Particles P and Q of mass 20 g and 40 g respectively are simultaneously projected from points A and

Q: Why do we assume they collide at the highest point?

We calculate the range $R = \frac{u^2 \sin 2\theta}{g}$ and find it is exactly $245\text{ m}$, which is the distance between $A$ and $B$. Since they are fired with the same speed and complementary angles ($45^\circ$ and $135^\circ$), their paths are perfectly symmetric. Therefore, they must meet exactly in the middle, which corresponds to the peak of their parabolic trajectories.

Q: Why is the vertical velocity zero just before the collision?

Because the collision happens exactly at the midpoint of their symmetric trajectories, which is the maximum height (the peak). At the peak of any projectile's flight, the vertical component of velocity becomes zero for an instant before it starts descending.

Q: What does "retraces its path" physically mean for particle P?

To retrace a parabolic path backwards, the particle must have the exact same speed but in the perfectly opposite direction. Since it's at the peak where vertical velocity is zero, its horizontal velocity must simply flip its sign. So, $v_{Px}$ becomes $-v_{Px}$.

Q: Why don't we consider vertical momentum during the collision?

We could, but it's trivial here. Just before the collision, both particles are at the peak, so their vertical velocities are zero. The total vertical momentum is zero. After the collision, $P$ retraces its path, so its vertical velocity is still zero. By conservation of momentum, $Q$'s vertical velocity must also be zero immediately after the crash.

Q: Why can't we use energy conservation to find Q's final velocity?

The problem does not state that the collision is perfectly elastic. In fact, because the particles change their velocities drastically and $Q$ comes to a dead stop, kinetic energy is definitely lost (converted to heat, sound, or deformation). Momentum, however, is always conserved in the absence of external forces, regardless of the type of collision.

Question

Particles $P$ and $Q$ of mass $20 	ext{ g}$ and $40 	ext{ g}$ respectively are simultaneously projected from points $A$ and $B$ on the ground. The initial velocities of $P$ and $Q$ make $45^\circ$ and $135^\circ$ angles respectively with the horizontal $AB$ as shown in the figure. Each particle has an initial speed of $49 	ext{ m/s}$. The separation $AB$ is $245 	ext{ m}$.

Both particles travel in the same vertical plane and undergo a collision. After the collision, $P$ retraces its path. (a) Determine the position $Q$ where it hits the ground. (b) How much time after the collision does the particle $Q$ take to reach the ground?
(Take $g = 9.8 	ext{ m/s}^2$).

Sigma Percentile · Accepted Answer

The collision of two projectiles in mid-air is one of the most visually stunning and conceptually rich problems in classical mechanics. It forces us to weave together the kinematics of parabolic motion with the rigid laws of momentum conservation. Let's embark on a thrilling journey to dissect this exact scenario.

### Analyzing the Setup

Imagine a vast, open field. Two points, $A$ and $B$, lie exactly $245	ext{ m}$ apart. From point $A$, a $20	ext{ g}$ particle named $P$ is launched into the sky at a $45^\circ$ angle. Simultaneously, from point $B$, a heavier $40	ext{ g}$ particle named $Q$ is fired back towards $A$ at a $135^\circ$ angle (which is simply $45^\circ$ measured from the negative x-axis).

Both particles are launched with an identical, blazing speed of $49	ext{ m/s}$. Our primary objective is to determine the exact location where these two particles will inevitably crash into each other, and what happens to particle $Q$ immediately after the impact.

### The Master Equation: Finding the Range

To understand where they will meet, we must first understand their individual trajectories. If they were to fly unimpeded, where would they land? This requires us to calculate their horizontal range, $R$.

The formula for the horizontal range of a projectile is given by:

$$R = \frac{u^2 \sin(2	heta)}{g}$$

Let's substitute our known values into this equation. The initial speed $u$ is $49	ext{ m/s}$, the angle $	heta$ is $45^\circ$, and the acceleration due to gravity $g$ is $9.8	ext{ m/s}^2$.

$$R = \frac{(49)^2 \sin(2 	imes 45^\circ)}{9.8}$$

Since $\sin(90^\circ) = 1$, the calculation simplifies beautifully:

$$R = \frac{2401 	imes 1}{9.8} = 245	ext{ m}$$

### The Geometric Revelation

Look closely at that result. The range of each projectile is exactly $245	ext{ m}$. This is not a coincidence; it is exactly equal to the initial separation distance between points $A$ and $B$!

Because both particles are fired with the exact same speed and at perfectly symmetric angles, their parabolic paths are mirror images of each other. Since their total range spans the entire distance between them, they are destined to meet exactly halfway.

This midpoint corresponds to the absolute highest point—the peak—of their parabolic trajectories.

### The Physics of the Peak

Let's freeze time just a fraction of a millisecond before the collision. At the peak of a projectile's flight, gravity has momentarily exhausted its upward momentum. The vertical component of velocity is exactly zero.

The particles are only moving horizontally. Particle $P$ is moving to the right with a velocity of $u \cos 45^\circ$, and particle $Q$ is moving to the left with a velocity of $-u \cos 45^\circ$. They are on a direct, head-on collision course.

### The Retracing Condition

The problem provides a massive, critical clue: after the collision, particle $P$ "retraces its path".

What does this mean physically? To travel perfectly backwards along the exact same parabola it just carved through the air, particle $P$'s velocity vector must be instantly flipped by $180^\circ$. Therefore, immediately after the crash, $P$'s new horizontal velocity becomes $-u \cos 45^\circ$.

### Conservation of Linear Momentum

During the intense mid-air collision, the forces the particles exert on each other are massive, but they are internal to the system. In the horizontal direction, there are absolutely no external forces acting on them (gravity only acts downwards).

This is our cue to deploy the Conservation of Linear Momentum. The total horizontal momentum just before the crash must equal the total horizontal momentum just after.

$$m_P v_{Px} + m_Q v_{Qx} = m_P v'_{Px} + m_Q v'_Q$$

Let's carefully substitute our masses and velocities into this master equation. The mass of $P$ is $20	ext{ g}$ and $Q$ is $40	ext{ g}$.

$$20(u \cos 45^\circ) + 40(-u \cos 45^\circ) = 20(-u \cos 45^\circ) + 40(v'_Q)$$

Notice the negative sign for $Q$'s initial velocity, as it is moving to the left. Now, let's simplify the algebra.

$$-20 u \cos 45^\circ = -20 u \cos 45^\circ + 40 v'_Q$$

When we cancel the identical $-20 u \cos 45^\circ$ terms from both sides, we are left with a stunningly simple result:

$$0 = 40 v'_Q \implies v'_Q = 0$$

### The Fate of Particle Q

The math has spoken. Immediately after the collision, particle $Q$'s horizontal velocity is exactly zero. Since the collision occurred at the peak, its vertical velocity was already zero.

Particle $Q$ has been completely stopped dead in its tracks. It hangs in the air for a fleeting moment, devoid of all kinetic energy, and then begins to fall straight down under the influence of gravity. Because the collision happened exactly halfway between $A$ and $B$, particle $Q$ will land exactly at the midpoint.

### The Final Descent

We know where it lands, but how long does the fall take? To find the time of descent, we first need to calculate the maximum height, $H$, from which it falls.

$$H = \frac{u^2 \sin^2	heta}{2g}$$

Substituting our launch parameters:

$$H = \frac{(49)^2 \sin^2 45^\circ}{2 	imes 9.8}$$

$$H = \frac{2401 	imes (\frac{1}{\sqrt{2}})^2}{19.6} = \frac{2401 	imes 0.5}{19.6} = 61.25	ext{ m}$$

Particle $Q$ is now in free fall from a height of $61.25	ext{ m}$. Using the standard kinematic equation for a drop from rest ($S = \frac{1}{2}gt^2$), we can solve for the time $t$:

$$t = \sqrt{\frac{2H}{g}}$$

$$t = \sqrt{\frac{2 	imes 61.25}{9.8}} = \sqrt{\frac{122.5}{9.8}} = \sqrt{12.5}$$

$$t \approx 3.53	ext{ s}$$

And there we have it! Particle $Q$ will strike the ground exactly midway between the launch points, $3.53	ext{ s}$ after the epic mid-air collision. The elegance of physics lies in how perfectly these independent concepts—kinematics and momentum—interlock to reveal the final truth.

Visualized Solution

$Visualizing the Setup$

$The Range of Projectiles$

$Calculating the Range$

$Deducing the Collision Point$

$Velocities Just Before Collision$

$The Retracing Condition$

$Conservation of Linear Momentum$

$Setting Up the Momentum Equation$

$Solving for Q’s Final Velocity$

$The Fate of Particle Q$

$Calculating Maximum Height$

$Computing the Height$

$Time of Free Fall$

$Final Calculation$

The Sigma Insight: Conservation of Linear Momentum

Analyzing the Setup

The Master Equation

The Geometric Revelation

The Physics of the Peak

The Retracing Condition

Conservation of Linear Momentum

The Fate of Particle Q

The Final Descent

Similar Questions

Three particles $A, B$ and $C$ of equal mass move with equal speed $v$ along the medians of an equilateral triangle as shown in figure. They collide at the centroid $G$ of the triangle. After the collision, $A$ comes to rest, $B$ retraces its path with the speed $v$ . What is the velocity of $C$ ?

Visualized Solution

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

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

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

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

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

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

.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; } Setting Up the Momentum Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for Q’s Final Velocity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Fate of Particle Q

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

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

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

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