Question

A ball of mass m is dropped from a height h on a smooth elastic floor, such that it rebounds with
same speed, then, the change in magnitude of momentum of ball before and after striking the floor
is: (Take vertically downward direction as positive)​

Answer 1

Given:

A ball of mass m is dropped from a height h on a smooth elastic floor, such that it rebounds with

same speed.

To find:

Change in magnitude of momentum

Calculation:

Considering downward direction as positive ;

Initial momentum just before hitting the ground:

$\therefore \: P1 = mv$

$= > \: P1 = m \sqrt{ {u}^{2} + 2gh}$

$= > \: P1 = m \sqrt{ {(0)}^{2} + 2gh}$

$= > \: P1= m \sqrt{ 2gh}$

Final momentum after rebounding :

$\therefore \: P2 = m ( -v )$

$= > \: P2= - m \sqrt{ 2gh}$

So, change in momentum :

$\Delta P = - m \sqrt{2gh} - m \sqrt{2gh}$

$= > \Delta P = - 2m \sqrt{2gh}$

$= > \Delta P = - m \sqrt{8gh}$

So, final answer is

$\boxed{ \large{ \bold{\Delta P = - m \sqrt{8gh} }}}$