  Question

# A body is projected up with a speed V along the line of greatest slope of an inclined plane of angle of inclination β. If the body collides elastically perpendicular to the inclined plane, find the time after which it crosses through its point of projection after collision ?  2Vg(sinβ)2Vg(tanβ)2Vg(√1+3sin2β)2Vg(√1+3tan2β)

Solution

## The correct option is C 2Vg(√1+3sin2β) Taking x axis along the inclined plane and y axis perpendicular to plane and resolving gravity in that direction On striking the plane perpendicularly, the velocity along x axis should be zero. So using vx=ux + axt we get  0 = Vcosα − gsinβt ⇒ t=Vcosαgsinβ. Also sy = 0 from A to B. So using sy = uyt+ ayt22 we get, 0 = 2V(sinα)t − g(cosβ)t22 ⇒ t = 2Vsinαgcosβ Equating t from above two equation , we get  Vcosαgsinβ = 2Vsinαgcosβ ⇒ tanα = cotβ2  ⇒ cosα = 2√4 + cot2β = 2sinβ√1 + 3 sin2β Putting value of cosα in t we get t = 2Vg√1 + 3sin2β  . Since collision is elastic at perpendicular to inclined plane , so it will trace back same path and will take same time t as mentioned above to get back to point of projection .  Suggest corrections   