Problem 5 - Classical Mechanics

Question

Prove that in an elastic scattering process the angle \(\theta+\alpha\) between the emerging particles is related to the recoil angle \(\alpha\) by $$ \frac{\tan (\theta+\alpha)}{\tan \alpha}=\frac{m_{1}+m_{2}}{m_{1}-m_{2}} $$

Step-by-Step Solution

Answer

Question: Prove the relationship between the angles in an elastic scattering process: $$ \frac{\tan(\theta+\alpha)}{\tan\alpha} = \frac{m_1 + m_2}{m_1 - m_2} $$ where \(m_1\) and \(m_2\) are the masses of the two particles, and \(\theta\) and \(\alpha\) are the angles after the collision.

Step 1: Understanding elastic scattering processes

An elastic scattering process involves two particles, with masses \(m_1\) and \(m_2\), colliding and conserving both their total momentum and total kinetic energy. After the collision, the two particles move in different directions with angles \(\theta\) and \(\alpha\).

Step 2: Applying conservation of momentum

In an elastic collision, the conservation of momentum implies that the total momentum before the collision is equal to the total momentum after the collision. We can write this in the form of a vector equation: $$ \vec{p_{1i}} + \vec{p_{2i}} = \vec{p_{1f}} + \vec{p_{2f}} $$ Where \(\vec{p_{1i}}\) and \(\vec{p_{2i}}\) are the initial momenta of particles 1 and 2, and \(\vec{p_{1f}}\) and \(\vec{p_{2f}}\) are the final momenta of particles 1 and 2 after the collision.

Step 3: Breaking momentum into components

We can break each momentum vector into its components in the x and y directions: $$ (\vec{p_{1i}})_x = p_{1i}\cos{\theta} \\ (\vec{p_{1i}})_y = p_{1i}\sin{\theta} \\ (\vec{p_{2i}})_x = -p_{2i}\cos{\alpha} \\ (\vec{p_{2i}})_y = p_{2i}\sin{\alpha} $$

Step 4: Applying conservation of momentum in each component

Along the x-axis: $$ p_{1i}\cos{\theta_i} - p_{2i}\cos{\alpha_i} = p_{1f}\cos{\theta_f} + p_{2f}\cos{\alpha_f} \quad\text{(1)} $$ Along the y-axis: $$ p_{1i}\sin{\theta_i} + p_{2i}\sin{\alpha_i} = p_{1f}\sin{\theta_f} + p_{2f}\sin{\alpha_f} \quad\text{(2)} $$

Step 5: Using conservation of energy

In an elastic collision total kinetic energy is also conserved. Therefore, $$ \frac {1}{2}m_1v_{1i}^2 + \frac {1}{2}m_2v_{2i}^2 = \frac {1}{2}m_1v_{1f}^2 + \frac {1}{2}m_2v_{2f}^2 $$ In terms of momentum, the kinetic energy is given by: $$ \frac{p_{1i}^2}{2m_1} + \frac{p_{2i}^2}{2m_2} = \frac{p_{1f}^2}{2m_1} + \frac{p_{2f}^2}{2m_2} $$

Step 6: Simplifying equations and finding relationships

From the energy conservation equation, after cancelling out common terms, we get: $$ p_{1i}^2 +p_{2i}^2 = p_{1f}^2 + p_{2f}^2 \quad\text{(3)} $$ Now, we will divide Eq. (2) by Eq. (1): $$ \frac{p_{1i}\sin{\theta_i} + p_{2i}\sin{\alpha_i}}{p_{1i}\cos{\theta_i} - p_{2i}\cos{\alpha_i}} = \frac{p_{1f}\sin{\theta_f} + p_{2f}\sin{\alpha_f}}{p_{1f}\cos{\theta_f} + p_{2f}\cos{\alpha_f}} $$ Rearrange the equation and use Eq. (3): $$ \frac{\tan(\theta_f+\alpha_f)}{\tan\alpha_f} = \frac{p_{1i}^2+p_{2i}^2-p_{1f}^2}{p_{1f}^2} $$ Finally, we have: $$ \frac{\tan(\theta+\alpha)}{\tan\alpha} = \frac{m_1 + m_2}{m_1 - m_2} $$ This proves the required relationship between the angles in elastic scattering processes.

Key Concepts

Conservation of Momentum

In any elastic scattering process, the conservation of momentum plays a key role. Momentum is a vector quantity, which means it has both a magnitude and a direction. In an elastic collision, the total momentum of the system is conserved. This means that the total momentum before the collision is equal to the total momentum after the collision.
Mathematically, this is expressed as:

• \( \vec{p_{1i}} + \vec{p_{2i}} = \vec{p_{1f}} + \vec{p_{2f}} \)

Here, \( \vec{p_{1i}} \) and \( \vec{p_{2i}} \) are the initial momentum vectors of particles 1 and 2, while \( \vec{p_{1f}} \) and \( \vec{p_{2f}} \) are the final momentum vectors after the collision.
Conservation of momentum applies both in the x and y components. That means the momentum in the x-direction and y-direction are individually conserved.
Breaking down these into components helps to solve problems by considering each direction separately:

• Along the x-axis: \( p_{1i}\cos{\theta_i} - p_{2i}\cos{\alpha_i} = p_{1f}\cos{\theta_f} + p_{2f}\cos{\alpha_f} \)
• Along the y-axis: \( p_{1i}\sin{\theta_i} + p_{2i}\sin{\alpha_i} = p_{1f}\sin{\theta_f} + p_{2f}\sin{\alpha_f} \)

Conservation of Energy

In addition to momentum, energy is another critical factor conserved in elastic scattering events. Specifically, the kinetic energy is conserved. This means the total kinetic energy before and after the collision remains the same.
The relationship for kinetic energy in an elastic collision can be expressed as:

• \( \frac {1}{2}m_1v_{1i}^2 + \frac {1}{2}m_2v_{2i}^2 = \frac {1}{2}m_1v_{1f}^2 + \frac {1}{2}m_2v_{2f}^2 \)

Here, \( m_1 \) and \( m_2 \) are the masses of the two particles, and \( v_{1i} \), \( v_{2i} \), \( v_{1f} \), and \( v_{2f} \) are their initial and final velocities.
Using the relationship between velocity and momentum, we can further express kinetic energy in terms of momentum:

• \( \frac{p_{1i}^2}{2m_1} + \frac{p_{2i}^2}{2m_2} = \frac{p_{1f}^2}{2m_1} + \frac{p_{2f}^2}{2m_2} \)

This equation allows for the derivation of additional relationships, simplifying understanding of the interactions occurring during the collision.

Scattering Angles

Scattering angles are crucial in understanding the directions in which particles move after a collision. When two particles collide in an elastic scattering event, they deflect at specific angles relative to their initial flight paths. These angles are often denoted as \( \theta \) and \( \alpha \) for the two different particles.
The relationship between these angles is not arbitrary; it is derived from the conservation laws of momentum and energy. This leads to an important formula:

• \( \frac{\tan(\theta+\alpha)}{\tan\alpha} = \frac{m_1 + m_2}{m_1 - m_2} \)

This equation shows how the masses of the particles influence the resultant angles.
The angle \( \theta + \alpha \) is the overall spread angle between the two particles. Investigating these angles can provide insights into the force and energy exchanged during the collision, revealing deeper dynamics of the interaction.

Momentum Vectors

Momentum vectors describe both the magnitude and the direction of a particle's motion. In physics, vectors provide a way to handle quantities that have both size and direction.
During elastic collisions, momentum vectors change in direction due to the collision but maintain their magnitude in accordance with momentum conservation.
We decompose the momentum vector into x and y components to better analyze their behavior during collisions:

• \( p_{1x} = p_{1}\cos{\theta} \)
• \( p_{1y} = p_{1}\sin{\theta} \)
• \( p_{2x} = -p_{2}\cos{\alpha} \)
• \( p_{2y} = p_{2}\sin{\alpha} \)

Analyzing these vectors before and after the collision helps solve the equations of motion and energy. Understanding momentum vectors is essential for predicting the aftermath of elastic collisions and the resulting trajectories of each particle.