Question

A system consists of two idential cubes each of mass $m$ linked together by a massless spring of spring constant $k$. The spring is compressed by $x$ connecting cubes by thread. Find minimum value of $x$ for which lower cube will bounce up after the thread has been burnt.

• A. $\frac{2mg}{k}$
• B. $\frac{3mg}{k}$
• C. $\frac{3mg}{2k}$
• D. $\frac{mg}{2k}$

Answer 1

The correct option is B $\frac{3mg}{k}$

The initial compression in the spring $x$ must be such that after burning of the thread, the upper cube rises to a height that produces a tension in the spring that is atleast equal to the weight of the lower cube.
Actually, the spring will first go from its compressed state to its natural length and then get elongated beyond this natural length.
Let $x^{\prime }$ be the maximum elongation produced under these circumstances.
$k{x}^{\prime }=mg$. (for which lower cube bounces off from the ground.)

Applying energy conservation,
As at the maximum elongation velocity of top most block is $u_{block}=0$.
$\frac{1}{2}k{x}^2=mg(x+x^{\prime })+\frac{1}{2}k{x}^{\prime 2}$
$x^2-\frac{2mgx}{k}-2\frac{mg}{k}{x}^{\prime }-x^{\prime 2}$
$x^2-\frac{2mg}{k}x-\frac{2mg}{k}\left(\frac{mg}{k}\right)-{\left(\frac{mg}{k}\right)}^2=0$
$x=\frac{3mg}{k}$