A particle of mass m is projected from the ground with an initial speed u0 at an angle alpha with th

Q: Wait, why is the vertical velocity of the first particle zero at the collision?

At the highest point of a projectile's trajectory, it momentarily stops moving upwards before starting to fall back down. Therefore, its vertical velocity component is exactly zero at that instant, leaving only the constant horizontal component.

Q: Why can we use conservation of momentum here if gravity is acting on the particles?

Gravity is indeed an external force, but the collision happens in an infinitesimally small time interval ($\Delta t \to 0$). The impulse due to gravity ($mg \Delta t$) during this tiny window is negligible compared to the massive internal impact forces. Thus, momentum is conserved just before and just after the collision.

Q: Why is the final mass $2m$?

The problem explicitly states it is a "completely inelastic collision". In such collisions, the colliding bodies stick together and move as a single composite mass. Since both particles have mass $m$, the total mass becomes $m + m = 2m$.

Q: What if the second particle was thrown with a different speed?

If the second particle had a different initial speed, its velocity at height $H$ would not perfectly match the horizontal velocity of the first particle. The $x$ and $y$ components of the final momentum would be different, resulting in an angle other than $\frac{\pi}{4}$.

Q: Does the angle $\alpha$ affect the final angle of the composite mass?

Surprisingly, no! As long as the second particle is thrown with the same initial speed $u_0$, the math shows that the $x$ and $y$ components of the final velocity will always be equal, regardless of $\alpha$. The angle $\alpha$ only changes the magnitude of the final velocity, not its direction.

Question

A particle of mass $m$ is projected from the ground with an initial speed $u_0$ at an angle $\alpha$ with the horizontal. At the highest point of its trajectory, it makes a completely inelastic collision with another identical particle, which was thrown vertically upward from the ground with the same initial speed $u_0$. The angle that the composite system makes with the horizontal immediately after the collision is

Options:

(A) $\frac{\pi}{4}$

(B) $\frac{\pi}{4} + \alpha$

(C) $\frac{\pi}{4} - \alpha$

(D) $\frac{\pi}{2}$

Sigma Percentile · Accepted Answer

The beauty of physics often lies in its hidden symmetries. In this problem, we are presented with a perfectly choreographed dance between two particles. One is launched as a graceful projectile, tracing a parabolic arc across the sky. The other is fired straight up, a vertical ascent designed to intercept the first particle at the exact apex of its flight.

When they meet, they don't just bounce off each other; they merge in a completely inelastic collision. Our goal is to determine the trajectory of this newly formed composite body immediately after the impact. To solve this, we must break the problem down into two distinct phases: understanding the state of each particle just before the collision, and then applying the fundamental laws of physics to the collision itself.

### The Projectile's Journey

Let's begin by analyzing the first particle, which is projected from the ground with an initial speed $u_0$ at an angle $\alpha$ with the horizontal.

As it travels along its parabolic path, gravity constantly pulls it downward, reducing its vertical velocity. However, in the absence of air resistance, there is no force acting in the horizontal direction. This means its horizontal velocity remains perfectly constant throughout its flight.

At the highest point of its trajectory, the particle momentarily stops moving upward before it begins its descent. At this exact instant, its vertical velocity is zero. Therefore, its entire velocity is purely horizontal. We can write the velocity vector of the first particle at the peak as:

$$ \mathbf{v}_1 = u_0 \cos\alpha \, \hat{\mathbf{i}} $$

Before we move on to the second particle, we need to know exactly how high this peak is. Using the standard kinematic equations for projectile motion, the maximum height $H$ reached by the particle is given by:

$$ H = \frac{u_0^2 \sin^2\alpha}{2g} $$

This height $H$ is the crucial meeting point where the two particles will collide.

### The Vertical Ascent

Now, let's turn our attention to the second particle. It is thrown vertically upward from the ground with the exact same initial speed $u_0$.

As it rises, gravity slows it down. We need to find its velocity exactly when it reaches the height $H$, because that is where the collision occurs. We can determine this using the third equation of kinematics:

$$ v^2 = u^2 - 2gS $$

Here, the initial velocity $u$ is $u_0$, and the displacement $S$ is the height $H$. Substituting the expression for $H$ that we found earlier, we get:

$$ v_{2y}^2 = u_0^2 - 2g \left( \frac{u_0^2 \sin^2\alpha}{2g} ight) $$

Notice how beautifully the $2g$ terms cancel out. This leaves us with:

$$ v_{2y}^2 = u_0^2 - u_0^2 \sin^2\alpha $$

By factoring out $u_0^2$, we reveal a fundamental trigonometric identity:

$$ v_{2y}^2 = u_0^2 (1 - \sin^2\alpha) = u_0^2 \cos^2\alpha $$

Taking the square root, we find the velocity of the second particle just before the collision:

$$ \mathbf{v}_2 = u_0 \cos\alpha \, \hat{\mathbf{j}} $$

This is a stunning result! The vertical velocity of the second particle at the collision point is exactly equal in magnitude to the horizontal velocity of the first particle.

### The Moment of Impact

We have now arrived at the climax of the problem. The two particles collide at the peak. The problem states that this is a **completely inelastic collision**, which means the two particles stick together and move as a single composite mass of $2m$.

During the infinitesimally short duration of the collision, the internal forces between the particles are overwhelmingly large compared to any external forces like gravity. Because the net external impulsive force is zero, we can safely apply the principle of **Conservation of Linear Momentum**.

The total momentum of the system just before the collision must equal the total momentum just after the collision:

$$ \mathbf{p}_i = \mathbf{p}_f $$

Let's construct the initial momentum vector. It is simply the vector sum of the momenta of the two individual particles:

$$ \mathbf{p}_i = m(u_0 \cos\alpha)\hat{\mathbf{i}} + m(u_0 \cos\alpha)\hat{\mathbf{j}} $$

Let the final velocity of the composite mass be $\mathbf{v}$. The final momentum vector is:

$$ \mathbf{p}_f = (2m)\mathbf{v} $$

Equating the initial and final momenta, we get our master equation:

$$ m(u_0 \cos\alpha)\hat{\mathbf{i}} + m(u_0 \cos\alpha)\hat{\mathbf{j}} = 2m\mathbf{v} $$

### The Aftermath

Solving for the final velocity $\mathbf{v}$ is now just a matter of simple algebra. We divide both sides of the equation by the total mass $2m$:

$$ \mathbf{v} = \frac{u_0 \cos\alpha}{2}\hat{\mathbf{i}} + \frac{u_0 \cos\alpha}{2}\hat{\mathbf{j}} $$

Take a close look at this final velocity vector. The $x$-component and the $y$-component are exactly identical!

The problem asks for the angle $	heta$ that this composite system makes with the horizontal immediately after the collision. The tangent of this angle is the ratio of the vertical velocity component to the horizontal velocity component:

$$ 	an	heta = \frac{v_y}{v_x} $$

Substituting our components into this ratio:

$$ 	an	heta = \frac{\frac{u_0 \cos\alpha}{2}}{\frac{u_0 \cos\alpha}{2}} = 1 $$

The only angle in the first quadrant whose tangent is $1$ is $45^\circ$, or in radians, $\frac{\pi}{4}$.

$$ 	heta = \frac{\pi}{4} $$

And there we have it. Despite the complex setup involving different launch angles and trajectories, the perfect symmetry of the initial speeds and the collision point results in the composite mass flying off at a perfect $45^\circ$ angle. This elegant result is a testament to the power and beauty of momentum conservation in physics.

Select Answer:

Visualized Solution

Visualizing the Setup

Velocity of Projectile 1 at Peak

Maximum Height of Projectile 1

Velocity of Particle 2 at Height $H$

Substituting $H$ for Particle 2

Simplifying Particle 2's Velocity

Conservation of Linear Momentum

Setting up the Momentum Equation

Solving for Final Velocity

Angle of the Composite System

The Sigma Insight: Oblique Collision

The Projectile's Journey

The Vertical Ascent

The Moment of Impact

The Aftermath

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