Question

A block of mass $m$ is dropped onto a spring of spring constant $k$ from a height $h$. The second end of the spring is attached to a second block of mass $M$ as shown in figure. Find the minimum value of $h$ so that the block $M$ bounces off the ground, if the block of mass $m$ sticks to the spring immediately after it comes into contact with it.

• A. $\frac{(M^2+2mM)g}{2kM}$
• B. $\frac{2(M^2+2mM)g}{km}$
• C. $\frac{2(M^2+2mM)g}{kM}$
• D. $\frac{(M^2+2mM)g}{2km}$

Answer 1

The correct option is D $\frac{(M^2+2mM)g}{2km}$

If the block $m$ dropped from a height $h$, it compresses the spring from its unstretched position $\text{A}$ to position as shown in figure, and after that it start elongated to position $\text{B}$ at a height $x$ above the initial level of spring.


Since, all the forces present here are conservative in nature, hence using conservation of mechanical energy, we get (assuming position $\text{A}$ as datum)

$mgh=\frac{1}{2}k{x}^2+mgx$

$h=\frac{1}{2}\frac{k{x}^2}{mg}+x........\left(1\right)$


Now, for the second block $M$ when it bounces off from the ground, it loses the contact of ground. From the FBD of the block $M$ we get,

$N+kx=Mg[\because N=0]$

$\Rightarrow x=\frac{Mg}{k}$

From $\left(1\right)$ we get,

$h=\frac{1}{2}\frac{k}{mg}\left(\frac{M^2{g}^2}{k^2}\right)+\frac{Mg}{k}$

$\Rightarrow h=\frac{M^{2}g}{2mk}+\frac{Mg}{k}$

$\therefore h=\frac{(M^2+2mM)g}{2km}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(D\right)$ is the correct answer.