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--- 183_notes:examples:mit_water_balloon_fight [2014/10/04 17:42] – pwirving
+++ 183_notes:examples:mit_water_balloon_fight [2014/10/11 06:03] (current) – pwirving
@@ Line 39: / Line 39: @@
 Surroundings: Surgical hose slingshot
-{{course_planning:|}}
+{{course_planning:projects:water_balloon_fight.jpg?300}}
 === Solution ===
+The first step to solving this problem is to define your system and surroundings which we have previously defined in our representations as system being the water balloon and the surroundings being the surgical hose slingshot.
+We are going to apply the work-kinetic energy theorem to the problem.
+According to this theorem the change in energy of the system is equal to the work done on the system.
  $\Delta K_{balloon} = W_{hose}$
+Kinetic final - kinetic initial is equal to the work of the hose.
  $K_{balloon,f} - K_{balloon,i} = W_{hose}$
+Sub in  $\dfrac{1}{2}mv^2$  for kinetic energy.
  $\dfrac{1}{2}mv_{f}^2 - \dfrac{1}{2}mv_{i}^2 = W$
+Initial kinetic energy of the balloon is zero. Therefore final energy of the balloon is equal to the work by the hose.
  $\dfrac{1}{2}mv_{i}^2 (0, v_{i} = 0)$
+Rearrange to solve for  $v_{f}^2$
  $\dfrac{1}{2}mv_{f}^2 = W \longrightarrow v_{f}^2 = \dfrac{2W}{m}$
-If we calculate the work done by the hose, we can find  $v_{f}$
+If we can calculate the work done by the hose, we can find  $v_{f}$
  $v_{f} = \sqrt {\dfrac{2W}{m}}$
-The force of the hose is not constant
+However the force of the hose is not constant
 But remember:  $F = -kx$
-$$W_{F} = \sum_{i} \vec{F}_i\cdot\Delta\vec{r}_i = \int_{i}^{f} \vec{F} d\vec {r}$$
+Remember that the work done by a non constant force is:
+$$W = \sum_{i} \vec{F}_i\cdot\Delta\vec{r}_i = \int_{i}^{f} \vec{F} d\vec {r}$$
+And the initial and final conditions are:
+Initial  =  $S_{max} \longrightarrow$  max stretch
+Final =  $0 \longrightarrow$  relaxed
+Sub in the force for hose into the equation for work done by a non constant force:
+ $W = \int_{S_{max}}^{0} (-kx)(dx) = -\dfrac{1}{2}kx^2$
+Solve the resultant:
+ $W = -\dfrac{1}{2}k(0)^2 - [-\dfrac{1}{2}ks^2_{max}] = \dfrac{1}{2}k(s)^2_{max}$
+ $W > 0$  makes sense because  $\Delta K > 0$
+Going back a couple of steps we now have an equation for work which we can relate to our previous equation for v_{f}.
+ $V_{f} = \sqrt {\dfrac{2W}{m}}$
+$$W = \dfrac{1}{2}S_{max}^2$$
-Initial  = S_{max} \longrightarrow max stretch
+Substitute work in the equation for $v_{f}$
-Final = 0 \longrightarrow relaxed
+$$V_{f} = \sqrt {\dfrac{2}{m}(\dfrac{1}{2}ks^2_{max})}$$
- $W_{F} = \int_{S_{max}}^{0} (-kx)(dx) = -\dfrac{1}{2}kx^2$
+Simplify down to:
-$$W_{F} = -\dfrac{1}{2}k(0)^2 - [-\dfrac{1}{2}ks^2_{max}] = \dfrac{1}{2}k(s)^2_{max}$$
+$$ = \sqrt {\dfrac{k}{m}}S_{max}$$
+Sub in values for known variables and solve for  $v_{f}$
+ $v_{f} = \sqrt {\dfrac{100N/m}{0.5kg}}  (3.5m)^2 \longrightarrow 50m/s = 111mph$