Question

A tunnel is dug along the diameter of the earth. Two balls of mass $m$ and $2m$ are dropped from the opposite ends of the tunnel simultaneously. They collide in the tunnel and stick to each other. If the new amplitude of their oscillation is $\frac{R}{n}$ where $R$ is earth's radius, then the value of $n$ is

Answer 1

For particle $m$:

Initial energy,

$E_i=-\frac{GMm}{R}+0$

Final energy,

$E_f=\frac{-GMm}{R}\times \frac{3}{2}+\frac{1}{2}m{v}_1^2$

$[\text{Potential at the centre of earth = 3/2}\times \text{Potential at the surface of earth}]$

For particle $2m$:

$E_i=\frac{-GM\left(2m\right)}{R}+0$

$E_f=\frac{-GM\left(2m\right)}{R}\times \frac{3}{2}+\frac{1}{2}\times \left(2m\right)\times (v_2{)}^2$

Using energy conservation for mass $m$,

$\frac{1}{2}m{v}_1^2-\frac{3}{2}\frac{GMm}{R}=-\frac{GMm}{R}$

$\Rightarrow \frac{v_1^2}{2}=\frac{3}{2}\frac{GM}{R}-\frac{GM}{R}$

$\Rightarrow \frac{v_1^2}{2}=\frac{GM}{2R}$

$\Rightarrow v_1=\sqrt{\frac{GM}{R}}$

Similarly, for mass $2m$,

$v_2=\sqrt{\frac{GM}{R}}$

So, $v_1=v_2=v=\sqrt{\frac{GM}{R}}$

From momentum conservation,

$mv-2mv=(m+2m)v_o$

$\Rightarrow v_o=-\frac{v}{3}=-\frac{1}{3}\sqrt{\frac{GM}{R}}$

Using energy conservation after collision between the mean position and extreme position,

$-\frac{3}{2}\times \frac{GM}{R}\times 3m+\frac{1}{2}\times 3m\times {(-\frac{1}{3}\sqrt{\frac{GM}{R}})}^2=\frac{-GM}{2R^3}[3R^2-r^2]\times 3m$

On solving this, we get,

$r=\frac{R}{3}$

$\therefore n=3$