Particle A of mass m_A = m/2 moving along the X-axis with velocity v_0 collides elastically with ano

Q: Why is momentum conserved in this collision?

Momentum is conserved because there are no external forces acting on the two-particle system in the horizontal direction. The forces they exert on each other during the collision are internal.

Q: What does an 'elastic collision' mean mathematically?

An elastic collision means that the total kinetic energy of the system is conserved. Mathematically, this simplifies to the relative velocity of approach being equal to the relative velocity of separation, which means the coefficient of restitution $e$ is equal to 1.

Q: Why do we only solve for particle A's velocity?

The question specifically asks for the change in the de-Broglie wavelength of particle A. Therefore, we only need to find the final velocity of particle A to calculate its new momentum and wavelength.

Q: How does velocity affect the de-Broglie wavelength?

The de-Broglie wavelength is inversely proportional to momentum ($p = mv$). If a particle's velocity decreases, its momentum decreases, which causes its de-Broglie wavelength to increase.

Q: Can the change in wavelength be negative?

Yes, if a particle speeds up, its momentum increases, and its wavelength decreases, resulting in a negative change. In this problem, particle A slows down, so its wavelength increases, making the change positive.

Question

Particle A of mass $m_A = m/2$ moving along the X-axis with velocity $v_0$ collides elastically with another particle B at rest having mass $m_B = m/3$. If both particles move along the X-axis after the collision, the change $\Delta \lambda$ in de-Broglie wavelength of particle A, in terms of its de-Broglie wavelength $\lambda_0$ before collision is

Options:

(A) $\Delta \lambda = \frac{5}{2}\lambda_0$

(B) $\Delta \lambda = 4\lambda_0$

(C) $\Delta \lambda = 2\lambda_0$

(D) $\Delta \lambda = \frac{3}{2}\lambda_0$

Sigma Percentile · Accepted Answer

This problem is a beautiful intersection of classical mechanics and modern physics. We are tasked with finding the change in the de-Broglie wavelength of a particle after it undergoes a perfectly elastic 1D collision. Let's break down the journey step-by-step.

### Analyzing the Setup
Imagine you are observing a microscopic billiards game. We have Particle A, with a mass of $m_A = \frac{m}{2}$, cruising along the X-axis with an initial velocity $v_0$. Waiting for it at rest is Particle B, which has a mass of $m_B = \frac{m}{3}$.

Because there are no external forces acting on our two-particle system in the horizontal direction, we can confidently rely on the **Conservation of Linear Momentum**.

### The Master Equations
The total momentum before the collision must equal the total momentum after the collision. Mathematically, this is expressed as:

$$m_A u_A + m_B u_B = m_A v_A + m_B v_B$$

Substituting our known values into this equation gives us:

$$\frac{m}{2} v_0 + \frac{m}{3} (0) = \frac{m}{2} v_A + \frac{m}{3} v_B$$

By canceling out the common mass $m$ and multiplying the entire equation by 6 to clear the denominators, we arrive at our first crucial relationship between the final velocities:

$$3v_A + 2v_B = 3v_0 \quad 	ext{--- (i)}$$

Now, we need a second equation because we have two unknowns ($v_A$ and $v_B$). The problem states that the collision is **elastically perfect**. This means kinetic energy is conserved, which elegantly simplifies to the condition that the coefficient of restitution $e$ is exactly 1.

The coefficient of restitution is the ratio of the relative velocity of separation to the relative velocity of approach:

$$e = \frac{v_B - v_A}{u_A - u_B}$$

Setting $e = 1$ and plugging in our initial velocities, we get:

$$1 = \frac{v_B - v_A}{v_0 - 0}$$

$$v_B = v_0 + v_A \quad 	ext{--- (ii)}$$

### Solving for the Final Velocity
We are specifically interested in Particle A, so let's substitute equation (ii) into equation (i) to eliminate $v_B$:

$$3v_A + 2(v_0 + v_A) = 3v_0$$

Expanding and solving for $v_A$:

$$5v_A + 2v_0 = 3v_0$$

$$5v_A = v_0 \Rightarrow v_A = \frac{v_0}{5}$$

Particle A has significantly slowed down after bouncing off Particle B.

### The Quantum Connection
Now we transition from classical mechanics to quantum mechanics using the **de-Broglie wavelength** formula, $\lambda = \frac{h}{p} = \frac{h}{mv}$.

The initial wavelength of Particle A, $\lambda_0$, is based on its initial momentum:

$$\lambda_0 = \frac{h}{m_A u_A} = \frac{h}{(\frac{m}{2}) v_0} = \frac{2h}{m v_0}$$

After the collision, Particle A's new wavelength, $\lambda_f$, is based on its new, slower velocity:

$$\lambda_f = \frac{h}{m_A v_A} = \frac{h}{(\frac{m}{2}) (\frac{v_0}{5})} = \frac{10h}{m v_0}$$

Notice the relationship? The final wavelength is exactly 5 times the initial wavelength:

$$\lambda_f = 5 \left( \frac{2h}{m v_0} ight) = 5\lambda_0$$

### Final Calculation
The question asks for the *change* in the de-Broglie wavelength, $\Delta \lambda$. This is simply the final wavelength minus the initial wavelength:

$$\Delta \lambda = \lambda_f - \lambda_0 = 5\lambda_0 - \lambda_0 = 4\lambda_0$$

Because Particle A lost momentum during the collision, its quantum wavelength stretched out significantly. The correct option is (b).

Select Answer:

Visualized Solution

Visualizing the Setup

Conservation of Linear Momentum

Substituting Values

Simplifying Momentum Equation

Coefficient of Restitution

Applying Elastic Collision Condition

Solving for Final Velocity of A

de-Broglie Wavelength Formula

Initial Wavelength of Particle A

Final Wavelength of Particle A

Change in Wavelength

The Way Forward

The Sigma Insight: Matter Waves and de Broglie Relation

Analyzing the Setup Imagine you are observing a microscopic billiards game

The Master Equations The total momentum before the collision must equal the total momentum after the collision

Solving for the Final Velocity

The Quantum Connection

Final Calculation The question asks for the change in the de-Broglie wavelength, $Δ λ$

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