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Question

A massless platform is kept on a light elastic spring as shown in the figure. When a particle of mass 0.1 kg is dropped on the pan from a height of 0.24m, the particle strikes the pan and the spring is compressed by 0.01m. From what height should the particle be dropped to cause a compression of0.04m?

Moderate

A

0.96 m
B

2.96 m
C

3.96 m
D

0.48 m

Solution

Let the particle be dropped from a height h and the spring be compressed by y. According to the conservation of mechanical energy, loss in PE of the particle = gain in elastic potential energy of the spring
$m g (h + y) = \frac{1}{2} k y^{2}$ , here m is mass of particle, g is acceleration due to gravity, h is height, y is extent of compression, k is force constant
Now, as the particle and spring are same in the second case also,
$\frac{m g (h_{1} + y_{1})}{m g (h_{2} + y_{2})} = \frac{\frac{1}{2} k {y_{1}}^{2}}{\frac{1}{2} k {y_{2}}^{2}}$ or
$(\frac{0.24 + 0.01}{h_{2} + 0.04}) = {(\frac{0.01}{0.04})}^{2}$
Solving we get $h_{2} = 3.96 m$

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