MT2ReviewSol - Problem 1 Let us choose two coordinate...

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Unformatted text preview: Problem 1 Let us choose two coordinate frames as follows: the snake is at rest in frame S' with its tail at the origin x' = 0 and its head at x' = 100cm. the two hatchets are at rest in frame S, the left one at origin x = 0 and the Γright one at x = 100cm. A. The graduate student's view in frame S at time t = 0 is below. The snake is moving to the right at speed v = .6c Fig. 1 B. As observed in frame S, the two hatchets bounce simultaneously at t = 0. At this time the snake's tail is at x = 0 and his head must therefore be at x = 80cm. [ You can check this using the transformation x' = Γ( x - vt ); with x = 80cm and t = 0, you will find that x' = 100cm, as required. ] Thus, as observed in S, the experiment is as show in Fig1. In particular, the student's prediction is correct and the snake is unharmed. Therefore, the snake's argument must be wrong. To understand what is wrong with the snake's argument, we must examine the coordinates , especially the times, at which the two hatchets bounce, as observed in the frame S'. The left hatchet falls at Event A ( x = 0 ). According to a Lorentz transformation, the coordinates of this event, as seen in S' are t 'A Γ tA x 'A Γ xA vx A c2 =0 vt A = 0 As expected, the left hatchet falls immediately beside the snake's tail, at the time t ' A 0. On the other hand, the right hatchet falls at tB =0 and x A = 100cm. Thus, as seen in S', it falls at a time given by the Lorentz transformation as t 'B Γ tB vxB c2 = 5 4 0 0.6 c 100 cm c2 = -2.5 ns. We see that, as measured in S', the two hatchets do not fail simultaneously. Since the right hatchet falls before the left one, it does not necessarily have to hit the snake, even though they were only 80cm apart ( in this frame ). In fact , the position at which the right hatchet falls is given by the Lorentz transformation as x 'B Γ xB vtB 5 4 100 cm 0 = 125 cm. and, indeed , the hatchet misses the snake. t 'A Γ tA x 'A Γ xA prob2.nb 2 vx A c2 =0 vt A = 0 As expected, the left hatchet falls immediately beside the snake's tail, at the time t ' A 0. On the other hand, the right hatchet falls at tB =0 and x A = 100cm. Thus, as seen in S', it falls at a time given by the Lorentz transformation as t 'B Γ tB vxB c2 = 5 4 0 0.6 c 100 cm c2 = -2.5 ns. We see that, as measured in S', the two hatchets do not fail simultaneously. Since the right hatchet falls before the left one, it does not necessarily have to hit the snake, even though they were only 80cm apart ( in this frame ). In fact , the position at which the right hatchet falls is given by the Lorentz transformation as x 'B Γ xB vtB 5 4 100 cm 0 = 125 cm. and, indeed , the hatchet misses the snake. Problem 3 Problem 5 a) The total energy : 100 Watts = 100 Joules / sec * 1000 sec = 105 Joules For photons p = E / c. So the total momentum p = 105 3 108 = 0.33 x 10 3 kg . m/s b) Since momentum is conserved we can set mblackbody vbb = pphotons = 0.33 x 10 Plugging in mblackbody = 1 kg 1000 3 kg . m/s. we solve for vbb = 0.33 m/s . ( We can intuit by looking at the physical situation that the black body will not get to relativistic speeds so we'll use the nonrelativistic formula for momentum here. ) 1 c) Kinetic energy = 2 mv2 = 1 2000 0.33 m s Almost all of the energy produces heat. 2 = 5.5x 10 5 Joules ...
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This note was uploaded on 10/05/2009 for the course PHYSICS 7C taught by Professor Lin during the Spring '08 term at University of California, Berkeley.

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