Question

A ball of mass $m$ hits the floor with a speed $v$ making an angle of incidence $\theta$ with the normal. The coefficient of restitution is $e$. Find the speed of the reflected ball and the angle of reflection of the ball.

• A. $v\sqrt{ecos\theta }$, $co{s}^{-1}\left(\frac{cos\theta }{e}\right)$
• B. $v\sqrt{ecos\theta }$, $ta{n}^{-1}\left(\frac{tan\theta }{e}\right)$
• C. $v\sqrt{si{n}^2\theta +eco{s}^2\theta }$, $ta{n}^{-1}\left(\frac{tan\theta }{e}\right)$
• D. $v\sqrt{e^{2}si{n}^2\theta +co{s}^2\theta }$, $ta{n}^{-1}\left(\frac{cot\theta }{e}\right)$

Answer 1

The correct option is C

$v\sqrt{si{n}^2\theta +eco{s}^2\theta }$, $ta{n}^{-1}\left(\frac{tan\theta }{e}\right)$

Suppose the angle the reflection is ${\theta }^{\prime }$ and the speed after the collision is $v^{\prime }$. The floor exerts a force on the ball along the normal during the collision. There is no force parallel to the surface. Thus, the parallel component of the velocity of the ball remains unchanged. This gives

$v^{\prime }sin{\theta }^{\prime }$ = $vsin\theta$ . . . (i)

For the components normal to the floor, the velocity of separation = $v^{\prime }cos{\theta }^{\prime }$ and the velocity of approach = $vsin\theta$.

Hence, $v^{\prime }cos{\theta }^{\prime }$ = $evsin\theta$ . . . (ii)

From (i) and (ii),

Note:For elastic collision, $e$ = 1 so that ${\theta }^{\prime }$ = $\theta$ and $v^{\prime }$ = $v$.