91Ó°ÊÓ

Given that asteel ball with a mass 40.0 g is dropped from a height of 2.00 m onto a horizontal steel lab. The ball rebounds to a height of 1.60 m

The mass, $\mathrm{m}=40\mathrm{g}$

Initial velocity, $\mathrm{u}=0$

The height, ${\mathrm{h}}_1=2\mathrm{m}$

The height, ${\mathrm{h}}_2=1.6\mathrm{m}$

Energy of a system is always conserved.

Impulse-Momentum Theorem: The impulse of the net force on a particle during a time interval equals the change in momentum of that particle during that interval

Thus Impulse is define by,

$\mathrm{J}={\mathrm{p}}_2-{\mathrm{p}}_1$

Where p₁ is initial momentum and p₂ is final momentum.

Also, according to definition of impulse, it is product of the net force and the time interval.

Let the velocity of striking the ground v₁ and velocity of bouncing back is v₂.

The kinetic and potential energy is the same. Therefore,

$\begin{array}{rcl}\frac{1}{2}{\mathrm{mv}}^2& =& \mathrm{mgh}\\ \mathrm{v}& =& Â\pm \sqrt{2\mathrm{gh}}\end{array}$

Define the velocity of striking the ground as beow.

$\begin{array}{rcl}{\mathrm{v}}_1& =& \sqrt{\left(2\right)\left(9.8\right)\left(2\right)}\\ & =& 6.26\mathrm{m}/\mathrm{s}\end{array}$

And the velocity of bouncing back is,

$\begin{array}{rcl}{\mathrm{v}}_2& =& -\sqrt{\left(2\right)\left(9.8\right)\left(1.6\right)}\\ & =& -5.6\mathrm{m}/\mathrm{s}\end{array}$

So Impulse, J is equal to the change in momentum, that is,

$\begin{array}{rcl}\mathrm{J}& =& \mathrm{m}\left({\mathrm{v}}_2-{\mathrm{v}}_1\right)\\ & =& \left(40×{10}^{-3}\right)\left(-5.6-6.26\right)\\ & =& -0.474\mathrm{kg}.\mathrm{m}/\mathrm{s}\end{array}$

Hence Impulse is data-custom-editor="chemistry" $0.474\mathrm{kg}.\mathrm{m}/\mathrm{s}$upwards.

Ball is in contact with the ground for $2\mathrm{ms}=0.002\mathrm{s}$.

Average force

$\begin{array}{rcl}\mathrm{F}& =& \frac{\mathrm{J}}{∆\mathrm{t}}\\ \mathrm{F}& =& \frac{-0.474}{0.002}\\ \mathrm{F}& =& -237\mathrm{N}\end{array}$

Force is $237\mathrm{N}$upwards.