Answer:
13.759 % of the initial mechanical energy is lost as thermal energy.
Explanation:
By the First Law of Thermodynamics we know that increase in internal energy of the object ([tex]U[/tex]), in joules, is equal to the lost amount of the change in gravitational potential energy ([tex]U_{g}[/tex]), in joules:
[tex]\frac{x}{100} \cdot \Delta U_{g} = \Delta U[/tex] (1)
Where [tex]x[/tex] is the percentage of the energy loss, no unit.
By definition of the gravitational potential energy and internal energy, we expand this equation:
[tex]\frac{x\cdot m \cdot g \cdot h}{100} = m\cdot c\cdot \Delta T[/tex] (1b)
Where:
[tex]m[/tex] - Mass of the object, in kilograms.
[tex]g[/tex] - Gravitational acceleration, in meters per square second.
[tex]h[/tex] - Initial height of the object above the lunar ground, in meters.
[tex]c[/tex] - Specific heat of aluminium, in joules per degree Celsius-kilogram.
[tex]\Delta T[/tex] - Temperature increase due to collision, in degree Celsius.
If we know that [tex]m = 0.95\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]h = 75\,m[/tex], [tex]c = 920\,\frac{J}{kg\cdot ^{\circ}C}[/tex] and [tex]\Delta T = 0.11\,^{\circ}C[/tex], then the percentage of energy loss due to collision is:
[tex]x = \frac{100\cdot c\cdot \Delta T}{g\cdot h}[/tex]
[tex]x = \frac{100\cdot \left(920\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (0.11\,^{\circ}C)}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot (75\,m)}[/tex]
[tex]x = 13.759\,\%[/tex]
13.759 % of the initial mechanical energy is lost as thermal energy.
