Question

A marble of mass ${}^{\prime }{m}^{\prime }$ is dropped from a height ${}^{\prime }{h}^{\prime }$ on a hard floor. If it rebounds to the same height the change in momentum of marble is :

• A. $m\sqrt{2gh}$
• B. $2m\sqrt{2gh}$
• C. $3m\sqrt{2gh}$
• D. $4m\sqrt{2gh}$

Answer 1

Answer: (B)

$2m\sqrt{2gh}$

The marble is dropped from height$h$then the velocity of the marble on reaching the ground is$\sqrt{2gh}$

Since it is given that it rebounds to the same height after striking the ground

Hence the change in velocity will be given as

$\Delta v=\sqrt{2gh}-\left(-\sqrt{2gh}\right)$

Hence the change is velocity is$2\sqrt{2gh}$

The change is momentum becomes

$2m\sqrt{2gh}$