Spaceship A moves past the earth at 0.80 c to the west....

Spaceship $A$ moves past the earth at $0.80 c$ to the west. Spaceship $B$ approaches $A,$ moving to the east. Both spaceship crews measure their relative speed of approach to be $0.98 c .$ What mass would the crews of both spaceships measure for the standard kilogram, kept at rest on the earth, (a) according to classical physics and (b) according to the special theory of relativity?

Key Concepts

Relativistic Velocity Addition

Unlike the simple arithmetic addition of velocities in classical mechanics, special relativity requires that velocities be combined using the relativistic velocity addition formula. This ensures that the resulting relative speed does not exceed the speed of light and properly accounts for the effects of high-speed motion, which is essential when analyzing scenarios where objects are moving at significant fractions of the speed of light.

Classical Mass Invariance

In classical mechanics, mass is considered an intrinsic and unchanging property of an object regardless of its state of motion. Thus, if an object has a mass of one kilogram at rest, it is always measured to be one kilogram in all inertial reference frames according to classical physics, independent of its speed relative to any observer.

Lorentz Factor and Relativistic Mass

Special relativity introduces the concept of the Lorentz factor, which accounts for time dilation and length contraction effects at speeds close to the speed of light. When an object is moving relative to an observer, its measured mass (often called the relativistic mass) increases by a factor of gamma (? = 1/?(1 - v²/c²)). This means that the mass of an object in motion will appear larger than its rest mass, reflecting the energy required to accelerate it.

Transcript

00:01 Hello, and in this question here we're going to be looking at this effect in special relativity where the mass measured increases when there's a relative velocity between the observer making the measurement and the object being measured.

00:15 This mass measured by the observer is equal to gamma times m0, where m0 was the rest mass, that is the mass and observer would measure if there was no relative velocity between the two objects, and gamma is equal to 1 divided by the square root of 1 minus v squared or the c squared, where v is the relative velocity between the observer and the object.

00:39 So we're going to set up the question, and in the question we have an observer on earth.

00:44 We're going to call this s prime.

00:46 And we have a spaceship which we're going to be, which is moving away from earth, and it's moving to the left.

00:53 And there's a relative velocity between the second spaceship and earth of 0 .8c, and this coordinate system we're going to call, s prime.

01:04 Now s prime is the coordinate system where a is at rest, so the spaceship a is at rest, and s is the coordinate system where the observer on earth is at rest.

01:15 According to an observer on spaceship a, there is a spaceship b traveling towards the earth, and this has a relative velocity v -prime is equal to 0 .98c.

01:29 So in order to find out that the mass of which both spaceships, a person on both spaceships would measure the one kilogram on earth.

01:41 We need to determine the relative velocities between earth and the spaceships.

01:47 So to do this, well, first of all, we already know the relative velocity between, so u is the relative velocity between earth and spaceship a, which i'm just going to say a.

02:12 And we're going to say v is the relative velocity between earth and the spaceship b.

02:27 Okay, cool.

02:28 So we know what u is.

02:30 We're given in the question, that's minus 0 .8c.

02:33 But in order to find v, well, v, so we have to find, we have a velocity in s prime.

02:41 So we're given v prime, and we need to find out what that velocity is in s prime.

02:46 And now from the book, there's a formula that relates velocities from different coordinate frames.

02:52 And this is, so this is v is equal to v prime plus u divided by 1 plus u times v prime over c squared.

03:02 Okay, so u is the relative velocity between frames between s and s prime...

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