Question

A solid sphere of mass $ 2 \,kg $ rolls on a smooth horizontal surface at $ 10 \,m/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

• A. $ 700 \, cm $
• B. $ 701 \, cm $
• C. $ 7.1 \, m $
• D. None of these

Answer 1

Answer: (C)

Given that,
mass of solid sphere, $m = 2\, kg$
velocity, $v = 10 \,m/ s$

Let the sphere attained a height $h$
When the sphere is at point $A$, it possesses kinetic energy and rotational kinetic energy, and when it is at point $B$ it possesses only potential energy. So, from law of conservation of energy
$\frac{1}{2} mv^{2}+\frac{1}{2} I\omega^{2}=mgh$
or $\frac{1}{2}mv^{2}+\frac{1}{2}mk^{2}\times\frac{V^{2}}{R^{2}}=mgh$
or $ \frac{1}{2}mv^{2}\left[1+\frac{K^{2}}{R^{2}}\right]=mgh$
or $\frac{1}{2}\times\left(10\right)^{2} \left[1+\frac{2}{5}\right]=9.8\times h$
[for solid sphere, $\frac{K^{2}}{R^{2}}=\frac{2}{5}$]
or $\frac{1}{2}\times100\times\frac{7}{5}=9.8\,h$
or $ h=7.1\,m$