( ) Physics 1401 Homework Solutions - Walker, Chapter 9

Transcription

1 Phyic 40 Conceptual Quetion CQ No Fo exaple, ey likely thee will be oe peanent deoation o the ca In thi cae, oe o the kinetic enegy that the two ca had beoe the colliion goe into wok that each ca doe on the othe ca to deo it Thi enegy i theeoe not peent a kinetic enegy ate the colliion Theeoe, the colliion i not elatic (The kinetic enegy o the yte i not coneed) On the othe hand, i thee wee not peanent deoation o the ca, then the colliion ay be ey nealy elatic So the act that they don t tick togethe iply doen t tell you whethe the colliion i elatic o inelatic it jut tell you that it wan t copletely inelatic Poble and Conceptual Execie 4 (a) We want: 0 i + i Sole thi o : ( ) i 035 / 069 / 06 i i The inu ign tell u that Cat oe in a diection oppoite o the way Cat i oing So the peed that Cat ut hae in ode o the total oentu to be zeo i 069 / (b) No, the kinetic enegy o the yte i not zeo kinetic enegie don t cancel jut becaue the two cat ae oing in oppoite diection (Thee no uch thing a a negatie kinetic enegy it alway geate than o equal to zeo o the kinetic enegie o the two cat cannot ee add to zeo unle each o the kinetic enegie i zeo) In act, the total kinetic enegy o the yte i: Ktotal i + i ( 035)( ) + ( 06)( 069) 040 J (c) See the calculation iediately aboe 4 Ipule-oentu theoe ay: ( ) Δt Δp F a In thi cae, the net oce on the ball i jut the oce exeted on it by the club (any oce that ai eitance exet on the ball i cetainly negligible copaed to the oce the club exet on the ball) So i F i the oce that the club exet on the ball, then we hae: F a Δ t Δp I we want only the agnitude o the aeage oce, we can take the agnitude o both ide o the equation iediately aboe to get: F a Δ t Δp Soling o the agnitude o the aeage oce gie:

2 Phyic 40 F a a Δp ball balli ball Δt Δt Δt F N ( 0045 kg)( 67 /) (a) We need to ind out what the change in the able oentu i while it in contact with the loo Then, by the ipule-oentu theoe, thi change in the able oentu will be equal to the ipule delieed to the able by the loo So what i the able oentu jut beoe it hit the loo? Well, to anwe thi quetion, we need to know the elocity o the able jut beoe it hit the loo You could do thi by aying, Well, the able i in ee-all, theeoe I can ue the ee-all equation I think I will do it a dieent way I will ay, Well, the only oce acting on the able while it all i gaity, theeoe the total enegy o the able ut be coneed while it alling The initial enegy i: E gy i i g ( 44 ), taking the potential enegy to be zeo at the loo When it eache the loo, all thi enegy i in the o o kinetic enegy, o: E The enegy i coneed, o: g So the peed o the able jut beoe it hit the loo i: ( 44 ) 53 / g I I take the poitie diection o y to be upwad, then the elocity o the able jut beoe it hit the gound i 53 / So the able oentu jut beoe it hit the gound i: kg p beoe beoe ( 0050 kg)( 53 /) Now how do we ind the able oentu jut ate it hit the loo? Well jut ate it hit the loo, it ll be oing up with a elocity I ll call ate So i we could ind ate, we d know what the oentu o the able i jut ate it hit the loo So how to ind ate? Well, on the way up, once again, the enegy will be coneed (ate the able leae the loo) Thi lead to: ( 44 ) ( 0640 ) ate g i: g( 0640 ) 354 / So ate ate, and thi tie I can leae the plu ign, becaue ate it hit the loo, the able i oing up So the elocity o the able jut ate it hit the loo i 354 / So the oentu o the able jut ate it hit the gound i:

3 Phyic 40 So the change in oentu o the able i: kg p ate ate ( 0050 kg)( 354 /) 0053 kg Δp p ate pbeoe 0053 ( 00797) 033 And thi change in oentu o the able i equal to the ipule delieed to the able by the loo Recall that the ipule i the poduct o ( F ) and Δ t : a ( F ) t I a Δ By the ipule-oentu theoe, then: I Δp So: kg I N (Note that it cutoay to gie the ipule in unit o N jut a a einde that ipule i a kg oce tie a tie Thee unit ae copletely equialent to, a you can poe o youel) So the agnitude o the ipule i 033 N The diection o the ipule i upwad (b) I the able had bounced to a geate height, the ipule delieed to it would hae been geate, becaue the change in oentu o the able would hae been geate 4 Let Piece # be the piece that ha the lage kinetic enegy ate the exploion Then: K K, That i, o:, () Now, we alo know that the oentu i coneed (no net extenal oce on the object), o: ( p 0 object wa at et initially) i 0 () + Sole Eq () o one o the inal elocitie I ll ole it o : 3

4 Phyic 40 Now quae thi: And ubtitute thi into Eq (): The cancel, and I wind up with an equation elating the ae: Let ole thi o the atio : Caying out the quaing on the RHS, I get: Now ipliy a little: So: Piece # ha twice the a o Piece # So the piece with the lage a i the piece that had the alle kinetic enegy 5 The oentu o the yte coniting o the atellite and the atonaut i coneed becaue thee i no extenal oce acting on thi yte (The only oce inoled ae the oce the atonaut and atellite exet on one anothe, and thee ae intenal to the yte o the atellite plu atonaut) The atellite and atonaut ae at et initially, o the initial oentu o the yte i zeo: p 0 i Theeoe the oentu o thi yte jut ate the atonaut puhe on the atellite ut alo be zeo: 0 atonaut atonaut + atellite atellite Let the poitie x axi be towad the pace huttle Then the elocity o the atellite ate the atonaut puhe on it i negatie: 4

5 Phyic / atellite So the atonaut ecoil towad the pace huttle with a elocity: atonaut atellite atonaut atellite atonaut 00 ( 04 /) 8 / 9 So the atonaut oe away o the atellite and towad the pace huttle with thi elocity Futheoe, the atonaut oe towad the huttle with contant peed becaue once he i no longe in contact with the atellite, thee nothing exeting a oce on hi So hi peed i 8 / the whole way to the pace huttle I it take hi 75 to get to the pace huttle, then the ditance he tael i: ( )( ) d atonaut t 8 / 75 4 Thi i the atonaut initial ditance o the pace huttle 8 Let Cat # be the cat that i oing with elocity initially and Cat # be the one that initially tationay Then coneation o oentu ay: ( ) + But the ae ae identical jut call each one o: ( ) So the tuck-togethe cobination o the two cat oe o ate the colliion with a elocity that one-hal o the initial elocity o Cat # Now the kinetic enegy o the yte ate the colliion i: K ( ) 4 So the kinetic enegy ate the colliion i one-hal the initial kinetic enegy o Cat # (The initial kinetic enegy o Cat # wa, o coue, ) So the kinetic enegy i not coneed, jut a we expect, becaue the colliion i not elatic 34 Thi i a poble ey uch like the poble o the ballitic pendulu dicued in Exaple 9-5 Think o thi poble in two pat: Fit, thee the colliion itel Second, thee what happen ate the colliion In the colliion, the oentu o the yte i coneed Ate the colliion, a the tuck-togethe 5

6 Phyic 40 cobination o putty and block copee the ping, the total echanical enegy i coneed, becaue thee ae no non-coneatie oce doing any wok (The oce in the poble ae the noal oce, which i nonconeatie but doe no wok, gaity, and the ping oce Gaity a coneatie oce, but it doen t atte anyway, becaue it doe no wok The only oce that doe any wok i the ping oce, and it coneatie) Let think about Pat # (the colliion itel) The oentu i coneed, and it a copletely inelatic head-on colliion Let the wad o putty be called Object # and the block be called Object # Then coneation o oentu ay: ( ) i +, in which i the elocity o the tuck-togethe cobination o putty and block jut ate the colliion Let ind out what thi elocity i: i ( 30 /) 040 / Now o Pat # o the otion (putty+block copee ping) The enegy i coneed What the initial enegy o the yte? Jut the kinetic enegy that putty+block ha iediately ate the colliion (The ping han t begun to be copeed yet, o thee no enegy toed in it yet) So: E ( kg kg)( 040 /) i What the inal enegy o the yte (ie, the enegy o the yte when the ping i at it axiu copeion)? Well, when the ping ha been copeed all it going to be copeed, the putty-block yte i oentaily at et So thee no kinetic enegy Thee jut the enegy that been toed in the ping So: E kx ax The act that the total enegy o the yte i coneed ean: E E i kx So: ( )( ) ax kg kg 040 / Sole o x ax : x ax ( kg kg)( 040 /) k ( kg kg)( 040 /) x ax c 00 N/ 6

7 Phyic (a) Since it a ubbe ball, the colliion between the ball and the elephant will be elatic (Thee no peanent deoation o ball o elephant) Let the ball be called Object # and the elephant be called Object # Alo, let take the poitie diection o the x axi to be in the diection o otion o the elephant Futheoe, let aue that the ball and elephant hae an elatic head-on colliion (not a glancing colliion) Then coneation o oentu ay: i i + + () Alo, we know one othe act about elatic head-on colliion: i i + + () And we d like to ole o : To do o, let eliinate the othe unkown, naely Sole () o i + i Plug thi into (): + ( + ) i i i i + Now do a little algeba to ole o paenthee by : I ll tat by caying out the ultiplication o eeything in i i + i + i + Now I ll gathe te inoling on one ide o the equation: i i + i + i + Now I ll do a little ipliying: ( ) i + i ( + ) So inally: i + i + + Actually, it a little bit illy to include the in the denoinato o the acto in paenthee I ean, 050 kg and i uch geate: 540 kg So kg, which i o cloe to a to be inditinguihable o, at leat o the kind o peciion we need So I ll jut ake the appoxiation that + Futheoe, 050 kg 540 kg kg, which i o cloe to that we can ay that, to a ey good appoxiation, So I ll ake thi appoxiation, too Then the expeion o look alot iple: 7

8 Phyic Plugging in nube: i i i i + + ( ) ( ) 78 / / 69 / 69 / (keeping 3 ig ig) Thi i the peed o the ball when it bounce back towad you (b) How do you account o the act that the ball kinetic enegy ha inceaed? Well, the ball kinetic enegy inceaed becaue thee wa oe net wok done on it While the ball and elephant wee in contact, the elephant exeted a oce on the ball By the wok-enegy theoe, thi eulted in an inceae in the kinetic enegy o the ball (a) I the a o the bullet i the ae in each cae, and i the bullet hit the block with the ae peed in each cae, then the block o wood will be oe likely to be knocked oe i the bullet i ubbe and bounce o The eaon i that the bullet ipat oe oentu to the block i it bounce o than i it ebed itel into the block Conide the change in the block oentu o thee two cae: Bullet tick in block (copletely inelatic head-on colliion): BEFORE COLLISION: AFTER COLLISION: Object # a Object # a Fo thi ituation, coneation o oentu tell u that: + The inal oentu o the block i: but o the coneation o oentu equation, So the inal oentu o the block i: ( ) i ( p ) block +, i ( p ) block i + 8

9 Phyic 40 Now i the bullet i ubbe, intead o haing a copletely inelatic head-on colliion with the block, it will hae an elatic head-on colliion So we ll hae: BEFORE COLLISION: AFTER COLLISION: Fo thi ituation, coneation o oentu tell u that: + () i But we alo know ( othe equation deied in cla coe o coneation o kinetic enegy and coneation o oentu): + + i i But i 0, o thi becoe: Sole () o : And ubtitute thi into (): Sole thi o : + () i i ( ) + i i i + So o the cae o an elatic colliion (ubbe bullet), the inal oentu o the block i: ( p ) block i + Thi i exactly twice the oentu ipated to the block o the cae whee the bullet tick in the block So oe oentu i ipated to the block i the bullet ae ubbe and bounce o Theeoe the block will be oe likely to be knocked oe i the bullet ae ubbe (a long a the a and the initial peed o the bullet i the ae) (b) The bet explanation i explanation I 9