Question

A particle strikes a horizontal smooth floor with a velocity $a$ making an angle $\theta$ with the floor and rebounds with velocity $v$ making angle $\varphi$ with the floor. The coefficient of restitution between the particle and the floor is $e$.

• A. The impulse delivered by the floor to the body is $(1+e)\sin \theta$
• B. $\tan \varphi =e\tan \theta$
• C. $v=u\sqrt{1-(1-e^2){\sin }^2\theta }$
• D. The ratio of final kinetic energy to the initial kinetic energy is ${\cos }^2\theta +e^2{\sin }^2\theta$

Answer 1

The correct option is B $\tan \varphi =e\tan \theta$

$\begin{array}{c}\alpha \cos \theta =vcos\varphi ---\left(1\right)\\ e=\frac{v\sin \varphi }{\alpha \sin \theta }\\ e=\frac{\cos \theta in\varphi }{\cos \varphi \sin \theta }\\ e=\frac{\tan \varphi }{\tan \theta }\\ \therefore \tan \varphi =\tan \theta \\ Hence,theoptionBiscorrect.\end{array}$