Question

A body falls from a height 'h' on a horizontal surface and rebounds. Then it falls again and again rebounds and so on. If the restitution coefficient is $\frac{1}{3}$, the total distance covered by the body before it comes to rest is

• A. $\frac{h}{4}$
• B. $\frac{5h}{4}$
• C. $2h$
• D. $3h$

Answer 1

The correct option is B $\frac{5h}{4}$

If a ball is dropped from a height h on a horizontal floor, then it strikes with the floor with a speed.
$v_0=\sqrt{2g{h}_0}[From{v}^2=u^2+2gh]$
and it rebounds from the floor with a speed

$v_1=e{v}_0=e\sqrt{2g{h}_0}$$[Ase=\frac{velocityaftercollision}{velocitybeforecollision}]$
$\left(1\right)Firstheightofrebound:h_1=\frac{v_1^2}{2g}=e^2{h}_0\therefore h_1=e^2{h}_0\left(2\right)Heightoftheballafter{n}^{th}rebound:Obviously,thevelocityofball{n}^{th}reboundwillbeThereforetheheightafter{n}^{th}reboundwillbe{h}_n=\frac{v_n^2}{2g}=e^{2n}{h}_0\therefore h_n=e^{2n}{h}_0$
$Totaldistance=h+2e^{2}h+2e^{4}h+\dots \dots =h+2e^{2}h(1+e^2+\dots \dots )=h+\frac{2e^{2}h}{1-e^2}=h\left(\frac{1+e^2}{1-e^2}\right)=h\left[\frac{1+{\left(\frac{1}{3}\right)}^2}{1-{\left(\frac{1}{3}\right)}^2}\right]=\frac{5h}{4}Hencethecorrectchoiceis\left(b\right).$