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, B, C

Explanation

(a) `u` is the relative velocity of the particle with respect to the box . Resolve `u` .
` u_(x) is the relative velocity of particle with respect to the box in ` x- direction` .
`u_(y)` is the relative velocity with respect to the box in ` y- direction ` .
Since , there is no velocity of the box in the ` y- direction` , therefore this is the vertical velocity of the particle with respect to ground also.
`y`- direction motion (Taking relative terms w.r.t. box)
`u_(y) = + u sin alpha`
` a_(y) = - g cos theta`
`s_(y) = 0 ( activity is taken till the time the particles comes back to the box .)
` t_(y) = t `
` s_(y) = u_(y)t + 1/2 a_(y) t^(2) rArr 0 = ( u sin alpha )t -1/2 g cos theta xx t^(2) `
`rArr t = 0 or t = (2 u sin alpha )/(g cos theta) `
`X` - direction motion ( Taking relative terms w.r.t. box)
`u_(x) = + u cos alpha , a_(x) = 0 , t_(x) = t , s_(x) = s_(x)`
`s_(x) = u_(x) t + (1)/(2) a_(x) t^(2) rArr s_(x) = u cos alpha xx ( 2u sin alpha)/( g cos theta) = (u^(2) sin 2 alpha)/( g cos theta)`
(b) For the observer ( on ground ) to see the horizontal displacement to be zero , the distance travelled by the box in time `(( 2u sin alpha )/(g cos theta))` should be equal to the range of the particle .
Let the speed of the box at the time of projection of particle be ` U` . Then for the motion of box with respect to ground.
` u_(x) = -U , a_(x) = -g sin theta , t_(y) = ( 2u sin alpha)/( g cos theta ), s_(x) = ( -u^(2) sin 2 alpha )/ ( gcos theta)`
` s_(x) = u_(x)t + 1/2 a_(x)t^(2) `
` -(u^(2) sin 2alpha)/( g cos theta ) = -U( ( 2u sin alpha )/(gcos theta)) -1/2 g sin theta (( 2u sin alpha)/( g cos theta))^(2)`
On solving we get ` U = ( ucos (alpha + theta ))/(cos theta)`