Question

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?

Short answer

a. Speed of the player is $1.916\mathrm{m}/\mathrm{s}.$

b. Speed of the player is $1.09\mathrm{m}/\mathrm{s}.$

Step-by-step solution

Part (a) Step 1: Given information

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.

Simplify

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,

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

Part (b) Step 1: Given information,

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.

Simplify

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)

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