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.

Step 1: Analyze the acceleration of the box.

The box slides down the smooth incline of angle $\theta$ with an acceleration:

$$a_{\text{box}} = g \sin\theta$$

Step 2: Determine the effective acceleration of the particle relative to the box.

In the non-inertial frame of the box, a pseudo-acceleration of $g \sin\theta$ acts up the incline, which cancels the component of gravity parallel to the incline.

The remaining effective acceleration is perpendicular to the bottom of the box:

$$g' = g \cos\theta$$

Step 3: Calculate the time of flight of the particle.

Using the perpendicular component of the initial velocity, $u \sin\alpha$, the time of flight $T$ is:

$$T = \frac{2u \sin\alpha}{g'}$$ $$T = \frac{2u \sin\alpha}{g \cos\theta}$$

Step 4: Find the distance $PQ$ along the bottom of the box.

The velocity component parallel to the bottom of the box is $u \cos\alpha$ and remains constant in this frame.

$$PQ = (u \cos\alpha) T$$ $$PQ = \frac{2u^2 \sin\alpha \cos\alpha}{g \cos\theta}$$ $$PQ = \frac{u^2 \sin 2\alpha}{g \cos\theta}$$

Step 5: Set up the horizontal velocity of the particle in the ground frame.

Let $v$ be the speed of the box down the incline at the instant of projection.

The horizontal component of the particle's velocity relative to the ground is:

$$v_{Px} = u \cos(\alpha + \theta) - v \cos\theta$$

Step 6: Solve for the speed of the box $v$.

For the horizontal displacement of the particle to be zero, its horizontal velocity relative to the ground must be zero.

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

Answer:

(a) The distance along the bottom of the box is $PQ = \dfrac{u^2 \sin 2\alpha}{g \cos\theta}$.

(b) The speed of the box with respect to the ground is $v = \dfrac{u \cos(\alpha + \theta)}{\cos\theta}$.