Problem 3 - University Physics with Modern Physics

Question

A particle of rest mass \(m_{0}\) travels at a speed \(v=0.20 c\) How fast must the particle travel in order for its momentum to increase to twice its original momentum? a) \(0.40 c\) c) \(0.38 c\) e) \(0.99 c\) b) \(0.10 c\) d) \(0.42 c\)

Step-by-Step Solution

Answer

Answer: (c) \(0.38c\)

Step 1: Relativistic momentum formula

Using the formula for relativistic momentum, we have: \(p = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}}\) Where - \(p\) is the momentum of the particle - \(m_0\) is the rest mass of the particle - \(v\) is the speed of the particle - \(c\) is the speed of light

Step 2: Calculate initial momentum

We are given the initial speed of the particle, \(v = 0.20c\). Let's calculate its initial momentum, \(p_0\): \(p_0 = \frac{m_0 (0.20c)}{\sqrt{1 - \frac{(0.20c)^2}{c^2}}} = \frac{0.20m_0c}{\sqrt{1 - 0.04}} = \frac{0.20m_0c}{\sqrt{0.96}}\)

Step 3: Set up the equation to find the final speed

We want to find the speed at which the momentum of the particle becomes twice its initial momentum. So, the equation we need to solve is: \(2p_0 = \frac{m_0v_f}{\sqrt{1 - \frac{v_f^2}{c^2}}}\) Where \(v_f\) is the final speed of the particle.

Step 4: Solve the equation

Now, we can substitute the initial momentum (\(p_0\)) into the equation: \(2\left(\frac{0.20m_0c}{\sqrt{0.96}}\right) = \frac{m_0v_f}{\sqrt{1 - \frac{v_f^2}{c^2}}}\) We can simplify and solve for \(v_f\): \(\frac{0.40m_0c}{\sqrt{0.96}} = \frac{m_0v_f}{\sqrt{1 - \frac{v_f^2}{c^2}}}\) Notice that \(m_0c\) can be canceled from both sides: \(\frac{0.40}{\sqrt{0.96}} = \frac{v_f}{\sqrt{1 - \frac{v_f^2}{c^2}}}\) Now, we can square both sides: \(0.40^2 \cdot 0.96 = (1 - \frac{v_f^2}{c^2})v_f^2\) Simplify and solve for \(v_f\): \(0.1536c^2 = v_f^2 - \frac{v_f^4}{c^2}\) \(v_f^4 - 0.8464c^2v_f^2 + 0.9536c^4 = 0\) This is a quadratic equation in terms of \(v_f^2\). We can use any method to solve it (e.g., factoring, completing the square, or quadratic formula). The solution is: \(v_f^2 = 0.146c^2\) \(v_f = \sqrt{0.146c^2} = 0.382c\)

Step 5: Select the correct answer

From the given choices, the speed at which the particle's momentum becomes twice its original value is closest to \(0.38c\). The correct answer is (c) \(0.38c\).

Key Concepts

Special Relativity

Special relativity is a fundamental theory in physics introduced by Albert Einstein in 1905. It revolutionized our understanding of space, time, and energy. Unlike classical mechanics, which operate under the assumption that speeds are much less than the speed of light, special relativity accounts for phenomena that occur at or near the speed of light.
Key points of special relativity include:

• Time Dilation: Time appears to move slower for an object in motion relative to a stationary observer.
• Length Contraction: An object in motion is measured to be shorter in the direction of motion as observed from a stationary frame.
• Relativity of Simultaneity: Two events that occur simultaneously in one frame of reference may not be simultaneous in another moving frame of reference.

These effects are particularly pronounced and must be taken into account when dealing with speeds close to the speed of light, such as in the calculations of relativistic momentum.

Lorentz Factor

The Lorentz factor, commonly denoted by the Greek letter gamma (\(\gamma\)), is a key component in the equations of special relativity. It represents how much time, length, and relativistic mass increase by a factor as the object’s speed approaches the speed of light.

Formula:

The Lorentz factor is calculated as:\[\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\]Here,

• \(v\) is the velocity of the object.
• \(c\) is the speed of light.

As velocity \(v\) increases towards the speed of light \(c\), the Lorentz factor grows significantly, affecting time dilation and relativistic mass.Considerations involving the Lorentz factor are crucial for describing and analyzing phenomena when speeds approach \(c\), especially for calculating relativistic momentum as done in the original exercise.

Speed of Light

The speed of light, denoted as \(c\), is a fundamental constant in physics, valued at approximately 299,792,458 meters per second. It is the maximum speed at which all energy, matter, and information in the universe can travel. The exact value of the speed of light is crucial in calculations concerning special relativity.
Significance includes:

• As an upper limit, it ensures that no object with mass can ever reach or exceed this speed.
• It serves as a foundational element in Einstein's theory of special relativity, affecting the perception of time and space.
• The speed of light is used in the determination of the Lorentz factor and thus influences calculations like momentum in relativistic speeds.

Understanding the speed of light is essential for solving problems involving relativistic momentum because it defines both the framework and limitations for particle velocities and interactions at high speeds.