Question

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

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

Answer 1

Answer: (A), (B), (C), (D)

The ratio of the final kinetic energy to the initial kinetic energy is $({\cos }^2\theta +e^2\sin \theta )$


Impulse $\left(J\right)=\Delta P$ = $mv\sin \varphi -m(-u\sin \theta )$
$=m(v\sin \varphi +u\sin \theta )=m(V_{sep}+V_{app})=m(e{V}_{app}+V_{app})$
$J=m\left(u\sin \theta \right)(1+e)$
In horizontal direction, momentum is conserved:
$u\cos \theta =v\cos \varphi v=\frac{u\cos \theta }{cos\varphi }$
$e=\frac{V_{sep}}{V_{app}}=\frac{v\sin \varphi }{u\sin \theta }=\frac{\tan \varphi }{tan\theta }$
$\tan \varphi =e\tan \theta$
In vertical direction ,
$e=\frac{v\sin \varphi }{u\sin \theta }$
$v\sin \varphi =eusin\theta$
$v=\sqrt{(eu\sin \theta {)}^2+(u\cos \theta {)}^2}=u\sqrt{e^2{\sin }^2\theta +{\cos }^2\theta }$
$v=u\sqrt{1-(1-e^2){\sin }^2\theta }$
Final kinetic energy = $\frac{1}{2}m{v}^2$
Initial kinetic energy = $\frac{1}{2}m{u}^2$
Ratio =$\frac{v^2}{u^2}=e^2{\sin }^2\theta +co{s}^2\theta$