Question

A ball is dropped from a height $h_0$ on a horizontal floor and keeps rebounding after hitting the floor. Find: (i) speed after $n$ rebounds

(ii) Total distance travelled by the ball before stopping

(iii) Height it rises after nth rebound

(iv) Total time before stopping

(coefficient of restitution is $e$)

• A. (i) , (ii), (iii) , (iv)
• B. (i) , (ii), (iii) , (iv)
• C. (i), (ii), (iii)infinite (iv) infinite
• D. None of these

Answer 1

The correct option is A

(i) , (ii), (iii) , (iv)

If a ball is dropped from a height $h_0$ 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

$v_1$ = $e{v}_0$ = $e\sqrt{2g{h}_0}$ [As $e$ = $\frac{velocityaftercollision}{velocitybeforecollision}$]

first height of rebound: $h_1$ = $\frac{v_1^2}{2g}$ = $e^2{h}_0$

$\therefore$ $h_n$ = $e^{2n}{h}_0$

Height of the ball after nth rebound: Obviously, the velocity of ball after nth rebound will be

$V_n$ = $e^n{V}_0$

therefore the height after nth rebound will be $h_n$ = $\frac{v_n^2}{2g}$ = $e^{2n}{h}_0$

$\therefore$ $h_n$ = $e^{2n}{h}_0$

Total distance travelled by the ball before it stops bouncing

$H$ = $h_0+2h_1+2h_2+2h_3+......$ = $h_0+2e^2{h}_0+2e^4{h}_0+2e^6{h}_0+.......$

$H$ = $h_0[1+2e^2+2e^4+2e^6+.......]$

$=h_0[1+2e^2\left(\frac{1}{1-e^2}\right)]$ [As $1+e^2+e^4+.......$= $\frac{1}{1-e^2}$]

$\therefore$ $H$ = $h_0\left[\frac{1+e^2}{1-e^2}\right]$

Total time taken by the ball to stop bouncing

$T$ = $t_0+2t_1+2t_2+2t_3+.......$ = $\sqrt{\frac{2h_0}{g}}+2\sqrt{\frac{2h_1}{g}}+2\sqrt{\frac{2h_2}{g}}+.......$

= $\sqrt{\frac{2h_0}{g}}[1+2e+2e^2+.......]$ [As $h_1$ = $e^2{h}_0$; $h_2$ = $e^4{h}_0$]

= $\sqrt{\frac{2h_0}{g}}[1+2e(1+e+e^2+e^3+........\left)\right]$ = $\sqrt{\frac{2h_0}{g}}[1+2e\left(\frac{1}{1-e}\right)]$ = $\sqrt{\frac{2h_0}{g}}\left(\frac{1}{1-e}\right)$

$\therefore$ $T$ = $\left(\frac{1+e}{1-e}\right)\sqrt{\frac{2h_0}{g}}$