* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
1 Relativity 1.) If you were on a spaceship traveling at 0.5c away from the earth, at what speed would a radio signal from the earth pass you? 2.) A car, at rest, has a length of 5 m. A garage has a length of 4m. How fast must the car be moving in order to fit in the garage (according to an observer standing next to the garage)? 3.) A certain star is 75.0 light years away. How long would it take a spacecraft traveling 0.95c to reach the star from earth, as measured by observers(a) on the earth; (b) on the spacecraft? 4.) When a uranium nucleus at rest breaks apart in the process of fission, the resulting fragments have a total kinetic energy of 200 MeV. How much mass was lost in the process? 5.) An observer on earth sees an alien vessel approach at a speed of 0.6c. The Enterprise comes to the rescue, overtaking the aliens from behind at a speed of 0.90c relative to the earth. At what speed do the aliens observe the Enterprise approaching? 6.) A spacecraft travels at 0.99c to a star 3.787 × 1013 kilometers away. star take from the point of view of someone on the earth? How long will a roundtrip to this 7.) With reference to the previous problem, how long will the roundtrip take for someone in the spaceship, according to someone measuring from the earth? 8.) Now in the reference frame of someone in the spaceship, what is the time taken for the roundtrip as observed by a passenger, and by someone on earth (ignoring the times when the spaceship is accelerating or decelerating). 9.) If one person stays on earth and one person travels to the distant star, who will age more during the trip and by what amount? 10.) Twin A floats freely in outer space. Twin B flies past in a spaceship at speed v0. Just as they pass each other they both start timers at t = 0. At the instant of passing B also turns on his engines so as to decelerate at g. This causes B to slow down and eventually to stop and accelerate back towards A so that when the twins pass each other again B is traveling at speed v0 again. If they compare their clocks, who is younger? 11.) If observer Bill, who is on a train moving with speed 0.6c, waves to Julie at four second intervals as measured in Bill’s frame, how long will Julie measure between waves? √ 12.) Bill and Julie are both now on identical trains. Bill’s train is moving to the right with velocity ( 3/2)c with respect to Julie’s train. Julie measures her train to be 100 meters long. How long does Julie measure Bill’s train to be? How long does Bill measure Julie’s train to be? 13.) Consider the following scenario: two meter sticks, call the SA and SB are oriented parallel to the y axis, some distance apart. The travel towards one another along the x-direction: that is, SA one moves in the positive x-direction and SB moves in the negative x-direction (see Figure). SA has paint brushes on its ends, pointing towards SB such that if SB is longer than SA, for example, it will leave paint marks on SB. Show that there is no length contraction in the y-direction (that is, the sticks both appear 1 meter long to each other)? (Hint: assume this is not the case and derive a contradiction). 14.) Imagine a train going through a tunnel. The train and the tunnel both have length l in their own frame. The train moves through the tunnel with speed v. There is a bomb at the front of the train which is designed to explode when the front of the train passes out the far end of the tunnel. However, there is a disarming sensor located on the back of the train which will disarm the bomb just as the back of the train enters the near end of the tunnel. Will the bomb explode? 2 Solution 6 6.) If we calculate the number of seconds in a year it turns out that 3.787 × 10 16 meters is about 4 light years (the distance light travels in one year at c). The spacecraft is traveling virtually at c, so the trip to the star takes 4 years of earth time. The roundtrip takes 8 years. 7.) According to an observer on the earth, since the spacecraft is moving, its passengers’ time is dilated. The factor by which this occurs isγ = 7.09. The passengers measure less time so, the roundtrip time is (1/7.09) × 8 = 0.14 × 8 = 1.1 years. 8.) The whole point of the twin paradox is that a passenger on the spaceship apparently measures the opposite: that is, that the trip takes 8 years for them, but only 1.1 years for those standing back on the earth. It turns out that this reasoning is incorrect and in fact the passengers measure the same times as an observer on the earth when the (General Relativistic) effects of acceleration and deceleration are taken into account. 9.) As we have seen, the reasoning of the passenger on the spaceship is erroneous because the spaceship is not in an inertial reference frame. The reasoning of the person on earth is correct (the earth is approximately inertial). They measure the passenger as aging less than themselves by an amount 8 − 1.1 = 6.9 years. 10.) This is just a variation of the same problem (that is, the twin paradox as stated earlier). Twin A is in an inertial reference frame so she can successfully apply the logic of Special Relativity to find that B’s time is dilated and hence that B is younger. B is not in an inertial reference frame so the opposite reasoning does not apply, and we conclude that when all the effects of the acceleration are accounted for he must agree with his twin that he is younger. 11.)Bill is in motion so we know that his seconds must be dilated (longer) with respect to Julie’s seconds, by a factor γ. Thus Julie will measure more seconds between waves. What is γ?γ = 5/4 Thus Julie measures 5/4×4 = 5s 12.)Bill’s train is in motion so we would expect it to appear contracted (shorter) by a factor γ to Julie. What is γ? γ = 2. Thus Julie will measure Bill’s train to be 50 meters long. We know that Bill’s train is identical, so because of the equivalence of frames and the symmetry of the situation, we can say that Bill must measure his own train to be 100 meters long and Julie’s to be 50 meters long. 13.) The crucial fact here is that if SA sees SB shorter than (or longer, or equal to) itself, then SB must also see SA as shorter than itself. This arises from the equivalence of all inertial reference frames. Moreover the factors by which each stick sees the other shorter or longer must be the same. First assume, then, that SA sees SB to be longer than itself. Then SA will paint marks on SB. But then, SB must see SA to be longer than itself, so its ends will miss SB and no marks will be painted. Hence we have a contradiction. If we assume that SA sees SB to be shorter than itself, then SA concludes no marks will be made, and SB concludes it will be painted. Again a contradiction. The only way out of this is if both sticks see each other to be the same length, in which case they both agree the brushes will just touch edges of SB. 14.)The answer is yes, the bomb will explode. In the frame of the train, it sees the tunnel as having length l /γ ¡ l so the front of the train will pass out of the tunnel before the rear enters the tunnel (the train has length l in its own frame). One might argue that in the frame of the tunnel, the train appears contracted by the same factor and so in the tunnel frame the train is shorter than the tunnel by a factor γ, so the rear of the train will enter the tunnel 3 before the front passes out, and the bomb will be disarmed. We appear to have a paradox. However, this second line of reasoning is false because it ignores the finite time that any disarming signal must take to move from the rear of the train to the bomb at the front. The fastest such a signal could move is at c. The bomb will be disarmed if and only if a signal traveling at c emitted from the rear of the tunnel at the instant the rear of the train passes, reaches the far end of the tunnel before the train does. Working still in the frame of the tunnel, the signal takes a time l length of the train) /c, and the train takes a time l−l/γ v , since the front of the train is already a distance l /γ (the p p through the tunnel. For the bomb not to explode we need: l/c < l−l/γ , which simplifies to 1 + v/c < 1 − v/c, v which is clearly false. The bomb explodes.