Question

A boy throws a ball with initial velocity u at an inclination of ${45}^o$. The ball strikes the smooth vertical wall at a horizontal distance $s$ from the boy and after rebounding returns to his hand. The coefficient of restitution between the wall and the ball is

• A. $\frac{u^2-gs}{gs}$
• B. $\frac{gs}{u^2-gs}$
• C. $\frac{u^2-2gs}{4gs}$
• D. $\frac{4gs}{u^2+gs}$

Answer 1

The correct option is B $\frac{gs}{u^2-gs}$

When the ball hits the wall at A, the vertical component of the impulse received is zero, because the wall is smooth. The vertical velocity component remains unaltered and consequently throughout both parts of flight and the height of the ball is the same at any time as it would be in the absence of the wall.
Total time of flight T is given by,

$0=\left(usin{45}^0\right)T-\frac{1}{2}g{T}^2$

$T=\frac{2usin{45}^0}{g}$

When the ball is rebounding from B to A, the horizontal velocity is constant at u cos ${45}^0$

$\Rightarrow$ time of flight t, from B to A$=\frac{s}{ucos{45}^0}$

Let the coefficient of restitution be e.
Horizontal velocity after impact $=$ eu cos ${45}^0$,
This is constant as it returns from A to B.

Time taken from A to B$=\frac{s}{eucos{45}^0}$

$\Rightarrow \frac{2usin{45}^0}{g}=\frac{s}{ucos{45}^0}+\frac{s}{eucos{45}^0}$

$\Rightarrow \frac{\sqrt{2}u}{g}=\frac{s\sqrt{2}}{u}+\frac{s\sqrt{2}}{eu}$

$\Rightarrow \frac{u}{g}=\frac{s}{u}(1+\frac{1}{e})$

$\Rightarrow \frac{u^2}{gs}=1+\frac{1}{e}$

$\Rightarrow e=\frac{gs}{u^2-gs}$