Question

A pinball launcher is a spring with spring constant k = 65.0 N/m. It is compressed a distance of 8.00 cm and a ball of mass 0.100 kg is placed in front of it. When the spring is released, the ball rolls up the incline of the pinball machine. Assuming that it does not hit anything and that no energy is lost to friction, what vertical height does the ball gain before coming momentarily to rest?

Answer 1

Answer:

The maximum height reached by the ball is 0.212 meters (21.2 centimeters).

Explanation:

In this case, the ball receives a potential energy from deformed spring and reaches its maximum height at the expense of such energy. By Principle of Energy Conservation, we define the phenomenon on the ball by the following model:

$U_{k,1}+U_{g,1} = U_{k,2}+U_{g,2}$ (1)

Where:

$U_{k,1}$, $U_{k,2}$ - Initial and final elastic potential energies, measured in joules.

$U_{g,1}$, $U_{g,2}$ - Initial and final gravitational potential energies, measured in joules.

By definitions of elastic and gravitational potential energies, we expand and simplify the equation above as follows:

$\frac{1}{2}\cdot k\cdot x^{2} = m\cdot g \cdot \Delta y$ (2)

Where:

$m$ - Mass of the ball, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second. .

$\Delta y$ - Height reached by the ball, measured in meters.

$k$ - Spring constant, measured in newtons per meter.

$x$ - Spring deformation, measured in meters.

If we know that $k = 65\,\frac{N}{m}$, $x = 0.08\,m$, $m = 0.1\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$, then the maximum height reached by the ball is:

$\Delta y = \frac{k\cdot x^{2}}{2\cdot m\cdot g}$ (2)

$\Delta y = \frac{\left(65\,\frac{N}{m} \right)\cdot \left(0.08\,m\right)^{2}}{2\cdot (0.1\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$\Delta y = 0.212\,m$

The maximum height reached by the ball is 0.212 meters (21.2 centimeters).