A ball whose mass is 0.2kg hits the floor with a speed of 8 m/s and rebounds upward with a speed of 7m/s. The ball was in contact with the floor for$\mathbf{0}.\mathbf{5}\mathbf{ms}\left(\mathbf{0}.\mathbf{5}\times \mathbf{10}-\mathbf{3}\mathbf{s}\right)$.

(a) What was the average magnitude of the force exerted on the ball by the floor? (b) Calculate the magnitude of the gravitational force that the Earth exerts on the ball.

(c) In a collision, for a brief time there are forces between the colliding objects that are much greater than external forces. Compare the magnitudes of the forces found in parts (a) and (b).

Mass of the ball is$m=0.2\text{kg}$.

Speed is$v_{i,y}=8\text{m}/\text{s}$.

Upward speed is$v_{f,y}=7\text{m}/\text{s}$.

Contact time with the floor is,$\Delta F=0.5\text{ms(}0.5\times {10}^{-3}\text{s)}$.

Acceleration due to gravity is $g=9.8\text{m}/{\text{s}}^2$.

The change in momentum is given by:

$$\begin{gathered} \mathit{\Delta }\mathit{P}\mathbf{=}{\mathit{P}}_{\mathbf{f}}\mathbf{-}{\mathit{P}}_{\mathbf{i}} \\ \mathit{\Delta }\mathit{P}\mathbf{=}\mathit{m}{\mathit{v}}_{\mathbf{f}\mathbf{,}\mathbf{y}}\mathbf{-}\mathit{m}{\mathit{v}}_{\mathbf{i}\mathbf{,}\mathbf{y}} \\ \mathit{\Delta }\mathit{P}\mathbf{=}\mathit{m}\left(v_{f,y}-v_{i,y}\right) \end{gathered}$$

The average force exerted on the ball can be calculated by dividing the change in momentum by the time interval.

${\mathit{F}}_{\mathbf{a}\mathbf{v}\mathbf{g}}\mathbf{=}\frac{\mathbf{\Delta }\mathbf{P}}{\mathbf{\Delta }\mathbf{F}}$

(a)

Substitute the values to the first formula to find for the change in momentum.

$\begin{array}{rcl}\Delta P& =& m\left(v_{f,y}-v_{i,y}\right)\\ & =& (0.2kg)(-7\frac{m}{\sec }-8\frac{m}{\sec })\\ & =& 3\text{kg.m}/\text{s}\end{array}$

Use the second formula and substitute the values.

$$\begin{gathered} F_{avg}=\frac{\Delta P}{\Delta F} \\ {F}_{avg}=\frac{3\text{kg}·\text{m}/\text{s}}{5\times {10}^{-3}\text{s}} \\ {F}_{avg}=6000\text{N} \end{gathered}$$

Therefore, the average force is $F_{avg}=6000\text{N}$.

(b)

The magnitude of the gravitational force that the Earth exerts on the ball.

Ball's mass:$m=0.2kg$

Gravity strength on earth:$g=9.8\text{m}/{\text{s}}^2$

The gravitational force$F_{grav}=mg$

Substitute the values I the above formula.

$$\begin{gathered} F_{grav}=0.2\times 9.8 \\ {F}_{grav}=2\text{N} \end{gathered}$$

Therefore, the gravitational force is 2 N.

(c)

From part (a) and (b):

$\begin{array}{rcl}\frac{\left|{\overrightarrow{F}}_{\text{floor}}\right|}{\left|{\overrightarrow{F}}_{\text{grav}}\right|}& =& \frac{6002}{2}\\ & =& 3000\end{array}$

So, the average magnitude of the force exerted on the ball by the floor is 3000 times the magnitude of the force exerted on the ball by the Earth.