question_answer

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            [EAMCET 2001]

A)          \[h\left( \frac{1+{{e}^{2}}}{1-{{e}^{2}}} \right)\]

B)                       \[h\left( \frac{1-{{e}^{2}}}{1+{{e}^{2}}} \right)\]

C)          \[\frac{h}{2}\left( \frac{1-{{e}^{2}}}{1+{{e}^{2}}} \right)\]

D)                       \[\frac{h}{2}\left( \frac{1+{{e}^{2}}}{1-{{e}^{2}}} \right)\]

Correct Answer: A

Solution :

Particle falls from height h then formula for height covered by it in nth rebound is given by \[{{h}_{n}}=h{{e}^{2n}}\] where e = coefficient of restitution, n = No. of rebound Total distance travelled by particle before rebounding has stopped \[H=h+2{{h}_{1}}+2{{h}_{2}}+2{{h}_{3}}+2{{h}_{n}}+........\] \[=h+2h{{e}^{2}}+2h{{e}^{4}}+2h{{e}^{6}}+2h{{e}^{8}}+.........\] \[=h+2h({{e}^{2}}+{{e}^{4}}+{{e}^{6}}+{{e}^{8}}+.......)\] \[=h+2h\left[ \frac{{{e}^{2}}}{1-{{e}^{2}}} \right]=h\,\left[ 1+\frac{2{{e}^{2}}}{1-{{e}^{2}}} \right]=h\,\left( \frac{1+{{e}^{2}}}{1-{{e}^{2}}} \right)\]