A ball falls vertically onto a floor with momentum p, and then bounces repeatedly. If the coefficient of restitution is e, then the total momentum imparted by the ball on the floor till the ball comes to rest is

Question

A ball falls vertically onto a floor with momentum p, and then bounces repeatedly. If the coefficient of restitution is e, then the total momentum imparted by the ball on the floor till the ball comes to rest is (A) p(1+e) (B) (p/1-e) (C) p(1+(1/e)) (D) p((1+e/1-e))

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/c3e99b7511bb611fbe0053126afdb796-.png" /> Δ p =(p+e p)+(e p+e2 p)+(e2 p+e3 p)+⋅s =p(1+e)[1+e+e2+⋅s]=(p(1+e)/1-e)

Q. A ball falls vertically onto a floor with momentum $p$ , and then bounces repeatedly. If the coefficient of restitution is $e$ , then the total momentum imparted by the ball on the floor till the ball comes to rest is

$p (1 + e)$

$\frac{p}{1 - e}$

$p (1 + \frac{1}{e})$

$p (\frac{1 + e}{1 - e})$

$Δ p = (p + e p) + (e p + e^{2} p) + (e^{2} p + e^{3} p) + \dots$
$= p (1 + e) [1 + e + e^{2} + \dots] = \frac{p ( 1 + e )}{1 - e}$

Q. A ball falls vertically onto a floor with momentum p, and then bounces repeatedly. If the coefficient of restitution is e, then the total momentum imparted by the ball on the floor till the ball comes to rest is

p(1+e)

1−ep​

p(1+e1​)

p(1−e1+e​)