Question

A ball is dropped from height h on a floor where Coefficient of restitution is e. Find the time required by the ball to stop rebounding.
(a) √(2h/g) (1 + e/1 - e)
(b) √(2h/g) (1 + e)
(c) √(2h/g) (1 - e/1 + e)
(d) none of the above



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Answer 1

Solution :-

Given :-

height = h

coefficient of restitution = e

Now :-

$v_0 = \sqrt{2gh}$

$v_1 = e \sqrt{2gh} \: \tiny{\left( \because e = \dfrac{relative \: velocity \: after\: collision}{relative \: velocity \: before\: collision}\right)}$

Similarly

v₂ = e² v₀... and so on .

Now

h₀ = h

$h_1 = \dfrac{v_{1}^{2} }{2g}= e^2. \dfrac{v_{0}^{2} }{2g} \\ \\ = e^2.h_0$

$h_2 = e^2.h_1 = e^4.h_0$

Now

$t_0 = \sqrt{\dfrac{2h_0}{g}}$

$t_1 = \sqrt{\dfrac{2h_1}{g}} =\sqrt{e^2.\dfrac{2h_0}{g}} \\= e.t_0$

$t_2 = e^2.t_0$

Now total time =

$= t_0 + 2t_1 + 2t_2. ...$

$= t_0 + 2e .t_0 + 2e^2.t_0 ...$

$= t_0( 1 + 2e + 2e^2. ... )$

$= t_0 ( 1 + 2e( 1 + e ... ))$

$= t_0 \left( 1 + 2e \left( \dfrac{1}{1-e} \right)\right)$

$= t_0 \left( 1 + \dfrac{2e}{1-e} \right)$

$= t_0 \left( \dfrac{1 + e}{1 -e}\right)$

$= \sqrt{\dfrac{2h_0}{g}}\left( \dfrac{1 + e}{1 -e}\right)$

Answer 2

QUESTION:-

A ball is dropped from height h on a floor where Coefficient of restitution is e. Find the time required by the ball to stop rebounding.

(a) √(2h/g) (1 + e/1 - e)

(b) √(2h/g) (1 + e)

(c) √(2h/g) (1 - e/1 + e)

(d) none of the above

ANSWER:-

A ball is dropped from height h on a floor where Coefficient of restitution is e. Find the time required by the ball to stop rebounding.

(a) √(2h/g) (1 + e/1 - e) [✏]

(b) √(2h/g) (1 + e) [✂]

(c) √(2h/g) (1 - e/1 + e) [✂]

(d) none of the above. [✂]