Physics 121 Hour Exam #5 Solution

Transcription

1 Physics 2 Hou xam # Solution This exam consists of a five poblems on five pages. Point values ae given with each poblem. They add up to 99 points; you will get fee point to make a total of. In any given poblem, points ae not necessaily evenly divided between the pats. Two pages ae supplied at the end of the booklet in case you need exta wokspace.. [2 points] In a supenova explosion, the inne coe of a sta collapses while the oute layes of the sta ae blown away at high speed in all diections, as suggested by the dawing below. Centuies afte such an explosion, the debis fom the oute laye can still be seen to be flying away fom the emaining coe of the sta. Befoe xplosion Afte xplosion Conside (i) the pe-explosion mass of the oiginal sta and (ii) the post-explosion mass consisting of the mass of the coe that was left behind plus the mass of the debis that was blown away afte the explosion. Which of these is geate, pe-explosion mass o post-explosion mass, o ae they the same? Biefly justify you answe. negy (the fist component of fou-momentum) is conseved. Mass is one fom that enegy can take. Befoe the explosion, all of the enegy is in the est mass of the sta. Afte the explosion, some of this enegy has been tansfomed into kinetic enegy of the debis. Thee is less enegy available fo the mass of the coe and the debis. The post-explosion mass is less.

2 2. [2 points] Can a photon (paticle of light) decay into a single paticle which has mass (m > )? Biefly justify you answe. No. This would violate the consevation of momentum and enegy. Hee s one way to look at it. Let s say that the photon is in the Home Fame, moving to the ight, and it decays into a single massive paticle. The photon had momentum, so this paticle must have momentum; it is moving to the ight at some speed v. Now conside the est fame of this paticle, which is moving at speed v elative to the Home Fame. The paticle has zeo momentum in this fame. Theefoe, by consevation of momentum, the photon must have had zeo momentum befoe the decay as obseved in this fame. But this is impossible: photons always have momentum. Thus it is impossible fo a photon to decay into a single paticle. Hee s anothe way to look at it. Let i be the enegy of the photon. Its momentum is p i = i. Let m f, v f, and γ f be the mass, velocity, and Loentz facto of the massive paticle. Fou-momentum consevation gives: v f The fist component gives i = and the second component gives i = v f. Compaing these, we find v f =. Howeve, as we have discussed in class, v f = is impossible fo a paticle with mass. Thus it is impossible fo a photon to decay into a single paticle. Going beyond this poblem. A staightfowad extension of the above aguments can be used to show that in addition to it being impossible fo a photon to decay into a single massive paticle, it also cannot decay into a goup of moe than one massive paticles. Convesely, neithe a single massive paticle no a goup of paticles can tansfom into a single photon. In any such tansfomation, thee must eithe be at least two initial photons, o at least two final photons, o at least one initial and one final photon. i i 2

3 . [2 points] A paticle with mass M and speed collides with a paticle of mass 7 M at est. They coalesce to fom a single new paticle. What is the mass and speed of this new paticle? v i = γ i = ( ) = 2 v 2i = γ 2i = v f γ i m i γ i m i v i γ 2i m 2i γ 2i m 2i v 2i M M 7 M M M The fist component gives The second component gives = M. v f = M v f = M v f = v f = =.2 M M = This gives Loentz facto So the mass is γ f = ( ) = 2 6 = =. = M m f = M γ f = M. m f = 2.9 M Altenative method. Afte setting up the momentum fou-vectos as above, use m 2 = 2 p 2 to calculate m f = ( ) 2 f 2 p2 f (M = ) 2 M = 2.9 M. Then solve = M to find γ f =., fom which you can calculate v f =.2.

4 . [2 points] A paticle of initial mass m i moves in the +x diection at speed v i, coesponding to Loenz facto γ i. A photon of enegy moves in the +y diection. The photon collides with the paticle and is absobed by it. (This is like two things hitting and sticking togethe, except one of those things is a photon of light). Afte absobing the photon, the paticle moves off at some angle θ, as shown. What is θ? (You answe should be a fomula which depends on, m i, v i, and/o γ i,) Befoe Absoption Afte Absoption v f v i θ photon Wite out consevation of fou-momentum. As will become evident when we do the tigonomety below, the calculation is easiest if we wite the x- and y-components of momentum in tems of v fx and v fy athe than substituting v fx = v f cos θ and v fy = v f sin θ. f p fx p fy The second and thid components give Tigonomety gives γ i m i γ i m i v i p fx = γ i m i v i p fy =. tan θ = p fy = ( ) θ = tan p fx γ i m i v i γ i m i v i Note: this same easoning woks if you wite the second and thid components of the final fou-momentum as v fx and v fy and then use tan θ = v fy /v fx. The factos of cancel out and the math is simila to that done above. θ p xf p yf

5 . [2 points] A spaceship is moving at speed v =.866, which coesponds to γ = 2, towad the ath. A lase beam is emitted by the spaceship in the diection of ath. The beam consists of photons each of which has enegy as measued in the spaceship efeence fame. v=.866 (a) What is the enegy of one of these photons in the ath efeence fame? (You answe should be a fomula which depends on.) (b) In quantum mechanics, it is claimed that the fequency of a photon is popotional to its enegy. This can be witten = hf, whee is the photon enegy, f is its fequency, and h is a physical constant which is the same numbe fo all photons. Conside the Dopple shift of the fequency of the a photon emitted in the spaceship efeence fame and eceived in the ath efeence fame. Is you answe to (a) consistent with this Dopple shift? Justify you answe using fomulas o a calculation. (Don t just answe yes o no ). (a) Fo a photon, momentum equals enegy, so the fou-vecto of this photon in the ocket ship is ath p = [,,, ]. The key point is that the x-component of momentum equals the enegy, p x =. fou-momentum tansfom equations to tansfom these to the eath fame: Use the = γ ` + βp x = 2( ) = 2( +.866) =.72 Incidentally, since = p x fo a ightwad-moving photon in the Home Fame, you could also use the tansfomation equation fo p x to get the same esult: p x = γ `β + p x = 2( ) = 2( ) =.72 = p x =.72 (b) Fo a signal emitted at speed v = β (spaceship) moving towad the eceive (ath), the fequencies ae elated by v f R = + v f If = hf in geneal, then in this poblem we have = hf R (eceived signal, ath fame) and = hf (emitted signal, Spaceship fame). Re-witing these as f R = /h and f = /h and plugging them into the Dopple fomula gives v h = v + v h = + v. Fo emitte moving towad eceive, v is negative, so v =.866: = s (.866) + (.866) =.72 The answe to (a) is consistent with the Dopple shift. They both give the same elation between and. It is stunning that two vey diffeent appoaches to this poblem, one in which the light is teated as a single paticle caying a cetain amount of enegy, and the othe in which the light is teated as a wave of a cetain fequency, give exactly the same esult.