A $70\mathrm{kg}$football player is gliding across very smooth ice at $2.00\mathrm{m}/\mathrm{s}$. He throws a $0.450\mathrm{kg}$football straight forward. What is the player’s speed afterward if the ball is thrown at

a. $15.0\mathrm{m}/\mathrm{s}$relative to the ground?

b. $15.0\mathrm{m}/\mathrm{s}$relative to the player?

We have given,

Mass of football player=$70\mathrm{kg}$

Speed of player =$2\mathrm{m}/\mathrm{s}$

mass of football=$0.45\mathrm{kg}$

Seed of ball =$15\mathrm{m}/\mathrm{s}$

We have to find the player’s speed afterward the ball is thrown.

Using the conservation of the momentum in the inelastic collision,

Then,

the momentum before the Collison will be,

$$\begin{gathered} {\mathrm{P}}_{\mathrm{i}}={\mathrm{P}}_{\mathrm{player}}+{\mathrm{P}}_{\mathrm{BALL}} \\ {\mathrm{P}}_{\mathrm{i}}=(70\mathrm{kg})(2\mathrm{m}/\mathrm{s})+(0.45)(2\mathrm{m}/\mathrm{s}) \\ {\mathrm{P}}_{\mathrm{i}}=\left(140\right)+(0.9)=140.9\mathrm{kg}\mathrm{m}/\mathrm{s} \end{gathered}$$

The momentum after the collision will be,

$$\begin{gathered} {\mathrm{P}}_{\mathrm{f}}={\mathrm{P}}_{\mathrm{player}}+{\mathrm{P}}_{\mathrm{ball}} \\ {\mathrm{P}}_{\mathrm{f}}=(70\mathrm{kg})\mathrm{v}+(0.45\mathrm{kg})(15\mathrm{m}/\mathrm{s}) \\ {\mathrm{P}}_{\mathrm{f}}=70\mathrm{v}+6.75={\mathrm{P}}_{\mathrm{i}} \\ 70\mathrm{v}+6.75=140.9 \\ \mathrm{v}=1.916\mathrm{m}/\mathrm{s} \end{gathered}$$

We have given,

Mass of player = $70\mathrm{kg}$

Mass of football =$0.45\mathrm{kg}$

Speed of player = $2\mathrm{m}/\mathrm{s}$

Speed of football with respect to player =$15\mathrm{m}/\mathrm{s}$

We have to find the player’s speed afterward the ball is thrown.

Since the speed of the football is given with respect to the player. then the observe speed will be less than origin speed of the player.

The original speed of the football is given by,

$$\begin{gathered} \mathrm{Original}\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{ball}=\mathrm{speed}\mathrm{of}\mathrm{player}\mathrm{after}\mathrm{he}\mathrm{kick}\mathrm{the}\mathrm{ball}-\mathrm{football}\mathrm{sped}\mathrm{with}\mathrm{respec}\mathrm{to}\mathrm{player} \\ {\mathrm{V}}_{\mathrm{O}}={\mathrm{v}}_{\mathrm{p}}-15\mathrm{m}/\mathrm{s}.............\left(1\right) \end{gathered}$$

Using the conservation of momentum in inelastic collision is given by,

$$\begin{gathered} {\mathrm{P}}_{\mathrm{i}}={\mathrm{P}}_{\mathrm{f}} \\ (70+0.45)\mathrm{kg}(2\mathrm{m}/\mathrm{s})=(70\hspace{0.17em}\mathrm{kg}){\mathrm{v}}_{\mathrm{p}}+(0.45\mathrm{kg}){\mathrm{v}}_{\mathrm{o}} \end{gathered}$$

Here put the value from equation (1)

$$\begin{gathered} (7.45)\left(2\right)=70\left({\mathrm{v}}_{\mathrm{p}}\right)+(0.45)({\mathrm{v}}_{\mathrm{p}}-15) \\ {\mathrm{v}}_{\mathrm{p}}=1.09\mathrm{m}/\mathrm{s} \end{gathered}$$