Question
A particle falls from a height $$h$$ on a fixed horizontal plane and rebounds. If $$e$$ is the coefficient of restitution, the total distance travelled by the particle before it stops rebounding is
A.
$$\frac{h}{2}\frac{{\left[ {1 - {e^2}} \right]}}{{\left[ {1 + {e^2}} \right]}}$$
B.
$$\frac{{h\left[ {1 - {e^2}} \right]}}{{\left[ {1 + {e^2}} \right]}}$$
C.
$$\frac{h}{2}\frac{{\left[ {1 + {e^2}} \right]}}{{\left[ {1 - {e^2}} \right]}}$$
D.
$$\frac{{h\left[ {1 + {e^2}} \right]}}{{\left[ {1 - {e^2}} \right]}}$$
Answer :
$$\frac{{h\left[ {1 + {e^2}} \right]}}{{\left[ {1 - {e^2}} \right]}}$$
Solution :
The velocity of particle after falling through height $$h$$
$$u = \sqrt {2gh} \,......\left( {\text{i}} \right)$$
After first rebounding, the velocity of ball is $$eu$$ and after attaining maximum height it will come to the ground with same velocity $$eu.$$ So, after second rebounding its velocity will be $${e^2}u.$$ Similarly, after third fourth ... etc reboundings its velocities will be $${e^2}u,{e^4}u,....$$ etc.
Since, it first rebounds with velocity $$eu$$ so if it attains height $$h$$ then from
$$\eqalign{
& {v^2} = {u^2} - 2gh \cr
& \therefore 0 = {e^2}{u^2} - 2g{h_1} \cr
& {\text{or}}\,\,{h_1} = \frac{{{e^2}{u^2}}}{{2g}} = \frac{{{e^2}2gh}}{{2g}} = {e^2}h\,\,\left[ {{\text{from}}\,{\text{Eq}}{\text{.}}\left( {\text{i}} \right)} \right] \cr} $$
The same height the ball travels while approaching ground. Now, it rebounds with velocity $${e^2}u$$ so if it attains a height $${h_2}$$ then
$$\eqalign{
& 0 = {e^4}{u^2} - 2g{h_2} \cr
& {\text{or}}\,\,{h_2} = {e^4}h \cr} $$
The similar process will follow for further reboundings
Hence, the total distance travelled by the practice before it stops rebounding.
$$\eqalign{
& = h + 2{h_1} + 2{h_2} + ...\infty = h + 2{e^2}h + 2{e^4}h + ...\infty \cr
& = h + 2{e^2}h\left( {1 + {e^2} + {e^4} + ...\infty } \right) = h + 2{e^2}h\left( {\frac{1}{{1 - {e^2}}}} \right) \cr
& = h\left( {1 + \frac{{2{e^2}}}{{1 - e}}} \right) \cr
& = \left( {\frac{{1 + {e^2}}}{{1 - {e^2}}}} \right)h \cr} $$