Question

A particle of mass m 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. The coefficient of restitution between the particle and the floor is e. Then

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

Answer 1

Answer: (B), (D)

The correct options are
B $tan\varphi =etan\theta$
D the ratio of the final kinetic energy to the initial kinetic energy is $co{s}^2\theta +esi{n}^2\theta$

$ucos\theta =vcos\varphi$.....(i)
$vsin\varphi =eusin\theta$.....(ii)
From equations (i) and (ii), we can see that,
$tan\varphi =etan\theta$
Momentum or velocity changes only in vertical direction.

$\therefore \mid Impulse\mid =\mid \Delta P\mid$

$=m(usin\theta +eusin\theta )$

$=m(1+e)usin\theta$

$v=\sqrt{(vcos\varphi {)}^2+(vsin\varphi {)}^2}$

$=\sqrt{(ucos\theta {)}^2+(eusin\theta {)}^2}$

$=\sqrt{u^2(co{s}^2\theta +e^{2}si{n}^2\theta )}$

$=u\sqrt{1-(1-e^2)si{n}^2\theta }$

$\frac{K_f}{K_i}=\frac{\frac{1}{2}m{v}^2}{\frac{1}{2}m{u}^2}=\frac{v^2}{u^2}=co{s}^2\theta +e^{2}si{n}^2\theta$