In a scattering experiment, a particle of mass 2m collides with another particle of mass m, which is

Q: Why do we have to split the momentum into x and y components?

Momentum is a vector quantity, meaning it has both magnitude and direction. In a 2D collision, the particles move off at angles, so the momentum is not just along a single line. To apply the conservation law correctly, we must ensure momentum is conserved independently in every spatial dimension (x and y).

Q: How do we know kinetic energy is conserved here? Isn't energy usually lost in collisions?

In macroscopic real-world collisions (like car crashes), kinetic energy is indeed lost to heat, sound, and deformation. However, the problem explicitly states the collision is 'perfectly elastic'. In physics, 'elastic' is the specific keyword that guarantees zero loss of kinetic energy.

Q: Why did we choose to eliminate the angle $\phi$ instead of $\theta$?

The question specifically asks for the maximum angular deviation $\theta$ of the heavier particle. The angle $\phi$ of the lighter particle is an intermediate variable that we don't need for our final answer. Eliminating it simplifies our system of equations to focus only on the variables we care about.

Q: How did a simple collision problem turn into a quadratic equation?

This is a classic mathematical pattern in physics! When you combine linear equations (like momentum, which depends on $v$) with quadratic equations (like kinetic energy, which depends on $v^2$), substituting one into the other naturally generates a quadratic polynomial. The roots of this polynomial represent the physically possible final states.

Q: I don't understand why the discriminant ($b^2 - 4ac$) must be greater than or equal to zero.

Think about what the variable $v_{1f}$ represents: it's the final speed of the particle. Speed must be a real, physical number (you can't have an imaginary speed like $3i$ m/s). The quadratic formula uses the square root of the discriminant. If the discriminant is negative, the square root is imaginary, meaning the physical scenario is impossible. Therefore, for the collision to actually happen, the discriminant must be positive or zero.

Question

In a scattering experiment, a particle of mass $2m$ collides with another particle of mass $m$, which is initially at rest. Assuming the collision to be perfectly elastic, the maximum angular deviation $	heta$ of the heavier particle, as shown in the figure, in radians is:

Options:

(A) $\pi$

(B) $	an^{-1}\left(\frac{1}{2}ight)$

(C) $\frac{\pi}{3}$

(D) $\frac{\pi}{6}$

Sigma Percentile · Accepted Answer

### The Anatomy of a Collision

Imagine a cosmic game of billiards. A heavy particle, boasting a mass of $2m$, is hurtling through space with an initial velocity $v_1$. Lying perfectly still in its path is a lighter particle of mass $m$. When they collide, it is a perfectly elastic event—no energy is lost to heat or sound; it is a pure exchange of momentum and kinetic energy.

Our mission is to determine the maximum angle $	heta$ by which the heavier particle can be deflected from its original path. To solve this, we must translate the physical reality of the collision into the rigorous language of mathematics.

### The Laws of the Universe: Momentum and Energy

In the absence of external forces, the universe demands that the total linear momentum of the system remains constant. Because the particles scatter in two dimensions, we must enforce this conservation law along both the horizontal ($x$) and vertical ($y$) axes.

Let the final velocity of the heavier particle be $v_{1f}$ at an angle $	heta$, and the final velocity of the lighter particle be $v_{2f}$ at an angle $\phi$.

Along the $x$-axis, the initial momentum must equal the sum of the final horizontal momentum components:
$$2m v_1 = 2m v_{1f} \cos	heta + m v_{2f} \cos\phi$$

Along the $y$-axis, the system starts with zero momentum. Therefore, the upward momentum of the heavier particle must perfectly cancel the downward momentum of the lighter one:
$$0 = 2m v_{1f} \sin	heta - m v_{2f} \sin\phi$$

Furthermore, the 'perfectly elastic' nature of the collision gives us a third powerful constraint: the conservation of kinetic energy. The initial kinetic energy of the heavy particle is redistributed between the two particles after the impact:
$$\frac{1}{2}(2m)v_1^2 = \frac{1}{2}(2m)v_{1f}^2 + \frac{1}{2}m v_{2f}^2$$

Simplifying this energy equation by dividing out the common mass terms, we get:
$$2v_1^2 = 2v_{1f}^2 + v_{2f}^2$$

### The Art of Elimination

We now possess a system of three equations, but we are burdened with an unwanted variable: the angle $\phi$ of the lighter particle. In physics, elegance often comes from eliminating what we do not need to observe.

We can isolate the terms containing $\phi$ from our momentum equations:
$$v_{2f} \cos\phi = 2(v_1 - v_{1f} \cos	heta)$$
$$v_{2f} \sin\phi = 2v_{1f} \sin	heta$$

By squaring both equations and adding them together, we exploit the fundamental trigonometric identity $\sin^2\phi + \cos^2\phi = 1$, causing $\phi$ to vanish entirely!
$$v_{2f}^2 = 4(v_1 - v_{1f} \cos	heta)^2 + 4v_{1f}^2 \sin^2	heta$$
$$v_{2f}^2 = 4(v_1^2 + v_{1f}^2 \cos^2	heta - 2v_1 v_{1f} \cos	heta) + 4v_{1f}^2 \sin^2	heta$$
$$v_{2f}^2 = 4(v_1^2 + v_{1f}^2 - 2v_1 v_{1f} \cos	heta)$$

### The Hidden Quadratic Constraint

Now, we substitute this beautiful expression for $v_{2f}^2$ back into our simplified kinetic energy equation:
$$2v_1^2 = 2v_{1f}^2 + 4(v_1^2 + v_{1f}^2 - 2v_1 v_{1f} \cos	heta)$$

Dividing the entire equation by 2 and rearranging the terms, a profound mathematical structure emerges:
$$3v_{1f}^2 - (4v_1 \cos	heta)v_{1f} + v_1^2 = 0$$

This is a standard quadratic equation in terms of $v_{1f}$, the final velocity of the heavier particle.

Here lies the crux of the problem: for this collision to be a physical reality, the final velocity $v_{1f}$ must be a real, measurable number. Algebra dictates that for a quadratic equation to possess real roots, its discriminant ($D = b^2 - 4ac$) must be greater than or equal to zero.

### The Final Verdict

Let us enforce this reality condition by calculating the discriminant:
$$(-4v_1 \cos	heta)^2 - 4(3)(v_1^2) \ge 0$$
$$16v_1^2 \cos^2	heta - 12v_1^2 \ge 0$$

Since the initial velocity $v_1$ is non-zero, we can safely divide the inequality by $4v_1^2$:
$$4 \cos^2	heta - 3 \ge 0$$
$$\cos^2	heta \ge \frac{3}{4}$$

Taking the square root (and knowing that the deflection angle $	heta$ must be acute in this forward-scattering scenario), we find:
$$\cos	heta \ge \frac{\sqrt{3}}{2}$$

Because the cosine function decreases as the angle increases from $0$ to $90^\circ$, this inequality implies that the angle $	heta$ cannot exceed the angle whose cosine is $\frac{\sqrt{3}}{2}$.

$$	heta \le \frac{\pi}{6}$$

Thus, the maximum angular deviation the heavier particle can experience is exactly **$\frac{\pi}{6}$ radians**. The universe's strict laws of momentum and energy simply will not allow it to turn any further!

Select Answer:

Visualized Solution

Visualizing the Collision

Conservation of Linear Momentum

Momentum along X-axis

Momentum along Y-axis

Conservation of Kinetic Energy

Eliminating $ϕ$

Substituting into Energy Equation

Forming the Quadratic Equation

Condition for Real Roots

Solving the Inequality

Finding Maximum Angle $θ$

The Way Forward

The Sigma Insight: Oblique Collision

The Anatomy of a Collision

The Laws of the Universe

The Art of Elimination

The Hidden Quadratic Constraint

The Final Verdict

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