Question

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

• A. 0.96 m
• B. 2.96 m
• C. 3.96 m
• D. 0.48 m

Answer 1

Answer: (C)

Let the particle be dropped from a height $h$ and the spring be compressed by $y$. According to conservation of mechanical energy, Loss in potential energy of the particle = gain in elastic potential energy of the spring
$\therefore \quad m g(h+y)=\frac{1}{2} k y^{2}$
Now, as the particle and spring remain same, hence
$\frac{h_{1}+y_{1}}{h_{2}+y_{2}}=\left(\frac{y_{1}}{y_{2}}\right)^{2} $
$\Rightarrow \left(\frac{0.24+0.01}{h_{2}+0.04}\right)=\left(\frac{0.01}{0.04}\right)^{2}$
Solving we get, $h_{2}=3.96\, m$