NTA Abhyas JEE Main2020PhysicsMotion in Two DimensionsPractice
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 the particle with respect to the box is u and the direction of projection makes an angle α with the bottom as shown in the figure: Find the distance along the bottom of the box between the point of projection P and the point Q where
Options
- Au 2 sin 2 α g
- Bu sin 2 α g cos θ
- Cu 2 sin α g
- Du 2 sin 2 α g cos θ
Correct answer
D. u 2 sin 2 α g cos θ
Step-by-step solution
u is the relative velocity of the particle with respect to the box. Resolve u . u x is the relative velocity of the particle with respect to the box in x – direction. u y is the relative velocity of the particle 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 α a y = − g cos θ S y = 0 (activity is taken till t