Two students are running to make it to class. They turn a corner and collide; coming to a complete stop. What force did they exert on each other.

Average mass of a person 68kg.

Students come to complete stop.

Choose student 1 as our system.

Collision time $ \Delta t$

Force of collision on student 1 in x-direction.

Approximate speed of person 5m/s^-1 (half speed of olympic sprinter)

On average student 1's body compresses a couple of cm on average - assume compression of 2.5cm.

$\Delta \vec{p}_{sys} = \vec{F}_{ext} \Delta t$

$\vec{p}_{sys,f} = \vec{p}_{sys,i} + \vec{F}_{ext} \Delta t$

$\vec{v}_{avg} = \dfrac{\vec{v}_{f} + \vec{v}_{i}}{2} = \dfrac{\Delta \vec{r}}{\Delta t}$

We use the momentum principle to relate the momentum of the students to the force applied. We have chosen student 1 as our system.

${p}_{fx} = {p}_{ix} + {F}_{x,coll} \Delta t$

We are told that the students come to a complete stop and so ${p}_{fx}$ = 0.

$0 = M{V}_{ix} + {F}_{x,coll} \Delta t$

$ {F}_{x,coll} = \dfrac{-M{V}_{ix}}{\Delta t}$

negative means $-\hat{x}$ direction

Need to find collision time ${\Delta t}$

$\vec{v}_{avg} = \dfrac{\vec{v}_{f} + \vec{v}_{i}}{2} = \dfrac{\Delta \vec{r}}{\Delta t}$

In 1D: $\vec{v}_{avg} = \dfrac{\Delta x}{\Delta t} = \dfrac{\vec{v}_{f} + \vec{v}_{i}}{2}$

$ {\Delta t} = \dfrac{\Delta x}{{V}_{avg}} = \dfrac{\Delta x}{\dfrac{\vec{v}_{f} + \vec{v}_{i}}{2}} $

$ \dfrac{0.025}{\dfrac{5m/s + 0m/s}{2}} = 0.01s $

$ {F}_{x,coll} = \dfrac{-M{V}_{ix}}{\Delta t} = - {\dfrac{(68kg)(5m/s)}{0.01s}}$

$ = -34,000 \,N$