(Figure (a))applies to the spring in a cork gun (Figure (b)); it shows the spring force as a function of the stretch or compression of the spring. The spring is compressed by 5.5 cm and used to propel a 3.8 g cork from the gun. (a) What is the speed of the cork if it is released as the spring passes through its relaxed position? (b) Suppose, instead, that the cork sticks to the spring and stretches it 1.5 cm before separation occurs. What now is the speed of the cork at the time of release?

1. The compression of the spring is,$x=5.5\mathrm{cm}=0.055\mathrm{m}$
2. The mass of the cork is,$m=3.8\mathrm{g}=3.8\times {10}^{-3}\mathrm{kg}$
3. The spring is stretched through a distance of $d=1.5\mathrm{cm}=1.5\times {10}^{-2}\mathrm{m}$

Use the concept of energy conservation law and elastic potential energy of the spring. According to the law of energy conservation, energy can neither be created, nor be destroyed.

Formulae:

$$\begin{gathered} U\left(x\right)=\frac{1}{2}k{x}^2 \\ K=\frac{1}{2}m{v}^2 \end{gathered}$$

where, K is kinetic energy, U(x)is potential energy, m is mass, v is velocity, x is displacement and kis spring constant.

From the slope of the graph, find the value of the force constant,

$$\begin{gathered} k=\frac{∆F}{∆x} \\ k=\frac{0.4-0.2}{4-2} \\ k=0.10\mathrm{N}/\mathrm{cm} \\ k=10\mathrm{N}/\mathrm{m} \end{gathered}$$

Now, the speed of the cork when the spring passes through its relaxed position can be find as follow:

When the cork is released, the potential energy of the compressed spring is converted to kinetic energy of the cork.

$$\begin{gathered} \frac{1}{2}k{x}^2=\frac{1}{2}m{v}^2 \\ {v}^2=\frac{k{x}^2}{m} \\ v=\sqrt{\frac{k}{m}}x \\ v=\sqrt{\frac{10\mathrm{N}/\mathrm{m}}{0.0038\mathrm{kg}}}\times 0.055\mathrm{m} \\ v=2.8\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed of the cork when the spring passes through its relaxed position is, $v=2.8\mathrm{m}/\mathrm{s}$.

In this case, the spring stretches by d. Hence, potential energy of the compressed spring is converted into kinetic energy and potential energy of the cork,

$$\begin{gathered} \frac{1}{2}k{x}^2=\frac{1}{2}m{v}^2+\frac{1}{2}k{d}^2 \\ \frac{1}{2}m{v}^2=\frac{1}{2}k{x}^2-\frac{1}{2}k{d}^2 \\ m{v}^2=k\left(x^2-d^2\right) \\ v=\sqrt{\frac{k\left(x^2-d^2\right)}{m}} \\ v=\sqrt{\frac{10\mathrm{N}/\mathrm{m}\left[{\left(0.055\mathrm{m}\right)}^2-{\left(0.015\mathrm{m}\right)}^2\right]}{3.8\times {10}^{-3}\mathrm{kg}}} \\ v=2.7\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed of the cork at the time of release is, $v=2.7\mathrm{m}/\mathrm{s}$.

Therefore, thevelocity of the cork can be found by using the conservation of energy.