Animated Solution for Physics - System of Particles: A simple pendulum is made of a string of length $l$ and a bob of mass $m$, is released from a small angle ${\theta }_0$. It strikes a block of mass $M$, kept on a horizontal surface at its lowest point of oscillations, elastically. It bounces back and goes up to an angle ${\theta }_1$. Then, $M$ is given by

The Pendulum's Journey

Imagine a simple pendulum, a classic setup in physics, consisting of a bob of mass $m$ suspended by a string of length $l$. It is pulled back to a small angle ${\theta }_0$ and released. As it swings down, gravity does work, converting its stored potential energy into kinetic energy.

Waiting patiently at the exact lowest point of the swing is a block of mass $M$, resting on a frictionless horizontal surface. The pendulum bob is on a collision course with this block. The problem states that this collision is perfectly elastic, meaning no kinetic energy is lost in the form of heat or sound. After the impact, the bob bounces back, retreating to a new, smaller angle ${\theta }_1$. Our mission is to find the mass $M$ of the block in terms of the given parameters.

Energy Conservation

Let's break the problem down chronologically. First, we need to determine the velocity of the pendulum bob just before it strikes the block. We invoke the powerful principle of conservation of mechanical energy.

At the release point, the bob is at a height $h$ relative to its lowest point. Using basic geometry, this height is $h=l-l\cos {\theta }_0=l(1-\cos {\theta }_0)$. Equating the initial potential energy to the kinetic energy at the bottom, we get:

Substituting the expression for $h$, we find the velocity $v$ just before impact:

Tracing the Steps Backwards

Similarly, after the collision, the bob rebounds with a new velocity, let's call it $v_1$, and swings up to a maximum angle ${\theta }_1$. Applying energy conservation again for this upward journey:

Where $h_1=l(1-\cos {\theta }_1)$. This gives us the rebound velocity:

A Dance of Momentum and Restitution

Now, let's analyze the split-second of the collision. Since there are no external horizontal forces acting on the system (the tension in the string is purely vertical at the lowest point), the linear momentum in the horizontal direction is conserved.

Let $v_2$ be the velocity of the block $M$ after the collision. Taking the initial direction of the bob as positive, the momentum equation is:

Notice the negative sign! The bob has bounced back, so its velocity vector points in the opposite direction. Rearranging this, we get:

We have two unknowns ($M$ and $v_2$), so we need another equation. This is where the nature of the collision comes in. For a perfectly elastic collision, the coefficient of restitution $e$ is $1$. This means the relative velocity of separation equals the relative velocity of approach:

Trigonometry Meets Physics

Now, we substitute our expression for $v_2$ back into the momentum equation:

Isolating the mass ratio, we get:

Let's plug in the ugly square root expressions we derived earlier for $v$ and $v_1$. The $\sqrt{2gl}$ terms beautifully cancel out from the numerator and denominator, leaving us with:

This looks intimidating, but trigonometry comes to the rescue! Recall the half-angle identity: $1-\cos \theta =2{\sin }^2\left(\frac{\theta }{2}\right)$. Substituting this transforms our equation into:

The Small Angle Approximation

The problem explicitly states that the pendulum is released from a small angle. In physics, when angles are small (typically less than 10 degrees), we can use the small-angle approximation: $\sin \theta \approx \theta$ (when $\theta$ is in radians).

Applying this approximation to our half-angles, $\sin (\theta /2)\approx \theta /2$. Our equation simplifies dramatically:

Multiplying the numerator and denominator by 2, we arrive at our final, elegant result:

This problem is a fantastic example of how JEE combines core mechanical principles (energy and momentum) with mathematical tools (trigonometry and approximations) to test your analytical depth.