1. Understanding the Setup and Accelerations

We consider two balls, A and B, released simultaneously from rest on two frictionless inclined planes with equal inclination $\theta$.

The acceleration of any object sliding down a frictionless incline is $g\sin\theta$. However, the directions are different:

• Ball A (Left Incline): Moves down and to the right.
• Ball B (Right Incline): Moves down and to the left.

Let’s analyze the Relative Acceleration ($\vec{a}_{rel} = \vec{a}_A – \vec{a}_B$).

When we subtract the acceleration vectors $\vec{a}_A – \vec{a}_B$, the vertical components (both pointing downward) cancel out. The horizontal components (one pointing right, the other left) add up.

$$ \vec{a}_{rel} = (g\sin\theta\cos\theta) \hat{i} – (-g\sin\theta\cos\theta) \hat{i} $$ $$ \vec{a}_{rel} = 2g\sin\theta\cos\theta \, \hat{i} $$

2. Displacements and the Common Level

We are given the times taken to reach a specific horizontal level: $t_A = 12\,\text{s}$ and $t_B = 4\,\text{s}$.

The distance traveled along the incline ($S$) is given by $S = \frac{1}{2}at^2$.

• $S_A = \frac{1}{2} (g\sin\theta) (12)^2$
• $S_B = \frac{1}{2} (g\sin\theta) (4)^2$

Let $\Delta S$ be the difference in distance along the incline. This creates a relative separation that needs to be “closed” for the balls to be closest.

$$ \Delta S = S_A – S_B = \frac{1}{2}g\sin\theta (12^2 – 4^2) $$

3. Solving for Time (t)

We model the relative motion. The horizontal component of the separation corresponds to the projection of the path difference $\Delta S$.

$$ \text{Relative Displacement (x)} = \Delta S \cos\theta $$ $$ \text{Relative Acceleration (x)} = a_{rel} = 2g\sin\theta\cos\theta $$

Using the kinematic equation for relative motion starting from rest ($u_{rel}=0$): $$ \text{Displacement} = \frac{1}{2} (a_{rel}) t^2 $$

Substitute our values: $$ \Delta S \cos\theta = \frac{1}{2} (2g\sin\theta\cos\theta) t^2 $$

Canceling $\cos\theta$ and the $\frac{1}{2} \times 2$ factor: $$ \Delta S = (g\sin\theta) t^2 $$

Now, substitute the expression for $\Delta S$ from Step 2: $$ \frac{1}{2}g\sin\theta (12^2 – 4^2) = g\sin\theta t^2 $$

Cancel $g\sin\theta$ from both sides: $$ \frac{1}{2} (144 – 16) = t^2 $$ $$ \frac{128}{2} = t^2 $$ $$ 64 = t^2 $$ $$ t = 8\,\text{s} $$