Question

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

Answer 1

The marble is dropped from the height 'h', so it has no initial velocity.

The velocity of the marble when it reaches the ground is, by third equation of motion,

$\longrightarrow\sf{(v_1)^2=0^2+2gh}$

$\longrightarrow\sf{v_1=\sqrt{0^2+2gh}}$

$\longrightarrow\sf{v_1=\sqrt{2gh}\ m\,s^{-1}}$

As the marble rebounds to the same height, its velocity at the highest point is zero.

Then by third equation of motion, the velocity of marble when it rises from the ground is given by,

$\longrightarrow\sf{0^2=(v_2)^2-2gh}$

$\sf{-2gh}$ is because marble is moving upwards during this time. Downward motion is considered as positive.

$\longrightarrow\sf{(v_2)^2=2gh}$

$\longrightarrow\sf{v_2=-\sqrt{2gh}\ m\,s^{-1}}$

Hence the change in momentum is,

$\longrightarrow\sf{\Delta p=m(v_2-v_1)}$

$\longrightarrow\sf{\Delta p=m(-\sqrt{2gh}-\sqrt{2gh})}$

$\longrightarrow\sf{\underline{\underline{\Delta p=-2m\sqrt{2gh}\ \ kg\,m\,s^{-1}}}}$

Negative sign implies $\sf{\Delta p}$ is acting upwards.