Question

A ball (m = 1 kg) is dropped from height 20 m. If
coefficient of a restitution in 0.9, the impulse
imparted to the ground (by the ball) is
(1) 38 NS
(2) 40 NS
(3) 57 NS
(4) 81 Ns​

Answer 1

Answer:

Option (1) 38 Ns

Explanation:

The velocity when the ball is about to hit the ground

using the third equation of motion

$v^2=u^2+2gh$

$v^2=0^2+2\times 10\times 20$

$v^2=400$

$v=\sqrt{400}$

$\implies v=20$ m/s

Coefficient of restitution = 0.9

Thus, after hitting the ground the velocity of the ball in the upward direction

$v'=0.9\times 20$

$\implies v'=18$ m/s

The impulse imparted = change in momentum

$=mv-(-mv')$

$=mv+mv'$

$=m(v+v')$

$=1\times (20+18)$ Ns

$=38$ Ns

Hope this helps.