Particle A of mass m1 moving with velocity (√3 i^ + j^) ms^-1 collides with another particle B of ma

Q: Why is linear momentum conserved in this collision?

Because there are no external forces (like friction or gravity acting horizontally) on the two-particle system. The forces during collision are purely internal.

Q: Can I solve this using components instead of vectors?

Absolutely! You can write separate momentum conservation equations for the x and y directions. It's mathematically equivalent to the vector approach.

Q: Why did we use the $\tan\theta$ method instead of the dot product?

The dot product method involves calculating magnitudes and simplifying a complex fraction, which can be prone to algebraic errors. The $\tan\theta$ method is much faster here because the components yield standard angles ($60^\circ$ and $-45^\circ$).

Q: What does a negative angle mean for $\mathbf{v}_2$?

A negative angle simply means the vector is pointing below the positive x-axis (in the fourth quadrant). It's a clockwise rotation from the reference line.

Q: Is this an elastic collision?

To find out, you would need to calculate the total kinetic energy before and after the collision. If they are equal, it's elastic. In this specific problem, we only needed momentum conservation to find the velocities.

Question

Particle A of mass $m_1$ moving with velocity $(\sqrt{3}\hat{i} + \hat{j})	ext{ms}^{-1}$ collides with another particle B of mass $m_2$ which is at rest initially. Let $\mathbf{v}_1$ and $\mathbf{v}_2$ be the velocities of particles A and B after collision, respectively. If $m_1 = 2m_2$ and after collision $\mathbf{v}_1 = (\hat{i} + \sqrt{3}\hat{j})	ext{ms}^{-1}$, then the angle between $\mathbf{v}_1$ and $\mathbf{v}_2$ is

Options:

(A) $15^\circ$

(B) $60^\circ$

(C) $-45^\circ$

(D) $105^\circ$

Sigma Percentile · Accepted Answer

The problem of finding the angle between two particles after a collision might seem daunting at first glance, especially when vectors are involved. However, by anchoring our thoughts to the fundamental laws of physics, we can unravel the mystery with surprising elegance. Let's embark on this journey together!

### Analyzing the Setup

Imagine you are observing a cosmic game of billiards. Particle A, which is twice as massive as Particle B, is cruising through space with a velocity of $\mathbf{u}_1 = (\sqrt{3}\hat{i} + \hat{j}) 	ext{ m/s}$. Particle B is just chilling there, completely at rest ($\mathbf{u}_2 = 0$).

Suddenly, BAM! They collide. After the collision, Particle A bounces off with a new velocity $\mathbf{v}_1 = (\hat{i} + \sqrt{3}\hat{j}) 	ext{ m/s}$. Our mission is to find the exact angle between the paths of Particle A and Particle B after they part ways.

### The Master Equation

In the absence of any external forces—like friction or a giant space hand pushing them—the total linear momentum of our two-particle system must remain perfectly conserved. This is our master key.

Mathematically, we write this as:
$$ \mathbf{p}_i = \mathbf{p}_f $$
$$ m_1\mathbf{u}_1 + m_2\mathbf{u}_2 = m_1\mathbf{v}_1 + m_2\mathbf{v}_2 $$

Now, let's bring in the specifics. We know $m_1 = 2m_2$ and $\mathbf{u}_2 = 0$. Substituting these into our master equation gives us the raw setup:
$$ 2m_2(\sqrt{3}\hat{i} + \hat{j}) + 0 = 2m_2(\hat{i} + \sqrt{3}\hat{j}) + m_2\mathbf{v}_2 $$

### Unveiling the Final Velocities

Notice how $m_2$ is a common factor in every single term? This is a beautiful moment in physics where the absolute mass doesn't matter, only the ratio does! Let's cancel $m_2$ out and rearrange the equation to isolate $\mathbf{v}_2$:
$$ \mathbf{v}_2 = 2(\sqrt{3}\hat{i} + \hat{j}) - 2(\hat{i} + \sqrt{3}\hat{j}) $$

By grouping the $\hat{i}$ and $\hat{j}$ components together, we get:
$$ \mathbf{v}_2 = 2(\sqrt{3} - 1)\hat{i} + 2(1 - \sqrt{3})\hat{j} $$
$$ \mathbf{v}_2 = 2(\sqrt{3} - 1)(\hat{i} - \hat{j}) 	ext{ m/s} $$

### The Geometric Elegance

Now we have both final velocity vectors. We *could* use the dot product formula ($\cos	heta = \frac{\mathbf{v}_1 \cdot \mathbf{v}_2}{|\mathbf{v}_1| |\mathbf{v}_2|}$) to find the angle between them. But let's be smart about this. Let's look at the geometry!

For Particle A's final velocity $\mathbf{v}_1 = \hat{i} + \sqrt{3}\hat{j}$, the angle it makes with the positive x-axis is:
$$ 	an	heta_1 = \frac{\sqrt{3}}{1} \Rightarrow 	heta_1 = 60^\circ $$

For Particle B's final velocity $\mathbf{v}_2 = 2(\sqrt{3} - 1)(\hat{i} - \hat{j})$, the x and y components are equal in magnitude but opposite in sign. This means:
$$ 	an	heta_2 = \frac{-1}{1} \Rightarrow 	heta_2 = -45^\circ $$

**The total angle between the two vectors is simply the difference between their individual angles:**
$$ 	heta = 	heta_1 - 	heta_2 = 60^\circ - (-45^\circ) = \mathbf{105^\circ} $$

And there we have it! By combining the raw power of momentum conservation with a touch of geometric intuition, we've elegantly solved the problem.

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Visualized Solution

$Initial Setup$

$Conservation of Linear Momentum$

$Substituting Values$

$Solving for v_{2}$

$Angles with X-axis$

$Angle Between Vectors$

$The Way Forward$

The Sigma Insight: Oblique Collision

Analyzing the Setup

The Master Equation

Unveiling the Final Velocities

The Geometric Elegance

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Visualized Solution

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

.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; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for v2​

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

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

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