Question

Fred (mass $60\mathrm{kg}$) is running with the football at a speed of $6.0\mathrm{m}/\mathrm{s}$when he is met head-on by Brutus (mass $120\mathrm{kg}$), who is moving at $4.0\mathrm{m}/\mathrm{s}$. Brutus grabs Fred in a tight grip, and they fall to the ground. Which way do they slide, and how far? The coefficient of kinetic friction between football uniforms and Astroturf is $0.30$.

Short answer

They slide $7.6\mathrm{cm}$distance in the direction of Brutus motion.

Step-by-step solution

Given information

We have given,

Fred mass =$60\mathrm{kg}$

Fred speed =$6\mathrm{m}/\mathrm{s}$

Brutus mass =$120\mathrm{kg}$

Speed =$4\mathrm{m}/\mathrm{s}$

coefficient of kinetic friction =$0.3$

We have to find how far they slide and which way.

Simplify

In an Collison the momentum will be always conserved. then,

${\mathrm{P}}_{\mathrm{I}}={\mathrm{P}}_{\mathrm{F}}$

Where the momentum is given by,

$\mathrm{P}=\mathrm{mv}$

Then,

$$\begin{gathered} {\mathrm{m}}_1{\mathrm{v}}_1+{\mathrm{m}}_2{\mathrm{v}}_2=({\mathrm{m}}_1+{\mathrm{m}}_2)\mathrm{V} \\ \mathrm{V}=\frac{{\mathrm{m}}_1{\mathrm{v}}_1+{\mathrm{m}}_2{\mathrm{v}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2} \\ \mathrm{V}=\frac{60\mathrm{kg}\times -6\mathrm{m}/\mathrm{s}+120\mathrm{kg}\times 4\mathrm{m}/\mathrm{s}}{60\mathrm{kg}+120\mathrm{kg}} \\ \mathrm{V}=0.67\mathrm{m}/\mathrm{s} \end{gathered}$$

This positive sign is indicates that the motion will be in BRUTUS side.

Simplify

Now we have to find the distance of slide.

The force after Collison will be equal to the force of friction.

$$\begin{gathered} {\mathrm{F}}_{\mathrm{net}}={\mathrm{f}}_{\mathrm{k}} \\ ({\mathrm{m}}_1+{\mathrm{m}}_2)\mathrm{a}=-{\mathrm{\mu }}_{\mathrm{k}}({\mathrm{m}}_1+{\mathrm{m}}_2)\mathrm{g} \\ \mathrm{a}=-{\mathrm{\mu }}_{\mathrm{k}}\mathrm{g} \\ \mathrm{a}=-0.3\times 9.8\mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{a}=-2.94\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Then using the equation of kinematics.

$$\begin{gathered} {\mathrm{v}}_2^2={\mathrm{v}}_1^2+2\mathrm{as} \\ \mathrm{s}=\frac{{\mathrm{v}}_2^2-{\mathrm{v}}_1^2}{2\mathrm{a}} \\ \mathrm{s}=\frac{0-{(0.67)}^2}{-2\times 2.94} \\ \mathrm{s}=7.6\mathrm{cm} \end{gathered}$$