Problem 22 - College Physics

Question

At what speed is the momentum of a particle three times as great as the result obtained from the nonrelativistic expression \(m v ?\)

Step-by-Step Solution

Answer

The speed is \( \frac{2c\sqrt{2}}{3} \).

Step 1: Understand the Problem

We're being asked to determine the speed at which the momentum of a particle is three times its nonrelativistic momentum, given by the formula \( m v \). This involves considering both nonrelativistic and relativistic momentum expressions.

Step 2: Write Down Nonrelativistic Momentum

The nonrelativistic momentum for a particle is given by \( p = m v \), where \( m \) is the mass of the particle, and \( v \) is its velocity.

Step 3: Write Down Relativistic Momentum

The relativistic momentum is given by \( p = \frac{m v}{\sqrt{1 - \frac{v^2}{c^2}}} \), where \( c \) is the speed of light. We need this to find when it is three times the nonrelativistic momentum.

Step 4: Set Up Equation for Momentum

According to the problem, the relativistic momentum is three times the nonrelativistic momentum: \( \frac{m v}{\sqrt{1 - \frac{v^2}{c^2}}} = 3 m v \). We can simplify this equation by canceling \( m \) and \( v \) (assuming \( v eq 0 \)).

Step 5: Simplify and Solve for Velocity

After canceling common terms, we get \( \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = 3 \). Solving for \( v \), we first square both sides to get \( \frac{1}{1 - \frac{v^2}{c^2}} = 9 \). Transfer \( 1 \) to the other side: \( 1 - \frac{v^2}{c^2} = \frac{1}{9} \).

Step 6: Final Calculation

Rearrange to find \( v^2 \): \( \frac{v^2}{c^2} = 1 - \frac{1}{9} = \frac{8}{9} \). Then \( v^2 = \frac{8c^2}{9} \), and thus \( v = c \sqrt{\frac{8}{9}} = \frac{2c\sqrt{2}}{3} \).

Key Concepts

Nonrelativistic Momentum

In classical physics, momentum is a fundamental concept. Nonrelativistic momentum is the product of the mass of an object and its velocity. This simple formula is expressed as:

• \( p = m v \)

Here, \( p \) represents momentum, \( m \) is the mass, and \( v \) is the velocity. This equation is straightforward. However, it's only accurate when the speed is much less than the speed of light.
At higher speeds, like those approaching the speed of light, we need to use relativistic physics instead. This ensures our calculations remain accurate.
Nonrelativistic momentum works well for most everyday applications. For example, calculating the momentum of a moving car or a baseball.
However, when dealing with particles moving at very high speeds, such as electrons in an accelerator, classical momentum doesn't suffice.

Relativity in Physics

Relativity in physics revolutionized our understanding of space and time. Introduced by Albert Einstein, it encompasses both the special and general theories of relativity.
Special relativity, which relates to our topic, describes how space and time are linked for objects moving at constant speeds close to the speed of light.
A key insight from special relativity is that the nonrelativistic expression for momentum must be adjusted to account for the effects of high velocity. This is where the concept of relativistic momentum comes in.

• Relativistic momentum is expressed as: \( p = \frac{m v}{\sqrt{1 - \frac{v^2}{c^2}}} \)
• This equation considers the increase in an object's mass as it approaches the speed of light.

These advanced calculations demonstrate that at high speeds, time appears to slow, and objects contract in length, according to an outside observer.
The relativistic effects ensure that as speeds increase, the laws of physics remain consistent and the same for all observers, a fundamental principle of relativity.

Speed of Light Calculation

The speed of light, denoted as \( c \), is a constant value of approximately 299,792 kilometers per second (or about 186,282 miles per second).
It's not just any speed but a universal speed limit in the universe, as described by the theory of relativity. When we calculate speeds approaching the speed of light, we use relativistic equations.
In our exercise, we sought the speed at which a particle's momentum is three times its nonrelativistic value. By using the formula for relativistic momentum:

• \( \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = 3 \)

We discovered the sought velocity to be \( \frac{2c\sqrt{2}}{3} \). This value is a specific fraction of the speed of light, showcasing how relativistic effects play a significant role as velocities approach the speed of light.
As particles move faster, calculations incorporating the speed of light emphasize the nonlinear nature of velocity and momentum in these extreme conditions.