Q32P - University Physics with Modern Physics

Question

 You have just landed on Planet X. You release a $\mathbf{100}\mathbf{ }\mathbf{g}$ ball from rest from a height of $\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{m}$ and measures that it takes $\mathbf{3}\mathbf{.}\mathbf{40}\mathbf{ }\mathbf{s}$ to reach the ground. Ignore any force on the ball from the atmosphere of the planet. How much does the $\mathbf{100}\mathbf{ }\mathbf{g}$ ball weigh on the surface of Planet X?

Step-by-Step Solution

Answer

The weight of the ball on the surface of Planet X is $0.173 \mathrm{N}$.

Step 1: Identification of the given data:

The given data can be listed below as,

The mass of a ball is, $\mathrm{m} = 100 \mathrm{g}$.

The initial height of the ball is $\mathrm{h} = 10.0 \mathrm{m}$. 

The time taken by the ball to reach the ground is $\mathrm{t} = 3.40 \mathrm{s}$.

The initial velocity of the ball is $\mathrm{u} = 0$. (Since the ball starts from rest).

Step 2: Significance of free-fall of an object:

Whenever an object initially lies at a specific altitude and releases from the altitude without any initial speed, the object will start moving towards the Earth's surface. The fall of the object is only under the action of gravitational acceleration.

Step 3: Determination of the gravitational acceleration of the planet X:

The relation of the gravitational acceleration of the planet X is expressed as,

$\mathrm{h}=\mathrm{ut}+\frac{1}{2}{\mathrm{gt}}^2$

Here, $\mathrm{g}$ is the gravitational acceleration of planet X.

Substitute all the known values in the above equation.

$\begin{array}{l}\left(1000\mathrm{m}\right)=\left(0\right)3340\mathrm{s})+\frac{1}{2}(9340{\mathrm{ss}}^2)\\ \left(100\mathrm{m}\right)=+\frac{1}{2}\mathrm{g}(3.40{\mathrm{s}}^2\end{array}$

$$\begin{gathered} \mathrm{g} = \left[\frac{2\left(10.0 \mathrm{m}\right)}{{\left(3.40 \mathrm{s}\right)}^2}\right] \\ =1.73 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Step 4: Determination of the weight of the ball on the surface of Planet X:

The relation of weight of the ball on the surface of Planet X is expressed as,

$\mathrm{W} = \mathrm{mg}$

Here, $\mathrm{W}$ is the weight of the ball on the surface of Planet X.

Substitute all the known values in the above equation.

$$\begin{gathered} \mathrm{W} = \left[\left(100 \mathrm{g}\right)\times \left(\frac{{10}^{-3} \mathrm{kg}}{1 \mathrm{g}}\right)\left(1.73\mathrm{m}/{\mathrm{s}}^2\right)\right] \\ =0.173 \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2 \\ =\left(0.173 \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\right)\times \left(\frac{1\mathrm{N}}{1 \mathrm{kg}.\mathrm{m}/\mathrm{s}}\right) \\ =0.173 \mathrm{N} \end{gathered}$$

Thus, the weight of the ball on the surface of Planet X is $=0.173 \mathrm{N}$.