Three space ships are seen flying past your classroom on a space station. When m

Question

Three space ships are seen flying past your classroom on a space station. When moving, Ship 1 is measured to be only 78% of its length compared to when it is at rest at the space station. Ship 2 is moving away from the space station at 0.6c and Ship 3 is moving toward the space station (in the opposite direction as Ship 2) at 0.8c.

1)How fast is Ship 1 moving relative to the space station?

a) 0c

            b) 0.23c

            c) 0.43c

*d) 0.63c

            e) 0.83c

2)After 3 hours pass on your watch, how much time do measure to have passed on Ship 2’s clocks?

a) 1.80 hours

*b) 2.40 hours

            c) 3.00 hours

            d) 3.75 hours

e) 5.00 hours

3)How fast does Ship 2 measure Ship 3 to be moving?

a) 0.20c

            b) 0.38c

*c) 0.95c

            d) 1.00c

            e) 1.40c

4)You send a radio signal (at the speed of light) to Ship 3. How fast does Ship 3 measure it approaching?

a) 0.2c

            b) 0.8c

*c) 1.0c

The answers are marked by * but would like an explanation on how these answers were obtained. Thanks in advance!

Explanation / Answer

(1)
L = Lo * sqrt(1 - v^2/c^2)
0.78*Lo = Lo * sqrt(1 - v^2/c^2)
0.78^2 = 1 - v^2/c^2
v = c * sqrt(1 - 0.78^2)
v = 0.625 c
v =0.63 c
Correct Option - (d)

(2)
Using Formula For Time Dilation -
T = To / ( sqrt(1 - v^2/c^2))
3hr = To / sqrt(1-0.6c^2/c^2)
3 hr = To / sqrt(1 -0.6^2)
To = 3hr * sqrt(1 -0.6^2)
To = 2.4 hr
Time Passed on Ship 2 Clock = 2.4 hr

(3)
Using Formula for Relative Velocity in Relativity -

v = w - u/ (1 - w*u/c^2)
w = 0.8 c
u = - 0.6 c
.
v = (0.8 c + 0.6 c)/(1 + 0.8*0.6 c^2/c^2)
v = 0.95 c
Fast does Ship 2 measure Ship 3 to be moving, v = 0.95 c

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