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--- 183_notes:examples:two_students_colliding [2014/09/25 05:56] – pwirving
+++ 183_notes:examples:two_students_colliding [2014/10/02 15:50] (current) – caballero
@@ Line 1: / Line 1: @@
-===== Example: Calculating the gravitational force exerted by the Earth on the Moon. =====
+===== 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.
@@ Line 30: / Line 30: @@
 $\vec{v}_{avg} = \dfrac{\vec{v}_{f} + \vec{v}_{i}}{2} = \dfrac{\Delta \vec{r}}{\Delta t}$
+{{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}$
-negative means $-\hat{x}$ direction
+The negative sign means that the force is in $-\hat{x}$ direction.
-Need to find collision time ${\Delta t}$
+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: $\vec{v}_{avg} = \dfrac{\Delta x}{\Delta t} = \dfrac{\vec{v}_{f} + \vec{v}_{i}}{2}$
+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$
+$ = -34,000 \,N$