HW10_soln[1]

HW10_soln[1] - 29-9: Picture the Problem: Due to time...

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29-9: Picture the Problem : Due to time dilation, as a proper time of 1.00 seconds pass on the ship, 1.00 minute passes on Earth. Strategy: The two events to be considered in this case are successive ticks of the clock that is in your spaceship. From your point of view, these events occur at the same place, so you observe the proper time interval t 0 and an observer on Earth measures the dilated time interval t . Solve equation 29-2 for the speed of the ship, where and 0 1.0000 s t  1.0000 m = 60.000 s. t   Solution : 1. Solve equation 29-2 for the speed: 0 0 22 1 1 t tv c vc t t  2. Insert the two time intervals: 2 1.0000 s 1 0.99986 60.000 s c     Insight: If 1.0000 hour had passed according to an Earth observer, instead of 1.0000 minute, the speed would have to be much closer to the speed of light. The speed would have been 0.9999999614 c . 29-26: Picture the Problem : The image shows a rectangular painting as viewed in its rest frame. The painting travels at sufficient speed that a stationary observer views the painting as square. Strategy: The height of the painting will not contract because that dimension is perpendicular to the motion. Therefore, use equation 29-3 to calculate the painting’s speed at which its length will be measured as 80.5 cm. Solution: 1. Solve equation 29-3 for the speed v : 00 1 1 LL vc LL 2. Substitute numerical values:  2 1 80.5 cm 124 cm 0.761 c Insight: It is not possible to make the painting square by moving the painting in the vertical direction, because length contraction would only shorten that dimension. 29-41: Picture the Problem : The image shows two ships (1 and 2) heading toward Earth (3) at speeds and relative to Earth. 13 0.80 23 0.80 v c Strategy: We want to calculate the speed of the one ship relative to the other. Use equation 29-4 to calculate the speed of Ship A relative to Ship B. 12 , v Solution: 1. Multiply both sides of equation 29-4 by the denominator: 12 23 13 2 12 23 2 13 12 23 12 23 1 1 vv v vv c v c v v  2. Isolate the unknown variable and solve: 12 v   2 12 23 13 13 23 13 23 12 23 13 1 0.80 0.80 1 1 0.80 0.80 v c v v cc v c c c   
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Insight: Because the two ships are traveling toward each other, their relative speed is greater than their speed toward Earth, but less than the speed of light. 29-52: Picture the Problem : The classical and relativistic momenta are essentially equal at very low speeds. As the speed increases, the relativistic momentum increases more rapidly than the classical momentum.
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This note was uploaded on 10/13/2011 for the course PHY 102 taught by Professor Alexandrakis during the Fall '06 term at FIU.

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HW10_soln[1] - 29-9: Picture the Problem: Due to time...

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