At the moment `t=0` the force `F=at` is applied to a small body of mass `m` resting on a smooth horizontal plane (a is constant).
The permanent direction of this force forms an angle `alpha` with the horizontal (figure). Find:
(a) the velocity of the body at the moment of its breaking off the plane,
(b) the distance traversed by the body up to this moment.

Options

A. `((mg^(2)costheta)/(2c sin^(2)theta))m//s,((mg^(3)contheta)/(6c^(2)sin^(3)theta))m`
B. `((mg^(2)costheta)/(2c sin^(2)theta))m//s,((m^(2)g^(3)costheta)/(6c^(2)sin^(3)theta))m`
C. `((mgcostheta)/(2c sin^(2)theta))m//s,((m^(2)g^(3)sintheta)/(6c^(2)cos^(3)theta))m`
D. `((mg^(2)costheta)/(2c sin^(2)theta))m//s,((m^(2)g^(3)sintheta)/(6c^(2)cos^(3)theta))m`

Answer

B

Explanation

Since `F` `cos theta =ma` or ct `cos theta = m (dv)/(dt)`
`underset(0)overset(v)intdv=(c"cos"theta)/(m)underset(0)overset(t)inttdt or v=((c"cos"theta)/(2m))t^(2)`
But velocity at the time of breaking off or at
`t =(mg)/(csintheta)`
`:.v=(c costheta)/(2m)x((mg)/(csintheta))^(2)=(mg^(2)costheta)/(2csin^(2)theta`
Now, `v = (dS)/(dt)=(c"cos"theta)/(2m)t^(2)`
`underset(0)overset(t)inttdtorS=((c"cos"theta)/(6m))t^(3)`
So the distance traversed till the time of break off i.e in the time interval `t = (mg)/(c"sin"theta)` is given by
`S=(c "cos"theta)/(6m)((mg)/(c"sin"theta))^(3)=(m^(2)g^(2)"cos"theta)/(6c^(2)"sin"^(3)theta)` .