A ball dropped on a large inclined plane, bounces repeatedly. Every bounce is perfectly elastic i.e.

Q: Why does the x-component of velocity remain unchanged during the bounce?

During a perfectly elastic collision with a smooth surface, the impulsive normal force acts only perpendicular to the surface. Since there is no force parallel to the incline during the impact, the velocity component in that direction is conserved.

Q: Why is the time of flight the same for every bounce?

The time of flight depends only on the initial y-component of velocity after the bounce and the y-component of acceleration. Since the collision is elastic, the ball rebounds with the same y-velocity magnitude every time, making the time of flight constant.

Q: Does the angle of the incline affect the ratio of the distances?

No! While the individual distances $R_{12}$ and $R_{23}$ depend on $\theta$, their ratio cancels out the $\sin\theta$ term, making the ratio $1:2$ independent of the incline's angle.

Q: What happens to the x-velocity over time?

The x-velocity continuously increases because there is a constant acceleration component ($g \sin\theta$) acting down the incline. It increases during the flight and remains unchanged during the bounces.

Q: Can we solve this using energy conservation?

Energy conservation holds, but it's not sufficient to find the distances directly because it doesn't give us the time of flight or the separation of velocity components easily. Kinematics is the most direct approach here.

Question

A ball dropped on a large inclined plane, bounces repeatedly. Every bounce is perfectly elastic i.e. there is no loss of speed and lines of motion make equal angles with the incline plane before and after the bounce. Find ratio of distance $R_{12}$ between the first and the second bounce to distance $R_{23}$ between the second and the third bounce.

Sigma Percentile · Accepted Answer

The problem of a bouncing ball on an inclined plane is a classic test of your ability to decouple motion into independent axes. At first glance, the continuous bouncing might seem chaotic, but by choosing the right coordinate system, a beautiful mathematical pattern emerges.

### Tilting Our Perspective

The secret to solving this problem elegantly lies in our choice of axes. Instead of using the standard horizontal and vertical axes, we align our coordinate system with the inclined plane. Let the $x$-axis point down the incline and the $y$-axis point perpendicular to it, away from the surface.

By doing this, the acceleration due to gravity, $g$, splits into two constant components:
- Down the incline: $a_x = g \sin	heta$
- Perpendicular to the incline: $a_y = -g \cos	heta$

When the ball is dropped and hits the incline for the first time, it has a purely vertical velocity, let's call it $v_0$. In our tilted coordinate system, this initial velocity has components $v_{0x} = v_0 \sin	heta$ and $v_{0y} = -v_0 \cos	heta$.

### The First Bounce: A Fresh Start

The problem states that the bounces are perfectly elastic. In the language of physics, this means the impulsive normal force from the plane only affects the perpendicular component of velocity. The parallel component remains completely untouched!

So, immediately after the first bounce, the velocity components are:
- $u_{1x} = v_{0x} = v_0 \sin	heta$
- $u_{1y} = -v_{0y} = v_0 \cos	heta$

Now, the ball is in the air. The time it spends in the air before hitting the plane again depends **only** on the $y$-motion. Setting the net $y$-displacement to zero, we find the time of flight $T_1$:
$$T_1 = \frac{2 u_{1y}}{|a_y|} = \frac{2 v_0 \cos	heta}{g \cos	heta} = \frac{2 v_0}{g}$$

Notice something incredible? The time of flight is completely independent of the incline's angle $	heta$!

During this time $T_1$, the ball is accelerating down the incline. The distance it covers, $R_{12}$, is given by the second equation of motion:
$$R_{12} = u_{1x} T_1 + \frac{1}{2} a_x T_1^2$$
Substituting our values:
$$R_{12} = (v_0 \sin	heta) \left(\frac{2 v_0}{g}ight) + \frac{1}{2} (g \sin	heta) \left(\frac{2 v_0}{g}ight)^2 = \frac{4 v_0^2 \sin	heta}{g}$$

### The Second Bounce: Gaining Momentum

While the ball was in the air, gravity was constantly pulling it down the incline, increasing its $x$-velocity. Just before the second bounce, its $x$-velocity is:
$$v_{1x} = u_{1x} + a_x T_1 = v_0 \sin	heta + (g \sin	heta)\left(\frac{2 v_0}{g}ight) = 3 v_0 \sin	heta$$

The $y$-velocity, right before impact, is back to $-v_0 \cos	heta$.

Upon the second elastic bounce, the $y$-velocity reverses again to $v_0 \cos	heta$, while the $x$-velocity remains $3 v_0 \sin	heta$. Because the initial $y$-velocity for this second jump is identical to the first jump, the time of flight $T_2$ is exactly the same: $T_2 = \frac{2 v_0}{g}$.

Now, let's calculate the distance of the second jump, $R_{23}$:
$$R_{23} = u_{2x} T_2 + \frac{1}{2} a_x T_2^2$$
$$R_{23} = (3 v_0 \sin	heta) \left(\frac{2 v_0}{g}ight) + \frac{1}{2} (g \sin	heta) \left(\frac{2 v_0}{g}ight)^2 = \frac{8 v_0^2 \sin	heta}{g}$$

### The Grand Finale: The Ratio

We have our two distances. The question asks for the ratio of $R_{12}$ to $R_{23}$. 
$$\frac{R_{12}}{R_{23}} = \frac{\frac{4 v_0^2 \sin	heta}{g}}{\frac{8 v_0^2 \sin	heta}{g}} = \frac{4}{8} = \frac{1}{2}$$

**The ratio is 1:2.**

If we were to calculate the distance of the third jump, $R_{34}$, we would find it to be $\frac{12 v_0^2 \sin	heta}{g}$. The distances between successive bounces form a beautiful arithmetic progression: $1 : 2 : 3 : 4 \dots$ This is a powerful generalized result that can save you immense time in competitive exams!

Visualized Solution

Coordinate System Setup

Velocity After First Bounce

Time of Flight ( $T_{1}$ )

Distance of First Jump ( $R_{12}$ )

Velocity After Second Bounce

Distance of Second Jump ( $R_{23}$ )

Ratio of Distances

Generalization for $n^{th}$ Bounce

The Sigma Insight: Projectile Motion

Tilting Our Perspective

The First Bounce

The Second Bounce

The Grand Finale

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