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
$\Rightarrow \frac{1}{2}kh^{2}=mg \left(h+x\right)+\frac{1}{2}kx^{3}$
$\Rightarrow \frac{1}{2}kh^{2}=mg\left(h+\frac{mg}{k}\right)+\frac{1}{2}k\left(\frac{mg}{k}\right)^{2}$
$\Rightarrow \frac{1}{2}kh^{2}=mgh+\frac{m^{2}g^{2}}{k}+\frac{m^{2}g^{2}}{2k}$
$\Rightarrow \frac{kh^2}{2} = mgh + \frac{3m^2g^2}{2k}$
$\Rightarrow k^{2}h^{2}- (2mgk)h -3m^2g^2 =0 $
$h = \frac {2mgk \pm \sqrt{4m^{2}g^{2}k^{2}-4(k^{2})(-3m^{2}g^{2})}}{2k^{2}}$
$=\frac{2mgk\pm4mgk }{2k^{2}}=\frac{3mg}{k}$
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 $