Question

As shown in the figure a ball of mass $m$ hits a floor with a speed $u=10\sqrt{2}\text{m/s}$ making an angle of incidence $\alpha ={45}^{\circ }$ with the normal. The coefficient of restitution is $e=0.5$. Find the speed of the reflected ball and the angle of the reflection.

• A. $10\sqrt{2},{\tan }^{-1}\left(1\right)$
• B. $5\sqrt{5},{\tan }^{-1}\left(2\right)$
• C. $10\sqrt{5},{\tan }^{-1}\left(1\right)$
• D. $5\sqrt{2},{\tan }^{-1}\left(2\right)$

Answer 1

The correct option is B $5\sqrt{5},{\tan }^{-1}\left(2\right)$

Given, speed of ball before collision $u=10\sqrt{2}\text{m/s}$
angle of incidence, $\alpha ={45}^{\circ }$
$e=0.5$
As we know,
$(e=\frac{V_{sep}}{V_{app}})$ along line of impact.
Here,


$(e=\frac{v}{u\cos {45}^{\circ }})$ along line of impact
$\Rightarrow 0.5\times u\times \frac{1}{\sqrt{2}}=v$
$\Rightarrow v_y=\frac{u}{2\sqrt{2}}$ along line of impact.
As we know,
Linear momentum of individual particles remains unchanged along perpendicular to common normal.
Hence, $v_x=u\sin {45}^{\circ }=\frac{u}{\sqrt{2}}$ along common tangent.
Hence, we have
$v=\sqrt{v_y^2+v_x^2}=\sqrt{{\left(\frac{u}{2\sqrt{2}}\right)}^2+{\left(\frac{u}{\sqrt{2}}\right)}^2}$
$\Rightarrow v=\frac{\sqrt{5}u}{2\sqrt{2}}=\frac{\sqrt{5}\times 10\sqrt{2}}{2\sqrt{2}}=5\sqrt{5}\text{m/s}$
Angle of the reflection is given by,
$\tan \beta =\frac{v_x}{v_y}$
$\Rightarrow \tan \beta =\left(\frac{\frac{u}{\sqrt{2}}}{\frac{u}{2\sqrt{2}}}\right)=2$
$\Rightarrow \tan \beta =2$
$\Rightarrow \beta ={\tan }^{-1}\left(2\right)$