ps02_solns

ps02_solns - Physics 1311 Spring 2010 PS #2 Solutions Page...

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Physics 1311 Spring 2010 PS #2 Solutions Page 1 of 9 Physics 1311 Problem Set #2 Solutions Problem 1: You slide a coin across the foor, and observe that it slows down and eventually stops. A sensitive thermometer shows that its temperature increased. What can we conclude? As your textbook explains in detail, interactions cause changes. In this example, both the temperature and the motion of the coin changed. (Its speed changed even if its direction remained the same.) Both of these changes indicate that the coin interacted with another object or set of objects. Problem 2: In which situations is it reasonable to use the approximate Formula For the momen- tum oF an object, instead oF the Full relativistically correct Formula? The criterion for determining whether we can approximate the momentum as ± p = m ± v is how close γ is to being 1. This in turn depends on how fast the object is moving compared to the speed of light. Well, clearly for highway travel and jet planes, the speed is negligible, so the approximation works. A neutron traveling at about 3000 m/s is still traveling at 0 . 001% the speed of light, which is pretty darn slow. What about a vacuum tube electron traveling at 0 . 02 c ? That sounds pretty fast, but γ =1 / 1 - 0 . 02 2 . 0002, so it’s still a very good approximation to use Newtonian momentum. In fact, γ < 1 . 01 even for speeds up to 14% of light speed. So it’s really only the outer-space proton that requires the full relativistic momentum formula in this case. Problem 3: An object oF mass 1 . 34 mg is traveling at a speed oF 0 . 480 c . (a) Calculate the Newtonian momentum oF the object. (b) Calculate the correct relativistic momentum oF the object. (c) Using the technique oF Taylor expansion, ±nd an approximate expression For the relativistic Factor γ that is correct to order v 2 . (d) Calculate the percentage error (including sign) between the momentum calculated using your approximate expression For γ , and the actual momentum oF the object. (Just enter a number, without the percent sign.) (a) The Newtonian momentum is simply p = mv = (1 . 34 × 10 - 6 kg)(0 . 480)(3 . 00 × 10 8 m/s) = 192 . 96 kg m/s . = 193 kg m/s . (b) The relativistic momentum has an extra factor γ compared with the Newtonian mo- mentum. For the given numbers, γ = 1 ± 1 - v 2 /c 2 = 1 1 - 0 . 480 2 . 140 . Using the result of part (a), we compute the relativistic momentum to be p = γmv = (1 . 140)(192 . 96) = 219 . 96 kg m/s . = 220 kg m/s .
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Physics 1311 Spring 2010 PS #2 Solutions Page 2 of 9 (c) There are several ways to compute the small-speed Taylor series (polynomial) expan- sion for the relativistic factor γ . The most straightforward way is to write γ ( v ) . = γ (0) + γ ± (0) v + γ ± (0) 2 v 2 + . . . , for which we need the Frst and second derivatives of γ with respect to v . Using the chain rule, we obtain: γ ( v )= 1 ± 1 - v 2 /c 2 γ ± ( v v c 2 ² 1 - v 2 c 2 ³ 3 / 2 γ ± ( v 1 c 2 ² 1 - v 2 c 2 ³ 3 / 2 + 3 v 2 c 4 ² 1 - v 2 c 2 ³ 5 / 2 .
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ps02_solns - Physics 1311 Spring 2010 PS #2 Solutions Page...

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