A solid sphere is projected up along an inclined plane of inclination `30^(@)` with the horizontal with a speed of `4ms^(-1)` . If it rolls without slipping , the maximum distance traversed by it is . .. . . . `(g = 10 ms^(-2))`

Options

A. `2 . 24 m`
B. 112 m
C. 1 . 12 m
D. 22.4 m

Answer

A

Explanation

Given, v = `4ms^(-1) Also v = r omega`
Hence,
KE at bottom
`= K_("transletion") + K_("rotation")`
`=(1)/(2) mv^(2) + (1)/(2) Iomega^(2)`
`= (1)/(2) mv^(2) + (1)/(2) [(2)/(5) mr^(2) (v^(2))/(r^(2))]`
`= (1)/(2) mv^(2) + (1)/(2) xx (2)/(5) mv^(2)`
`= (1)/(2) ((7)/(5)mv^(2))`
At hight h, energy conservation gives
`(1)/(2) ((7)/(5)) mv^(2) = mgh`
`rArr h = (7)/(10) xx (v^(2))/(g)`
`= (7)/(10) xx (16)/(10)`
= 1 . 12m
As, ` l = h// sin 30^(@)`
`rArr l = 2 xx h`
= 224 m