Animated Solution for Physics - Laws of Motion: Two fixed frictionless inclined plane making the angles $30^{\circ }$ and $60^{\circ }$ with the vertical are shown in the figure. Two blocks $A$ and $B$ are placed on the two planes. What is the relative vertical acceleration of $A$ with respect to $B$?

This is a classic kinematics problem that beautifully tests your ability to resolve vectors and understand relative motion. At first glance, it might seem like a straightforward application of Newton's laws on an inclined plane, but there is a subtle geometric trap waiting for the unwary student. Let's break it down step-by-step and master the physics behind it.

The problem presents two frictionless inclined planes with blocks $A$ and $B$ resting on them. The crucial detail here is how the angles are defined. The problem states that the planes make angles of $60^{\circ }$ and $30^{\circ }$ with the vertical.

In physics, our standard formulas for inclined planes (like $a=g\sin \theta$) are almost always derived assuming $\theta$ is the angle the plane makes with the horizontal. To avoid confusion and potential sign errors, our first step should be to convert these vertical angles into horizontal angles.

Since the vertical and horizontal axes are perpendicular ($90^{\circ }$), we can easily find the horizontal angles by subtracting the given angles from $90^{\circ }$: - For Block $A$: ${\theta }_A=90^{\circ }-60^{\circ }=30^{\circ }$ - For Block $B$: ${\theta }_B=90^{\circ }-30^{\circ }=60^{\circ }$

Now, let's consider a generic block of mass $m$ sliding down a frictionless incline of angle $\theta$. What is its acceleration?

The only forces acting on the block are gravity ($mg$) pointing straight down, and the normal force ($N$) pointing perpendicular to the surface. To analyze the motion, we resolve gravity into two components: 1. Perpendicular to the plane: $mg\cos \theta$ (balanced by the normal force $N$). 2. Parallel to the plane: $mg\sin \theta$ (the driving force causing acceleration).

Applying Newton's Second Law ($F_{net}=ma$) along the plane, we get:

Notice how the mass $m$ beautifully cancels out! This means the acceleration along the incline is purely a geometric consequence of gravity, independent of how heavy the block is.

The problem doesn't ask for the acceleration along the plane; it specifically asks for the relative vertical acceleration. This means we need to take our acceleration vector $\vec{a}$ (which points down the incline) and resolve it into its vertical component.

The incline itself is at an angle $\theta$ below the horizontal. Therefore, the acceleration vector $\vec{a}$ also points at an angle $\theta$ below the horizontal. Using basic trigonometry, the vertical component $a_y$ is:

Substituting our expression for $a$ ($a=g\sin \theta$), we get a powerful derived formula for the vertical acceleration of any object sliding down a frictionless incline:

Armed with our derived formula, calculating the vertical acceleration for each block becomes a breeze.

For Block $A$: The angle with the horizontal is ${\theta }_A=30^{\circ }$. Since $\sin 30^{\circ }=\frac{1}{2}$, we have: This acceleration is directed downwards.

For Block $B$: The angle with the horizontal is ${\theta }_B=60^{\circ }$. Since $\sin 60^{\circ }=\frac{\sqrt{3}}{2}$, we have: This acceleration is also directed downwards.

Finally, we need to find the relative vertical acceleration of $A$ with respect to $B$. Relative acceleration is defined as the vector difference:

Since we are only looking at the vertical axis, and both accelerations are pointing in the exact same direction (downwards), we can simply subtract their magnitudes to find the relative difference:

Taking the standard value of acceleration due to gravity as $g=9.8{\text{ m/s}}^2$, we get:

The magnitude of the relative vertical acceleration is $4.9{\text{ m/s}}^2$. This perfectly matches option (d). This problem is a fantastic reminder to always read the given angles carefully and to trust in the systematic resolution of vectors!