Question

A solid spherical ball rolls on a horizontal surface at $10\, m / s$ and continues to roll up on an inclined surface as shown in the figure. If the mass of the ball is $11\, kg$ and frictional losses are negligible, the value of $h$ where the ball stop and starts rolling down the inclination is (Assume, $g=10\, m / s ^{2}$ )

• A. 8 m
• B. 6 m
• C. 7 m
• D. 10 m

Answer 1

Answer: (C)

For rolling motion, the rotational kinetic energy is given by expression $k$ $$ KE =\frac{1}{2} m v^{2}\left(1+\frac{k^{2}}{R^{2}}\right) $$
Given, rolling velocity on horizontal surface,
$v_{H}=10\, m / s$, mass of ball, $m=11\, kg$ and $g=10\, m / s ^{2}$
Kinetic energy of rotation,
$KE =\frac{1}{2} \times 11 \times(10)^{2}\left(1+\frac{\frac{2}{5} R^{2}}{R^{2}}\right)$
($\because$ For solid spherical ball, $k=\sqrt{\frac{2}{5}} R$)
$\Rightarrow =11 \times 50 \times \frac{7}{5} J$
By applying the law of energy conservation,
$KE _{i}+v_{i}= KE _{f}+v_{f}$
$\Rightarrow 11 \times 50 \times \frac{7}{5}+ 0 =0+ mgh$
$\Rightarrow h=\frac{50}{10} \times \frac{7}{5}=7\, m$