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A solid sphere of mass $2 \mathrm{~kg}$ rolls on a smooth horizontal surface at $10 \mathrm{~m} / \mathrm{s}$. It then rolls up a smooth inclined plane of inclination $30^{\circ}$ with the horizontal. The height attained by the sphere before it stops is [take $g=10 \mathrm{~m} / \mathrm{s}^2$ ]
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The correct answer is:
$7.0 \mathrm{~m}$
The given situation is shown in the following figure
If $h$ be height attained by the sphere before it stopped, then according conservation of energy,
$K_{\text {rot }}+K_{\text {Trans }}=m g h$
$\Rightarrow \quad \frac{1}{2} I \omega^2+\frac{1}{2} m v_{\mathrm{CM}}^2=m g h$
$\Rightarrow \quad \frac{1}{2} \times\left(\frac{2}{5} m R^2\right)\left(\frac{v_{\mathrm{CM}}}{R}\right)^2+\frac{1}{2} m v_{\mathrm{CM}}^2=m g h$
$\left[\therefore I=\frac{2}{5} m R^2 \Rightarrow \omega=\frac{v_{\mathrm{CM}}}{R}\right]$
$\Rightarrow \quad \frac{v_{\mathrm{CM}}^2}{5}+\frac{v_{\mathrm{CM}}^2}{2}=g h \Rightarrow \frac{10^2}{5}+\frac{10^2}{2}=10 \times h$
$\Rightarrow \quad 20+50=10 h \Rightarrow 70=10 h$
$\Rightarrow \quad h=\frac{70}{10}=7 \mathrm{~m}$
If $h$ be height attained by the sphere before it stopped, then according conservation of energy,
$K_{\text {rot }}+K_{\text {Trans }}=m g h$
$\Rightarrow \quad \frac{1}{2} I \omega^2+\frac{1}{2} m v_{\mathrm{CM}}^2=m g h$
$\Rightarrow \quad \frac{1}{2} \times\left(\frac{2}{5} m R^2\right)\left(\frac{v_{\mathrm{CM}}}{R}\right)^2+\frac{1}{2} m v_{\mathrm{CM}}^2=m g h$
$\left[\therefore I=\frac{2}{5} m R^2 \Rightarrow \omega=\frac{v_{\mathrm{CM}}}{R}\right]$
$\Rightarrow \quad \frac{v_{\mathrm{CM}}^2}{5}+\frac{v_{\mathrm{CM}}^2}{2}=g h \Rightarrow \frac{10^2}{5}+\frac{10^2}{2}=10 \times h$
$\Rightarrow \quad 20+50=10 h \Rightarrow 70=10 h$
$\Rightarrow \quad h=\frac{70}{10}=7 \mathrm{~m}$
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