A marble of mass ‘m’ is dropped from a height ‘h’ on a hard floor. If it rebounds to the same height, the change in momentum of marble is

Given that a marble of mass m is dropped from a height = h

The velocity of marble before striking the hard floor $u=-\sqrt{2gh}$

$\left(\because \text{final}\text{velocity}\text{of}\text{body}\text{dropped}\text{from}\text{height}\text{h}\text{is}\text{equal}\text{to}v=\sqrt{2gh}\right)$

Initial momentum before striking ${\text{p}}_{\text{i}}=-\text{mu}$  $\Rightarrow -m\sqrt{2gh}$

$\left(\begin{array}{c}\text{by}\text{sign}\text{convertion}\text{we}\text{considered}\text{quantities}\text{in}\text{downward}\\ \text{direction}\text{as}-\text{ve}\text{and}\text{in}\text{upward}\text{direction}\text{as}+\text{ve}\end{array}\right)$

After striking the hard floor

Let the velocity of marble = v

$\therefore \text{h}=\frac{{\text{v}}^{\text{2}}}{\text{2g}}$  $\left(\because \text{We}\text{know}{\text{h}}_{\text{max}}=\frac{{\text{u}}^{\text{2}}}{\text{2g}}\text{for}\text{a}\text{body}\text{thrownup}\text{with}\text{initial}\text{velocity}\text{u}\right)$

$\therefore \text{v}=+\sqrt{2\text{gh}}$

The momentum of marble after striking

${\text{p}}_{\text{f}}=\text{mv}=\text{m}\sqrt{2\text{gh}}$

Hence change in momentum ${\text{p}}_{\text{f}}-{\text{p}}_{\text{i}}=\text{m}\sqrt{2\text{gh}}-\left(-\text{m}\sqrt{2\text{gh}}\right)$

$\Delta \text{p}=2\text{m}\sqrt{2\text{gh}}$  in upward direction.