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?

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.

Velocity of balloon leaving window.

No energy is lost to other forms in the system.

$$\Delta K = W$$

$$K_E = \dfrac{1}{2}mv^2$$

$$F = -kx$$

$$W_{F} = \sum_{i} \vec{F}_i\cdot\Delta\vec{r}_i = \int_{i}^{f} \vec{F} d\vec {r}$$

$$W = \dfrac{1}{2}mv_{f}^2 - \dfrac{1}{2}mv_{i}^2$$

System: Water balloon (model as a point particle)

Surroundings: Surgical hose slingshot

course_planning

$$\Delta K_{balloon} = W_{hose}$$

$$K_{balloon,f} - K_{balloon,i} = W_{hose}$$

$$\dfrac{1}{2}mv_{f}^2 - \dfrac{1}{2}mv_{i}^2 = W$$

$$\dfrac{1}{2}mv_{i}^2 (0, v_{i} = 0)$$

$$\dfrac{1}{2}mv_{f}^2 = W \longrightarrow v_{f}^2 = \dfrac{2W}{m}$$

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

$$W_{F} = \int_{S_{max}}^{0} (-kx)(dx) = -\dfrac{1}{2}kx^2$$

$$W_{F} = -\dfrac{1}{2}k(0)^2 - [-\dfrac{1}{2}ks^2_{max}] = \dfrac{1}{2}k(s)^2_{max}$$

$$W_{f} > 0$$ makes sense because $$\Delta K > 0$$

Work back

$$V_{f} = \sqrt {\dfrac{2W}{m}}$$ ; $$W = \dfrac{1}{2}S_{max}^2$$

$$V_{f} = \sqrt {\dfrac{2}{m}(\dfrac{1}{2}ks^2_{max})}$$

$$= \sqrt {\dfrac{k}{m}}S_{max}$$