Q71P - University Physics with Modern Physics

Question

A certain volcano on earth can eject rocks vertically to a maximum height H. (a) How high (in terms of H) would these rocks go if a volcano on Mars ejected them with the same initial velocity? The acceleration due to gravity on Mars is 3.71 m/s2; ignore air resistance on both planets. (b) If the rocks are in the air for a time T on earth, for how long (in terms of T) would they be in the air on Mars?

Step-by-Step Solution

Answer

(a) So the volcano can throw 2.64 times the height on the earth.

(b) So the rock will stay 2.64 times time on earth.

Step 1: identification of given data

${\mathrm{g}}_{\mathrm{m}}=3.71\mathrm{m}/{\mathrm{s}}^2$

Step 2: Calculation of height

Let us consider mass of a stone is m. So the energy of the stone with which it was thrown is

 ${\mathrm{E}}_{\mathrm{p}}=\mathrm{mgH}$

Where, $\mathrm{g}=9.8\mathrm{m}/{\mathrm{s}}^2$

Now if the same stone is thrown in the mars, if it reaches the height H_m then the potential energy is

 ${\mathrm{E}}_{\mathrm{p}.\mathrm{m}}={\mathrm{mg}}_{\mathrm{m}}{\mathrm{H}}_{\mathrm{m}}$

Now equating the above energies, we can write that

$$\begin{gathered} \mathrm{mgH}={\mathrm{mg}}_{\mathrm{m}}{\mathrm{H}}_{\mathrm{m}} \\ \frac{\mathrm{g}}{{\mathrm{g}}_{\mathrm{m}}}=\frac{{\mathrm{H}}_{\mathrm{m}}}{\mathrm{H}} \\ {\mathrm{H}}_{\mathrm{m}}=\frac{\mathrm{g}}{{\mathrm{g}}_{\mathrm{m}}}\mathrm{H} \end{gathered}$$ 

Now putting the given values,

$$\begin{gathered} {\mathrm{H}}_{\mathrm{m}}=\frac{9.8}{3.71}\mathrm{H} \\ {\mathrm{H}}_{\mathrm{m}}=2.64 \mathrm{H} \end{gathered}$$  

Step 3: Calculation of time

Consider the time taken for the rock to reach the top is T.

By the symmetry it becomes $\frac{\mathrm{T}}{2}$

Now using kinematic equation 

 $$\begin{gathered} \mathrm{v}=\mathrm{u}+\mathrm{at} \\ \mathrm{v}=0 \end{gathered}$$

Also

$$\begin{gathered} \mathrm{v}=\mathrm{u}+\mathrm{g}\frac{\mathrm{T}}{2} \\ {\mathrm{v}}_{\mathrm{m}}=\mathrm{u}+{\mathrm{g}}_{\mathrm{m}}\frac{{\mathrm{T}}_{\mathrm{m}}}{2} \end{gathered}$$ 

Equating both equation

 $$\begin{gathered} \mathrm{g}\frac{\mathrm{T}}{2}={\mathrm{g}}_{\mathrm{m}}\frac{{\mathrm{T}}_{\mathrm{m}}}{2} \\ {\mathrm{T}}_{\mathrm{m}}=2.64\mathrm{T} \end{gathered}$$