A large heavy box is sliding without friction down a smooth plane of inclination theta. From a point

Q: Why do we use relative acceleration in part (a)?

The box is accelerating, making it a non-inertial frame. To use standard kinematics equations (like the projectile range formula) inside the box, we must account for this by finding the particle's acceleration *relative* to the box.

Q: Why does the effective gravity become $g \cos \theta$?

The particle falls with absolute acceleration $g$ downwards. The box slides with acceleration $g \sin \theta$ down the incline. When you subtract the box's acceleration from the particle's, the components parallel to the incline cancel out exactly! You are left only with the perpendicular component, $g \cos \theta$.

Q: In part (b), why does zero horizontal displacement mean zero horizontal velocity?

The particle is in free fall, meaning the only force acting on it is gravity (which is purely vertical). Since there is no horizontal force, there is no horizontal acceleration. A constant horizontal velocity that results in zero displacement must be exactly zero.

Q: Why is the horizontal velocity of the box $-v \cos \theta$?

The box is sliding *down* the incline. If we take the positive x-axis to the right, the box is moving to the left and downwards. The angle of its velocity vector is $\theta$ below the negative x-axis, giving a horizontal component of $-v \cos \theta$.

Question

A large heavy box is sliding without friction down a smooth plane of inclination $	heta$. 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.
(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 the particle was projected.

Sigma Percentile · Accepted Answer

### The Setup: A Box on a Slide

Imagine standing on a smooth, frictionless inclined plane tilted at an angle $	heta$. A heavy box is sliding down this plane. Inside this moving box, a tiny particle is projected from the bottom corner $P$ towards the top corner $Q$ with an initial speed $u$ at an angle $\alpha$ relative to the floor of the box.

This setup might seem incredibly complex at first glance because both the box and the particle are moving simultaneously. However, physics offers us a beautiful tool to simplify such problems: the concept of relative frames of reference.

### The Magic of Relative Frames

This is where the true elegance of physics shines. Instead of trying to track the particle from the stationary ground, let's mentally step inside the sliding box.

The box is sliding down a smooth incline, meaning its acceleration is purely due to gravity along the plane: $a_B = g \sin 	heta$. Meanwhile, the particle, once projected, is in free fall. Its absolute acceleration is simply $g$ acting vertically downwards.

To find the particle's acceleration *relative* to the box, we subtract the box's acceleration from the particle's absolute acceleration: $$\vec{a}_{pB} = \vec{a}_p - \vec{a}_B$$

If we resolve the true gravity $g$ into components parallel and perpendicular to the incline, we get $g \sin 	heta$ down the plane and $g \cos 	heta$ perpendicular to it. Notice what happens when we subtract the box's acceleration! The parallel component ($g \sin 	heta$) cancels out entirely.

We are left with only the perpendicular component: $g \cos 	heta$. This means that inside the box, the particle feels a constant "effective gravity" pulling it straight towards the floor. The problem has magically transformed into a standard projectile motion over a flat surface!

Using the standard range formula $R = \frac{u^2 \sin 2\alpha}{g_{eff}}$, we substitute our effective gravity to find the distance $PQ$:

$$PQ = \frac{u^2 \sin 2\alpha}{g \cos 	heta}$$

### The Zero Displacement Mystery

Moving to the second part of the problem, we are given a fascinating condition: an observer on the ground sees the particle's horizontal displacement as exactly zero.

What does this imply? Since the particle is in free fall, the only force acting on it is gravity, which is purely vertical. There is absolutely no horizontal force, and therefore, no horizontal acceleration. If a particle has zero horizontal acceleration and its horizontal displacement is always zero, its horizontal velocity must be exactly zero at all times.

### The Final Calculation

To enforce this condition, we need to find the absolute horizontal velocity of the particle and set it to zero. The absolute velocity is the vector sum of the particle's velocity relative to the box and the box's velocity relative to the ground.

Let the box's speed down the incline be $v$. Since it moves down an incline of angle $	heta$, its horizontal velocity component is $-v \cos 	heta$ (assuming the positive x-axis points to the right and the box slides to the left).

The particle is projected at an angle $\alpha$ relative to the box's floor, which is already tilted at an angle $	heta$ to the horizontal. Therefore, the particle's total angle of projection with respect to the horizontal is $(\alpha + 	heta)$. Its horizontal velocity component relative to the box is $u \cos(\alpha + 	heta)$.

Adding these together gives the absolute horizontal velocity:

$$v_{px} = u \cos(\alpha + 	heta) - v \cos 	heta = 0$$

Rearranging this elegant equation to solve for the box's speed $v$, we arrive at our final answer:

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

Visualized Solution

Visualizing the System

Acceleration of the Box

Absolute Acceleration of the Particle

Relative Acceleration (Effective Gravity)

Projectile Motion in the Box Frame

Calculating the Distance PQ

Part (b): Zero Horizontal Displacement

Velocity Vectors in Ground Frame

Absolute Horizontal Velocity

Solving for Box Velocity

The Sigma Insight: Relative Velocity

The Setup

The Magic of Relative Frames

The Zero Displacement Mystery

The Final Calculation

Similar Questions

Comprehension Passage

When and for how long can the particle C fulfil the given condition?

What is speed of the particle C at the instant $t = 5 s$ ?

What is modulus of acceleration of the particle C at the instant $t = 5 s$ ?

At the instant, when the line joining locations of A and B is perpendicular to the line joining locations of B and C, what are the magnitudes of velocities of C relative to A and B respectively?

List I describes four systems, each with two particles A and B in relative motion as shown in figure. List II gives possible magnitudes of their relative velocities (in $ms^{- 1}$ ) at time $t = \frac{π}{3} s$ .

List-I

List-II