Using mass and speed data and assuming that the football player catches the ball with his feet off the ground with both of them moving horizontally, calculate: (a) the final velocity if the ball and player are going in the same direction and (b) the loss of kinetic energy in this case. (c) Repeat parts (a) and (b) for the situation in which the ball and the player are going in opposite directions. Might the loss of kinetic energy be related to how much it hurts to catch the pass?

Kinetic energy is the energy that an item or particle has as a result of its movement. When work is done on an object by exerting a net force, the object accelerates and gains kinetic energy as a result.

The mass of the first object before collision$={\mathrm{m}}_{\mathrm{a}}=0.410\mathrm{kg}$

The mass of the second object before collision$={\mathrm{m}}_{\mathrm{b}}=110\mathrm{kg}$

The velocity of the first object before the collision $={\mathrm{V}}_{\mathrm{a}}=25\mathrm{m}/\mathrm{s}$

The velocity of the second object before the collision$={\mathrm{V}}_{\mathrm{b}}=8\mathrm{m}$

Mass of the objects when they are together after the collision$=\mathrm{M}=60+75=135\mathrm{kg}$

(a) By putting all the value into the equation we get

$$\begin{gathered} {\mathrm{M}}_{\mathrm{a}}{\mathrm{V}}_{\mathrm{a}}+{\mathrm{M}}_{\mathrm{b}}{\mathrm{V}}_{\mathrm{b}}=\left({\mathrm{M}}_{\mathrm{a}}+{\mathrm{M}}_{\mathrm{b}}\right){\mathrm{V}}_{\mathrm{x}} \\ {\mathrm{V}}_{\mathrm{x}}=\frac{{\mathrm{M}}_{\mathrm{a}}{\mathrm{V}}_{\mathrm{a}}+{\mathrm{M}}_{\mathrm{b}}{\mathrm{V}}_{\mathrm{b}}}{\left({\mathrm{M}}_{\mathrm{a}}+{\mathrm{M}}_{\mathrm{b}}\right)} \\ {\mathrm{V}}_{\mathrm{x}}=\frac{\left(0.41\right)\left(25\right)+\left(110\right)\left(8\right)}{\left(110.410\right)} \\ {\mathrm{V}}_{\mathrm{x}}=8.06 \mathrm{m}/\mathrm{s} \end{gathered}$$

Velocity of the system after the collision is$8.06 \mathrm{m}/\mathrm{s}$.

(b) The kinetic energy lost will be the difference of the initial kinetic energy to the final kinetic energy.

$$\begin{gathered} {\mathrm{KE}}_{\mathrm{lost}}={\mathrm{KE}}_{\mathrm{i}}-{\mathrm{KE}}_{\mathrm{f}} \\ {\mathrm{KE}}_{\mathrm{lost}}=\left[\frac{1}{2}{\mathrm{m}}_{\mathrm{a}}{{\mathrm{v}}_{\mathrm{a}}}^2+\frac{1}{2}{\mathrm{m}}_{\mathrm{b}}{{\mathrm{v}}_{\mathrm{b}}}^2\right]-\frac{1}{2}{\mathrm{m}}_{\mathrm{x}}{{\mathrm{v}}_{\mathrm{x}}}^2 \\ {\mathrm{KE}}_{\mathrm{lost}}=\frac{1}{2}\left[{\mathrm{m}}_{\mathrm{a}}{{\mathrm{v}}_{\mathrm{a}}}^2+{\mathrm{m}}_{\mathrm{b}}{{\mathrm{v}}_{\mathrm{b}}}^2-{\mathrm{m}}_{\mathrm{x}}{{\mathrm{v}}_{\mathrm{x}}}^2\right] \\ {\mathrm{KE}}_{\mathrm{lost}}=\frac{1}{2}\left[\left(0.41\right){\left(25\right)}^2+\left(110\right){\left(8\right)}^2-\left(110.410\right){\left(8.06\right)}^2\right] \end{gathered}$$

${\mathrm{KE}}_{\mathrm{lost}}=61.8 \mathrm{J}$

The kinetic energy lost is$61.8 \mathrm{J}$.

(c) Let’s say the football player is moving in opposite side.

The velocity of the second object before the collision$={\mathrm{V}}_{\mathrm{b}}=8\mathrm{m}/\mathrm{s}$

All other value will be same

$$\begin{gathered} {\mathrm{M}}_{\mathrm{a}}{\mathrm{V}}_{\mathrm{a}}+{\mathrm{M}}_{\mathrm{b}}{\mathrm{V}}_{\mathrm{b}}=\left({\mathrm{M}}_{\mathrm{a}}+{\mathrm{M}}_{\mathrm{b}}\right){\mathrm{V}}_{\mathrm{x}} \\ {\mathrm{V}}_{\mathrm{x}}=\frac{{\mathrm{M}}_{\mathrm{a}}{\mathrm{V}}_{\mathrm{a}}+{\mathrm{M}}_{\mathrm{b}}{\mathrm{V}}_{\mathrm{b}}}{\left({\mathrm{M}}_{\mathrm{a}}+{\mathrm{M}}_{\mathrm{b}}\right)} \\ {\mathrm{V}}_{\mathrm{x}}=\frac{\left(0.41\right)\left(25\right)+\left(110\right)\left(-8\right)}{\left(110.410\right)} \\ {\mathrm{V}}_{\mathrm{x}}=-7.88 \mathrm{m}/\mathrm{s} \end{gathered}$$

The final velocity if the ball and the player are moving in opposite direction is$-7.88 \mathrm{m}/\mathrm{s}$

The kinetic energy lost will be the difference of the initial kinetic energy to the final kinetic energy.

$$\begin{gathered} {\mathrm{KE}}_{\mathrm{lost}}={\mathrm{KE}}_{\mathrm{i}}-{\mathrm{KE}}_{\mathrm{f}} \\ {\mathrm{KE}}_{\mathrm{lost}}=\left[\frac{1}{2}{\mathrm{m}}_{\mathrm{a}}{{\mathrm{v}}_{\mathrm{a}}}^2+\frac{1}{2}{\mathrm{m}}_{\mathrm{b}}{{\mathrm{v}}_{\mathrm{b}}}^2\right]-\frac{1}{2}{\mathrm{m}}_{\mathrm{x}}{{\mathrm{v}}_{\mathrm{x}}}^2 \\ {\mathrm{KE}}_{\mathrm{lost}}=\frac{1}{2}\left[{\mathrm{m}}_{\mathrm{a}}{{\mathrm{v}}_{\mathrm{a}}}^2+{\mathrm{m}}_{\mathrm{b}}{{\mathrm{v}}_{\mathrm{b}}}^2-{\mathrm{m}}_{\mathrm{x}}{{\mathrm{v}}_{\mathrm{x}}}^2\right] \\ {\mathrm{KE}}_{\mathrm{lost}}=\frac{1}{2}\left[\left(0.41\right){\left(25\right)}^2+\left(110\right){\left(-8\right)}^2-\left(110.410\right){\left(-7.88\right)}^2\right] \end{gathered}$$

${\mathrm{KE}}_{\mathrm{lost}}=220 \mathrm{J}$

The kinetic energy lost is$220 \mathrm{J}$ if initially object is moving opposite direction.