Problem 81 - Fundamentals of Physics

Question

A particle with mass \(m\) has speed \(c / 2\) relative to inertial frame S. The particle collides with an identical particle at rest relative to frame \(S\). Relative to \(S\), what is the speed of a frame \(S^{\prime}\) in which the total momentum of these particles is zero? This frame is called the center of momentum frame.

Step-by-Step Solution

Answer

The speed of frame \( S' \) is \( \frac{c}{4} \).

Step 1: Define the system

Two particles are involved. One has mass \( m \) and speed \( \frac{c}{2} \) relative to frame \( S \), and the second identical particle is at rest in the same frame.

Step 2: Calculate total momentum in frame S

The total momentum in frame \( S \) is the sum of the momentum of the first particle and the second particle. Since momentum \( p \) is given by \( p = mv \), the momentum of the first particle is \( m \cdot \frac{c}{2} = \frac{mc}{2} \) and the momentum of the second particle is \( 0 \). Thus, total momentum \( p_{total} = \frac{mc}{2} + 0 = \frac{mc}{2} \).

Step 3: Determine velocity of center of momentum frame

In the center of momentum frame, the total momentum must be zero. To find the speed \( v \) of this frame relative to \( S \), the velocity of each particle in this new frame must be adjusted such that their total momentum is zero. This requires solving the equation:\[ m \left( \frac{c}{2} - v \right) + m \left( 0 - v \right) = 0 \] Solving for \( v \) gives \( \frac{c}{2} - v = v \), so, \( \frac{c}{2} = 2v \), therefore \( v = \frac{c}{4} \).

Step 4: Verify the solution

Verify that the calculated speed of the center of momentum frame indeed results in zero momentum. In frame \( S' \), the velocity of the moving particle is \( \frac{c}{2} - \frac{c}{4} = \frac{c}{4} \), and the velocity of the particle initially at rest is \( -\frac{c}{4} \). Thus, the total momentum in frame \( S' \) is \( m \cdot \frac{c}{4} + m \cdot \left(-\frac{c}{4}\right) = 0 \), verifying the solution.

Key Concepts

Momentum Calculation

Momentum is a crucial concept in both classical and relativistic mechanics. It helps us understand motion and interactions. For any object with mass, momentum is calculated by multiplying its mass by its velocity. This is represented mathematically as \( p = mv \). Here, \( p \) is momentum, \( m \) is mass, and \( v \) is velocity. In the given exercise, we have two particles. The first has momentum because it is moving with velocity \( \frac{c}{2} \) relative to a reference frame \( S \). Its momentum is calculated as \( p_1 = m \times \frac{c}{2} = \frac{mc}{2} \). The second particle is at rest, meaning it has zero momentum unless acted upon. Hence, its momentum is \( p_2 = m \times 0 = 0 \). The total momentum in this system, considering both particles, is simply \( p_{total} = \frac{mc}{2} + 0 = \frac{mc}{2} \).

Relativistic Mechanics

Relativistic mechanics accounts for the effects of traveling at speeds close to that of light. When dealing with high speeds, such as a fraction of the speed of light \( c \), classical mechanics might not suffice as it doesn't account for certain effects like time dilation and mass increase. In this context, special relativity becomes relevant. It allows us to understand how physics behaves under such conditions. The formula for momentum slightly adjusts in the relativistic regime. While classical momentum uses \( p = mv \), relativistic momentum incorporates the Lorentz factor, adjusted for high-speed effects:

• \( p = \gamma mv \)
• \( \gamma = \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} \)

However, since the problem given is within manageable speeds (\( c/2 \), and \( c/4 \)), the classical calculation serves well for clarity and ease.

Reference Frames

Reference frames are essential for observing and measuring motion. They provide a background to compare velocities and accelerations.In this exercise, frame \( S \) is the inertial frame where one particle moves and the other is stationary. When calculating momentum, it's crucial to determine it relative to a specific frame. Here, \( S \) serves that purpose.To find the center of momentum frame, denoted \( S' \), means finding a reference where the total momentum is zero. This requires adjusting that frame's velocity so that both particles' momentum cancel each other out. The exercise shows that in \( S' \), the new frame moves at \( \frac{c}{4} \) relative to \( S \). Hence, the center of momentum frame gives a true balance point for analysis where forces seem symmetrical.

Conservation of Momentum

The principle of conservation of momentum suggests that the total momentum in a closed system remains constant if no external forces are applied.In the problem, despite the collision, if no other forces act, momentum is conserved. That means we calculate pre-collision momentum, then adjust velocities to ensure the total remains unchanged post-collision, albeit possibly redistributed between particles.In our setup:

• Initial Total: \( \frac{mc}{2} \)
• After accounting for the center of momentum frame: each particle has half the speed in opposite directions, resulting in zero total momentum.

This validates the conservation law and the correctness of the center of momentum velocity calculated.