Question

A rocket is traveling at speed \(0.9 c\) along the \(x\) axis of frame \(\delta\). It shoots a bullet whose velocity \(\mathbf{v}^{\prime}\){ measured in the rocket's rest frame } \(\mathcal{S}^{\prime} )\) is \(0.9 c\) along the \(y^{\prime}\) axis of \(\mathcal{S}^{\prime} .\) What is the bullet's velocity (magnitude and direction) as measured in \(\mathcal{S}\) ?

Short answer

The bullet's velocity magnitude is approximately \(0.982c\), and direction is \(23.33^\circ\) from the x-axis.

Step-by-step solution

Identify the Known Variables

We have the speed of the rocket in frame \( \delta \): \( v = 0.9c \). The speed of the bullet in the rocket's rest frame \( \mathcal{S}^{\prime} \) is \( v' = 0.9c \) in the \( y' \) direction. We need to find the velocity of the bullet in the stationary frame \( \mathcal{S} \).

Relativistic Velocity Transformation for Y-Component

Use the relativistic velocity transformation formula for the y-component:\[v_y = \frac{v'_y}{\gamma(1 + \frac{vv'_x}{c^2})}\]where \( v'_y = 0.9c \), \( v'_x = 0 \) (since it's purely in \( y' \) axis), and \( \gamma = \frac{1}{\sqrt{1-(v/c)^2}} \). Calculate \( \gamma \).

Calculating Gamma

The Lorentz factor \( \gamma \) is calculated as:\[\gamma = \frac{1}{\sqrt{1-(0.9c/c)^2}} = \frac{1}{\sqrt{1-0.81}} = \frac{1}{\sqrt{0.19}} \approx 2.294\]

Compute the Transformed Y-Component

Substitute \( v'_y = 0.9c \), \( v'_x = 0 \), \( v = 0.9c \), and \( \gamma \approx 2.294 \) into the transformation:\[v_y = \frac{0.9c}{2.294(1 + 0)} = \frac{0.9c}{2.294} \approx 0.392c\]

Determine the X-Component of Velocity

The x-component of the bullet's velocity in the stationary frame \( \mathcal{S} \) is simply the velocity of the rocket along the x-axis: \( v_x = 0.9c \).

Calculate the Magnitude of the Velocity

Use the Pythagorean theorem to find the magnitude of the velocity:\[v = \sqrt{v_x^2 + v_y^2} = \sqrt{(0.9c)^2 + (0.392c)^2} = c\sqrt{0.81 + 0.153664} = c\sqrt{0.963664}\]Compute the result to find \( v \approx 0.9816c \).

Compute the Direction of the Velocity

The direction \( \theta \) can be found using the tangent function:\[\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) = \tan^{-1}\left(\frac{0.392c}{0.9c}\right) = \tan^{-1}(0.435)\]Calculate \( \theta \approx 23.33^\circ \) relative to the x-axis.

Key concepts

Lorentz Factor

The Lorentz Factor, denoted as \( \gamma \), plays a crucial role in the realm of relativity as it accounts for the effects of moving near the speed of light. It's used for adjusting various measured quantities, such as time, length, and mass, between different inertial reference frames. The factor is calculated using the formula:

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

where \( v \) is the velocity of the moving object, and \( c \) is the speed of light in a vacuum.
In the given exercise, since both the rocket and the bullet travel at \( 0.9c \), the Lorentz factor becomes vital in transforming velocities between the rocket's frame, \( \mathcal{S}' \), and the stationary frame, \( \mathcal{S} \). Calculating \( \gamma \) gives us approximately 2.294, indicating that the relativistic effects are substantial. The Lorentz factor essentially tells us how much the measurement of velocities, or indeed any measurements, actually change for observers who are in different frames of reference moving relative to one another.

Pythagorean Theorem

The Pythagorean Theorem provides a simple yet incredibly useful method to calculate resultant magnitudes in physics problems, particularly when dealing with components in orthogonal directions. The theorem states:

• For a right-angled triangle with sides \( a \) and \( b \), and hypotenuse \( c \): \( c^2 = a^2 + b^2 \)

In the context of velocity calculations, this is used to determine the magnitude of an object's velocity when its components are known.
For our problem with the bullet's velocity, by identifying the x-component as \( 0.9c \) and the transformed y-component as \( 0.392c \), we use the theorem:

• \( v = \sqrt{(0.9c)^2 + (0.392c)^2} = 0.9816c \)

This step illustrates how velocities in different directions can combine to give a resultant velocity in space. It's essential to know how to apply such geometric principles to comprehend the full picture of multicomponent movements.

Tangential Direction Calculation

Understanding the direction of motion is as important as knowing the speed or magnitude, especially in physics problems involving multiple velocity components. The tangential direction of an object's velocity can be determined using basic trigonometry. Specifically, for a right triangle formed with velocity components, we use the inverse tangent function:

• \( \theta = \tan^{-1}(\frac{v_y}{v_x}) \)

This equation determines the angle \( \theta \), which describes the direction of movement relative to one of the axes.
In our scenario, we find the direction of the bullet. By substituting the y-component \( v_y = 0.392c \) and x-component \( v_x = 0.9c \), we calculate:

• \( \theta \approx \tan^{-1}(0.435) \approx 23.33^\circ \)

relative to the x-axis. This angle gives insightful information about how much the bullet deviates from traveling directly along the x-axis. It's crucial for constructing a complete and accurate path of motion in relativistic contexts.