Animated Solution for Physics - System of Particles: Comprehension Passage

Analyzing the Projectile

Let's embark on a thrilling journey through a classic JEE physics problem involving projectile motion and the brilliant concept of the Center of Mass. We start with a projectile launched from the origin $O$. The problem states the angle of projection is $45^{\circ }$ from the vertical.

However, since the total angle between the horizontal and vertical is $90^{\circ }$, the angle with the horizontal is also $90^{\circ }-45^{\circ }=45^{\circ }$. This symmetry makes our initial calculations incredibly elegant.

Given the initial speed $u=5\sqrt{2}\text{ m/s}$, we can resolve this velocity into its horizontal and vertical components.

With these components, we can easily determine the fundamental parameters of the projectile's trajectory. The time of flight $T$ is given by $\frac{2u_y}{g}$, which evaluates to exactly $1\text{ s}$. The total horizontal range $R$ is $u_x\times T=5\text{ m}$. Furthermore, the maximum height reached by the projectile is $H_{\text{max}}=\frac{u_y^2}{2g}=1.25\text{ m}$.

A Mid-Air Explosion

Imagine the projectile reaching the absolute peak of its parabolic path. At this exact moment, $t=0.5\text{ s}$, an internal explosion occurs, splitting the projectile into two equal masses, each of mass $m$.

At the highest point, the vertical velocity of the projectile is momentarily zero. Its entire velocity is purely horizontal, $\vec{v}=5\hat{i}\text{ m/s}$.

The problem presents a fascinating clue: one of the pieces falls vertically downwards and hits the ground exactly $0.5\text{ s}$ after the explosion. Let's analyze this carefully. If an object is dropped from a height of $1.25\text{ m}$ with zero initial vertical velocity, the time it takes to hit the ground is $t=\sqrt{\frac{2H}{g}}=\sqrt{\frac{2.5}{10}}=0.5\text{ s}$.

This perfectly matches the given time! This mathematical alignment proves that the first piece was essentially "dropped" from the highest point, meaning its initial vertical velocity right after the explosion was exactly zero.

Conservation of Momentum

Now, we apply the sacred law of Conservation of Linear Momentum. Since the explosion is caused by internal forces, the total momentum of the system must remain conserved in all directions.

Let's look at the vertical direction. Before the explosion, the total vertical momentum was zero. After the explosion, we just proved that the first piece has zero vertical velocity, meaning its vertical momentum is also zero.

This is a profound realization! The second piece must also have an initial vertical velocity of zero. Because it starts from the same height ($1.25\text{ m}$) with the same zero initial vertical velocity, it will take the exact same time to free-fall to the ground. Therefore, the time $t$ for the second part is $0.5\text{ s}$.

Finding the Final Position

We must now determine where the second piece lands. We turn our attention to the horizontal direction and apply momentum conservation once again.

Before the explosion, the entire mass $2m$ was moving horizontally at $5\text{ m/s}$. After the explosion, the first piece falls straight down, meaning its horizontal velocity has become zero.

Solving this simple equation reveals that the second piece shoots forward with a horizontal velocity $v_{2x}=10\text{ m/s}$.

Since it travels for $0.5\text{ s}$ before hitting the ground, the additional horizontal distance it covers is $10\times 0.5=5\text{ m}$. The explosion happened at the midpoint of the range, which is $x=2.5\text{ m}$ from the origin.

Adding these distances together, the final position of the second piece from the origin is $x=2.5+5=7.5\text{ m}$.

A Masterstroke

While the kinematic approach is beautiful, there is a much faster, more elegant way to solve for $x$ using the Center of Mass (CM) concept.

Because the explosion is an internal event, it cannot alter the trajectory of the system's Center of Mass. The CM is completely oblivious to the explosion and continues its original parabolic path, landing exactly at the original range $R=5\text{ m}$.

We know the landing positions of the CM ($x_{cm}=5\text{ m}$) and the first piece ($x_1=2.5\text{ m}$). We can simply use the Center of Mass coordinate formula to find the landing position of the second piece ($x_2=x$).

This brilliant shortcut bypasses the need to calculate the individual velocities entirely, showcasing the true power of physics principles in simplifying complex problems!