Question

A glass marble dropped from a certain height above the horizontal surface reaches the surface in time t and then continues to bounce up and down. The time in which the marble finally comes to rest will be (coefficient of restitution is e):
A) ${e^n}t$
B) ${e^2}t$
C) $t\left[ {\dfrac{{1 + e}}{{1 - e}}} \right]$
D) $t\left[ {\dfrac{{1 - e}}{{1 + e}}} \right]$

Answer 1

Here when the ball is dropped the initial velocity of the ball is zero and it is at maximum height and after the ball hits the ground and bounces upward the max height of the bounced ball would be less than the height when the ball was dropped, similarly the time and velocity to reach the ground varies by each bounce the ball makes.

Complete step by step solution:
Here, the kinetic energy is equal to the potential energy as the ball reaches its max height:
$\dfrac{1}{2}m{v^2} = mgh$ ;
$\Rightarrow {v^2} = 2gh$;
Now, we apply the equation of kinematics:
$S = ut + \dfrac{1}{2}g{t^2}$ ;
Here, S = height; u = initial velocity =0 ; g = acceleration due to gravity and t = time taken;
$h = 0 + \dfrac{1}{2}g{t^2}$;
$h = \dfrac{1}{2}g{t^2}$;
Put the value of h in the equation ${v^2} = 2gh$:
$v = \sqrt {2g \times \dfrac{1}{2}g{t^2}}$;
$\Rightarrow v = \sqrt {{g^2}{t^2}}$;
Write the above equation in terms of time:
$\Rightarrow v = gt$;
$\Rightarrow \dfrac{v}{g} = t$;
Here, the velocity of separation $v'$ is the first velocity after the ball bounces upwards;
Now, we know that the coefficient of restitution is the velocity of separation upon the velocity of approach:
$e = \dfrac{{v'}}{v}$ ;
$ev = v'$;
Here, the ball after touching the ground first time will again go in the upward direction and would reach its maximum height:
${v^2} - {u^2} = 2as$;
Now, at max height of the ball the final velocity would be zero:
$0 - {\left( {v'} \right)^2} = - 2g{h_1}$;
$\Rightarrow {\left( {v'} \right)^2} = 2g{h_1}$;
Now, put $ev = v'$in the above relation:
${\left( {ev} \right)^2} = 2g{h_1}$;
$\Rightarrow \dfrac{{{{\left( {ev} \right)}^2}}}{{2g}} = {h_1}$;
Now, the time required by the ball to reach ${h_1}$:
Now, we know that:
${h_1} = \dfrac{1}{2}gt_1^2$;
Put the above relation in the equation:
$\Rightarrow \dfrac{{{{\left( {ev} \right)}^2}}}{{2g}} = \dfrac{1}{2}gt_1^2$
$\Rightarrow \dfrac{{ev}}{g} = {t_1}$;
Similarly, for the ball to come down:
$\Rightarrow \dfrac{{2ev}}{g} = {t_2}$;
Similarly, we can have n number time period for n number of bounces:
$t = {t_1} + {t_2} + {t_3} + {t_4}.........$
$\Rightarrow t = \dfrac{v}{g} + \dfrac{{2ev}}{g} + \dfrac{{{e^2}{v^2}}}{g} + .........$;
Take the common out:
$\Rightarrow t = \dfrac{v}{g}\left( {e + 2e + 2{e^2} + .....} \right)$;
Now, $\dfrac{v}{g}$= t;
$\Rightarrow t = t\left( {1 + 2e + 2{e^2} + .....} \right)$;
$\Rightarrow t = t\left( {1 + 2e\left( {1 + e + {e^2}} \right).....} \right)$;
Apply property of geometric progression:
$\Rightarrow t = t\left( {1 + 2e\left( {\dfrac{1}{{1 - e}}} \right)} \right)$;
Take, the LCM:
$\Rightarrow t = t\left( {\dfrac{{1 - e + 2e}}{{1 - e}}} \right)$;
$\Rightarrow t = t\left( {\dfrac{{1 + e}}{{1 - e}}} \right)$;

Final Answer: Option “C” is correct.

Note: The process is very lengthy so, go step by step and carefully make relations. In the first case when the ball is dropped the K.E = P.E (vice-versa is also true), now similarly for the second time the ball bounces make an equation for v and t and so on. Do this for three bounces and then make a G.P and apply the property of G.P (Geometric Progression) including coefficient of restitution after the first bounce.