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` )

Options

A. 8 m
B. 6 m
C. 7 m
D. 10 m

Answer

C

Explanation

Key Idea for rolling motion, the rotational kinetic energy is given by expression k
`KE =1/2 mv^2 (1+(k^2)/(R ^2))`
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 =1/2 xx 11 xx (10)^2 (1+(2/5R^2)/(R^2))`
`(because " For solid spherical ball " , k = sqrt(2/5 R))`
`rArr = 11 xx 50 xx 7/5 J`
By applying the law of energy conservation,
`KE _i +V_i =KE _f +V_f`
`rArr 11 xx 50 xx 7/5 +0 =0 + mgh `
`rArr h = (50)/(10) xx 7/5 =7 m`