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 $\varphi$ with vertical. The coefficient of restitution between the particle and floor is $e$. The magnitude of $v$ is

• A. $eu$
• 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

The correct option is D $u\sqrt{{\sin }^2\theta +e^2{\cos }^2\theta }$

As the floor is frictionless and there is no horizontal force, therefore, momentum must be conserved in the horizontal direction.
i.e. $mu\sin \theta =mv\sin \varphi$
or $v\sin \varphi =u\sin \theta$ ....(1)
And in vertical direction, $\frac{v\cos \varphi }{u\cos \theta }=e$
or $v\cos \varphi =eu\cos \theta$ ....(2)
Squaring and adding (1) and (2),
$v^2({\sin }^2\varphi +{\cos }^2\varphi )=u^2{\sin }^2\theta +e^2{u}^2{\cos }^2\theta$
$v=u\sqrt{{\sin }^2\theta +e^2{\cos }^2\theta }$.