A body of mass m falls from height h on ground. If e be the coefficient of restitution of collision between the body and ground, then the distance travelled by body before it comes to rest is

Question

A body of mass m falls from height h on ground. If e be the coefficient of restitution of collision between the body and ground, then the distance travelled by body before it comes to rest is (A) h (1+e2/1-e2)  (B) h (1-e2/1+e2)  (C) (2 e h/1+e2) (D) (2 e h/1-e2)

Tardigrade Admin · Accepted Answer

S=h+2 e2 h+2 e4 h+ ldots S=h+2 h[e2+e4+e6+ ldots .] S=h+2 h[(e2/1-e2)] Solving S=(h(1+e2)/(1-e2))

Q. A body of mass $m$ falls from height $h$ on ground. If $e$ be the coefficient of restitution of collision between the body and ground, then the distance travelled by body before it comes to rest is

$h {\frac{1 + e ^{2}}{1 - e ^{2}}}$

$h {\frac{1 - e ^{2}}{1 + e ^{2}}}$

$\frac{2 e h}{1 + e ^{2}}$

$\frac{2 e h}{1 - e ^{2}}$

$S = h + 2 e^{2} h + 2 e^{4} h + \dots$
$S = h + 2 h [e^{2} + e^{4} + e^{6} + \dots .]$
$S = h + 2 h [\frac{e ^{2}}{1 - e ^{2}}]$
Solving $S = \frac{h ( 1 + e ^{2} )}{( 1 - e ^{2} )}$

Q. A body of mass m falls from height h on ground. If e be the coefficient of restitution of collision between the body and ground, then the distance travelled by body before it comes to rest is

h{1−e21+e2​}

h{1+e21−e2​}

1+e22eh​

1−e22eh​