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1. Homework Statement [/B]
This is a problem that was in my Physics HW.
Two powerless rockets are on a collision course. The rockets are moving with speeds of 0.800c and 0.600c and are initially ## 2.52 × 10^{12} ## m apart as measured by Liz, an Earth
observer, as shown in Figure P1.59. Both rockets are 50.0 m in length as measured by
Liz. (a) What are their respective proper lengths? (b) What is the length of each rocket as
measured by an observer in the other rocket? (c) According to Liz, how long before the
rockets collide? (d) According to rocket 1, how long before they collide? (e) According
to rocket 2, how long before they collide? (f) If both rocket crews are capable of total
evacuation within 90 minutes (their own time), will there be any casualties?
My doubt is on letters (d) and (e). I don't know if I am supposed to apply the time Lorentz transformation using the value obtained in (c) or if I should calculate this time based on the speed each rocket sees the other approaching and the distance using length contraction. I found two answers on the internet.
## L = L_{0}\sqrt {1  \frac {v^2} {c^2}} ##
## \Delta t' = \frac 1 {\sqrt {1  \frac {v^2} {c^2}}} \Delta t ##
## V' = \frac {u  V_x} {1  \frac {uV_x} {c^2}} ##
By using the mentioned equations, I obtained that (a) ## L_1 = 83.3 m ## and ## L_2 = 62.5 m ## . (b) ## L_1 = 27.0 m ## in the frame of rocket 2 and ## L_2 = 21.0 m ## in the frame of rocket 1. (c) ## \frac {\Delta S} {v_1 + v_2} = 6000 sec = 100 min ## .
When it comes to letter (d) that something goes wrong. my first approach to it was to use the length contraction observed by 1 and divide it by the speed 1 sees 2 approaching. ## L = L_{0}\sqrt {1  \frac {v^2} {c^2}} = 2.52 \times 10^{12} \times 0.6 = 1.512 \times 10^{12} ## and ## V' = \frac {u  V_x} {1  \frac {uV_x} {c^2}} = \frac { 0.8c  (  0.6c)} {1  \frac { ( 0.48c^2)} { c^2 }} = 0.945c ## . Dividing these results we have ## \frac {L} {V'} = 5,333 sec = 88.9 min ## . Although, using ## \Delta t' = \frac 1 {\sqrt {1  \frac {v^2} {c^2}}} \Delta t ## , where t' is Liz's time of 100 min, we obtain ## 100 min = 1.6666 \times \Delta t ## and ## \Delta t = 60 min ##. This same problem happens when I try to solve (e), and I've taken a look at several solutions on the internet, being half of them solved in the first way, and half in the second. Shouldn't these results agree? If not, why?
This is a problem that was in my Physics HW.
Two powerless rockets are on a collision course. The rockets are moving with speeds of 0.800c and 0.600c and are initially ## 2.52 × 10^{12} ## m apart as measured by Liz, an Earth
observer, as shown in Figure P1.59. Both rockets are 50.0 m in length as measured by
Liz. (a) What are their respective proper lengths? (b) What is the length of each rocket as
measured by an observer in the other rocket? (c) According to Liz, how long before the
rockets collide? (d) According to rocket 1, how long before they collide? (e) According
to rocket 2, how long before they collide? (f) If both rocket crews are capable of total
evacuation within 90 minutes (their own time), will there be any casualties?
My doubt is on letters (d) and (e). I don't know if I am supposed to apply the time Lorentz transformation using the value obtained in (c) or if I should calculate this time based on the speed each rocket sees the other approaching and the distance using length contraction. I found two answers on the internet.
Homework Equations
## L = L_{0}\sqrt {1  \frac {v^2} {c^2}} ##
## \Delta t' = \frac 1 {\sqrt {1  \frac {v^2} {c^2}}} \Delta t ##
## V' = \frac {u  V_x} {1  \frac {uV_x} {c^2}} ##
The Attempt at a Solution
By using the mentioned equations, I obtained that (a) ## L_1 = 83.3 m ## and ## L_2 = 62.5 m ## . (b) ## L_1 = 27.0 m ## in the frame of rocket 2 and ## L_2 = 21.0 m ## in the frame of rocket 1. (c) ## \frac {\Delta S} {v_1 + v_2} = 6000 sec = 100 min ## .
When it comes to letter (d) that something goes wrong. my first approach to it was to use the length contraction observed by 1 and divide it by the speed 1 sees 2 approaching. ## L = L_{0}\sqrt {1  \frac {v^2} {c^2}} = 2.52 \times 10^{12} \times 0.6 = 1.512 \times 10^{12} ## and ## V' = \frac {u  V_x} {1  \frac {uV_x} {c^2}} = \frac { 0.8c  (  0.6c)} {1  \frac { ( 0.48c^2)} { c^2 }} = 0.945c ## . Dividing these results we have ## \frac {L} {V'} = 5,333 sec = 88.9 min ## . Although, using ## \Delta t' = \frac 1 {\sqrt {1  \frac {v^2} {c^2}}} \Delta t ## , where t' is Liz's time of 100 min, we obtain ## 100 min = 1.6666 \times \Delta t ## and ## \Delta t = 60 min ##. This same problem happens when I try to solve (e), and I've taken a look at several solutions on the internet, being half of them solved in the first way, and half in the second. Shouldn't these results agree? If not, why?
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