Question

A pair of identical light elastic springs support a mass less platform as shown in the figure. When a piece of clay of mass $0.1kg$ falls on the platform from a height of $0.24m$ and sticks to it, the compression in the spring is $0.01m$. The height in meters from which the same piece of clay has to fall to create a compression of $0.04m$ is:

• A. $4.25m$
• B. $3.96m$
• C. $6m$
• D. $7.5m$

Answer 1

Answer: (B)

$3.96m$

Let k be the spring constant of each spring. So spring constant of this system $k_1=2k$. Here the potential energy of the clay piece is equal to potential energy due to both springs compression.

$mg(h_1+x_1)=\frac{1}{2}{k}_1{x}_1^2$

$0.1\times 10(0.24+0.01)=\frac{1}{2}{k}_1(0.01{)}^2$

$0.5=k_1\times {10}^{-4}$

$k_1=5000$

Let $h_2$ be the required height to create compression $x_2=0.04m$.

Thus, $mg(h_2+x_2)=\frac{1}{2}{k}_1{x}_2^2$

$0.1\times 10(h_2+0.04)=\frac{1}{2}\times 5000\times (0.04{)}^2$

$h_2+0.04=\frac{1}{2}\times 5000\times \frac{16}{10000}=4$

$h_2=4-0.04=3.96m$