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Example: MIT Water Balloon Fight
During the spring semester at MIT, residents of the parallel buildings of the East Campus Dorms battle one another with large sling-shots made from surgical hose mounted to window frames. Water balloons (with a mass of about 0.5kg) are placed in a pouch attached to the hose, which is then stretched nearly the width of the room (about 3.5 meters). If the hose obeys Hooke's Law, with a spring constant of 100N/m, how fast is the balloon traveling when it leaves the dorm room window?
Facts
Mass of water balloons = 0.5kg
Hose stretched = 3.5m
Spring Constant = 100N/m
Initial State: Hose stretched 3.5m; no velocity
Final State: Hose relaxed; non zero v.
Lacking
Velocity of balloon leaving window.
Approximations & Assumptions
No energy is lost to other forms in the system.
Representations
$$\Delta K = W$$
$$K_E = \dfrac{1}{2}mv^2$$
$$ F = -kx$$
$$W_{F} = \sum_{i} \vec{F}_i\cdot\Delta\vec{r}_i = \int_{i}^{f} \vec{F} d\vec {r}$$
$$W = \dfrac{1}{2}mv_{f}^2 - \dfrac{1}{2}mv_{i}^2$$
System: Water balloon (model as a point particle)
Surroundings: Surgical hose slingshot
Solution
$$\Delta K_{balloon} = W_{hose}$$
$$K_{balloon,f} - K_{balloon,i} = W_{hose}$$
$$ \dfrac{1}{2}mv_{f}^2 - \dfrac{1}{2}mv_{i}^2 = W$$
$$ \dfrac{1}{2}mv_{i}^2 (0, v_{i} = 0)$$
$$ \dfrac{1}{2}mv_{f}^2 = W \longrightarrow v_{f}^2 = \dfrac{2W}{m}$$
If we calculate the work done by the hose, we can find $v_{f}$
$$v_{f} = \sqrt {\dfrac{2W}{m}}$$
The force of the hose is not constant
But remember: $$F = -kx$$
$$W_{F} = \sum_{i} \vec{F}_i\cdot\Delta\vec{r}_i = \int_{i}^{f} \vec{F} d\vec {r}$$
Initial = S_{max} \longrightarrow max stretch
Final = 0 \longrightarrow relaxed
$$W_{F} = \int_{S_{max}}^{0} (-kx)(dx) = \dfrac-{1}{2}kx^2 $$