Question

A smooth tunnel is dug along the radius of earth that ends at centre. A ball is released from the surface of earth along tunnel. Coefficient of restitution for collision between soil at centre and ball is $0.5$. Calculate the distance travelled by ball just before second collision at centre. Given mass of the earth is $M$ and radius of the earth is $R$.

• A. $\frac{3R}{2}$
• B. $2R$
• C. $R$
• D. $4R$

Answer 1

The correct option is B $2R$

Given,
Co-efficient of restitution, $e=0.5=\frac{1}{2}$

Since, initially the ball is at rest at point $\text{A}$, So $u=0$.

Let, $v$ be the speed of the ball just before hitting $\text{B}$.

If, $V_A$ and $V_B$ are the potentials at points $\text{A}$ and $\text{B}$. Applying the conservation of energy, we get


$\frac{1}{2}m{v}^2=m(V_A-V_B)$

Substituting the values of $V_A$ and $V_B$,

$\Rightarrow \frac{1}{2}m{v}^2=m[-\frac{GM}{R}-(-1.5\frac{GM}{R})]$

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

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

Velocity of ball just after collision,

$v_f=ev=\frac{1}{2}\sqrt{\frac{GM}{R}}$

Let $r$ be the distance from the centre upto where the ball reaches after collision. Then, on applying the conservation of principle, we get

$\frac{1}{2}m{v}_f^2=m(V_r-V_{\left(centre\right)})$

$\Rightarrow \frac{1}{2}m{v}_f^2=m[\frac{-GM}{R^3}(1.5R^2-0.5r^2)-\frac{-3GM}{2R}]$

Substituting the value of $v_f$,

$\Rightarrow \frac{1}{2}m{\left(\frac{1}{2}\sqrt{\frac{GM}{R}}\right)}^2=m[\frac{3GM}{2R}-\frac{GM}{R^3}(1.5R^2-0.5r^2)]$

$\Rightarrow \frac{1}{8}=+\frac{3}{2}-\frac{3}{2}+\frac{r^2}{2R^2}$

$\Rightarrow \frac{r^2}{R^2}=\frac{1}{4}$

$\therefore r=\frac{R}{2}$

So, the distance covered by the ball just before the second collision will be,
$=R+2r$
$=R+2\times \frac{R}{2}$
$=2R$

Hence, option (b) is the correct answer.