Q. A particle falls from a height $h$ upon a fixed horizontal plane and rebounds. If $e$ is the coefficient of restitution, the total distance travelled before rebounding has stopped is:-
NTA AbhyasNTA Abhyas 2022
Solution:
After every collision, the velocity of the ball becomes $e$ times of the initial velocity, therefore, the height will become $e^{2}$ times of initial height.
where $e=$ the coefficient of restitution,
$n=$ The number of rebounds.
Total distance travelled by particle before rebounding has stopped
$H=h+2h_{1}+2h_{2}+2h_{3}+2h_{4}+\ldots \ldots $
$H=h+2he^{2}+2he^{4}+2he^{6}+2he^{8}+\ldots \ldots \ldots $
$H=h+2h\left(e^{2} + e^{4} + e^{6} + e^{8} + \ldots \ldots \right)$
$H=h+2h\left[\frac{e^{2}}{1 - e^{2}}\right]=h\left[1 + \frac{2 e^{2}}{1 - e^{2}}\right]$
$H=h\left(\frac{1 + e^{2}}{1 - e^{2}}\right)$ .
After every collision, the velocity of the ball becomes $e$ times of the initial velocity, therefore, the height will become $e^{2}$ times of initial height.
where $e=$ the coefficient of restitution,
$n=$ The number of rebounds.
Total distance travelled by particle before rebounding has stopped
$H=h+2h_{1}+2h_{2}+2h_{3}+2h_{4}+\ldots \ldots $
$H=h+2he^{2}+2he^{4}+2he^{6}+2he^{8}+\ldots \ldots \ldots $
$H=h+2h\left(e^{2} + e^{4} + e^{6} + e^{8} + \ldots \ldots \right)$
$H=h+2h\left[\frac{e^{2}}{1 - e^{2}}\right]=h\left[1 + \frac{2 e^{2}}{1 - e^{2}}\right]$
$H=h\left(\frac{1 + e^{2}}{1 - e^{2}}\right)$ .