Question

A tunnel is dug n the earth across one of its diameter. Two masses 'm' & '2m' are dropped from the ends of the tunnel. The masses collide and stick to each other and performs S.H.M. Then amplitude of S.H.M. will be : [R=radius of the earth]

• A. R
• B. R/2
• C. $\frac{\sqrt{2R}}{3}$
• D. 2R/3

Answer 1

Answer: (C)

$\frac{\sqrt{2R}}{3}$

For mass ${}^{\prime }{m}^{\prime }$

$\begin{array}{c}ui+ki=uf+kf\\ \Rightarrow \frac{-GMm}{R}+0=\frac{-GM}{2R^3}\left(3R^2-0\right)m+\frac{1}{2}m{v}^2\\ \Rightarrow \frac{-GMm}{R}=\frac{-3GM}{2R}+\frac{V^2}{2}\\ \Rightarrow \frac{-GM}{R}=\frac{V^2}{2}\\ \Rightarrow V^2=\frac{2GM}{R}\\ \Rightarrow V=\sqrt{\frac{2GM}{R}}\end{array}$

Conserving momentum before and after collinear.

Let $V_{cm}$ is the velocity of combines mass.

$V_{cm}=\frac{2mv-mv}{3m}=\frac{v}{3}$

Let the amplitude is $x$.

$\begin{array}{c}ui+ki=uf+kf\\ \Rightarrow \frac{-3GM\left(3m\right)}{2R}+\frac{1}{2}\times 3m\times \frac{v^2}{9}=\frac{-GM}{2R^3}\left(3R^2-x^2\right)\left(3m\right)+0\\ \Rightarrow x=\frac{\sqrt{2R}}{3}\end{array}$

$\therefore$ AMplitude is $\frac{\sqrt{2R}}{3}$

$\therefore$ Option $C$ is correct answer.