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. $tan\varphi =etan\theta$
• C. $v=u\sqrt{1-(1-e{)}^{2}si{n}^2\theta }$
• D. the ratio of final kinetic energy to the initial kinetic energy is $(co{s}^2\theta +e^{2}si{n}^2\theta )$

Answer 1

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

The correct options are
A the impulse delivered by the floor to the body is $mu(1+e)sin\theta$
B $tan\varphi =etan\theta$
D the ratio of final kinetic energy to the initial kinetic energy is $(co{s}^2\theta +e^{2}si{n}^2\theta )$

Let velocity of particle in y direction after the collision be $v^{\prime }$.

Initial velocity of the particle in x direction remains the same but its velocity in y direction changes.

$\therefore$ $ucos\theta =vcos\varphi$ ...............(1)

Also $\frac{v^{\prime }-0}{-usin\theta -0}=-e$

$⟹$ $v^{\prime }=usin\theta$

$\therefore$ $eusin\theta =vsin\varphi$ ................(2) $(\because v^{\prime }=vsin\varphi )$

Dividing (1) from (2) we get $\frac{vsin\varphi }{vcos\varphi }=\frac{eusin\theta }{ucos\theta }$

$⟹tan\varphi =e$ $tan\theta$

Change in momentum of the particle $\Delta P=m\left(vsin\varphi \right)-m(-usin\theta )$

$\Delta P=m\left(eusin\theta \right)+m\left(usin\theta \right)=mu(1+e)sin\theta$

Impulse delivered $I=\Delta P=mu(1+e)sin\theta$

Squaring and adding (1) and (2), $u^{2}co{s}^2\theta +e^2{u}^{2}si{n}^{\theta }=v^2(co{s}^2\varphi +si{n}^2\varphi )$

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

Ratio of final to initial kinetic energy $\frac{K.E_f}{K.E_i}=\frac{\frac{1}{2}m{v}^2}{\frac{1}{2}m{u}^2}=co{s}^2\theta +e^{2}si{n}^{\theta }$