A uniform solid ball rolls smoothly along a floor, then up a ramp inclined at 15.0. It momentarily stops when it has rolled 1.50 malong the ramp. What was its initial speed?

$$\begin{gathered} \mathrm{\theta }=15.0^{°} \\ \mathrm{x}=1.50\mathrm{m} \end{gathered}$$

To find the initial velocity, use the law of conservation of energy. According to the law of conservation of energy, energy can neither be created, nor be destroyed.

Formula is as follow:

Conservation of mechanical energy,

${\mathbf{PE}}_{\mathbf{1}}\mathbf{+}{\mathbf{KE}}_{\mathbf{1}}\mathbf{=}{\mathbf{PE}}_{\mathbf{2}}\mathbf{+}{\mathbf{KE}}_{\mathbf{2}}$

Where, PE is potential energy and KEis kinetic energy.

Now, to find the initial speed, use conservation of energy. At height h, solid ball has only potential energy, whereas at the bottom, there is translational kinetic energy and rotational kinetic energy.

From the diagram,

$\begin{array}{rcl}\sin 15& =& \frac{\mathrm{h}}{1.5}\\ \mathrm{h}& =& 1.5\sin 15\\ \mathrm{h}& =& 0.3882\mathrm{m}\end{array}$

Now, the solid ball has inertia about center as given below,

$\mathrm{I}=\frac{2}{5}{\mathrm{mr}}^2$

So, conservation of energy,

$\begin{array}{rcl}\mathrm{mgh}& =& 0.5\times \mathrm{m}\times {\mathrm{v}}_{\mathrm{i}}^2+0.5\times \mathrm{I}\times {\mathrm{w}}_{\mathrm{i}}^2\\ \mathrm{mgh}& =& 0.5\times \mathrm{m}\times {\mathrm{v}}_{\mathrm{i}}^2+0.5\times \frac{2}{5}\times \mathrm{m}\times {\mathrm{r}}^2\times {\mathrm{w}}_{\mathrm{i}}^2\\ \mathrm{mgh}& =& \frac{7}{10}{\mathrm{mV}}_{\mathrm{i}}^2\\ \mathrm{gh}& =& \frac{7}{10}{\mathrm{V}}_{\mathrm{i}}^2\\ 9.81\times 0.3882& =& \frac{7}{10}{\mathrm{V}}_{\mathrm{i}}^2\end{array}$

So,

data-custom-editor="chemistry" ${\mathrm{V}}_{\mathrm{i}}=2.33\mathrm{m}/\mathrm{s}$

Hence,initial speed of the ball is data-custom-editor="chemistry" $\mathrm{V}=2.33\mathrm{m}/\mathrm{s}$.

Therefore, to find the initial velocity, conservation of energy can be used along with the formula for potential energy, linear kinetic energy, and rotational kinetic energy.