Question

A particle is projected on a frictionless inclined plane of inclination ${37}^{\circ }$ from the horizontal, with the projection angle ${45}^{\circ }$ from the inclined plane as shown in the figure. After one collision from the plane, it reaches to its initial point of projection. Coefficient of restitution between particle and plane is

• A. $\frac{3}{25}$
• B. $\frac{2}{3}$
• C. None of these
• D. $\frac{1}{3}$
  

Answer 1

The correct option is D $\frac{1}{3}$

Hint: “Coefficient of restitution”

Formula used:

$e=\frac{\text{Velocity of seperation}}{\text{Velocity of approach}}$

$s=ut+\frac{1}{2}a{t}^2$

Solution:

$v\sin \theta =u\sin {45}^{\circ }$

$t_2=\frac{2ev\sin \theta }{g\cos {37}^{\circ }}$

$=\frac{5\sqrt{2}eu}{4g}$

During upward motion;

$R=u\cos {45}^{\circ }{t}_1-\frac{1}{2}g\sin {37}^{\circ }{t}_1^2$

$=\frac{4}{\sqrt{2}}\frac{5\sqrt{2u}}{4g}-\frac{1}{2}g\frac{3}{5}\frac{25\times 2u^2}{16g^2}=\frac{5}{16}\frac{u^2}{g}$

$v\cos \theta =u\cos {45}^{\circ }-g\sin {37}^{\circ }{t}_1$

$=\frac{u}{\sqrt{2}}-\frac{3\sqrt{2}}{4}u$

During downward motion;

$R=-v\cos \theta t_2+\frac{1}{2}g\sin {37}^{\circ }.t_2^2$

$\frac{5}{16}\frac{u^2}{g}=-(\frac{u}{\sqrt{2}}-\frac{3\sqrt{2}u}{4})\frac{5\sqrt{2}eu}{4g}+\frac{1}{2}g\frac{3}{5}\left(\frac{25e^2\times 2u^2}{16g^2}\right)$

$3e^2+2e-1=0$

$e=\frac{-2+\sqrt{4+12}}{6}=\frac{-2+4}{6}=\frac{1}{3}$

Final answer : $\left(b\right)$.