First slide

Coefficient of restitution

Question

A girl throws a ball with initial velocity v at an inclination of 45°. The ball strikes the smooth vertical wall at a horizontal distance d from girl and after rebounding returns to her hand. What is the coefficient of restitution between wall and ball ?

Difficult

A

$v^{2} - gd$
B

$gd / (v^{2} - gd)$
C

$gd / v^{2}$
D

$v^{2} / gd$

Solution

See fig.
As the wall is smooth, the vertical component of the impulse it receives is zero, The total time of flight is given by
$0 = (vsin 45^{\circ}) t - \frac{1}{2} {gt}^{2} or t = \frac{2 vsin 45^{\circ}}{g}$
The time of flight t₂ from G to A
$= \frac{d}{vcos 45^{\circ}}$
The time of flight t₂ from A to G
$= \frac{d}{evcos 45^{\circ}}$
where e = coefficient of restitution.
$So, \frac{2 v \sin 45^{\circ}}{g} = \frac{d}{Vcos 45^{\circ}} + \frac{d}{evcos 45^{\circ}}$
$\begin{aligned} or & \frac{\sqrt{2} v}{g} = \frac{d \sqrt{2}}{v} + \frac{d \sqrt{2}}{ev} \\ \frac{v}{g} = \frac{d}{V} (1 + \frac{1}{e}) or \frac{v^{2}}{gd} = 1 + \frac{1}{e} \end{aligned}$
$Solving, we get e = (\frac{gd}{v^{2} - gd}) .$

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