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Question: A large heavy box is sliding without friction down a smooth plane of inclination θ. From a point P o...

A large heavy box is sliding without friction down a smooth plane of inclination θ. From a point P on the bottom of the box, a particle is projected inside the box. The initial speed of particle with respect to the box is u and the direction of projection makes an angle a with the bottom as shown. Find the distance along the bottom of box between the 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)

A

u2sin⁡2αg\frac { \mathrm { u } ^ { 2 } \sin 2 \alpha } { \mathrm { g } }

B

u2sin⁡2α2gcos⁡θ\frac { u ^ { 2 } \sin ^ { 2 } \alpha } { 2 g \cos \theta }

C

u2sin⁡2αgcos⁡θ\frac { \mathrm { u } ^ { 2 } \sin 2 \alpha } { \mathrm { g } \cos \theta }

D

u2sin⁡αg\frac { u ^ { 2 } \sin \alpha } { g }

Answer

u2sin⁡2αgcos⁡θ\frac { \mathrm { u } ^ { 2 } \sin 2 \alpha } { \mathrm { g } \cos \theta }

Explanation

Solution

Acceleration of particle w.r.t. block = Acceleration of particle – acceleration of block

= (gsin⁡θi^+gcos⁡θj^)−gsin⁡θi^( g \sin \theta \hat { \mathrm { i } } + g \cos \theta \hat { \mathrm { j } } ) - \mathrm { g } \sin \theta \hat { \mathrm { i } } = g cos θ j^\hat { j }

Motion of particle with reference to block is parabolic

∴ PQ = range = u2sin⁡2αgcos⁡θ\frac { \mathrm { u } ^ { 2 } \sin 2 \alpha } { \mathrm { g } \cos \theta }