The spring in the muzzle of a child’s spring gun has a spring constant of 700 N/m. To shoot a ball from the gun, first, the spring is compressed and then the ball is placed on it. The gun’s trigger then releases the spring, which pushes the ball through the muzzle. The ball leaves the spring just as it leaves the outer end of the muzzle. When the gun is inclined upward by 30^oto the horizontal, a 57 gball is shot to a maximum height of 1.83 mabove the gun’s muzzle. Assume air drag on the ball is negligible. (a) At what speed does the spring launch the ball? (b) Assuming that friction on the ball within the gun can be neglected, find the spring’s initial compression distance.

The spring constant of the spring is, k = 700 N/m

The angle of inclination is, $\mathrm{\theta }=30°$

The mass of the ball is,$\mathrm{m}=57\mathrm{g}\left(\frac{{10}^{-3}\mathrm{kg}}{1\mathrm{g}}\right)=5.7\times {10}^{-2}\mathrm{kg}$

The maximum height of the muzzle is, h = 1.83 m

Initial velocity can be calculated by using a kinematic equation for motion under gravity. The spring’s compression can be calculated by using the law of conservation of energy.

Formulae:

The third equation of the kinematic motion,${\mathrm{v}}_{\mathrm{f}}^2={\mathrm{v}}_0^2-2\mathrm{gy}$ (1)

Applying the law of conservation of energy,

Initial total energy = Final total energy (2)

Using equation (1) for the case of launching the ball, we can get the vertical speed of the ball as given: $\left({\mathrm{v}}_{\mathrm{f}}=0\mathrm{m}/\mathrm{s}\right)$

$$\begin{gathered} 0\mathrm{m}/\mathrm{s}={\mathrm{v}}_{0\mathrm{y}}^2-\left(2\times 9.8\mathrm{m}/{\mathrm{s}}^2\times 1.83\mathrm{m}\right) \\ {\mathrm{v}}_{0\mathrm{y}}^2=35.87{\mathrm{m}}^2/{\mathrm{s}}^2 \\ {\mathrm{v}}_{0\mathrm{y}}^2=6\mathrm{m}/\mathrm{s} \end{gathered}$$

We have,

The vertical component of initial velocity will give the launching speed that is given as:

$$\begin{gathered} {\mathrm{v}}_{0\mathrm{y}}={\mathrm{v}}_0\sin \left(\mathrm{\theta }\right) \\ 6\mathrm{m}/\mathrm{s}={\mathrm{v}}_0\sin \left(30°\right) \\ {\mathrm{v}}_0=12\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the launching speed of the ball is 12 m/s .

Using equation (2), we can get the compressed distance as given:

$$\begin{gathered} \frac{1}{2}{\mathrm{kd}}^2=\frac{1}{2}{\mathrm{mv}}_0^2+\mathrm{mgdsin}\left(\mathrm{\theta }\right) \\ \frac{1}{2}\times 700\mathrm{N}/\mathrm{m}\times \left({\mathrm{d}}^2\right)=\frac{1}{2}\times \left(5.7\times {10}^{-2}\mathrm{kg}\right)\times {\left(12\mathrm{m}/\mathrm{s}\right)}^2+\left(5.7\times {10}^{-2}\mathrm{kg}\right)\times 9.8\mathrm{m}/{\mathrm{s}}^2\times \mathrm{d}\times \sin \left(30°\right) \\ \left(350\mathrm{N}/\mathrm{m}\right){\mathrm{d}}^2=4.104\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2+\left(0.2793\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\right)\mathrm{d} \\ \left(350\mathrm{N}/\mathrm{m}\right){\mathrm{d}}^2-\left(0.2793\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\right)\mathrm{d}-4.104\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2=0 \end{gathered}$$

By solving the above quadratic equation for d: we get,

d = 0.11 m

Hence, the value of the distance of compression is 0.11 m .