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. $R$
• B. $2R$
• C. $3R$
• D. $4R$

Answer 1

The correct option is B $2R$

Let mass of the ball be m.

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

$=m[-\frac{GM}{R}-(-1.5\frac{GM}{R})]$

$=\frac{GMm}{2R}$

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

Velocity of ball just after collision,

$v^{\prime }=ev=\frac{1}{2}\sqrt{\frac{GM}{R}}$

Let r be the distance from the centre upto where the ball reaches after collision. Then,

$\frac{1}{2}m{v}^{\prime 2}=m[V\left(r\right)-V\left(centre\right)]$

or

$\frac{1}{8}\frac{GMm}{R}=m[\frac{3GM}{2R}-\frac{GM}{R^3}(\frac{3R^2}{2}-\frac{r^2}{2})]$

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

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

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

$\therefore$ The desired distance,

$s=R+\frac{R}{2}+\frac{R}{2}=2R$