What is the velocity of car 2 when car 1 is going 1/2 the speed of light?

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Homework Help Overview

The discussion revolves around the relativistic effects of length contraction as two cars move relative to a stationary observer, specifically focusing on the scenario where car 1 is traveling at half the speed of light. Participants are attempting to determine the velocity of car 2 based on the lengths of the cars as perceived by an observer.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are exploring the relationship between the lengths of the cars as observed in different reference frames, questioning the application of the length contraction formula. There are attempts to equate the lengths of the cars while considering relativistic effects, leading to discussions about the correct placement of terms in the equations used.

Discussion Status

The discussion is ongoing, with participants providing feedback on each other's mathematical approaches and questioning the assumptions made regarding the reference frames. Some guidance has been offered regarding the correct formulation of the equations, but no consensus has been reached on the final velocity of car 2.

Contextual Notes

Participants are working under the constraints of relativistic physics, specifically addressing how length contraction affects measurements in different inertial frames. There is an emphasis on ensuring that the proper length is correctly identified and applied in calculations.

ehrenfest
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car 2 is twice as long as car 1 when they are at rest

a stationary policeman observes that car 2 is the same length as car 1 as car 2 passes car 1 going through a speed trap.

car 1 is going 1/2 the speed of light

I tried to solve this with

L 1/sqrt(1-(1/2)^2) = 2L/sqrt(1-v^2/c^2) and I got an imaginary anwer
 
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ehrenfest said:
car 2 is twice as long as car 1 when they are at rest

a stationary policeman observes that car 2 is the same length as car 1 as car 2 passes car 1 going through a speed trap.

car 1 is going 1/2 the speed of light

I tried to solve this with

L 1/sqrt(1-(1/2)^2) = 2L/sqrt(1-v^2/c^2) and I got an imaginary anwer

your equation is wrong. the sqrt(1-v^2/c^2) should be in the numerator on the RHS

[edit: and the LHS too]
 
Last edited:
Why? I took the stationary policeman as my intertial reference frame. To him L 1/sqrt(1-(1/2)^2) is the length of car 1 and 2L/sqrt(1-v^2/c^2) is the length of car 2. These quantities must be equal, right?
 
ehrenfest said:
Why? I took the stationary policeman as my intertial reference frame. To him L 1/sqrt(1-(1/2)^2) is the length of car 1 and 2L/sqrt(1-v^2/c^2) is the length of car 2. These quantities must be equal, right?

As olgran said that sqrt should be in the numerator on both sides. The length contracts... Can you explain the steps you took to get it in the denominator?
 
ehrenfest said:
Why? I took the stationary policeman as my intertial reference frame. To him L 1/sqrt(1-(1/2)^2) is the length of car 1 and 2L/sqrt(1-v^2/c^2) is the length of car 2. These quantities must be equal, right?

Nope. For example, to him the length of car one is

<br /> \frac{L}{\gamma(c/2)}\;.<br />

Which is *not* the same as what you have written.

Don't get your gammas upsidedown.
 
remember length contracts when moving and proper length is always biggest. and since \gamma \geq 1 always, you can quickly check whether your answer make sense or not... for \frac{L}{\gamma} \leq L
 
I see. So the velocity of car 2 must be sqrt(13/16)c, right?
 
ehrenfest said:
I see. So the velocity of car 2 must be sqrt(13/16)c, right?

yep.
 

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