First slide

Coefficient of restitution

Question

A particle collides with $15 {ms}^{- 1}$ at an angle 60⁰ from vertical on a smooth horizontal surface. The coefficient of restitution (e) is $\frac{1}{\sqrt{3}} .$ The range R, as shown in diagram will be

Moderate

A

12.5 m
B

12 m
C

11.25 m
D

12.75 m

Solution

$Given e = \frac{1}{\sqrt{3}}$
$\tan ϕ = \frac{1}{e} \tan θ$
$\tan ϕ = 3 \Rightarrow \sin ϕ = \frac{3}{\sqrt{10}}$
Along horizontal,
$m \times 15 \sin 60^{\circ} = m \times v \sin ϕ$
$v = \frac{15 \sqrt{5} \sqrt{10}}{2 \times 3} = 13.7 {ms}^{- 1}$
$x = \frac{v^{2} \sin 2 (\frac{π}{2} - ϕ)}{g} = \frac{v^{2} \sin 2 ϕ}{g}$
$= \frac{(13.7)^{2} \times \frac{2 \times 3}{10}}{10} = 11.25 m$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App