Rubber Physics (Part 2)

By increasing the width w of the rubber band, the externally applied force is divided to a greater length, so that the rubber band is stretched less. Instead of a rubber band of width $w = 4\,\text{cm}$ you could also use 4 rubber bands of width $w = 1\,\text{cm}$ , which are arranged parallel to each other and each experienced only a quarter of the applied force. The general force-distance curve is therefore $f_n(w) = \frac{w}{w_1} f_1(x) = n \cdot f_1(x)$ with the force $f_1(x)$ of rubber band of width $w_1 = 1 \,\text{cm}$ and the ratio $n = w/w_1$ . These force curves are outlined in the diagram below for the cases $n = 1,2,3$ and 4. To find out the maximum tensile length $x_n$ of the rubber band when the band is loaded with the force $f_\text{max}$ , we draw the horizontal in the diagram and graphically determine the points of intersection with $f_n(x_n) = f_\text{max}$ . This yields: $\begin{aligned} x_1 &= 1\,\text{m}, \\ x_2 &\approx 0.855 \,\text{m}, \\ x_3 &\approx 0.743 \,\text{m}, \\ x_4 & \approx 0.588\,\text{m} \end{aligned}$ In order to determine the elastic energy stored in the rubber, we must integrate the force $f(x)$ over the displacement: $\begin{aligned} U_n(x) &= - \int_{0}^{x} f_n(x) dx = - n \int_{0}^{x} f_1(x) dx \\ &= n \left(75 \,\frac{\text{J}}{\text{m}^2}\cdot x^2 - 75 \,\frac{\text{J}}{\text{m}^4}\cdot x^4 + 60 \,\frac{\text{J}}{\text{m}^6}\cdot x^6 \right) \\ \Rightarrow \quad W_1 &= U_1(x_1) \approx 60 \,\text{J}, \\ W_2 &= U_2(x_2) \approx 76.3 \,\text{J}, \\ W_3 &= U_3(x_3) \approx 86\,\text{J}, \\ W_4 &= U_4(x_4) \approx 77.7\,\text{J} \end{aligned}$ Because the rubber band with $w = 3\,\text{cm}$ stores the largest elastic energy, the fired ball in this case receives the highest kinetic energy and therefore the highest speed.