Question

Frame S' has a speed v = 0.6c relative to S. Clocks are adjusted so that t=t'=0 at x=x'=0.

Answer 1

An event occurs in S at \[t = 2 \times 10 ^{-7}s\] at a point for which x = 50m. At what time does the event occur in S'?

If a second event occurs at 10m and \[3 \times 10^{-7}s\] in S, what is the time interval between the events as measured in S'?

Answer 2

Bonus question:
Two observers are in different reference frames, moving at speed v relative to each other. It is true that,
a) both measure the speed of light as c
b) they disagree on the magnitude of v
c) they agree about which events are simultaneous d) all of the above
e) only a) and c) are correct

Answer 3

\(\gamma = \dfrac{1}{\sqrt{1-(v/c)^2}} =\dfrac{1}{\sqrt{1-(0.6)^2}} = 1.25 \)

Answer 4

perhaps this may help

Answer 5

4 am here, I will catch you later :) have fun!

Answer 6

well, good night :P

Answer 7

for the bonus question, assuming the frames are inertial,
the speed of light must be c in both frames according to speed of light postulate

they must agree on the magnitude of relative velocity : if A's velocity in B's frame is v, then B's velocity in A's frame would be -v.

simultaneity is broken, observers in different inertial frames don't agree about the simultaneity of events in general

Answer 8

I luv physics but cant wait to study this in the future......😊😝

Answer 9

@Astrophysics medal for your new profile pic <3

Answer 10

@baru Haha, thanks!

@ganeshie8 Yes! Also the math was sort of a distraction for the first problem, I wanted you to notice that time dilation is not applicable because since time dilation assumes both events occur at the same position in frame S note the x1=x2.

Answer 11

@YumYum247 It is fun :D

Answer 12

\(\Delta t' = (3.5-1.25)\times 10^{-7} = 2.25 \times 10^{-7} s\)

\(\Delta t = (3-2)\times 10^{-7} = 1 \times 10^{-7} s\)

looks the S frame is in slow motion when viewed from S' frame