A large , heavy box is sliding without friction down a smooth plane of inclination `theta` . From a point `P` on the bottom of the box , a particle is projected inside the box . The initial speed of the particle with respect to the box is `u` , and the direction of projection makes an angle `alpha` with the bottom as shown in Figure .
(a) Find the distance along the bottom of the box between the point of projection `p` and the point `Q` where the particle lands . ( Assume that the particle does not hit any other surface of the box . Neglect air resistance .)
(b) If the horizontal displacement of the particle as seen by an observer on the ground is zero , find the speed of the box with respect to the ground at the instant when particle was projected .

Answer

(a) `(u^(2)sin 2 alpha)/(g cos theta)`, (b) `v = (v cos (alpha+theta))/(cos theta)`

Explanation

(a) `u_(x) = ucos alpha, u_(x) = usin alpha`
`a_(x) = 0 a_(y) = g cos theta`
Time of Flight `T = (2usin alpha)/(g cos theta)`

`R = u cos alpha T`
`R = (u cosalpha.2u sin alpha)/(g cos theta) rArr R = (u^(2)sin 2 alpha)/(g cos theta)`
(b) `vecu_(PB) = vecu_(P) - vecu_(B) rArr vecu_(P) = vecu +vecv_(B)`
For horizontal displacement w.r.t. ground `0`.
`u_(p(x)) = u_(x) +u_(B(x))`
`0 = ucos (theta +alpha) - u_(B) cos theta`
`u_(B) = (ucos (theta +alpha))/(cos theta)`