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===== Example: Colliding Students ===== 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. === Facts === Average mass of a person 68kg. Students come to complete stop. Choose student 1 as our system. === Lacking === Collision time $ \Delta t$ Force of collision on student 1 in x-direction. === Approximations & Assumptions === 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. === Representations === $\Delta \vec{p}_{sys} = \vec{F}_{ext} \Delta t$ $\vec{p}_{sys,f} = \vec{p}_{sys,i} + \vec{F}_{ext} \Delta t$ $\vec{v}_{avg} = \dfrac{\vec{v}_{f} + \vec{v}_{i}}{2} = \dfrac{\Delta \vec{r}}{\Delta t}$ {{183_notes:untitled.jpg?400|}} === Solution === 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. $0 = M{V}_{ix} + {F}_{x,coll} \Delta t$ Rearranging the equation we relate ${F}_{x,coll}$ to the remaining variables. We are trying to find force. $ {F}_{x,coll} = \dfrac{-M{V}_{ix}}{\Delta t}$ The negative sign means that the force is in $-\hat{x}$ direction. Now that we have the above relationship we must find the missing variables in order to solve for $ {F}_{x,coll}$. First we need to find collision time ${\Delta t}$. We can relate the average velocity to displacement over time. $\vec{v}_{avg} = \dfrac{\vec{v}_{f} + \vec{v}_{i}}{2} = \dfrac{\Delta \vec{r}}{\Delta t}$ In 1D this looks like: $\vec{v}_{avg} = \dfrac{\Delta x}{\Delta t} = \dfrac{\vec{v}_{f} + \vec{v}_{i}}{2}$ Relate these 3 equations together to solve for ${\Delta t}$ $ {\Delta t} = \dfrac{\Delta x}{{V}_{avg}} = \dfrac{\Delta x}{\dfrac{\vec{v}_{f} + \vec{v}_{i}}{2}} $ Fill in the values for the variables from the assumptions and approximations you made previous. $ \dfrac{0.025}{\dfrac{5m/s + 0m/s}{2}} = 0.01s $ Having solved for $ {\Delta t}$ fill this value and the known value for mass and the approximated value for velocity into the equation that we arranged earlier to find $ {F}_{x,coll}$ $ {F}_{x,coll} = \dfrac{-M{V}_{ix}}{\Delta t} = - {\dfrac{(68kg)(5m/s)}{0.01s}}$ $ = -34,000 \,N$