Question

A tunnel is dug in the earth across one of its diameters. Two masses $m$ and $2m$ are dropped from the ends of the tunnel. If the masses collide and stick to each other and perform $\text{S.H.M.}$ then amplitude of $\text{S.H.M.}$ will be: $(R=$ radius of the earth$)$

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

Answer 1

The correct option is C $\frac{R}{3}$

Let velocity of the particle of mass $m$ at the centre be $u$. Applying energy conservation principle for the particle of mass $m$ between centre and the point on the surface of the tunnel.

$\Rightarrow \frac{1}{2}m{u}^2+\frac{-3GMm}{2R}=0+\frac{-GMm}{R}$

$\Rightarrow \frac{1}{2}m{u}^2=\frac{GMm}{2R}$

$\Rightarrow u=\sqrt{\frac{GM}{R}}$,

Maximum velocity of a particle at the centre of tunnel is independent of their masses and is given by, $$u=\sqrt{\frac{GM}{R}}$$Let, the speed of the two masses just before collision be $u$ and just after collision be $v$.

Applying conservation of linear momentum,

$2mu-mu=(2m+m)v$

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

Now, applying the conservation of energy principle after collision between the mean and the extreme points of $\text{SHM}$,

$K.E_{mean}+P.E_{mean}=K.E_{extreme}+P.E_{extreme}$

$\Rightarrow \frac{1}{2}(2m+m)v^2+\frac{-3GM(m+2m)}{2R}=0+\frac{-GM(m+2m)(1.5R^2-0.5A^2)}{R^3}$

Where, $A\to$ amplitude of the $\text{S.H.M.}$

$\Rightarrow \frac{1}{2}{v}^2+\frac{-3GM}{2R}=0+\frac{-GM(1.5R^2-0.5A^2)}{R^3}$

Substituting the value of $v$,

$\Rightarrow \frac{1}{2}{\left(\frac{1}{3}\sqrt{\frac{GM}{R}}\right)}^2+\frac{-3GM}{2R}=0+\frac{-GM(1.5R^2-0.5A^2)}{R^3}$

$\Rightarrow \frac{1}{2}\left(\frac{1}{9R}\right)+\frac{-3}{2R}=0-\frac{(1.5R^2-0.5A^2)}{R^3}$

$\Rightarrow -\frac{26}{18R}\times R^3=0-(1.5R^2-0.5A^2)$

$\Rightarrow \frac{A^2}{R^2}=\frac{1}{9}$

$\therefore A=\frac{R}{3}$

Hence, option (b) is the correct answer.

Why this Question:This is a multi-conceptual question that requires the concept of momentum conservation and energy conservation.