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 )

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

Tardigrade Admin · Accepted Answer

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 k g

and

g = 10 m / s^{2}

Kinetic energy of rotation,

K E = \frac{1}{2} \times 11 \times (10)^{2} (1 + \frac{\frac{2}{5} R ^{2}}{R ^{2}})

(

∵

For solid spherical ball,

k = \frac{2}{5} R

)

\Rightarrow= 11 \times 50 \times \frac{7}{5} J

By applying the law of energy conservation,

K E_{i} + v_{i} = K E_{f} + v_{f}

\Rightarrow 11 \times 50 \times \frac{7}{5} + 0 = 0 + m g h

\Rightarrow h = \frac{50}{10} \times \frac{7}{5} = 7 m