A perfectly elastic particle is projected with a velocity `v` on a vertical plane through the line of greatest slope of an inclined plane of elevation `alpha`.If after striking the plane, the particle rebounds vertically show that it will return to the point of projection at the end of time equal to

Options

A. `(6v)/(gsqrt(1+8sin^(2)alpha))`
B. `(6v)/(gsqrt(1+sin^(2)alpha))`
C. `v/(gsqrt(1+8sin^(2)alpha))`
D. `v/(gsqrt(1+sin^(2)alpha))`

Answer

A

Explanation

The path of the motion of particle is as shown in figure.
Let particle is projected at an angle `theta` with the plane.Its displacement along `y`-axis becomes zero in time `T`.Then we have
`y-u_(y)T+1/2a_(y)T^(2) or T=(2v sin theta)/(g cos alpha)`....(i)
Let `v` is the velocity with which particle strikes the plane `alpha` is the angle which it makes with this vertical.Then `alpha` is the angle which it makes with the vertical.
Then we have, `v sin alpha=v cos theta-g cos theta T`..(ii)
and `v cos alpha=v sin theta-g sin alphaxx(2v sin theta)/(g cos alpha)` or
`v=(v cos theta)/(sin alpha)-(2v sin theta)/(cos alpha)`...(iv)
Substituting the value of `T` in equation (ii) and (iii),
`v cos alpha=v sin theta-g cos alpha[(2v sin theta)/(g cos alpha)]`...(vi)
Dividing equation (v) by (vi), we have
`tan alpha=(cos theta2sin theta tan alpha)/(sin theta)=cot theta-2 tan alpha`
`:. cos theta=(3 tan alpha)/(sqrt(1+9 tan ^(2)) alpha)`
`sqrt(1+9 tan ^(2) alpha)`
Total time taken by particle is equal to the sum of time taken from `O` and `P` and `P` to `Q` and then from `Q` to `P` to `Q`.Thus total time
`=2T+2t=2(T+t)` For `t`, we have `0=v`
`-"gt" or t=v/g`
`:.` Total time `=2[(2v sin theta)/(g cos alpha)+v/g]`
`=2[(2v sin theta)/(g cos alpha)+1/g]((v cos theta)/(sin alpha)-(2v sin theta)/(cos alpha))]`
`=(2v cos theta)/(g sin alpha)=(2v[(3tan alpha)/(sqrt(1+9 tan^(2)alpha))]]/(g sin alpha)`
After solving, we get Total times `=(6v)/(gsqrt(1+8 sin ^(2) alpha))`
`T_(1)=(2v_(y))/a_(y)=(2v_(1)sin theta)/(g cos theta),T_(2)=(2v_(2))/( g cos theta)`