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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 |
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Answer» `(6v)/(gsqrt(1+8sin^(2)alpha))` 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)`  
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