A large heavy box is sliding without friction down a smooth inclined plane of inclination $θ$ .From a point P on bottom of box, a particle is projected inside the box. The initial speed of particle with respect to box is $u$ and direction of projection makes and angle $α$ with the bottom as shown in the figure
Find the distance along the bottom of box between point of projection P and point Q where the particle lands.(Assume that the particle does not hit any other surface of the box. Neglect air resistance)

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$P Q = \frac{u^{2} \sin 2 θ}{g \cos α}$

$P Q = \frac{u^{2} \sin 2 α}{g \cos θ}$

$P Q = \frac{u^{2} \sin 2 θ}{g \sin α}$

$P Q = \frac{u^{2} \sin 2 α}{g \sin θ}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Lets consider reference frame $x - y$ fixed with box with origin at one corner O and moving down with box. The acceleration of this frame down the incline is $g \sin θ$ .
Acceleration of the particle launched inside the box w.r.t ground is $g$ (vertically downwards) with components
a) $g \sin θ$ ;(along the negative x-axis) acting as retardation for particle’s motion along the incline
Question Image
$s_{y} = u_{y} t + \frac{1}{2} a_{y} t^{2} 0 = (u \sin α) t - \frac{1}{2} g \cos θ t^{2} t = \frac{2 u \sin α}{g \cos θ} P Q = (u \cos α) t = \frac{u^{2} \sin α}{g \cos θ}$