A particle of mass m is projected with a speed u from the ground at an angle theta = pi/3 w.r.t. hor

Q: Why is the vertical velocity of the first particle zero at the maximum height?

At the maximum height, the particle momentarily stops moving upward before it begins to fall back down. Therefore, its vertical velocity component is exactly zero at that instant, leaving only the constant horizontal component.

Q: Why do we only conserve momentum and not kinetic energy during the collision?

The problem states it is a 'completely inelastic collision', which means the particles stick together. In all inelastic collisions, kinetic energy is lost (usually as heat or sound), but linear momentum is always conserved if there are no external forces.

Q: Does the collision affect the time it takes for the combined mass to fall to the ground?

No, it doesn't. The collision is purely horizontal, meaning the combined mass still has zero initial vertical velocity right after the crash. The time of fall depends only on the height and gravity, which remain unchanged.

Q: Why is the initial velocity of the second particle given as $u\hat{i}$?

The $\hat{i}$ indicates that the second particle is moving purely in the positive x-direction (horizontally). This makes it easy to add its momentum to the horizontal momentum of the first particle at the peak.

Q: Can we use the standard range formula $R = \frac{u^2 \sin(2\theta)}{g}$ for the combined mass?

No, the standard range formula only applies to a projectile launched from the ground and landing on the ground. The combined mass is launched horizontally from a height $h_{max}$, so you must calculate the horizontal distance manually using $R' = v_x imes t$.

Question

A particle of mass $m$ is projected with a speed $u$ from the ground at an angle $	heta = \frac{\pi}{3}$ w.r.t. horizontal (X-axis). When it has reached its maximum height, it collides completely inelastically with another particle of the same mass and velocity $u\hat{i}$. The horizontal distance covered by the combined mass before reaching the ground is

Options:

(A) $\frac{3\sqrt{3}}{8}\frac{u^2}{g}$

(B) $\frac{3\sqrt{2}}{4}\frac{u^2}{g}$

(C) $\frac{5}{8}\frac{u^2}{g}$

(D) $2\sqrt{2}\frac{u^2}{g}$

Sigma Percentile · Accepted Answer

### The Ascent: Reaching the Peak

Imagine you are standing on the ground, and you launch a particle of mass $m$ into the air with an initial speed $u$ at an angle $	heta = \frac{\pi}{3}$ to the horizontal. This is a classic projectile motion scenario. As the particle ascends, gravity continuously pulls it downward, reducing its vertical velocity.

However, gravity has no effect on the horizontal motion (assuming no air resistance). Therefore, when the particle reaches its absolute highest point—the peak of its parabolic trajectory—its vertical velocity becomes exactly zero. At this precise moment, the particle is moving purely horizontally.

We can calculate this horizontal velocity, let's call it $\vec{v}_1$, using the horizontal component of the initial velocity:
$$ \vec{v}_1 = u \cos\left(\frac{\pi}{3}ight)\hat{i} = \frac{u}{2}\hat{i} $$

We also need to know how high this peak is. Using the standard formula for the maximum height of a projectile, we get:
$$ h_{max} = \frac{u^2 \sin^2\left(\frac{\pi}{3}ight)}{2g} = \frac{u^2 \left(\frac{\sqrt{3}}{2}ight)^2}{2g} = \frac{3u^2}{8g} $$

### The Mid-Air Collision: Momentum Conservation

Now, here is where the problem gets thrilling. Just as our particle is peacefully gliding horizontally at the peak, a second particle of the same mass $m$ comes hurtling in horizontally with a velocity of $u\hat{i}$.

They collide! And not just any collision—it's a **completely inelastic collision**. This means the two particles crash into each other and stick together, forming a single combined mass of $2m$.

Because there are no external horizontal forces acting on the system during this split-second crash, we can invoke the powerful **Principle of Conservation of Linear Momentum**. The total momentum just before the crash must equal the total momentum right after.

$$ \vec{p}_{initial} = \vec{p}_{final} $$
$$ m\vec{v}_1 + m\vec{v}_2 = (m + m)\vec{v}_{comb} $$

Let's plug in the velocities we know:
$$ m\left(\frac{u}{2}\hat{i}ight) + m(u\hat{i}) = 2m\vec{v}_{comb} $$
$$ \frac{3mu}{2}\hat{i} = 2m\vec{v}_{comb} $$

Solving for the velocity of the newly formed combined mass, we find:
$$ \vec{v}_{comb} = \frac{3u}{4}\hat{i} $$

### The Descent: A Horizontal Projectile

Our new combined mass of $2m$ is now at height $h_{max}$, moving horizontally with a speed of $\frac{3u}{4}$. From this point onward, it behaves exactly like a horizontal projectile dropped from a cliff.

To find out how far it travels horizontally before hitting the ground, we first need to know how long it stays in the air. The time of flight $t$ depends solely on the vertical height and gravity. Since its initial vertical velocity post-collision is zero, we use the second equation of motion:
$$ h_{max} = \frac{1}{2}gt^2 $$
$$ t = \sqrt{\frac{2h_{max}}{g}} $$

Let's substitute the value of $h_{max}$ we found earlier:
$$ t = \sqrt{\frac{2}{g} \cdot \frac{3u^2}{8g}} = \sqrt{\frac{3u^2}{4g^2}} = \frac{\sqrt{3}u}{2g} $$

### Bringing It All Together

We are in the final stretch! We know the horizontal speed of the combined mass, and we know exactly how long it will be flying. The horizontal distance covered, let's call it $R'$, is simply the product of this horizontal speed and the time of flight.

$$ R' = |\vec{v}_{comb}| 	imes t $$
$$ R' = \left(\frac{3u}{4}ight) 	imes \left(\frac{\sqrt{3}u}{2g}ight) $$
$$ R' = \frac{3\sqrt{3}u^2}{8g} $$

And there we have it! By carefully breaking down the problem into projectile motion, a momentum-conserving collision, and a final horizontal projection, we've arrived at the elegant final answer.

Select Answer:

Visualized Solution

$Initial Projection$

$Velocity at Maximum Height$

$Calculating v_{1} and h_{ma x}$

$The Inelastic Collision$

$Conservation of Linear Momentum$

$Velocity of Combined Mass$

$Time of Fall$

$Calculating Time t$

$Horizontal Distance R^{'}$

$Conclusion$

The Sigma Insight: Conservation of Linear Momentum

The Ascent

The Mid-Air Collision

The Descent

Bringing It All Together

Similar Questions

Two particles of equal mass $m$ have respective initial velocities $u \hat{i}$ and $u (\frac{i ^ + j ^}{2})$ . They collide completely inelastically. The energy lost in the process is

A wedge of mass $M = 4 m$ lies on a frictionless plane. A particle of mass $m$ approaches the wedge with speed $v$ . There is no friction between the particle and the plane or between the particle and the wedge. The maximum height climbed by the particle on the wedge is given by

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Projection

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Velocity at Maximum Height

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating v1​ and hmax​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Inelastic Collision

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conservation of Linear Momentum

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Velocity of Combined Mass

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Time of Fall

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Time t

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Horizontal Distance R′

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion