Question

A ball is thrown at a triangular wedge with an angle of inclination ${60}^o$ parallel to the ground as shown in the figure. After bouncing against the wedge, the velocity of the ball is perpendicular to the ground. What is the coefficient of restitution?

• A. $1$
• B. $\frac{1}{\sqrt{3}}$
• C. $\frac{1}{6}$
• D. $\frac{1}{3}$

Answer 1

The correct option is D $\frac{1}{3}$


Resolving $u$ and $v$ along and perpendicular to L.O.I.



We know,
Perpendicular to the L.O.I, component of initial and final velocity will remain same.
$\therefore u\cos {60}^{\circ }=v\sin {60}^{\circ }\Rightarrow \frac{v}{u}=\frac{\cos {60}^{\circ }}{\sin {60}^{\circ }}...\left(i\right)$

We know ,
$e=\frac{\text{Velocity of separation}}{\text{Velocity of approach}}\text{and along the L.O.I,}\text{Velocity of separation}=v\cos {60}^{\circ }\text{Velocity of approach}=u\sin {60}^{\circ }=\frac{v\cos {60}^{\circ }}{u\sin {60}^{\circ }}....\left(ii\right)$

From equation $\left(i\right)$ and $\left(ii\right)$

$e={\left(\frac{\cos {60}^{\circ }}{\sin {60}^{\circ }}\right)}^2={\cot }^2{60}^{\circ }=\frac{1}{3}$