Question

A ball is let fall from a height $h_0$. It makes $n$ collisions with the earth. After $n$ collisions it rebounds with a velocity '${\upsilon }_n$' and the ball rises to a height $h_n$. then coefficient of restitution is given by:

• A. ${e=\left[\frac{h_n}{h_0}\right]}^{1/2n}$
• B. ${e=\left[\frac{h_0}{h_n}\right]}^{1/2n}$
• C. $e=\frac{1}{n}\sqrt{\frac{h_n}{h_0}}$
• D. $e=\frac{1}{n}\sqrt{\frac{h_0}{h_n}}$

Answer 1

The correct option is A ${e=\left[\frac{h_n}{h_0}\right]}^{1/2n}$

From conservation of energy, potential energy at the top equals the kinetic energy at the bottom for each cycle.

For zeroth free-fall (before the first collision),

$mg{h}_o=\frac{1}{2}m{v}_o^2$

$v_o=\sqrt{2g{h}_o}...............\left(i\right)$

Similarly, after $n^{th}$ collision,,

$v_n=\sqrt{2g{h}_n}..........\left(ii\right)$

Let velocity of the ball just after $n^{th}$ collision be $v_n$.

Then applying conservation of energy, velocity of the ball just before the $(n+1{)}^{th}$ collision is given by, $\frac{1}{2}m{v}_n^{\prime 2}=\frac{1}{2}m{v}_n^2$

$\therefore v_n^{\prime }=v_n$

Let coefficient of restitution be $e$. By its definition,

$e=\frac{v_n}{v_{n-1}}=\frac{v_{n-1}}{v_{n-2}}=.....=\frac{v_1}{v_o}$

$⟹v_n=e^n{v}_o$

$e={\left(\frac{v_n}{v_o}\right)}^{1/n}..........\left(iii\right)$

From (i), (ii) and (iii),

$e={\left[\frac{h_n}{h_o}\right]}^{1/2n}$