Question

A particle strikes a horizontal frictionless floor with a speed $u$ at an angle $\theta$ with the vertical, and rebounds with a speed $v$ at an angle $\phi$ with vertical. The coefficient of restitution between the particle and floor is $e$. The magnitude of $v$ is

• A. $e u$
• B. $(1-e) u$
• C. $u \sqrt{e^{2} \sin ^{2} \theta+\cos ^{2} \theta}$
• D. $u \sqrt{\sin ^{2} \theta+e^{2} \cos ^{2} \theta}$

Answer 1

Answer: (D)

As the floor is frictionless and there is no horizontal force, therefore, momentum must be conserved in the horizontal direction.
i.e. $ m u \sin \theta=m v \sin \phi$ or $u \sin \theta=v \sin \phi \dots$(i)
And in vertical direction, $\frac{v \cos \phi}{u \cos \theta}=e$
or $ v \cos \phi=e u \cos \theta \dots$(ii)
Squaring and adding (i) and (ii), we get
$v^{2}\left(\sin ^{2} \phi+\cos ^{2} \phi\right)=u^{2} \sin ^{2} \theta+e^{2} u^{2} \cos ^{2} \theta$
or $ v=u \sqrt{\sin ^{2} \theta+e^{2} \cos ^{2} \theta}$