Question

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 the figure. 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 the particle was projected.

Answer 1

Answer:

$\: v = \frac{u \cos( \theta \: + \alpha)}{ \cos( \theta) }$

Explanation:

Let v be the speed of the box at the instant when the particle was projected.

Velocity components of the particle at the time of projection with respect to ground reference frame is attached. The velocities are resolved into vertical and horizontal component.

It is given that that particle has no horizaltal displacement w.r.t. ground frame. As such resultant initial velocity of the particle in horizontal direction must be zero. Hence

$v \cos( \theta) - u \cos( \theta + \alpha ) = 0$

$or \: v \cos( \theta) = u \cos( \theta + \alpha )$

$or \: v = \frac{u \cos( \theta \: + \alpha)}{ \cos( \theta) }$