Chapter 11, Problem 14P

To determine

To Calculate: The speed of the ball when it leaves the gun.

Answer to Problem 14P

  $\text{9}\text{.43 m/s}$

Explanation of Solution

Given info:

  $\text{F = 80 N}$

Mass, $\text{m = 0}\text{.180 kg}$

Distance, $\text{x = 0}\text{.2 m}$

Formula used:

From the available information, first find the spring constant.

  $\text{k = }\frac{\text{F}}{\text{x}}$

When the ball is in motion, the elastic potential energy in the loaded position would become kinetic energy of the ball and it is described by the following formula,

  $\begin{array}{l}{\text{E}}_{\text{i}}{\text{ = E}}_{\text{f}}\\ \frac{\text{1}}{\text{2}}\times \text{k}\times {\text{x}}^{\text{2}}\text{ = }\frac{\text{1}}{\text{2}}\times \text{m}\times {\text{v}}_{\text{max}}{}^{\text{2}}\\ {\text{v}}_{\text{max}}{}^{\text{2}}\text{ = }\frac{\text{k}\times {\text{x}}^2}{\text{m}}\end{array}$

Calculation:

Substitute the given values,

  $\text{k = }\frac{\text{F}}{\text{x}}\text{ = }\frac{\text{80}}{\text{0}\text{.2}}\text{ = 400 N/m}$

  $\begin{array}{l}{\text{v}}_{\text{max}}{}^{\text{2}}\text{ = }\frac{\text{k}}{\text{m}}{\text{x}}_{\text{max}}{}^{\text{2}}\\ {\text{v}}_{\text{max}}{}^{\text{2}}\text{ = }\frac{\text{400}}{\text{0}\text{.180}}\text{ (0}{\text{.2)}}^{\text{2}}\\ {\text{v}}_{\text{max}}{}^{\text{2}}\text{ = 88}{\text{.889 (m/s)}}^{\text{2}}\\ {\text{v}}_{\text{max}}\text{ = 9}\text{.43 m/s}\end{array}$

Conclusion:

Thus, the speed of the ball when it is leaving the gun is $\text{9}\text{.43 m/s}$ .