\frac{2mgk\pm 4mgk}{2k^2}=\frac{3mg}{k}$
or $hk=3mg$
Now, considering equilibrium in position 1,
Spring force = Weight of upper block +Weight of mass M
$\Rightarrow kh=mg+Mg$
$\Rightarrow 3mg=mg+Mg$
or $M=2m$