A simple pendulum is suspended from a peg on a vertical wall. The pendulum is pulled away from the w

Q: Wait, why does the velocity become $ev$ after the collision?

By definition, the coefficient of restitution $e$ is the ratio of the velocity of separation to the velocity of approach. Since the wall is stationary, the velocity of separation is simply the rebound velocity, making it $e$ times the approach velocity $v$.

Q: Why can we use energy conservation after the collision if it's inelastic?

We don't use energy conservation *during* the collision! Energy is indeed lost as heat and sound when the bob hits the wall. However, *between* collisions, as the bob swings up and down, the only forces doing work are gravity (conservative) and tension (does zero work). Thus, mechanical energy is conserved during the swing itself.

Q: How did we get the formula $h = l(1 - \cos\theta_n)$?

Draw a right triangle with the pendulum string at an angle $\theta_n$. The vertical distance from the pivot to the bob is $l \cos\theta_n$. Since the total length of the string is $l$, the height $h$ it has risen from the lowest point is the total length minus this vertical distance: $l - l \cos\theta_n$.

Q: Why does the inequality flip when dealing with cosine?

The cosine function decreases as the angle increases from $0^\circ$ to $90^\circ$. Therefore, if we want the angle $\theta_n$ to be *less* than $60^\circ$, its cosine value must be *greater* than $\cos 60^\circ$. It's a classic mathematical trap!

Q: Can we solve this using simple harmonic motion (SHM) concepts?

No, SHM approximations only work for very small angles (typically less than $5^\circ$ or $10^\circ$) where $\sin\theta \approx \theta$. Here, we are dealing with large angles like $90^\circ$ and $60^\circ$, so we must use exact energy conservation and geometry.

Question

A simple pendulum is suspended from a peg on a vertical wall. The pendulum is pulled away from the wall to a horizontal position (see fig.) and released. The ball hits the wall, the coefficient of restitution being $\frac{2}{\sqrt{5}}$.
What is the minimum number of collisions after which the amplitude of oscillations becomes less than $60^\circ$ ?

Sigma Percentile · Accepted Answer

### The Initial Drop: Trading Height for Speed

Imagine a pendulum pulled completely horizontal and then released. At this exact moment, it is momentarily at rest, meaning its kinetic energy is zero, but its potential energy is at its absolute maximum. If we take the lowest point of the swing as our reference level, this initial potential energy is simply $mgl$, where $l$ is the length of the string.

As the pendulum swings down, gravity does work, converting that potential energy entirely into kinetic energy. By applying the conservation of mechanical energy, we can easily find the velocity at the lowest point right before it hits the wall.

$$mgl = \frac{1}{2}mv^2$$

Solving for $v$, we get $v = \sqrt{2gl}$. This is the maximum speed the pendulum will ever achieve in its journey.

### The Wall: A Thief of Momentum

Now comes the main event. The pendulum smashes into the vertical wall. This isn't a perfectly elastic bounce; energy is lost as heat and sound. The coefficient of restitution, $e$, tells us exactly how much velocity is retained after the impact. Since the wall is stationary, the rebound velocity $v_1$ is simply $e$ times the impact velocity $v$.

$$v_1 = e\sqrt{2gl}$$

Think about what happens next. Every single time the bob hits the wall, its velocity gets multiplied by another factor of $e$. So, if we let it bounce $n$ times, the velocity right after the $n$-th collision will be $e$ raised to the power of $n$, multiplied by our original velocity.

$$v_n = e^n\sqrt{2gl}$$

Armed with this new, reduced velocity, the pendulum swings back up, fighting gravity until it stops at some new maximum height, let's call it $h$. We can use energy conservation once again. Equating the kinetic energy after $n$ bounces to the potential energy at height $h$, we find:

$$\frac{1}{2}mv_n^2 = mgh$$

Substituting our expression for $v_n$, we get $h = e^{2n}l$. The problem gives us a very specific value for $e$: $\frac{2}{\sqrt{5}}$. Let's substitute that into our height equation. When we square it, the square root disappears, and we are left with:

$$h = \left(\frac{4}{5}ight)^n l$$

### The Geometry of the Swing

Now, let's connect this height to the angle of the swing. Look closely at the geometry of the pendulum. If the string makes an angle $	heta_n$ with the vertical, the vertical drop from the pivot is $l \cos	heta_n$.

Therefore, the height $h$ from the bottom is simply the total length $l$ minus this vertical drop. This is a favorite concept for JEE!

$$h = l - l\cos	heta_n = l(1 - \cos	heta_n)$$

We now have two different ways to express the height $h$. Let's equate them. The length $l$ beautifully cancels out from both sides, leaving us with a clean, elegant relationship between the number of bounces $n$, and the angle $	heta_n$.

$$\left(\frac{4}{5}ight)^n = 1 - \cos	heta_n$$

### The Inequality: Racing to the Threshold

Here is where mistakes happen. We want the amplitude angle to drop below $60^\circ$. But remember, the cosine function is decreasing in the first quadrant! So, if the angle is less than $60^\circ$, its cosine must be strictly greater than $\cos 60^\circ$, which is $\frac{1}{2}$.

$$\cos	heta_n > \frac{1}{2}$$

Let's bring it all together. We substitute our expression for $\cos	heta_n$ into the inequality.

$$1 - \left(\frac{4}{5}ight)^n > \frac{1}{2}$$

After a quick rearrangement, we find that $\left(\frac{4}{5}ight)^n$ must be strictly less than $0.5$. Don't rush through this final step. Let's test integer values for $n$.

For $n=1$, the value is $0.8$. For $n=2$, it's $0.64$. For $n=3$, it's $0.512$. All of these are still greater than $0.5$.

But the moment we plug in $n=4$, we get $\left(\frac{4}{5}ight)^4 = \frac{256}{625} = 0.4096$, which finally breaks the threshold! So, it takes exactly **4** collisions for the amplitude to drop below $60^\circ$.

Enter Numerical Value:

Visualized Solution

Initial Energy Setup

Velocity at Lowest Point

First Rebound

Velocity after $n$ Rebounds

Height after $n$ Rebounds

Substituting Coefficient of Restitution

Geometry of the Pendulum

Equating the Heights

Condition for Amplitude

Solving for $n$

Finding Minimum $n$

The Sigma Insight: Head-on Collision

The Initial Drop

The Wall

The Geometry of the Swing

The Inequality

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