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 .

Options

A. `(u^(2) sin 2 alpha)/( g cos theta)`
B. `(u^(2) cos 2 alpha)/(g cos theta)`
C. `(u^(2) sin^(2) alpha)/(2g cos theta)`
D. `(u^(2) sin^(2) alpha)/(2g sin theta)`

Answer

A

Explanation

Take the x-axis along incline and y-axis perpendicular to the plane. `:.` Acceleration of particle `=gsinthetahati+gcosthetahatj` and acceleration of block `=gsinthetahati`
`:.` Acceleration of particle with respect to block
= acceleration of particle -acceleration of block
`=(gsinthetahati+gcosthetahatj)-(gsintheta)hati=gcosthetahatj`
Now motion of particle with respect to block will be a projectile as shown.
The only difference is, g will be replaced by g cos `theta`
`:.PQ="Range"(R)=(u^(2)sin2alpha)/(gcostheta)`
`PQ=(u^(2)sin2alpha)/(gcostheta)`