A small solid ball is released from rest while fully submerged in a liquid and then its kinetic energy is measured when it has moved $\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$in the liquid. Figure 14-40 gives the results after many liquids are used: The kinetic energy is plotted versus the liquid density, and sets the scale on the vertical axis. (a)What is the density?(b) What is the volume of the ball?

a) Height of ball in the liquid, $\mathrm{h}=4.0\mathrm{cm}$

b) Kinetic energy in the liquid, ${\mathrm{K}}_{\mathrm{s}}=1.6\mathrm{J}$

When the liquid and the object to be submerged are of the same density, then it will never accelerate inside the liquid. Thus, the point corresponding to zero kinetic energy on the graph must be the point where the density of the object is equal to the density of the liquid. With the help of the distance moved by the ball, we can find the kinetic energy and volume of the ball.

Formula:

The third equation of motion,${\mathrm{v}}^2={\mathrm{v}}_0^2-2\mathrm{ah}$(i)

Kinetic energy of a body,$\mathrm{KE}=\frac{1}{2}\times \mathrm{m}\times {\mathrm{V}}^2$(ii)

Density of a body, $\mathrm{Density}=\frac{\mathrm{Mass}}{\mathrm{Volume}}$ (iii)

As we know, there is no acceleration when the object and liquid have the same density. So the point showing zero kinetic energy suggests that point where the density of the object must equal the density of the liquid.

${\text{ρ}}_{\mathrm{liq}}={\mathrm{\rho }}_{\mathrm{ball}}$

$=1.5\mathrm{g}/{\mathrm{cm}}^3$

$=1500\raisebox{1ex}{$\mathrm{kg}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{m}}^3$}\right.$

Hence, the density of the ball is$1500\raisebox{1ex}{$\mathrm{kg}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{m}}^3$}\right.$

Now, the ball is falling from height from rest position; then the velocity using equation (i) is given by:

${\mathrm{v}}^2=0+2\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)(0.04\mathrm{s})$

$=0.784\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$

Now, substituting the value of ${\mathrm{v}}^2$and the given values in equation (ii), we get

$1.6=\frac{1}{2}\times \mathrm{m}\times 0.784$

$\mathrm{m}=4.082\mathrm{kg}$

Again, using equation (iii), we can get the volume as follows:

$\text{Volume}=\frac{4.082\mathrm{kg}}{1500\mathrm{kg}/{\mathrm{m}}^3}$

$=2.72\times {10}^{-3}{\mathrm{m}}^3$

Hence, the volume of the ball is $2.72\times {10}^{-3}{\mathrm{m}}^3$.