Question

A massless platform is kept on a light elastic spring, as shown in the figure. When a particle of mass $0.1\text{kg}$ is dropped on the pan from a height of $0.24\text{m}$, the spring gets compressed by $0.01\text{m}$. From what height should the same particle be dropped to cause a compression of $0.04\text{m}$ in the spring?
Neglect the effect of collision.

• A. $0.96\text{m}$
• B. $2.96\text{m}$
• C. $3.96\text{m}$
• D. $0.48\text{m}$

Answer 1

The correct option is C $3.96\text{m}$

Let the sand particle be dropped from a height $h$ and the spring be compressed by $y$.
Hence total height through which particle falls is $(h+y)$

$\Rightarrow$Applying conservation of mechanical energy,
Loss in PE of the sand particle = gain in elastic potential energy of the spring
$\Rightarrow mg(h+y)=\frac{1}{2}k{y}^2....\left(i\right)$
For the two different cases the height from which particle dropped and compression in spring be $(h_1,h_2)$ and $(y_1,y_2)$ respectively.

From Eq.$\left(i\right)$ taking ratio for both cases,
$\Rightarrow \frac{h_1+y_1}{h_2+y_2}={\left(\frac{y_1}{y_2}\right)}^2$
Given:$(h_1=0.24\text{m},y_1=0.01\text{m},y_2=0.04\text{m})$
$\Rightarrow \left(\frac{0.24+0.01}{h_2+0.04}\right)={\left(\frac{0.01}{0.04}\right)}^2$
$\Rightarrow h_2+0.04=4$
$\therefore h_2=3.96\text{m}$