Special Relativity Ship Problem

[harrietstowe]

Homework Statement

A ship (attached to reference frame S') passes us (we are standing in reference frame S) with velocity = 0.920c. A proton is fired at speed 0.975c relative to the ship from the front of the ship to the rear. The proper length of the ship is 775 m.
What is the temporal separation between the time the proton is fired and the time it hits the rear wall of the ship according to us?

Homework Equations

u'=(u-v)/(1-((uv)/c^2))
Length contraction:
L=L_o/gamma
t=x/v

The Attempt at a Solution

I thought to let u be .975c
Let v be -.920c
solve for u' and get 2.99684e8 m/s
Observers on Earth will see a shorter length for the ship that is 775m/gamma
gamma = 1/sqrt(1-(.920^2)) and so the contracted length is 303.737m
This contracted length is x and divide it by u' to get 1.014 μs
This though was not the right answer.
Thanks

 

[vela]

You're not accounting for the fact that the rear end of the ship is moving forward toward the proton, so in S', the proton doesn't actually have to travel the entire distance equal to the contracted length of the ship.

Are you familiar with the Lorentz transformations? I usually find problems like these are easier to do using the transformations (not that there's anything wrong with your approach either).

 

[harrietstowe]

If I am interested in the velocity of the proton relative to observers on Earth I am struggling to see why we would have to take a length contraction into account for the proton. If I could get the velocity of the proton for the Earth observers I think i would have this solved and yes I am very familiar with Lorentz transformation equations.

 

[vela]

That's not what I'm saying. Suppose you and I were 10 meters apart and you throw a ball to me at with a horizontal velocity of 10 m/s. If we were both standing still, I'd catch the ball at t=1 s. If I'm running toward you, however, I will catch the ball before 1 second has elapsed, right?

 

[rwisz]

for your submission, but I would like to provide a more thorough explanation for the solution to this problem.

First, we need to understand the concept of relative velocity in special relativity. In classical physics, the relative velocity between two objects is simply the difference between their individual velocities. However, in special relativity, the relative velocity is calculated using the relativistic velocity addition formula:

u' = (u + v) / (1 + (uv/c^2))

Where u' is the relative velocity between two frames of reference, u is the velocity of an object in one frame, v is the velocity of that frame relative to another frame, and c is the speed of light.

In this problem, we have a ship (frame S') moving with a velocity of 0.920c relative to us (frame S). A proton is fired from the front of the ship to the rear with a velocity of 0.975c relative to the ship. We are asked to find the temporal separation between the time the proton is fired and the time it hits the rear wall of the ship according to us.

To solve this problem, we need to use the time dilation formula, which states that the time interval between two events in one frame is longer than the time interval between the same two events in another frame if the frames are moving relative to each other. This can be expressed as:

t' = t / gamma

Where t' is the time interval in the moving frame, t is the time interval in the stationary frame, and gamma is the Lorentz factor, which is equal to 1/sqrt(1-(v^2/c^2)).

In this problem, the time interval t' between the firing of the proton and its impact on the rear wall of the ship is what we are looking for. The time interval t is the proper time interval, which is the time interval measured by an observer in the same frame as the events. We can calculate the proper time interval using the distance and velocity of the proton, which is given by the formula:

t = x / v

Where x is the distance traveled by the proton and v is its velocity relative to the ship. Since the ship is moving at a velocity of 0.920c, the velocity of the proton relative to us (frame S) is:

v = (0.975c - 0.920c) / (1 - (0.975c*0.920