230011 Christiaan Huygens

8/21/2019 230011 Christiaan Huygens

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Christiaan Huygens' The Motion of Colliding BodiesAuthor(s): Richard J. Blackwell and Christiaan Huygen

Source: Isis, Vol. 68, No. 4 (Dec., 1977), pp. 574-597Published by: The University of Chicago Presson behalf of The History of Science SocietyStable URL: http://www.jstor.org/stable/230011.

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hristiaan Huygens h e M o t i o n oColliding o d i e s

Translated by Richard J. Blackwell"lHypothesis I: Any body already in motion will continue to move perpetually withthe same speed and in a straight line unless it is impeded.2Hypothesis II: Whatever may be the cause of hard3 bodies rebounding frommutual contact when they collide with one another,4 let us suppose that when twobodies, equal to each other and having equal speed, directly collide with one another,each rebounds with the same speed which it had before the collision.Two bodies are said to directly collide when both their motion and their contactoccur in the same straight line which passes through the center of gravity of eachbody.Received July 1974: accepted October 1974.*Department of Philosophy, Saint Louis University, Saint Louis, Missouri 63103.'The De motu corporum ex percussione appears, along with an opposite-page French version, inOeuvres compltes de Christiaan Huygens publi&espar la Societe Hollandaise des Sciences (The Hague:Martinus Nijhoff, 1888-1950), Vol. XVI, pp. 29-91 (hereafter referred to as Oeuvres). A German

translation by F. Hausdorff appears in Ostwald's Klassiker der exakten Wissenschaften, No. 138 (1903),pp. 1-34. The De motu was first published in the Opusculapostuma in 1703, eight years after the death ofHuygens (1629-1695), printed from a manuscript copy written by someone else but corrected by Huygens.The exact date of the manuscript version of the De motu is unknown, although it is later than 1673.However, there are numerous indications that Huygens had established all the propositions and theirproofs by 1656 at the latest (see the Avertissement in Oeuvres, Vol. XVI, pp. 3-14, for the evidence), whichis thus the effective date of origin of the De motu even though minor modifications were clearly made overa long period of years (e.g., the reference to the Horologium oscillatorium of 1673 and the very lastparagraph of the treatise). That the mid-i 650s was the period of origin of the present treatise is attested toby Huygens in a letter of July 20, 1656, to G. P. Roberval: "IIy a quelques temps quej'ai quitdtoute autrespeculation pour m'attacher uniquement a cette matiere de la Percussion dont je pense vous avoir diteautrefois que des Cartes l'avoit traitee si malheureusement. J'ay acheve mon petit ouvrage depuis peu dejours, par lequel je pretens de faire veoir qu'il n'a pas este impeccable non plus dans la Phisique que dans laGeometrie" (Oeuvres, Vol. I, p. .457).2It is significant that this quite straightforwardstatement of what has come to be known as the principleof inertial motion is asserted here as axiomatic, indicating its active acceptance by the mid-i 650s. Huygenssays (Oeuvres, Vol. XVI, p. 140) that this notion is found in Galileo (which is incorrect), Descartes, andmany others. The main source for Huygens no doubt was Descartes, whom he met frequently in hisfather's house during his formative years in the 1640s and with whose writings he was intimatelyacquainted. The components of the inertial principle are contained quite clearly in the firstand second lawsof nature in Descartes's Principiaphilosophiae (1644), II, 37 and 39: "Prima lex naturae:quod unaquaequeres, quantum in se est, semper in eodem statu perseveret; sicque quod semel movetur, semper moveripergat." "Altera lex naturae: quod omnis motus ex se ipso sit rectus; et ideo quae circulariter moventur,tendere semper ut recedant a centro circuli quem describunt."

3Huygens uses this term (dura) to indicate that both here and throughout this treatise his analysis isrestricted to perfectly elastic collisions. As a result (to use modern parlance) both the principle of theconservation of momentum and the principle of the conservation of kinetic energy apply throughout.Regarding the former, see Proposition 6 with our notes; regarding the latter see Proposition 11. ForHuygens' later (c. 1667) contributions to the analysis of inelastic collisions, see Oeuvres, Vol. XVI, pp.161-168.4Concealed behind this phrase stand the reasons for Huygens' consistent refusal to publish the De motu,despite strong encouragement from several sources prompted by incomplete glimpses of his results. OnJan. 5, 1669, Huygens sent a letter to Henry Oldenburg at the Royal Society "containing the firstfour rulesISIS, 1977, 68 (No. 244) 574

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THE MOTION OF COLLIDING BODIES 575Hypothesis III: Both the motion of bodies and their equal or unequal speeds mustbe understood in relation to other bodies considered to be at rest, even if both sets ofbodies happen to be involved in some other common motion.5 As a result, when twobodies collide, then even if each of them is simultaneously subject to some otheradditional equal motion, they will in no way act on each other with respect to thecommon motion by which they are each moved. It is as if that additional motion weretotally absent.For example, someone who is carried along by a boat which progresses withuniform speed makes two equal balls collide with each other with equal speed asdetermined in relation to himself and to the parts of the boat. We say then that eachball ought to rebound with an equal speed in relation to the man carried along in theboat, as would clearly also happen if he had made the same balls collide with equalspeed while he was standing in a motionless boat or on the ground.Having made these suppositions concerning the collision of equal bodies, we willdemonstrate the laws by which they mutually collide; we will then proceed to infer inthe proper place6other hypotheses which we will need for the case of unequal bodies.Proposition 1: If a body collides with another equal body which is at rest, thenafter the contact the former is at rest and the latteracquires the same speed which wasin the body which struck it.

concerning the motion of bodies after mutual impulse, together with their demonstration" (PhilosophicalTransactions of the Royal Society, 1669, No., 46: 925), and on Mar. 18, 1669,he sent a letter to the Journalde Scavans containing a short synopsis of his theory in the form of seven rules without proofs (Oeuvres,Vol. XVI, pp. 179-181.) Ina letter of Feb. 4, 1669,Oldenburg "earnestlyintreated" Huygens to publish histheory, but the reply ignored the point. Huygens himself explains his refusal to publish. Writing a marginalcomment on a draft of an uncompleted and unpublished projected preface for the De motu c. 1690, hesays, "Me jam inde ab anno 1652-r54veras leges reperisse quae ad dura seu resistentia pertinent, sed de ijsin lucem edendis supersedisse, quod praeter eas leges superessent quaedam de motus natura nondumpenitus mihi nec satis liquido perspecta, quae longiorem meditationem requirebant"(Oeuvres, Vol. XVI,p. 204).What are these additional factors concerning the nature of motion which Huygens did not yet under-stand? Perhaps he was concerned to determine the essence and true cause of motion and forces, in whichcase he would be much closer to Descartes than to Newton in his conception of the goal of scientificunderstanding. Perhaps he was thinking of an unresolved problem left over by Galileo to which he oftenrefers. Just prior to Proposition 5 of the Fourth Day of Galileo's Dialogues Concerning Two New Sciences(Edizione Nazionale, Vol. VIII, pp. 292-293) Sagredo expresses amazement over the immense forceexerted by the blow of a hammer. Salviati replies that his Academician friend (Galileo) has spent a greatdeal of time on this question and has solved it with some remarkable new ideas. He promises to relate theexplanation at a later discussion, but the promise was not kept in the originally published four days ofdialogue. The so-called Sixth Day (first published in 1718)does return to the problem (see Vol. VIII, pp.319-346). Huygens frequently asserts that Galileo did not go beyond merely stating this problem, and hefelt also that his own De motu was inadequate to solve it. At any rate such concerns seem to stand behindthe first phrase of Hypothesis 11.The "nature of motion" and the "cause of hard bodies rebounding" arenot to be treated in the De motu, only the laws governing collision.5Hypothesis III is, in effect, Huygens' version of the principle of the relativity of motion. The basicnotion was of course well known at the time, but Huygens' extensive and detailed use of it as an instrumentof analysis is quite distinctive and contributed in large part to his successes in dealing with the problem ofcollision. His frequently used illustration of the principle in terms of a boat moving uniformly and parallelto the bank of a river may have been borrowed from Descartes (Principia philosophiae, 111,26), who wasalso clearly aware of the relativity of motion. Ironically Descartes' failure to use this notion in his analysisof collision is one of the main reasons for the erroneous character of his seven rules of impact (see R. J.Blackwell, "Descartes' Laws of Motion," Isis, 1966, 57:220-234). We might add that Huygens was alsoconcerned with the problem of whether the comparison of relative motions ultimately leads to an absolutemotion and thus to an absolute space (see Oeuvres, Vol. XVI, pp. 213-233.) However, this latter issue isirrelevant to the theory of collision formulated in the De motu, and Hypothesis III and its later use do notspeak to this question.6Hypothesis IV ff.



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576 RICHARD J. BLACKWELLImagine a boat which is floating down riverparallel to the bank and so close to thebank that a sailor standing in it can extend his hands to a friend standing on the

[Fig. I.]

C

bank. The sailor holds in his hands,A and B, [Fig. 1] two bodies, E andF, suspended on strings. The dis-tance EF is divided in half at thepoint G. As he moves his handstogether into contact with equal mo-tions as judged in relation to himselfand to the boat, at the same time hemakes the balls E and F collide withequal speed. Thus it is necessary thatthe balls rebound from mutual con-tact with equal speed [Hypothesis II]in relation to the sailor and the boat.Moreover it is supposed that theboat meanwhile has moved towardthe left with the speed GE which isthe same speed by which the lefthand A was moved toward the right.From this it is clear that the sai-lor's hand A was unmoved in rela-tion to the bank and to his friendstanding there, and also that thesailor's hand B was moved in respect to his friend with the speed FE which is doublethat of GE or FG. Accordingly let us suppose that the friend standing on the banktakes hold of, with his own hand C, the sailor's hand A together with the top of thestring from which the ball E is suspended; also he takes hold of, with his hand D, thesailor's hand B which holds the string from which F is suspended. It is apparent thenthat while the sailor made the balls E and F collide with equal speed as judged inrelation to himself and the boat, at the same time his friend standing on the bank

made the moving ball F collide with the motionless ball E with a speed FE asjudgedin relation to himself and to the bank. It is also clear that the sailor who moves theballs as described is not affected in any way by the fact that his friend standing on thebank has taken hold of his hands and the tops of the strings since the motion of thefriend's hands is only a companion motion and produces no hindrance for the sailor.For the same reason the friend on the bank who moves the ball F toward themotionless ball E is not affected by the fact that his hands and the sailor's hands arejoined, since both hands A and C are at rest in relation to the friend and the bank andboth hands B and D are moved with the same speed FE. However, as was said, aftermutual contact the balls E and F rebound with equal speed in relation to the sailorand the boat, the ball E with speed GE and the ball F with speed GF, and all the whilethe boat itself moves toward the left with the speed GE or FG. Hence it follows thatin relation to the bank and the friend standing there, the ball F comes to rest after thecollision and the other ball E moves toward the left with a speed of double BGE,hatis, the speed FE which is the same speed by which the ball F was originally movedtoward the ball E. Thus we have shown that when a man standing on the groundmakes a body collide with another equal motionless body, then after the contact theformer body loses all its motion and the latter acquires all of it. Q.E.D.



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THE MOTION OF COLLIDING BODIES 577[Fig. 2.]

.4I

Proposition 2: If two equalbodies moving with unequalspeeds collide with eachother, then after the contactthey move with their speedsreciprocally exchanged.Let a body E [Fig. 2] bemoved toward the right withthe speed EH, and an equalbody F be moved initially inthe opposite direction withthe smaller speed FH. Theythus collide at H. After theirmutual collision I say that thebody E is moved toward theleft with speed FH, and Ftoward the right with speedEH.For imagine a man stand-ing on the bank of the riverwho produces the above mo-tion of bodies by holding inhis hands C and D the top ofstrings from which E and F are suspended; he brings his hands together, as well as thebodies E and F, with the above-mentioned speeds EH and FH. The distance EF is cutin half at G. Further, imagine that a boat moves toward the right with the speed GH.

[Fig 3.]

c lXC l.:7 ,i

In the boat standsanother man in rela-tion to whom the ballE is moved only withthe speed EG and theball F with the speedFG. Therefore in re-lation to him the twoballs are moved totheir mutual collisionwith equal speed. Ac-cordingly if we sup-pose that with hisown hands A and Bhe has taken hold ofthe hands C and D ofhis friend standingon the bank, togetherwith the tops of thestrings from whichthe balls are sus-pended, then it happens simultaneously that the man standing on the bank makesthe balls collide with the speeds EH and FH and the man moving in the boat makes



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578 RICHARD J. BLACKWELLthe same balls collide with the equal speeds EG and FG. Thus it follows in the lattercase [Hypothesis II] that each ball rebounds from contact with an equal speed, thatis, E with the speed GE and F with the speed GF. Further all this time the boatcontinues to move with the speed GH. Thus in relation to the bank and to the manstanding there F will have a speed composed of both GF and GH, that is, equal toEH, while E will have the speed HF which is the difference between the speeds GEand GH. Therefore we have shown that, for a man standing on a bank who makes theballs E and F collide with the speeds EH and FH, after the collision E will reboundwith speed FH and F with speed EH. Q.E.D.Furthermore, let both bodies E and F move toward the right [Fig. 3], E with thespeed EH and F preceding it with the smaller speed FH. Thus E follows F, and theycollide at H. I say that after contact F will advance with the speed EH and E willfollow with the speed FH. The demonstration is the same as above.

Hypothesis IV: If a larger body collides with a smaller body at rest, it gives someof its motion to it and consequently loses some of its motion.Proposition 3: Any large body is moved by a collision with any small body havingany speed whatsoever.7[Fig. 4.]

C I'D

C~~~

Imagine a boat moving parallelto the bank of a river, and a sailorstanding in it holding the bodies Aand B [Fig. 4] suspended fromstrings. Let A on the left be larger,and B smaller. His right hand Dwhich holds the body B is motion-less in relation to himself and theboat. He then moves his hand Ctoward D, together with the bodyA, with the speed AB [HypothesisIV]. Therefore body A collides withB and loses some of its speed, thuscontinuing to move toward theright with a speed less than AB.Meanwhile, while this is happen-ing, let the boat move toward theleft with the speed BA. Thus ithappens that while the sailor movesthe body A with the speed AB inrelation to himself and to the shipwhich carries him, the same body Aas well as the sailor's hand C is motionless in relation to the bank and to an observer

standing there. Also the other hand D and the body B will be moved toward the leftwith the speed BA in relation to the observer, because we assumed that they weremotionless in relation to the boat which moves toward the left with the speed BA.Therefore if the observer standing on the bank uses his hands E and F to take hold ofthe sailor's hands C and D, then it is clear that while the ball A moves toward the7Proposition 3 is intended by Huygens to be a direct refutation of Descartes's fourth rule of impact:"Quarto, si corpus C plane quiesceret, essetque paulo majus quam B, quacunque cum celeritate Bmoveretur versus C, nunquam ipsum C moveret, sed ab eo repelleretur in contrariam partem" (Principiaphilosophiae, II, 49).



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THE MOTION OF COLLIDING BODIES 579motionless ball B in relation to the sailor, simultaneously B moves toward A which ismotionless and at rest in relation to the observer and the bank. But we said that afterthe collision the ball A, in relation to the sailor and the boat, moves toward the rightwith a speed smaller than AB, and that the boat moves toward the. eft with the speedBA. Therefore, in relation to the bank and to the observer standing there, it is clearthat after the collision A moves somewhat toward the left. Hence it has been shownfor an observer standing on the ground that when he makes any small body B havingsome speed BA collide with any large body A which is at rest, then the body A is putin motion. Q.E.D.Hypothesis V: If two hard bodies collide with each other, and if after the collisionone of them retains all the motion which it had, then the motion of the other isneither increased nor decreased.Proposition 4: Whenever two bodies collide with each other, the relative speed ofthe departing bodies is the same as the relative speed of the approaching bodies.For equal bodies this is clear from Proposition 2. Let us then consider unequalbodies. The first case to be proposed is that in which a small body B moving toward

[Fig-5.]B ?~~~~~IA. "% D C4~~ X CliI _ \A -

the right collideswith a large body Awhich is at rest [Fig.5]. I say that afterthe collision the twobodies move separ-ately with the samespeed BA such that if in one unit of time the body B crossed the space BA, then in alater equal unit of time the bodies are found to have separately crossed a space whichis equal to BA.For A clearly receives some speed from the collision of the body B. Let this speedbe AC, which must be less than the speed BA by which B itself was moved. For if Awere equal to B, then A would have acquired the whole speed BA from the collision[Proposition 1]. Next let AC be divided into two equal parts at D, and let AE equalAD. Therefore, if we imagine that these motions take place in a boat which floats

toward the left with the speed DA, then it is necessary that before the collision thebody A, which was at rest in the boat, moved toward the left with the speed DA inrelation to the bank. Furthermore, after the collision, since A was said to be moved inthe boat toward the right with the speed AC while the boat itself was moved in theopposite direction with the speed DA, then A will be moved in relation to the banktoward the right with the speed DC or AD. Therefore in relation to the bank the bodyA has the same speed before and after the collision. Therefore B also [Hypothesis V]in relation to the bank cannot lose any of its speed. Now before the collision B movedtoward the right with the speed BE in relation to the bank because in the boat it hadthe speed BA toward the right while the boat itself was moving in the oppositedirection with the speed DA or AE. Therefore after the collision B should be movedwith the speed BE in relation to the bank, but toward the left, because the slowermotion of the body A prevents it from moving toward the right. Therefore, since afterthe collision B is moved toward the left with the speed EB in relation to the bank, andA is moved toward the right with the speed AD or EA, it necessarily follows thatthese bodies depart from each other with a speed composed of both BE and EA, thatis, with the speed BA, in relation to both the bank and the boat; and indeed they domove away from each other with this speed. Moreover what happens to the colliding



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580 RICHARD J. BLACKWELLballs in the moving boat also clearly happens in the same way everywhere outside theboat.This having been proven, the rest follows easily. Four different cases remain: either

[Figr.6.]

A B~~

the small body is at rest, orboth bodies move in oppo-site directions, or the smallbody pursues the large onewith a faster motion, orvice versa. All of thesecases can be treated to-gether.Let there be, as before, abody A which is largerthan B [Fig. 6], and let Amove with the speed AC. Bis either completely at restor has the speed BC. Thus,when these bodies are somoved, they have the rela-tive speed AB, and I saythat after the collision theymove separately with the same speed.

For if we imagine again that these motions occur in a boat moving with the speedCA, that is, the same speed by which A moves but in the opposite direction, then it isevident that in relation to the bank A remains motionless and B in every case collideswith A with the speed BA. But A is larger than B. Thus we have the preceding casefrom which it is clear that after the collision the bodies must move away from each[Fig. 7.]

CD ]/ X

A D B

2 .I C e/_%I\.. I - D

tA_ _ D

other with thesame speed ABin relation toboth the bankand the boat.And it is evi-dent that thebodies do re-cede from eachother with thisspeed.Proposition5: If two bodiescollide a sec-ond time withthe same speedwith whichthey reboundedfrom a priorcollision, thenafter the sec-ond collision each acquires the same speed which it had before the first collision.



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THE MOTION OF COLLIDING BODIES 581Let us suppose that the body A is moved with the speed AC [Fig. 7] and B with thespeed BC so that they collide. After the collision A departs with the speed CD, and Bwith the speed CE. Afterward each of the bodies turns back toward another collisionwith the same speed, that is, A with the speed DC, and B with the speed EC. From

this I say that A will rebound with the speed CA, and B with the speed CB, which arethe speeds by which they originally moved toward the first collision. While the bodiesare progressing toward the second collision, A with the speed DC and B with thespeed EC, let us imagine that these motions occur in a boat moving with the speedAD. Then in relation to the bank A is now moving with the speed AC because itmoves with the speed DC in the boat which itself has the speed AD. And in relationto the bank B moves with the speed BC. For since DE is equal to AB [Proposition 4],then by subtracting the common segment DB, BE will equal AD. Therefore the boatmoves with the speed BE. But B has the speed EC in the boat. Hence in relation to thebank B will be moved with the speed BC, as we said. Therefore it is necessary that inrelation to the bank the bodies rebound from the first collision with A having thespeed CD and B having the speed CE. For it was granted above that if A approachesthe first collision with the speed AC and B with the speed BC, then after the collisionA rebounds with the speed CD and B with the speed CE. Therefore, while in relationto the bank A moves with the speed CD and the boat moves with the speed AD, itfollows that A moves in the boat with the speed CA. Further since in relation to thebank B moves with the speed CE and the boat with the speed AD or BE, it followsthat the speed of B in the boat is CB. Therefore it has been shown that the bodieswhich move toward the second collision in the boat with the speeds DC and ECrebound in the boat with the speeds CA and CB. And this same thing must alsohappen everywhere else. Thus the proposition has been proven.Proposition 6: When two bodies collide with each other, the quantity of motion8found in both of them taken together before collision is not always conserved afterthe collision, but may be either increased or decreased.The quantity of motion is estimated as follows: for unequal bodies having equalspeeds, the larger each body is the larger is its quantity of motion; for equal bodieshaving unequal speeds, the quantity of motion is estimated in terms of how muchfaster the one body moves than the other.9

IQuantitas motus was a technical phrase whose meaning evolved through several stages in the seven-teenth century. Its meaning finally became fixed in Definition II of Newton's Principia: "Quantitasmotusest mensura ejusdem orta ex velocitate et quantitate materiae conjunctim," which is the now familiarnotion of momentum as mass times velocity. However, in Descartes, quantitas motus (which Godconserves in the universe as a whole) means spatial volume times scalar speed (Principia philosophiae, II,36), since Descartes did not have a working notion either of mass or of velocity as a vector quantity (seeBlackwell, "Descartes' Laws of Motion"). The meaning of quantitas motus for Huygens lies between theCartesian and Newtonian definitions. Huygens has a clear notion of velocity as a vector quantity (althoughhe uses the term celeritas much more often than velocitas in the De motu). However, he is thinking of theweight rather than either the volume or the mass of a body when he talks about the magnitude of a body,one body being equal, larger, or smaller in relation to another, etc. Thus for Huygens quantitas motus isbest defined in the De motu in terms of weight and velocity, which is not quite the same as momentum.9The meaning of this paragraph is not clear. It could be interpreted to mean that the quantity of motionis measured either in terms of the weight of each body or the difference of their velocities (and apparentlyalso the combination of the two for the third and more common case not mentioned here;namely, unequalbodies having unequal velocities.) Or it could be measured in terms of the weight of each body and itsactual velocity in relation to a common frame of reference. Further, how are the two components of weightand velocity combined in the measurement of the quantity of motion? Huygens does not explicitly say herethat they form a product, i.e., weight times velocity, in this calculation. However, in a 1652 draft he givesan earlier version of Proposition 6 in which it is clear that magnitude and velocity are taken as a product:"Ex his demonstrari potest non semper post duroum corporum collisionem, tantundem motus remanere



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582 FRANK SPENCER

[Fig. 8.]Let the body A be larger thanB, while A is at rest and B movesto a collision with A with thespeed BA [Fig. 8]. A will be

moved [Proposition 3] and willacquire some speed, say AC. Bwill rebound with the speed ADsuch that the whole speed CD which the bodies have in relation to each other is equalto AB [Proposition 4]. If the body A were equal to B, the same quantity of motionwould exist after the collision as before. For it clearly makes no difference whethertwo bodies equal to B are moved, the one with the speed AD and the other with thespeed AC, or that only B is moved with the speed CD or BA. But the body A is largerthan B. Therefore it is apparent that there is a greater quantity of motion after thecollision when the body A moves with the speed AC and the body B with the speedAD, than before when only B had a speed of BA. Further, the fact that the quantityof motion can be decreased is shown as follows. If B with the speed BA collides withthe body A which is at rest, then A acquires the speed AC and B retains the speedAD. On the other hand if A advances with the speed CA and B with the speed DA inthe opposite direction, then after the collision A remains motionless and B reboundswith the speed AB [Proposition 5]. Therefore, from what was shown earlier, therewill be a smaller quantity of motion after the collision than there was before.Proposition 7: When a larger body collides with a smaller body at rest, it givesless than twice its speed to it.Let the body A with the speed AB collide with a smaller body B at rest [Fig. 9]. I

[Fig. 9.] say that B will acquire a speed lessthan the double of AB. For after thecollision the two bodies must departfrom each other with the same speedAB [Proposition 4]. Now if the speedof the body B were to become twicethe speed AB, it would be necessary that after the collision A would pursue B withthis same speed AB, which is impossible [Hypothesis IV]. Further, if the speed of Bwere more than twice AB, it would be necessary-that after the collision A wouldquantum erat antea, eo videlicet sensu, ut corporum magnitudines cum velocitatibus multiplicatae,eundum numerum producant quem prius producebant" (Oeuvres, Vol. XVI, p. 95). His thinking hereseems to be that since the weights are constant, the key question is how much velocity is found in eachbody after a collision and how does this compare to how much velocity was present in each body before thecollision. This comparison does not always give an equality according to Proposition 6, which is thus notthe same as the principle of the conservation of momentum. The latter is much more clearly anticipated byHuygens in his "admirable law of nature" announced in his letter of Mar. 18, 1669, to the Journal deSfavans (Oeuvres, Vol. XVI, p. 181):Au reste j'ay remarque une loy admirable de la Nature, laquelleje puis demonstrer en ce qui est descorps Spheriques, et qui semble estre generale en tous les autres tant durs que mols, soit que lerencontre soit directe ou oblique: C'estque le centre commun de gravit6de deux ou de trois ou de tantqu'on voudra de corps, avance touijoursegalement vers le meme coste en ligne droite devant et apresleur rencontre.The above-mentioned demonstration of this law is not to be found anywhere in the writings of Huygens.Also in this same letter (Rule 5) Huygens gives a corrected version of Proposition 6 which reads: "Laquantite du mouvement qu'ont deux corps, se peut augmenter ou diminuer par leur rencontre, mais il yreste touijoursla mesme quantit6 vers le mesme coste, en soustrayant la quantite du mouvement contraire."Since this modified rule was already known to Huygens by the mid-1650s (see Oeuvres, Vol. XVI, pp. 146-147), it is both surprising and disappointing that he did not use it in the De motu.



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THE MOTION OF COLLIDING BODIES 583continue to move with a speed greater than AB, which is also absurd. Therefore theproposition is proven.In regard to equal bodies it has been shown in all cases how one of them givesmotion to the other, on the supposition that when equal bodies collide with equalspeeds, then they also rebound equally. Likewise in regard to bodies of differentmagnitudes all cases can be determined, and there are many such cases, by grantingwhat follows. For if two unequal bodies collide with each other, and if their speedsare inversely proportional to their magnitudes, then they will each rebound from thecollision with the same speed which they had before.

[Fig. I o.]- -

For example, let A be three timeslarger than B, but the speed BC bywhich B moves is triple the speedAC by which A moves [Fig. 10].Having collided at C, each bodyrebounds with the same speedwhich it had before. However,since this is not as evident as whatwas granted for equal bodies (even though it is not contrary to reason and conformsquite well with experiments), we will try to prove it with a demonstration.It is quite well known that when two bodies fall downward with naturally acceler-ated motion, the ratio of the distances they have traversed is equal to the square ofthe ratio of the maximum grades of speed which they have acquired. This was

Rdemonstratedby Galileo in his third dialogue on motion'0 and is also observed ininnumerable and very exquisite experiments. What is also well known is that thespeed acquired by a falling body can restore it to the same height from which it fell.Demonstrations of both of these points are developed in our treatise on clocks."From this then the above-mentioned theorem can be demonstrated.Proposition 8: If two bodies whose speeds are inversely proportional to theirmagnitudesl2 collide with each other, then each rebounds with the same speed whichit had before the collision.Let the bodies A and B collide with each other [Fig. 11]. A is larger than B.'?In the passage to which Huygens refers, Galileo proves the theorem "The ratio of the spaces traversedis the same as the squared ratio of the time-intervals." He then immediately adds, "Evidently then the ratioof the distances is the square of the ratio of the final velocities" (Dialogues Concerning Two New Sciences,trans. H. Crew and A. de Salvio, New York: Dover, n.d., p. 175; Edizione Nazionale, Vol. VII, p. 210)."Propositions 3 and 4, Part II, of Horologium oscillatorium (1673) in Oeuvres, Vol. XVIII, pp.131-137.'2The Latin term magnitudo (as used in Proposition 8 and frequently hereafter) we will consistentlytranslate as "magnitude."This is the same term used by Descartes in his seven rules of impact (Principiaphilosophiae, II, 46-52), where it means spatial volume, a natural consequence of Descartes's identifica-tion of matter with extension. However, this is not Huygens' meaning. Rather he uses the term to refer tothe amount of weight in a body, as he asserts in his letter of Mar. 18, 1669, to the Journal de Scavans.

Referring specifically to his laws of collision, he says, "le considere en tout cecy des corps d'une mesmematiere, ou bien j'entends que leur grandeur soit estimee par le poids" (Oeuvres, Vol. XVI, p. 180). H.Crew has claimed (The Rise of Modern Physics, Baltimore: Williams and Wilkins, 1928, p. 121) that thefirst clear distinction between mass and weight appears in Huygens' De vi centrifuga (completed in 1659;conclusions published in 1673;full text not published until 1703).The relevant passage reads: "Unde etiamconcludemus vires centrifugas mobilium inaequalium sed in circulis aequalibus aequali velocitate latorumesse inter se sicut mobilium gravitates, seu quantitates solidas" (Oeuvres, Vol. XVI, p. 267). These "solidquantities" are certainly what we would now call "masses" expressed in the relation of F1:F2::mI:m2.However this notion may have been distinguished from that of weight in Huygens' mind at that time, whatwe have here is at least a major step toward a distinct notion of mass. But there is no textual evidence thatthis distinction is operative in the De motu, despite our being habituated after Newton to regard whatHuygens calls "magnitude" as "mass" rather than "weight."



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584 RICHARD J. BLACKWELLFurther, the ratio of the magnitude of A to the magnitude of B is the same as the ratioof the speed BC of the body B to the speed AC of the body A. What must be proven isthat after the collision each body rebounds with the same speed which it had before,

[Fig. I I.]

PX~~~

ar

ID cr

that is, A with thespeed CA and B withthe speed CB. Now itis clear that if A re-bounds with thespeed CA, then B isreflected with thespeed CB. Otherwisethe mutual speed ofrebounding wouldnot be the same asthe mutual speed ofthe bodies approach-ing the collision[Proposition 4].Therefore if body Adid not rebound withthe speed CA, thenin the first case letit rebound with asmaller speed CD, ifthat were possible,And thus B wouldrebound with thespeed CE which islarger than the speedit had before in such a way that DE is equal to AB [Proposition 4]. Let us assume thatthe body A had acquired its initial speed AC by which it approaches the collision by

falling from a height of HA such that after falling as far as A it changed itsperpendicular motion into a horizontal motion of the speed AC. Likewise let usassume that the body B acquired its speed BC by falling from the height KB.Therefore these heights are proportional to the squares of the speeds; that is, thesquare of AC is related to the square of CB as HA is related to KB. Next after thecollision let the bodies A and B convert their horizontal motions, whose speeds aremeasured by CD and DE, into perpendicular motions upward. It follows that thebody A will rise to the height AL such that AL is to AH as the squareof CD is to thesquare of CA. For when this is the ratio of AL to AH, it is certain that a body fallingfrom the height LA will acquire the speed CD. Likewise a body having the speed CDcan rise to the height AL, because of what was granted above. Moreover, when thebody B converts its speed CE into a perpendicular motion upward, it will rise to theheight BM such that MB is to KB as the square of CE is to the square of CB. Let usnext draw the lines HK and LM which necessarily intersect, let us say at the point P.Further let each of these lines be divided in the same way at N and 0 so that as themagnitude B is to the magnitude A, or as AC is to CB, so also is HN to NK and LO toOM. Thus when the center of gravity of the body A is located at H, and the body B'scenter of gravity at K, their composite center of gravity is at the point N. Now after



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THE MOTION OF COLLIDING BODIES 585they fall from H and K, and after they collide and rise again up to the points L andM, the composite center of gravity will be at 0. But this cannot occur. For, as we willsoon show, the point 0 is higher than N, and it is a most certain axiom in mechanicsthat, when bodies move because of their gravity, their common center of gravitycannot rise.I That the point 0 is higher than N is shown as follows. The excess of thesquare of EC over the square of BC is equal [Euclid, Book II, Proposition 414] to tworectangles CBE plus the square of BE, that is, is equal to the rectangle which isconstructed with the sum of EC and CB for one side and with BE as the other side.Likewise the excess of the square of AC over the square of CD is equal to a rectanglewhich is constructed with the sum of AC and CD for one side and with AD as theother side.'5 But AD is equal to BE since AB is equal to DE. It follows then that theratio of the excess of the squares of EC and CB to the excess of the squares of AC andCD is the same as the ratio of the sum of the lines EC and CB to the sum of the linesAC and CD. But the sum of the two lines EC and CB is more than double the lineCB, and the sum of the two lines AC and CD is less than double the line AC.Therefore the ratio of the sum of EC and CB to the sum of AC and CD will be greaterthan the ratio of CB to CA. And therefore the ratio of the excess of the square of ECover the square of CB to the excess of the square of AC over the square of CD isgreater than the ratio of BC to CA. Since the ratio of the square of EC to the squareof CB is the same as the ratio of the length MB to the length BK, then by division itfollows that the ratio of the excess of the square of EC over the square of CB to the

'3Thisaxiom is actually Huygens' translation into a context of dynamics of the principle from statics tothe effect that the center of gravity is always located in the lowest possible point. For the history of thislatter principle, see Pierre Duhem, Origines de la statique (Paris: Hermann, 1906), Vol. II, Ch. 15. As such,what we have here is the creative modification of a previous mechanical principle to deal with newproblems (see Oeuvres, Vol. XVI, pp. 21-24). In the demonstrations of the various cases under Proposi-tion 8 which follow, this axiom is central and is used quite skillfully. However, the proofs are rathercomplex and cumbersome because Huygens elects to formulate them in geometrical terms. For the simpleralgebraic forms of these as well as the other proofs in the De motu, see the editor's notes in Oeuvres, Vol.XVI, pp. 29-91.

'4Proposition 4, Book II, of Euclid's Elements reads as follows:"If a straight line be cut at random, tfie square on the whole is equalto the squares on.the segments and twice the rectangle contained bythe segments" (The Thirteen Books of Euclid's Elements, trans. T.L. Heath, 2nd ed., New York: Dover, 1956, Vol. I, p. 379). By wayof illustration, let the line AB be cut at any point C, as shown in thefigure. Put algebraically,(AB)2 = (CB)2 + (AC)2 + 2(CB X AC)

I

-A

A C B

A D C E~~~~~~~~~~~~~~~~~~~~~~~~~~~~'5Huygens' point in this and in the previoussentence can be illustrated as shown. Put alge-braically,(EC)2 - (BC)2 = [(EC + CB) >XBE]

(AC)2 - (CD)2 = [(AC + CD) X AD]



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586 RICHARD J. BLACKWELLsquare of CB will be the same as the ratio of MK to KB. Also the ratio of the squareof CB to the square of CA is the same as the ratio of the line KB to the line HA. Andthe square of CA has the same ratio to its excess over the square of CD as HA has toHL, for the square of AC is related to the square of CD as HA is related to AL. Thusby equality it follows that the ratio of the excess of the square of EC over the squareof CB to the excess of the square of AC over the square of CD is the same as the ratioof MK to HL. For this reason the ratio of MK to HL will also be larger than the ratioof BC to CA. But MK is to HL as MP is to PL, and also BC is to CA as MO is to OL.Therefore the ratio of MP to PL is greater than the ratio of MO to OL. And byaddition the ratio of ML to LP is greater than the ratio of ML to LO. Thus LO isgreater than LP. Hence it is necessary that the point 0 fall on that side of theintersection P which is toward M. But the line which joins the points 0 and N isparallel to the perpendiculars MB and HA, because these points divide the straightlines LM and HK by the same ratio. Thus as M is higher than K, so also is it clearthat 0 is higher than N. Q.E.D.

[Fig. In.] x

/C -S

Next let usassume that itwould be pos-sible for thebody A to re-bound from thecollision witha speed CDwhich is largerthan the speedCA by which itapproached thecollision [Fig.12]. But CDwill be less thanCB, which wasthe speed of thebody B beforethe collision.For if B werenot less than Abut equal to it,then indeed A

would rebound from the collision with the speed CB [Proposition 2], but B wouldrebound with the speed CE so that DE would be equal to AB [Proposition 4]. Nowlet all the other things done above be granted, and let the construction be completedas in the previous case. It follows then that L is higher than H because DC is greaterthan AC; also that M is lower than K because EC is smaller than CB. Next we willshow as before that the ratio of the difference of the squares of DC and CA to thedifference of the squares of BC and CE is the same ratio as the sum of AC and CD tothe sum of EC and CB. Since this last sum is less than twice CB, and since the formersum is greater than twice AC, the ratio of the sum of AC and CD to the sum of ECand CB will be greater than the ratio of AC to CB. Therefore, the ratio of thedifference of the squares of DC and CA to, the differenceof the squares of BC and CE



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THE MOTION OF COLLIDING BODIES 587will be greater than the ratio of AC to CB. But what was demonstrated about theratio of the one differenceto the other is also demonstrated of the ratio of LH to KM.Thus also the ratio of LH to KM, that is, of LP to PM, is greaterthan the ratio of ACto CB, that is, of LO to OM. Accordingly the point 0 falls on the side of theintersection P which is toward L. But as before ON is parallel to LH. Therefore as thepoint L is higher than H, also 0 will be higher than N. But this is absurd for the samereason which we gave in the previous case.Next let us assume that after the collision the body A [Fig. 13]remains motionlessand only B is rebounded. Then B will rebound with the speed AB [Proposition 4]

[Fig13.]

0

/ - IA$

I z~~~~~A

since the two bodies hadthe speed AB in respectto each other before thecollision. Now let us as-sume as before that thebody B acquires itsspeed BC by fallingfrom the height KB.Consequently if thesquare of CB is relatedto the square of AB asthe length BK is relatedto BM, then the body Bcould rise to the heightBM if it converted itshorizontal motion hav-ing the speed AB into aperpendicular motionupward. However, thebody A will remain onthe line AB, because af-ter the collision it hasno motion at all. Thus,if a line is drawn fromM to A and is dividedat 0 such that AO isto OM as AC is toCB, then the point 0will mark the heightto which the center ofgravity composed ofboth bodies will rise.Now if we suppose thatthe bodies had been lo-cated at H and K fromwhich they are presumed to have fallen, then their common center of gravity wouldhave been at the point N, which in a like manner divides the straight line HK by theratio of AC to CB. Thus if we show again in this case that the point 0 is higher thanthe point N, the demonstration will be reduced to the same absurdity as above. Nowthis is shown as follows. The square of AB is to the square of BC as the length MB is



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588 RICHARD J. BLACKWELLto BK. Then by division the ratio of the excess of the square of AB over the square ofBC to the square of BC is the same as the ratio of MK to KB. But the square of BC isto the square of CA as KB is to HA, for this is assumed as in the first case. Therefore,by equality the ratio of the excess of the square of AB over the square of BC to thesquare of CA will be the same as the ratio of MK to HA. But the ratio of this excessto the square of CA is clearly greater than the ratio of the straight line BC -to CA.Thus the ratio of MK to HA, that is, of MP to PA, is greater than the ratio of BC toCA, that is, greater than the ratio of MO to OA. By addition it follows that the ratioof MA to AP is greater than the ratio of MA to AO. Hence the point 0 must fall onthe side of the intersection P which is toward M. But M is higher than K. Thereforesince ON is necessarily parallel to MK, the point 0 will also be higher than N. Thiswas what required proof.

[Fig. 14.]aIf e

Finally let thebody A after thecollision continueto move in thesame directionwith the speed CF[Fig. 14]. CF willnot be greaterthan AC, which was A's speed before the collision. But the body B ought to run onahead of A with the speed CG whose excess FG over the speed CF is equal to AB

[Proposition 4]. It will be shown as follows that this is not possible. Let CD equal CF,and thus DE equals AB. Therefore CE is smaller than ED by the same quantity thatCG is larger than ED or FG. However, if we assume as in the first case that the bodyA rebounded from the collision with the speed CD, then it is clearly proven that thebody B cannot have the speed CE without falling into absurdity. For after convertingtheir horizontal motions into perpendicular motions, the composite gravity of thebodies will rise higher than the place from which they descended. This is even muchmore necessary if the body B acquires a speed CG which is much greater than CE,while A has a speed CF which is equal to CD. Therefore the body A does notcontinue to move in the same direction after the collision. It follows then that Arebounds after the collision with the same speed which it had before. And thus also Brebounds with the speed CB. Q.E.D.Proposition 9: Given two unequal bodies which directly collide, and given thateither both or only one of them is in motion, and given the speed of each or of onlyone if the other is at rest, find the speed by which each is moved after the collision.Let the body A move toward the right with the speed CD [Fig. 15]. And let B eithermove in the opposite direction, or let it precede A in the same direction with the speedBD, or finally let B be at rest, that is, let the point D fall in B. Therefore AB will bethe mutual speed of A and B.Next let AB be divided at C such that AC is to CB as the magnitude of B is to themagnitude of A. And let CD equal CE. I say that EA will be the speed of the body Aafter the collision, and EB the speed of the body B. The direction of the speeds isshown by the order of the points EA and EB. When the point E coincides with A,then the body A is brought to rest, and when E coincides with B, then the body B is atrest.When we have shown that this is what happens in a boat moving with equal speed,it follows that this will also happen in the same way for an observer standing on the



17/25

THE MOTION OF COLLIDING BODIES 589ground. Let us imagine then a boat moving parallel to the bank of a river. A sailorstands in the boat and holds in his hands F and G the balls A and B suspended from

[Fig. i5.]A -D Cf E;- ~~~I I


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590 RICHARD J. BLACKWELLthe ground, as we have said. Therefore the proposition is proven. From the construc-tion of this problem one can formulate the following rules for the purposes ofcalculation.Given two bodies A and B, both of which are moved, to find the speed of the bodyA after the collision, let the relative speed of A and B be related to another speed,called C, as the sum of the bodies is related to twice the body B. The differencebetween C and the speed of A before the collision (or in one case their sum when Aprecedes in motion) will give the speed by which A will be moved after the collision.A rebounds when C is greater than A's initial speed, and it continues forward when Cis smaller. If there is no difference, the body A remains at rest after the collision.When the speed of the body A has been found, the speed of the body B is alsoknown because the relative speed of the bodies must be the same before and after thecollision.Given that the body A is at rest and only B moves toward it, it is clear that thespeed of A after the collision will be equal to the speed C, found as we have alreadysaid. From this the following theorem is deduced.Proposition 10: The ratio of the speed which a large body gives to a small body atrest to the speed which a small body of the same speed gives to a large body at rest isthe same as the ratio of the magnitude of the large body to the magnitude of the smallbody.Let the body A be larger than B. Let us assume that the speed AC is given to A atrest when it is struck by the body B moving with the speed BA [Fig. 16]. Also assumethat the speed BD is given to the body B at rest when it is struck by the body A havingthe same speed AB. I say that the speed BD is related to the speed AC as themagnitude of A is related to the magnitude of B.

[Fig. i6;]C BC-{

For the speed BDis related to twice thespeed AB as the bodyA is related to both Band A taken together[Proposition 9]. Alsoboth B and A taken together are related to B as twice the speed AB is related to thespeed AC [Proposition 9]. By equality it follows that the speed BD will be related tothe speed AC as the body A is related to B. Q.E.D.Proposition 11: If two bodies collide with each other, and if the ratio of both theirmagnitudes and their speeds is given either in numbers or in lines, then the sum oftheir magnitudes multiplied by the squares of their respective velocities is equalbefore and after the collision.16Let us suppose two bodies A and B. A moves before the collision with the speedAD, and B with the speed BD [Fig. 17]. After the collision we find as before that Ahas the speed EA, and B the speed EB. Let AB be divided at C so that A is to B as BCis to CA, and let CE equal CD. Therefore the ratio of magnitude A to B is designatedby the ratio of the line CB to CA. What must be proven is that the solid figure

'6It is quite remarkable that Huygens was able to formulate the principle of the conservation of kineticenergy in perfectly elastic collisions as early as 1652. At that time he merely stated the principle as anaxiom and gave no indication of hoW he arrived at the notion. His axiom was, "Sed necesse est quadratavelocitatum ducta in magnitudinem corporum semper eundem numerum producere"(Oeuvres, Vol. XVI,p. 95).



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THE MOTION OF COLLIDING BODIES 591composed of the line CB multiplied by the square of AD plus the solid of CA timesthe square of BD is equal to the solid of CB times the square of EA plus the solid ofCA times the square of EB. Now if there are four magnitudes of which the first

tFig. 17.]

rvX~. c x~i_~~ ir

exceeds the secondas much as the thirdexceeds the fourth,or of which the firstis smaller than thesecond by as muchas the third issmaller than thefourth, then it iscertain that the firstplus the fourthequals the secondplus the third. Thusthe proposition willbe proven if weshow that the solidfigure of CB timesthe square of AD iseither larger orsmaller than thesolid of CB timesthe square of EA bythe same amountthat the solid of CAtimes the square ofEB is either largeror smaller than thesolid of CA times the square of BD. This is shown as follows.In every case either the point C falls between A and D or the point D falls betweenA and C. Whenever C is located between A and D, AD will equal the sum of AC andCD, and AE will equal the difference between AC and CD. For CE is equal to CD,and thus AD will always be largerthan AE. In these same cases BE will equal the sumof BC and CE, and BD will equal the difference between BC and CE. And thus BEwill always be larger than BD. Further, whenever D falls between A and C, AE willequal the sum of AC and CE, and AD will equal the difference between AC and CE.And thus AE is larger than AD. But in these same cases BD will be larger than BE,since BD is equal to the sum of BC and BD while BE is equal to the difference be-

tween BC and BD. Thus it is clear that whenever AD is larger than AE, then BE is alsolarger than BD, and whenever AE is larger than AD, then BD is also larger than BE.Next, since DE is divided in half at C, then wherever the point A is located, thedifference of the squares of AD and AE will always be equal to four times therectangle ACD or ACE, according to Euclid's Elements, Book II, Proposition 8.17In17Proposition 8, Book I1,of Euclid's Elements reads as follows: "If a straight line be cut at random, fourtimes the rectangle contained by the whole and one of the segments together with the square of the



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592 RICHARD J. BLACKWELLthe first and fifth cases, for the line cut in any way whatsoever, take the line AC whichis divided at E; in the second and eighth cases, take AC which is divided at D; in thethird and fourth cases, take EC which is divided at A. In the sixth and seventh cases,where there is no square of AE, it is clear that the above-mentioned difference is thesquare of AD which likewise is necessarily equal to four times the rectangle ACD orACE. For the same reason, since the line DE is bisected at C, then wherever the pointB is located, the difference of the squares of BE and BD will always be equal to fourtimes the rectangle BCD or BCE. However since the rectangles have a commonheight, four times the rectangle BCD is to four times the rectangle ACD, which wasequal to the difference of the squares of AD and AE, as BC is to AC. Therefore thedifference of the squares of BE and BD is to the difference of the squares of AD andAE as BC is to AC. Consequently the product of the straight line BC times thedifference of the squares of AD and AE, which is the same as the difference of thesolid figures composed of BC times the square of AD and of BC times the square ofAE, is equal to the product of the straight line AC times the difference of the squaresof BE and BC, that is, is equal to the difference of the solid figures composed of'ACtimes the square of BE and of AC times the square of BD.But when the square of AD is larger or smaller than the square of AE, then thesquare of BE will always be respectively larger or smaller than the square of'BD.Therefore it is clear that the solid composed of BC times the square of AD always islarger or smaller than the solid composed of BC times the square of AE by the sameamount that the solid composed of AC times the square of BE is respectively largerorsmaller than the solid composed of AC times the square of BD. Q.E.D.Lemma 1: Let the straight line AB be cut at C and D such that the segment AC issmaller than CD, and CD is smaller than BD [Fig. 18]. I say that the rectanglecomposed of AD and CB is less than twice the sum of the rectangles ACD and CDB.Let the square CGND be constructed on the segment CD. Extend CG up to E sothat GE is equal to CA. Construct the rectangle ECBF, and extend DN to K and GNto H. Thus since CG is equal to CD and GE is equal to AC, the entire line CE will beequal to AD. Hence the rectangle CF is the rectangle composed of AD and CB. Butthe rectangle EN is equal to the rectangle ACD, and the rectangle NB is equal to therectangle CDB. Therefore what must be shown is that the rectangle CF is less thantwice the sum of the rectangles EN and NB. Let GL be equal to GE, and draw the lineLM parallel to AB. Since GL is smaller than GC (for GE or AC is smaller than CD),LM falls between GH and CB.Now since CD is less than DB, the rectangle LD will be less than the rectangle DM.But LN is equal to the rectangle NE, and the rectangle NM is equal to the rectangle

remaining segment is equal to the square described on the wholeand the aforesaid segment as on one straight line"(Elements, trans.Heath, Vol. I, p. 389). By way of illustration, let the line AB be cutat any point C, and let BD = AC, as shown in the figure. Putalgebraically,4(AB X AC) + (CB)2 = (AB + AC)2

--I

"7c B D



21/25

THE MOTION OF COLLIDING BODIES 593[Fig. i8.]

46 x __ _

.XC C .0 .

NF. There-fore the sumof the rec-tangles LNand NF isequal to thesum of NEand NM.Thus if un-equals areadded toequals, thatis, the rec-tangle LD tothe rectanglesLN and NF,and the rec-tangle MD to the rectangles NE and NM, then the sum of the square CN and therectangle NF is less than the sum of the rectangles NB and NE. Hence the sum of allof these figures, that is, the rectangle CF, must be less than twice the sum of therectangles NB and NE. Q.E.D.Lemma 2: Let AB, AC, and AD be three proportional straight lines of which AB is

the largest, and let the same length AE be added to all three lines [Fig. 19]. I say that[Fig. '9.]_D Athe rectangle composed of BEand DE is larger than thesquare of CE.For since AB. AC. and ADare proportional, the excess BC is related to the excess CD as BA is related to AC, oras CA is related to AD. But the ratio of CA to AD is larger than the ratio of CE toED. Thus the ratio of BC to CD is also larger than the ratio of CE to ED. Bypermutation the ratio of BC to CE is larger than the ratio of CD to DE, and by

addition the ratio of BE to EC is larger than the ratio of CE to ED. Therefore therectangle composed of BE and ED is larger than the square of CE, as was proposed.Proposition 12: If a body moves toward another larger or smaller body at rest, thenit will give it a greaterspeed if there is another body of intermediate magnitude and atrest interposed between them than if they collide without any intermediary. And itwill confer the maximum speed when the interposed body is a middle proportionalbetween the extremes.Let the body A move toward C which is at rest [Fig. 20]. And let A be larger orsmaller than C, and let the motionless body B of intermediate magnitude be placed inbetween A and C. Thus when A first strikes B, and then B strikes C, I say that agreater motion is acquired by C than if A had collided directly with C.Let the straight lines DE, ELI, and HK have the same ratio to each other as thebodies A, B, and C. Let the speed of the body A be LP whose double is LQ. If we letLQ be related to MR as the sum of DE and ELI s related to DE, then MR will be thespeed acquired by the motionless body B when it is struck by A [Proposition 9]. LetMS be the double of MR. Thus again if we let MS be related to N as the sum of EHand H K is related to ELI,then N will be the speed of the body C after it is struck by Bhaving the speed MR [Proposition 9]. But if LQ is related to 0 as the sum of DE and



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594 RICHARD J. BLACKWELL[Fig. 20.]

@ 3

1- f

b4 I isa *, I a 0 __u- 4 --- I

HK is related to DE,then 0 will be the speedacquired by the body Cif it is struck by thebody A having thespeed LP. What mustbe proven is that thespeed N is larger than0. The ratio of LQ to Nis composed of the ra-tios of LQ to MR andof MR to N. But theratio of LQ to MR isthe same as the ratio ofHD to DE. And the ra-tio of MR to N is thesame as the ratio of KE to twice EH. For as KE is to EH, so also is SM to N. ThusKE is to twice EH as SM is to twice N, that is, as RM is to N. Therefore the ratio ofLQ to N is composed of the ratios of HD to DE and of KE to twice EH, and thuswill be the same as the ratio of the rectangle composed of HD and KE to twice therectangle DEH. But the ratio of LQ to 0 is the same as the ratio of the sum of DEand HK to DE by construction, that is, by taking EH as the common height, it is thesame as the ratio of the sum of the rectangles DEH and EHK to the rectangle DEH,or the same as twice that sum to twice the rectangle DEH. But the rectanglecomposed of HD and KE is less than twice the rectangles DEH and EHK [Lemma 1].Therefore the ratio of the rectangle composed of HD and KE to twice the rectangleDEH will be less than the ratio of the rectangles DEH and EHK, taken twice, to twicethis same rectangle DEH. But we have said that the ratio of the rectangle composedof HD and KE to twice the rectangle DEH is the same as the ratio of LQ to N. Andwe have also said that the ratio of twice the sum of the rectangles DEH and EHK totwice the rectangle DEH is the same as the ratio of LQ to 0. Therefore the ratio of

LQ to N will be less than the ratio of LQ to 0, and thus N is larger than 0.Now let B be a middle proportional between A and C. I say that with this ratio thegreatest speed of all will be given to the body C.For in place of B let the larger body X be interposed from the start such that Astrikes X and X strikes C [Fig. 21]. Then let it be granted, if that were possible, thatthe speed acquired by the body C is greater than if B were interposed. Thus A is to Xas DE to ET. Therefore ET is larger than EH, assuming as before that DE, EH, andHK are proportional in the same ratio as the bodies A, B, and C. But VE is the thirdproportional to the two lines TE and HE, and the speed N acquired by the body Cthrough the interposed body B is found as before. The speed which is acquired by Cthrough the interposed body X is found in the same way [Proposition 9]. Now if theratio of the sum of A and X to X, that is, of the sum of DE and ET to DE, is the sameas the ratio of LQ to IY, then IY will be the speed acquired by the body X whenstruck by A. Further, if the ratio of the sum of X and C to X, that is, of the sum of ETand HK to ET, is the same as the ratio of twice IY, that is, ZY, to G, then G will bethe speed acquired by the body C. From this it has been shown that N is largerthan G.As was shown before, the ratio of LQ to N is composed of the ratios of HD to DEand of KE to twice HE. But as KE is to twice HE, so is HD to twice ED since KH,



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THE MOTION OF COLLIDING BODIES 595[Fig. 2I.]

, a ~~~Z^ 41 I Ia~Yi .4 I af

HE, and ED are proportional. Therefore the ratio of LQ to N is now composed of theratios of HD to DE and of HD to twice DE, and because of this it will be the same asthe ratio of the square of HD to twice the square of DE. But the ratio of LQ to G iscomposed of the ratios of LQ to IY and of IY to G, of which the ratio of LQ to IY isthe same as the ratio of TD to DE by construction, and the ratio of IY to G is thesame as the ratio of the sum of KH and TE to twice TE. For by construction, the sumof KH and TE is related to TE as ZY is related to G. Therefore, by doubling thesecond terms the sum of KH and TE is related to twice TE as ZY is related to twice G,or as IY is related to G, as was said. Therefore the ratio of LQ to G is composed of theratios of TD to DE and of the sum of KH and TE to twice TE. But since DE, EH, andHK are proportional, the rectangle composed of DE and HK will be equal to thesquare of EH. However, the rectangle composed of EV and ET is also equal to thesquare of EH, since EV, EH, and ET are proportional. Therefore the rectanglecomposed of DE and HK is equal to the rectangle composed of EV and ET. Hence,as VE is to ED, HK is to ET. And by addition VD is to DE as the sum of KH and TE

[Fig. 22.]

9 ~ Iiw i~~i

Z,v II Ir,

is to TE. By doubling thesecond terms, VD is to twiceDE as the sum of KH and TEis to twice TE. Thus the ratioof LQ to G is composed ofthe ratios of TD to DE and ofVD to twice DE, and thus isthe same as the ratio of therectangle TDV to twice thesquare of DE. But the ratio ofLQ to N has been shown tobe the same as the ratio of thesquare of HD to twice thesquare of DE. Thus since therectangle TDV is larger thanthe square of HD by Lemma2 (for TE, HE, and VE areproportionals to which thelength ED was added), it follows that the ratio of LQ to G is larger than the ratio ofLQ to N. And therefore N is larger than G. Q.E.D.Next let it be assumed that the body C acquires a greater speed when a body X



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596 RICHARD J. BLACKWELLsmaller than B is interposed [Fig. 22]. Again let A be to X as DE is to ET. Then sinceit is assumed that X is smaller than B, ET will also be smaller than EH, for DE is toEH as A is to B. For the rest repeat the same construction and apply again thedemonstration which shows that the speed N is larger than G. Thus it is clear that themaximum speed is acquired by the motionless body C through the interposition ofthe body B, which is a middle proportional between A and C.Proposition 13: To the extent that many bodies are interposed between twounequal bodies, of which one is at rest and the other is in motion, to the same extent agreater motion can be given to the body at rest. The maximum motion is conferredthrough any one group of interposed bodies if the latter constitute a continuousproportional series with the extremes.Let A, B, and C be proportional bodies [Fig. 23]. A is in motion while the othertwo are at rest. The maximum motion which C can acquire through the interpositionof one body is produced by the interposition of B [Proposition 12]. The fact that astill larger motion can be produced by the interposition of two bodies will be clear

[Fig. 23.]

(~ /

from the following. If the middle proportionalD is interposed between A and B, then agreater motion is acquired by the body B thanif it had been struck directly by the body A. Tothe extent that the speed of B is greater, to thesame extent a greater speed is given to C.Therefore C will be moved more through theinterposition of bodies D and B than throughB alone. Further, if in the place of B anotherbody is introduced which is a middle propor-tional between D and C, it is clear that a stillgreater motion will be given to C than was the case when the bodies D and B wereinterposed. Next it is shown as follows that the maximum motion is given to the lastbody through the interposition of two bodies when A, D, B, and C are continuouslyproportional. First it is clear that the speed of the body C through the interposition oftwo bodies cannot increase indefinitely. For the speed of the body D will always be

less than twice the speed of A [Proposition 7]. In the same way the speed of B willalways be less than twice the speed of D, and the speed acquired by the body C willalways be less than twice the speed of B. Therefore in every case the speed of C will beless than eight times the speed of A. From this it is clear that there is a definite limitspeed which the body C cannot surpass through the interposition of two bodies. Letthis speed be E, and let us assume that the body C has acquired this speed when thebodies D and B are interposed between it and A. Now I say that A, D, B, and C arecontinuously proportional. For in the first place, if the three bodies A, D, and B arenot proportional, then by substituting for D another body which is the middleproportional between A and B, it will happen that a larger motion is given to B thanthrough the interposition of D, and also C will acquire a greater speed than throughthe interposition of D and B, that is, a speed greater than E. But this is absurd since itwas assumed that E is the maximum speed which the body C can acquire through theinterposition of two bodies. Likewise, if D, B, and C are not proportional, then it willbe possible to introduce in the place of B and between D and C another body which isa middle proportional such that it will happen again that C acquires a greater speedthan through the interposition of D and B, that is, a speed greater than E, which isabsurd for the same reason. Thus since both A, D, B and D, B, C are proportional, all



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THE MOTION OF COLLIDING BODIES 597the bodies A, D, B, C are in continuous proportion, which was the point to beproven. From this and with the same argument it can be shown that when threebodies are interposed between A and C, then C can acquire a still greatermotion thanwhen only two bodies were interposed, and so forth. Also with the same argument itis shown that C acquires the maximum motion when there is a continuous proportionof all the bodies. Q.E.D.Let there be a row of one hundred bodies in a proportion ot two to one, and let themotion begin from the largest body. Then, by a calculation, omitted here, accordingto the rule given in Proposition 9 but in an abridged form, one finds that the ratio ofthe speed of the smallest body to the speed by which the largest body was moved isapproximately 14,760,000,000 to 1. If the motion begins from the smallest body, thequantity of motion increases in general approximately according to the ratio of 1 to4,677,000,000,000.18

'8This last paragraph was added to the treatise at a later date, probably in about 1667. For Huygens'calculations of these values, see Oeuvres, Vol. XVI, pp. 156-158. The rule referredto is contained in thelast three paragraphs of Proposition 9.

230011 Christiaan Huygens

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