physics student... |
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Hi all, there's just a few questions I'm having trouble with regarding relativity and I was wondering if someone could help me out. 1. a) How much work W must be done on a particle with a mass of m to accelerate it from rest to a speed of 0.894c ? I thought this was just: E = (1/sqrt(1- (u/c)^2))(mc^2) as the total energy (and E = mc^2 is the rest energy). (u/c)^2 is .894^2, and when I followed through on that, it turns out I was incorrect (I got something like 0.49 J). b) How much work W must be done on a particle with a mass of m to accelerate it from a speed of 0.894c to a speed of 0.982c ? However, I figure I should understand part a first before doing part b. 2. Consider three galaxies, Alpha, Beta and Gamma. An observer in Beta sees the other two galaxies each moving away from him in opposite directions at speed 0.70c. At what speed would an observer in Alpha see the galaxy Beta moving? In this problem, I wanted to define my frames: s' as Alpha, s as Beta (Alpha is the moving observer). v' = (v-u)/(1-uv/c^2) In this case, using the Lorenz velocity transformation, I thought v = 0.7c (the velocity in frame s), u = -0.7c (velocity of s' relative to s): v' = (v-u)/(1-(uv)/c^2) However, obviously this has to be incorrect as v' gives something like 2.7c. Am I defining my frames incorrectly? 3. An electron has a speed of 0.837 c. Through what potential difference would the electron need to be accelerated (starting from rest) in order to reach this speed? (c = 3.00 × 10^8 m/s, e = 1.60 × 10^-19 C, m el = 9.11 × 10^-31 kg) I thought this was similar to problem 1, but I figure I should make sure I know how to do problem one. Is my potential difference, delta V, just the difference in energy from the final speed vs. the initial energy (which would just be mc^2)? 4. Assume that a certain city consumes electrical energy at an average rate of 2.0 × 10^9 W. What would be the mass change in producing enough energy to keep this city running for 12 weeks? (c = 3.00 × 10^8 m/s) I know Force/Area = Power. Would I just try to find something in terms of period (since I have a time number) and try to apply this? Not quite sure how to go about this one. Any help with any of these questions would be greatly appreciated. Thank you so much for looking! |

physics student... |
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Want to update: Solved #4 Looking at #3, my set up is: (1/2)(mass electron)(v^2) = q (V) I want to solve for V. I plug in all my numbers, and get V = 1.79 x 10^5 V, which doesn't work as it is not one of the answers listed (multiple choice question). |

physics student... |
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Update: Got #1 too.. |

physics student... |
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Still stuck on 2 and 3 though, so if anyone knows those, much appreciated! |

Bad Wigins |
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2 is the easiest one. If beta sees alpha moving away at 0.7c, then alpha sees beta moving away at 0.7c too. Maybe you meant how fast is gamma going from alpha's perspective. That's the question they usually ask. |

physics student... |
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Cripes, you're right! I accidentally read it as how fast is gamma going from alpha's perspective.. so it's my fault for reading it wrong. Thanks! In that case, it's only #3 that I'm not totally sure on... is my expression for E wrong (i.e. not 1/2 mv^2, but is it (gamma -1)mc^2 where gamma = 1/sqrt(1-(u/v)^2)? Also, here's one that I seem to be off on as well, any suggestions on this one? Suppose that the speed of light in a vacuum (c), instead of being a whooping 3x10^8m/s, was a rather sluggish 40mph. How would that affect everyday life? Throughout this problem we are going to assume that c=40mph and that time dilation is in full effect. Let's start by assuming that it is fairly easy to accelerate to speeds close to 40mph. We will also ignore gravity throughout this problem. Otherwise, the earth (with an escape velocity of 11 km/s) would have turned into a black hole long ago. a) Suppose that a bored student wants to go to a restaurant for lunch, but she only has an hour in which to go, eat, and get back in time for class. Considering that it usually takes about 30 minutes in most restaurants to get served and to eat, what is the farthest restaurant the student can go to without being late for class? Assume in this part that the student has a car that can accelerate to its top speed in a negligible amount of time. Also, the local speed limit is 30 mph and the student would not like to get a speeding ticket. This is easy, 7.5 miles. b) The restaurant the student likes to go to doesn't have any clocks. As a result, the only way that the student can keep track of the time so as not to be late is to keep an eye on her wristwatch. According to the student's watch, how much time does she actually have for lunch if she wants to go to the furthest restaurant (including travel time)? So she has 15 "real world" minutes... I need to find out how many minutes she has in this world. So is my t = 2d/c ? That doesn't seem to work, would I be using time dilation? Where t = (gamma)t_0, where t is 15 min and gamma is 1/sqrt(1 -(u/c)^2)? c) Now, suppose the student wishes to bring back some ice cream from the restaurant for her friends at school, but since it is such a hot day, the ice cream will melt away in the car in only 5 minutes. How fast will the student have to drive back to get the ice cream to her friends before it completely melts? Not sure how to go about doing this one... |

Bad Wigins |
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b. gamma = 1/sqrt(1-9/16) = 1/sqrt(5/16) = sqrt(1/(5/16)) = sqrt(16/5) = 4/sqrt5. So, she should figure that the there-and-back trip of 30 minutes by the office/restaurant perspective should correspond to 30/(2x4/sqrt5) = 30sqrt5/8 minutes In the restaurant, her watch is back in the office perspective. So she's got, by her watch, 30 + 30sqrt5/8 minutes. Less than 40 minutes. c. find v and t such that 5 minutes = t/sqrt(1-(v/40)^2) and vt = 7.5 |

the average male runner |
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woah. bump. we have the exact same hw. |

Bad Wigins |
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30/4sqrt5 = 7.5sqrt5 minutes and she's got, by her watch, 30 + 7.5sqrt5 minutes. Or 46.77 minutes. b. find v and t' such that vt' = 7.5 t' = gamma x t = t/(sqrt(1-(v/40)^2)) = 1/(12sqrt(1-(v/40)^2)) where t = 5 minutes = 1/12 hour, and t' is the corresponding dilated time in the non-traveler's perspective. ok dammit, now I'm pretty sure it's right. The answer should be 36.55 mph. |

DontFeedTheTroll |
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Your equation for #2 is incorrect. |

physics student... |
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Hmm, part b is actually 49.9 minutes. Is it perhaps because you aren't converting units correctly? |

Bad Wigins |
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no, it's because of this error
then the travel time is 7.5sqrt7 = 19.84 minutes. shouldn't affect c though. |

physics student... |
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Awesome, I actually solved it as well and got the same thing you did which looks like it's the correct answer. Do you mind taking a look at one last problem? I feel like I'm on the right path but my answer isn't working out. Suppose that you decide to look at a known binary star system. The system is too far away to resolve the individual stars, so it appears to be just one point of light. By looking at the spectrum of the system, though, you should be able to use the Doppler shift to determine some parameters of the two-star system. The first thing that one notices is that there appear to be two hydrogen spectra shifted relative to each other owing to the motion of the stars relative to each other. Furthermore, one notices that the position of these lines will shift over time as the stars orbit around each other. Consider a binary star system that has bright lines at 656.72 and 656.86 nm. Over the course of six months the 656.72-nm line moves to longer wavelength and the 656.86-nm line moves to shorter wavelength, until finally the two have swapped (i.e., the spectrum of the star system again shows bright lines at 656.72 and 656.86 nm). Assume that the stars are of roughly equal mass and moving in a circular orbit with axis perpendicular to the line connecting the star system to the earth as shown in the figure: http://imgur.com/UY6HSRR Determine the mass M of each star. My method: So using the relativistic Dopper shift equation, I got that the speed of the two stars are 119 km/s and 183 km/s... so the relative speed is 64 km/s (you just subtract the two). Next, I set my forces equal to each other: (Gmm)/r^2 = (mv^2)/r I know that v = (2pi r)/T, and T, my period, is 1 year. So I solve for r and plug it back into the initial equation and get: m = (v^3)T/(2pi G) T = 3.156 x 10^7 s, v = 6.4 x 10^4 m/s, G is the gravitational constant. I end up with something like 2 x 10^31 kg which is incorrect... but it seems like my method works/makes sense. I know that the speeds of each star I have are correct (119 and 183 km/s), because that was a previous part of the question that I got right. |

physics student... |
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bump |

physics student... |
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bump |

Bad Wigins |
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Doesn't your school have paid tutors? I mostly post on these boards to argue with people and occasionally insult them. Homework help's really not my thing. Sorry. |