Question

A solid sphere is projected up along an inclined plane from its bottom with a velocity $2.8m{s}^{-1}$ so that it rolls upward. If the inclined plane makes an angle of ${30}^o$ with the horizontal, the sphere can roll to a maximum height of : (Take $g=10m/s^2$)

• A. $0.35m$
• B. $0.55m$
• C. $0.7m$
• D. $0.84m$

Answer 1

The correct option is B $0.55m$

we know that,

loss in gravitational potential energy = gain in rotational kinetic energy +gain in translational kinetic energy

so,

$KE=\frac{1}{2}m{V}^2+\frac{1}{2}\times \frac{2}{5}m{V}^2$

$=\frac{1}{2}\times m{V}^2\times \frac{7}{5}$
which is equal to work done by gravity
$mgH=\frac{1}{2}\times \frac{7}{5}\times m{V}^2$

$H=\frac{7}{10\times 10}\times (2.8{)}^2$

$=0.5488m$