Answer
(a) $2.09 \space Kg.m/s$
(b) $F_{avg}=22N$ The direction of force is upward
Work Step by Step
(a) We can find the change in momentum as follows:
First, we will determine the velocity of the ball when it hits the ground, let it be $v$
$\implies v^2=v_{\circ}^2+2gh_1$
$\implies v^2=2gh_1$ ($v_{\circ}$=0 because initially the ball is at rest)
$\implies v=\sqrt{2gh_1}$
We plug in the known values to obtain:
$v=\sqrt{2(9.8)(2)}$
This simplifies to:
$\implies v=6.26m/s$
We take $v=-6.26m/s$ because the direction of the velocity is downward.
Now, let $v^{\prime}$ be the velocity of the ball when it rebounds
${v^{\prime}}^2=v_{\circ}^2+2gh_2$ ($v_{\circ}$=0)
The above equation simplifies to:
$v^{\prime}=\sqrt{2gh_2}$
We plug in the known values to obtain:
$v^{\prime}=\sqrt{2(9.8)(0.9)}$
$\implies v^{\prime}=4.2m/s$
The change of momentum is given as
$\Delta P=m(v^{\prime}-v)$
We plug in the known values to obtain:
$\Delta P=(0.2)(4.2-(-6.26))$
$\implies \Delta P=2.09\space Kg.m/s$
(b) The average force can be determined as follows:
$F_{avg}=\frac{\Delta P}{\Delta t}$
We plug in the known values to obtain:
$F_{avg}=\frac{2.09\space Kg.m/s}{0.095}$
$\implies F_{avg}=22N$
