Question

Consider a one-dimensional collision at relativistic speeds between two particles with masses \(m_{1}\) and \(m_{2}\). Particle 1 is initially moving with a speed of \(0.700 c\) and collides with particle \(2,\) which is initially at rest. After the collision, particle 1 recoils with a speed of \(0.500 c\), while particle 2 starts moving with a speed of \(0.200 c\). What is the ratio \(m_{2} / m_{1} ?\)

Short answer

Answer: The ratio of mass m2 to mass m1 is 4.

Step-by-step solution

Calculate initial and final momenta

In this step, we need to calculate the initial and final momenta for both particles. Since \(m_1\) and \(m_2\) are unknowns, we will express the momenta in terms of their masses and velocities only. Initial momenta: - For particle 1: \(p_{1i} = \gamma_{1i} m_{1} v_{1i}\) - For particle 2: \(p_{2i} = \gamma_{2i} m_{2} v_{2i}\) (particle 2 is at rest, so \(v_{2i}=0\) and \(p_{2i}=0\)) Final momenta: - For particle 1: \(p_{1f} = \gamma_{1f} m_{1} v_{1f}\) - For particle 2: \(p_{2f} = \gamma_{2f} m_{2} v_{2f}\)

Write the conservation of momentum equation

Using the conservation of momentum principle, we can write the following equation: \(p_{1i} + p_{2i} = p_{1f} + p_{2f}\) Now we can substitute the momenta expressions obtained in step 1: \(\gamma_{1i} m_{1} v_{1i} = \gamma_{1f} m_{1} v_{1f} + \gamma_{2f} m_{2} v_{2f}\)

Solve the equation for \(\frac{m_2}{m_1}\)

To solve this equation for \(\frac{m_2}{m_1}\), we can first divide both sides of the equation by \(m_1v_{1i}\): \(\gamma_{1i} = \frac{\gamma_{1f}v_{1f}}{v_{1i}} + \frac{\gamma_{2f}m_{2}v_{2f}}{m_1v_{1i}}\) Next, we can isolate the term with \(\frac{m_2}{m_1}\): \(\frac{m_2}{m_1} = \frac{\gamma_{1i}v_{1i} - \gamma_{1f}v_{1f}}{\gamma_{2f}v_{2f}}\) Now, we can substitute the velocities and calculate the relativistic factors: \(\gamma_{1i} = \frac{1}{\sqrt{1-\frac{(0.7c)^2}{c^2}}} = \frac{1}{\sqrt{1-0.49}} = 2\) \(\gamma_{1f} = \frac{1}{\sqrt{1-\frac{(0.5c)^2}{c^2}}} = \frac{1}{\sqrt{1-0.25}} = \frac{4}{3}\) \(\gamma_{2f} = \frac{1}{\sqrt{1-\frac{(0.2c)^2}{c^2}}} = \frac{1}{\sqrt{1-0.04}} = \frac{5}{3}\) Finally, substitute the values of the relativistic factors and velocities into the ratio equation: \(\frac{m_2}{m_1} = \frac{2\cdot0.7c - \frac{4}{3}\cdot0.5c}{\frac{5}{3}\cdot0.2c} = \frac{(1.4 - \frac{4}{3}\cdot0.5)c}{\frac{1}{3}c} = \frac{(1.4 - \frac{2}{3})c}{\frac{1}{3}c} = 4\) The ratio \(\frac{m_2}{m_1}\) is equal to 4.

Key concepts

One-Dimensional Collision

In physics, when we speak of a one-dimensional collision, we're describing a scenario where two particles interact along a single straight line. This type of collision allows for a simplified analysis since it reduces the variables to a minimum, focusing purely on the linear motion.

During a one-dimensional collision, two key quantities are often of interest: the velocities of the particles and their masses. It is also important to clarify whether the collision is elastic or inelastic, as this will affect how energy is conserved during the event. However, regardless of the collision's nature, the total momentum along the line of motion must be conserved.

The problem provided illustrates a one-dimensional collision at relativistic speeds between two particles. When dealing with such high velocities (significant fractions of the speed of light), we must rely on relativistic mechanics to accurately describe the motion, as classical mechanics would lead to incorrect results.

Relativistic Speeds

At relativistic speeds, close to the speed of light (denoted by the symbol c), the standard Newtonian mechanics learned at school begins to break down. Instead, we use Einstein's theory of relativity to analyse motion.

One crucial concept from this theory is the relativistic factor, Gamma (γ), which is given by the equation \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \), where v is the velocity of the particle. As the velocity increases towards the speed of light, gamma increases, which has significant implications for momentum and energy.

For example, at relativistic speeds, simple addition and subtraction of velocities no longer apply. Instead, velocities combine through a more complex formula, and momentum increases disproportionately, becoming significantly greater than the classical calculation \( p = mv \). The inclusion of the relativistic factor in the momentum calculation ensures that our description of the motion remains accurate at high velocities, as shown in the exercise solution.

Conservation of Momentum in Relativity

Momentum conservation is a foundational concept in both classical and relativistic physics. It states that the total momentum of a closed system remains constant if no external forces are acting upon it. However, in the realm of relativistic speeds, we must consider the relativistic momentum defined by \( p = \gamma mv \).

In our problem, we apply the conservation of momentum principle to a relativistic one-dimensional collision. To do so, we calculate the initial and final momentum for each particle, taking into account their respective relativistic factors. This allows us to set up an equation representing momentum conservation throughout the collision.

It's important to note that while the velocities and momenta are subject to relativistic effects, the conservation laws themselves are not altered by the high speeds. This enduring principle provides a crucial link between the familiar physics of everyday experiences and the less intuitive physics governing ultra-fast particles.