Question

A solid sphere of mass $2kg$ rolls on a smooth horizontal surface at $10m/s$ . 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. $700cm$
• B. $701cm$
• C. $7.1m$
• D. None of these

Answer 1

Answer: (C)

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

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
$\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2=mgh$
or $\frac{1}{2}m{v}^2+\frac{1}{2}m{k}^2\times \frac{V^2}{R^2}=mgh$
or $\frac{1}{2}m{v}^2\left[1+\frac{K^2}{R^2}\right]=mgh$
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 $\frac{1}{2}\times 100\times \frac{7}{5}=9.8h$
or $h=7.1m$