Question

A particle collides with smooth surface with speed u at an angle theta coefficient of restitution is e find ratio of horizontal range of two consecutive projectile motion

Answer 1

Check this attachment :-)

Answer 2

Answer: Ratio of horizontal range of two consecutive projectile motion is $=\frac { 1 }{ e }\quad or \ 1:e$

Given data states about coefficient of restitution to be e. therefore for the range of projectile the velocity components will be $v_y$ and $u_x$ which changes along the line of impact.

Therefore, for the first projectile,

                $v_y = e.u.cos \theta\quad and\quad u x = u.sin \theta$

Thereby the Range $1 =\frac { 2v_yu_x }{ g }$

                $\Rightarrow R_{ 1 }=\frac { 2eu. 2cos\theta sin\theta }{ g }$ .

And similarly, for the second projectile the $v_y^{'} = e.v_y = e.2 u.cos\theta\quad and\quad u x = u.sin \theta$

Thereby

                $Range_2 =\frac { 2vy^{'}ux }{ g }$

                $\Rightarrow R_{ 2 }=\frac { 2e.2u.2cos\theta sin\theta }{ g }.$

Hence, $\frac { R_{ 1 } }{ R_{ 2 } } =\frac { 1 }{ e }$