Two particles at a distance `5 m` apart, are thrown towards each other on an inclined smooth plane with equal speed 'v'. Inclined plane is inclined at an angle of `30^@` with the horizontal. It is known that both particle move along the same straight line. The particle colllide at the point from where the lower is thrown. Find the value of `["take " g = 10 m//s^2]`.

Options

A. `1 m//s`
B. `1.5 m//s`
C. `2 m//s`
D. `2.5 m//s`

Answer

D

Explanation

Down the plane,
`5 = v.t + (1)/(2) (g sin theta) t^2`
At the plane, `0 = v - g sin theta t_1 rArr t_1 = (v)/(g sin theta)`
`t = 2 t_1 = (2v)/(g sin theta)`
[time taken by lower particle coming back to initial position]
`5 = (2.v^2)/(g sin theta) +(1)/(2)(g sin theta. 4 v^2)/(g^2 sin^2 theta)`
`10 g sin theta = 8 v^2`
`v = sqrt((10 xx 10 xx(1)/(2))/(8)) = sqrt((100)/(16)) = (10)/(4) = 2.5 m//s`
Alternate sol.
`2vt = 5`
`(2v)/(g sin theta) = t rArr 2v xx (2v)/(g sin theta) = 5`
`rArr v = sqrt((5 g sin theta)/(4)) =(5)/(2) m//s`.