hwsol7 - Homework Solutions #7 Momentum and Inertial Forces...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Homework Solutions #7 Momentum and Inertial Forces Part A: Fmd . p (t) , the x component of the total momentum of the system at time . t TAe..0. +ato-I mome.ntc)m h' In e... t. is ihe..sIJfY\ ~ tn, lI, o? rhe. lnclil/{oJ.v~l !'H.(Jrt"e- I1 +o.. + .s~ f'Yle P (t) l-f) 1"" tn:J V~ (I:) Part B: Find the time derivativedp{t)jdt of the x component of the system's total momentum. ~ [p We.. l t) vi = rn ~ I V, (t) t- rn ~ V~ (t)] =- iJ..I1) ::- '((t I tJ j ~t) l' YY1;). cJ.. \/2 (t) J~ Y\D ==I'" W'I 0- : '" I ;"".. tZ:e ott _ J'l.e.-CA 11 ,. tl ~ Part C: The quantity ma(mass times acceleration) is dimensionally equivalent to which of the following? f;::.l"r\CA- FOl'ce f Aue-fa Acce/e.,~t-'G>n re- F:- 'fff {o,c,e- Part D: Acceleration is due to which of the following physical quantities? ,'s clue -10 Part E: Since we have assumed that the system composed of blocks 1 and 2 is closed, what could be the reason for the acceleration of block I? S/Y1ce.. -rh.e.. sy.{fem tnt farLe. w-OV leA611 is be otftt.r wis.e ~/ase.. cA / e...xert-e.cA... 1- 1, >' b/ 0 C J~ .J. bJocX Part F: What could be the reason for the acceleration of block 2? A 1-0 r c. e... (..X e r f-eo<. ~ y b Joe I{ ~ Part G: Let us denote the x component of the force exerted by block Ion block 2 by F12 ,and the x component of the force exerted by block 2 on block 1 by F21 Which of the following pairs equalities is a direct consequence of Newton's second law? r;~ .; ty1~ a.2. Ol/ F~l ~ ml Part H: Let us recall that we have denoted the force exerted by block 1 on block 2 by exerted by block 2 on block 1 by F21.If we suppose that m1is greater than statements about forces is true? n11.!, F12 , and the force which of the following rv e w fon bot-A Opposi '5 L 0. ~s Wa. I~1'1tJ vV .' e.l:fJ U a,/ a.vld.. Ff"OY'Yl for c e.s, n (>.II~ +e... YYle:tJV YJ,"r vc1e . Part I: Now recall the expression for the time derivative of the x component of the system's total ,f'. "h . momentum: dpjz(t)/dt = F;2:. ConSl.d. enng t h.e lnJormation t at you now h ave, c h oose th e b est a Iternative for an equlvalent expreSSlOn to . . dpz(t)/dt . Sin c he -I .L -I he10 r C SYS4-lUY) CA. 15 G~ c/o~e.cJ-.J tAe e.. c f-; n:J f he. s YeS f--en'l Is zero T),e,,.-e- t-Dre s6R oAt: =-0 Filling the Boat A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 3.00 m/s when it starts to rain. The rain is falling vertically, and it accumulates in the boat at the rate of 1O. 0 kg/hr. Part A: A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 3.00 m/s when it starts to rain. The rain is falling vertically, and it accumulates in the boat at the kgjhr rate of 10.0 . J,. Y\ C. e.. res;s I-t:A-Yl C e. is n efj}; e i b Ie. : we.. ~ /'\0 t4/ ,0; ;. PIyn; Vi :: fYl.p V.p J'Yl/;; m ,. ~ ~50 I-<'!j m; -+ 10 ... ~/hr .~ jhUS J.. r .= J 70 K::; J50 K:; :J. 7t> M E:J 3 ~ .:::-. V...e - :2. 78' r'1/s Part B: Now assume that the boat is subject to a drag force Fddue to water resistance. Is the component of the total momentum of the system parallel to the direction of motion still conserved? No. sa Just will ,..,,,,, e r (",'ch'on (or c.(4.v..se..s QY\o"'BY d,'s.sir>~1-""f/e nor +0 b ~ c.onse",ved... +r,-"f-t"O'Y'l t:l.I"IY J;"ce) (!ClvSe... th~ c."nse.rVtl.hGn of momeYlrlJ1'Y\ fo hrea...~ 0<..0\4/11" Part C: The drag is proportional to the square of the speed of the boat, in the form . What is the acceleration of the boat just after the rain starts? Take the positive xaxis along the direction of motion. I f -::. :!J.. -F., = O.5v2 V :z.. :::: m ex- Sol \I]Yl~ 0...:::::' \ ;).rn rt.e-o..1J Y\ -/. v".l, +,'Ofl Lfj 0'. r'" ~ h'li e tX J a.cc.e-lerc.... is -oJ O-C f-I"j In f-he.- d I' r ec.. fiaJl1 a. == g () X/O A~ Introduction to Collisions Let two particles of equal mass particle 2 is initially stationary. T1l collide. Particle 1has initial velocity 1" directed to the right, and Part A: If the collision is elastic, what are the final velocities 'f)land lfJ-l:of particles 1 and 2? Give the velocity v10f particle 1followed by the velocity 1>:1of particle 2, separated by a comma. Express each velocity in terms of 17. h \I S ",U. se eo. :::::) (v, t- y' .1)/'" .::: p( v V;.W'-I' ~ v, m= voN \/Yfl + fl Pi h6 we Part B: Now suppose that the collision is perfectly inelastic. What are the velocities vIand V:lofthe two particles after the collision? ri/ v"- ~ c V 'let' 1 v, 0./ . V.;l }::; V VI CJ vi . c) "., for Ctn i ne.lastj m.= col I '0' V.f (yY\.,.)'Y\ J VI::= ht V Part C: Now assume that the mass of particle 1 is 2m, while the mass of particle 2 remains m. If the collision is elastic, what are the final velocities vland 1].lof articles 1 and 2? p C"fle5ervo"h'otl ton$et"vat;ol1 So ~ y' ::= o! Of fiome,r!vlH: Jm v:l:::;:: V~ :J. ty\\/, or mv~ ~J...f- ~ ~ne.rr.oy: ~",)v)..::= v, -:J.. r)4 'V':l ~ ~m Yd.m v~ :Jv, +- v: (V , .: ~Uv v':l :; 1-3 Part D: Let the mass of particle 1 be mand the mass of particle 2 be 3m. lfthe collision is perfectly inelastic, what are the velocities of the two particles after the collision? COflser V().t-i (Jf' of mDWfe,t?tr.Jrn m v;:. (3 rn +- m) V, Tne- ve-io.-c,-+yof 'mhl borA po..rt,'c./es . v l/ - L/ , -- Part E: What qualitative change takes place as the ratio of the mass of the blue disk to the mass of the orange disk, TT4x, increases from 0.3 to 4.0? Set the elasticity to 1.0 for a perfectly elastic collision. rAe... fino..l velOCIty of rAe..- (!)f'a.,n[)e oLl'S J~ G han B <U;;.. d.. " r e.c.;ri a Vl . Part F: If the two particles with equal masses mcollide with elasticity E = 0.650, what are the final velocities of the particles? Assume that particle 1 has initial velocity vand particle 2 is initially at rest. Look at the applet to be sure that your answer is reasonable. LVe.. ~nO\V V~-~:::.i3.v (;\.1.50 rYlV, r~v9--;::"fY1V \/, :::.~ -/76 V v~.=:. o. 8~5 V Collision at an Angle Two cars, both of mass m, collide and stick together. Prior to the collision, one car had been traveling north at speed 2v , while the second was traveling at speed Vat an angle tP south of east (as indicated in the figure). After the collision, the two-car system travels at speed Vfintl1at angle an north. () I I I I ____________ ,,~ 1!f'S..~ I I 0 v linul I I east of Part A: Find the speedlJfiilAlofthe joined cars after the collision. LE v'COS.J. _ 'f.J ;2. BreaK vI int-O ;::.. ()\ V Cos " / c-o fI'lP 0 ne fI fs : P,)( ::. P""X f. x _y\ rp :::. J m v':l:X P";j" ==) v'--,?( _- I' +- h L at i(" e.cf;() : 11 PI Y ::;. Y \I ;;.. S .5 i ~ ntJ P,y::' J. P(V-I-p(v i,.., ~ = -;..,< V~y .::.) v~ y -~i n VI; 1\ 0. 1 - t-e::f'"<. v'( " IJ t ~ 5 -'1.s,'Y\f ! ;l. Part B: What is the angle 0with respect to north made by the velocity vector of the two cars after the collision? We- Kt'\o w V7...;< V c..OJ ~ :=. ~ v';).. y ::: \I ( ~ -:. "n f) ft:An e v v)( y -CJ.- ACos ~ i -/ /('J..-~ ,'1'\ .5 j"l'\ (j> - e =- rc t-CA.n r -cosf; L:l. J 1') Center of Mass and External Forces Part A: Find the x-coordinate of the center of mass of the system. m XI m2 1 X~.-." - - X, m, t- XQTYl;!., X2 m, .,.n1:z. Part B: If 1712 ~ m1, 1-0 Part C: If m2 t" e.. then the center of mass is located: Ie. f t- 0 P m~ : d X;L-X, = m1, then the center of mass is located: hOllf WfAy be,t~ee-Y} fYl, q-m;L Part D: Let us assume that the blocks are in motion, and the :J:-components of their velocities at a certain moment are viand 'L'2, respectively. Find the x-component of the velocity of the center of mass at that moment. Remember that, in general, dx V;r= - dt o' .1'1 Vc.m 3e..Y1 e r .= V CM (1.. l: ,;. Xc. 11'\ _ olt" vc.1'11 So. J... X em tit .::: ~ I v, ~, -;;If d1t, P\ I J.)(~ tm,Tm~ art="" ~ re..(,o.l' J. v ,- ~1(1 ckt V::::;:. otX'=-, :;l. ~ + m;. v~ rfl12,. Part E: Suppose that 1.hand ~have equal magnitudes. Also, "lis directed to the right andV2is directed to the left. The velocity of the center of mass is then: . .. ~ ;:. _ -; J.f lv, I.::. l\l~ 1 ()...f\tJ. +ht.y f tJ,'"f in ofJfJDS, t~ J.Jrec.tlo~sJ rJ..eYl V, ~ V c~ -::. fYI, V, - m:l v, :::) m f he.. Vc.n" de.-p e..r1 cls "n th e r ~'h'o , r~~ Part F: Let us assume that the blocks are in motion, and the x-components of their momenta at a certain moment are P1and Pl, respectively. Find the x-component of the velocity of the center of mass at that moment. ~, ""2.. p, ::: , m SO "'''t. VJ P:J. ::W1a V~ lie"!" c.tl'\l-~r- -".,. -~o...tS c..,'t-y IS. vC. PY\ ::. Part G: Suppose that 'V ern = O. Therefore, the following must be true: If,l~lf~l Part H: Let us assume that the blocks are accelerating, and the -x components of their accelerations at a certain moment are Gland a2, respectively. Find the x-component general, of the acceleration of the center of mass at that moment. Remember that, in Ilx = Cit dux we.- "" }'\0 W f(\ I v, r m~ V~ ~,,...m"J.. ..,... ~ rY'~ ~t- VC.M ::; at- at Vc,M - Wl, ~'!l ~t YY1, rm::l. ()... F .=. r1I 0'.... ,.. m;l rJ1'\ ~ 0.. '). c..mit', Part I: Consider the same system of two blocks. An externalforce is now acting on blocktn1 No forces are applied to block Find the acceleration of the center of mass of the system. F _tim m."2. 1 X2 frh Xl F== U sin Cj fYl, tt, ttA. ::: 0 H: +- h e. eCfJ lJ 0...t ;0 " C r-o..... fJ a r IlAc m =- m, t-~~ F . Part J: Consider the same system of two blocks. Now, there are two forces involved. An external force external force F lis acting on block mland another 1 _all~l 1 Xl 1 12 1112 1 i0.is acting on block n12. .1": Find the acceleration of the center of mass of the system. F, = + rn, ~c.W\ lA, ~ ')..:::. rrt ~ lA ~ F, -r F" Yl"t I t- it"::L Part K: Consider the previous situation. Under what condition would the acceleration of the center of mass be zero? Keep in mind that F1 and F2represent the components, not the magnitudes, of the corresponding forces. If P, ::::- -F"J.. LA.c. m - - F, - F, tl'l, t-rn"). - Q Part L: Consider the same system of two blocks. Now, there are two internal forces involved. An internal force 112is applied to block m1by block 'm2and another internal force block ~by block mI. F~l is applied to 1 xl 1 x2 Find the acceleration of the center of mass of the system. F ~/").. F-;I f rO~ JJ e. wto n-..s ~ d Io..u/ ~c.. ~ -.=~. tV\. A Rocket in Deep Space 1 A rocket is fired in deep space, where gravity is negligible. In the first second it ejects 160 of its mass as exhaust gas and has an acceleration of 14.6 mj'; . What is the speed "Vgosoftheexhaust gas relative to the rocket? The, rt)..te. <' f0 c.~~ m L!t e. of rn e. lJ W\M .s is : ot ~ 1= ra m _D ott = - moll ~o ,1'1 t.~v a.ti CL s n _ LA. - t\ t.3Cf -the- 1000/1) ) -rite.. o-c.c~/e('e.tt-"t'n i s cJ..t, ,. e..cA f n : eft - cAv - - ve-x &..m Mo CAt S i J'\ c..e. w~ I~ no vV e:tt= V (,x_ ~ ~Wl {)...:::. -f \J./ l. JI\ (-t ~II':-O) Ls V e, e..e.ol t-o Sol fo r V t.x . ()...(I b 0) ( Is)..: V e..x Ve.x::' {J Lt. t ~s~J (I' 0) ( IS) [Vex; ;;'-336 /5 ) Problem 8.56 A single-stage rocket is fired from rest from a deep-space platform, where gravity is negligible. If the rocket burns its fuel in a time of 50.0 s'and the relative speed of the exhaust gas is V"r = 2100 mIs, what must the mass ratio mfJJmbe for a final speed 'vof 8.00 km/s (about equal to the orbital speed of an earth satellite)? 1'0 1 Yl"0 solll~ rJ,is problem.; (I LCI'!.- nl.e.eA 1-0 f-()./~t. Jelll-"'~{ r-htnfJS c...o S i al-e ,o..f ,'-" n Q) In fo t'war ~ovj 00(' 'f"e.fe.r.f' e.e. frtArtt e. VltSe.~ ()..t'\ ~ -fA-L ('t>c}"\e.-r ,.,,,vlllj cJ.... w; th a.. S)O Le~ war~ V -/--h e.. rocXe t 'Is ex ho-V.s +- I\J 1\ b~Lt~ wit-A o....S)"eecJ.- V;Ue,/ .. L r Vex +h e.. ro c M.e...t e.XdUS+- be; '" 's I' ~f e..r LYle e..j e..e.teJ. 0..+ r()c/~ f ro.AJ e .J 0... i +- s e.e's fit. e. LDn.sro-Y't c.;oe.e,r). re...lo.. 1-i IIe.. +-D +- h e. d. c> Ve.XClU.s t ~ I" .., e.1: j c> \,Vt.. W f''' t-'t. w' Yl : -=- v - Vfue.1 ~ (1) fe.w 1'0 c ty1 /" r h (... ,.()~~et~ in t}.e frC/J'He JI\ ahseve.,.J" frct.m~ We. w'" 1/ use.. Y'tl 0 t- A is res v J I- ,. 0- v f-es 111 0"" Ar 0- c.. e,; r 0. i n ~().lf\ me" f ,. 1-'" " Nt e-) Me ..,.,'s e+1 rlJY"f iQ: -f he- p-=we. c..ht,.~"" s y s ru-n th ~ fA ()1 V 1CJ,.j..~Y"" se.conJ.. SfnceV ~ fhi"s i..s C\. C 10so e.';""" m 0 1"1 e..1"+v,." e..LO not " s c..(:J n S er v e.~ . 1~1-~r./ V ~ A & r oJ...V rn ~ m +olm ~ i 1\ C f e.tl.S e,..J.- ve.loG"t;' (). e.e"- (~s e.d. m~.0 () "-0 e ~ e + . f 1( cJ.m is,' n kift1\Hy fA e.. rOc,AL-r .rhe. ex1,()"II,s1- 1\0 W h(u.$}.... m 0 WI e-nrvW\ mop' -LYlftJrrt) O.p 'fif'o c~U-)::::: (fY\ t' rJ... m) (V t-c1 \I ) Bi &Ie" o../so Pfvel AO-S hy " ne 5lX.r ; lie, (,. '" o..s s u ~ P f-i 0 n ) +""s :;. -d..rY1 V~vt..1 I.S' c.~"rifllJUJl. SiJ'\c.e the ,rlO,;lent\1M of t~e. \I~l f'ocl-'\.e, ..-/v(.1 .sysrQ,.m "s ctmse.,.lIe.eJ.: P, :: f roc,f\ e.+ ... f t f fhi5 is .:= c fI''' = (~ I-d-M) ( \I + o\\I) - olI'M V {vel sobsl-"fvt,'o", fJ,. 0.. ~ooJ. p1o..cc. tD use. "ur- (J). (V,cuc.l) is vo..t' ,.. >IeI 1 It,, w e..11e." we.. c.~" ex f res S (r C4 s .. V;l' ue,l :;:; v - v't-~ 0n f n is j.5 0.. l) DO Q.. sub s f; fvf-i b~ c.o..lIS e.. we. ~ss e.'"h'",1 J. (II Qo' r t. d. lJ C e.. r I, . til)'" b t.r "f t-() 1. (v). ~ em t.rn be.r I v~ ~o: 'X p &-.1\ 11 tA" i '" b Ie s f r" '" lie-x) V;.oe..t) i.} 0- C" n sfa I'tf. m"~ (yn-l- olm)(V cJ.i 1'\ ~: rcJ.v) -elM (v - ~~ ~ +- J,.~+- /'lid-v 1"' J,. tAv - ,,(";1/ rd.'" 1Ie.x So w e. ho.tlt., Ie,r-l" "" rJ,. tI .::. - cJ..y>;\ v(., ' X - ol ~ J.. v o..no c..vl 1-11 e~ (-0 re. the ft.r h to n Co HI 01.,.. tJ.1I i5 vvy very f,.", Y e~' e.c re.ol- mtA\I ~ -cAW' \/.>" V _ - 1/ e, c.A"; r\ ~ rt,f'rn s '" j y\ -l .1 o..,v). m ...-...yV\ v _ cAm ., ~ X v r" te,.t*'c.... -rj "3 \AI-(.. ~o 11-11" r c>hr~;II\: v rJ.- \/ ::. - Ve.)tf f "" =: - lie-x [J."... L"'] "'0 t;- ~ t:.,..,,] J Q.,..ssvtn e "o'=' 0 ~ t.,;< = .J.,.... [~] F tf\ a, It y'" e.. xp L jL. 1::; Ve.Je ~ .:::. '-/5 1""\ I Problem 8.70 A rifle bullet with mass 8.00 g strikes and embeds itself in a block with mass 0.992 kg that rests on africtionless, horizontal suiface and is attached to a coil spring. The impact compresses the spring 15.0 cm. Calibration of the spring shows that aforce of 0.750 N is required to compress the spring 0.250 cm. , . ~15()Cnl-~ Part A: Find the magnitude of the block's velocity just after impact. 1 s.r .f " Y\ J. 1)., froW\ +of" t-h~ SjO Y"i nfj : 0 the. c <'0.1 " h ro..h' "l W= Y:l. 14. X ~::. F .. 1< :- K=- ~ X :1 Co. 7501.1) .QO~5..,., ,. oS 6oot:J. J'Y\ We.. 1-<'"a w the b lOG to the.. c.Ol'Y'pre.sSle:JJ1 /~ 'J in)' t ,. 0../ ~(. vel 0 G" +1' re.I (Jf.. +--e et ot,'sf-o..nCe- rAe.. sl't"""'j X'=~15m y~ m v ~ = v;)..= ~ I" x~ I~:: m 6 00 bJ. )'II) m.:: O. '1 9 ~ I~:J .,.. ~ :;. I I~j B ~x~ 1 ""'" v=- j X 3.67mj.s Part B: What was the initial speed of the bullet? -r~ e. )In U oS f- i ., i oJ-; o..l ~~ u ex. J mom ~n J.- cJn1 J't'1 0 f of h 0(., b ()J1 e..+ - h 10 c.., H s Y s f-e W\ f h e. II f,. n 0..1 a ".. ~11 f v W\ Pb II (. t' Qg ::: bull P f.. + + b Jo c..l~ (1 H ~ ) {3 . (, 1 "Ys) (~ 0 I~~ ) (Vb)'::; [Vb:::' l.j S 9. H "'/) } Problem 8.72 A movie stuntman (mass 80.0 kg) stands on a window ledge 5.0 m above the floor. Grabbing a rope attached to a chandelier, he swings down to grapple with the movie's villain (mass 70.0 kg), who is standing directly under the chandelier. (Assume that the stuntman's center of mass moves downward 5.0 m. He releases the rope just as he reaches the villain.) Part A: With what speed do the entwined foes start to slide across the floor? ,The. momellt" 01- po..t l'I1o.n ri[j '" t be. fore. he. hits +h~ v,/, CA-n is e'i Lio..l f-o the rna '" " II t v IY1 (' if:J r, +- cJ ref' f he! r, I r. (E 0 }"',,) II, ::;:. (E D /'I~ .,.. 70 K3) V", wno..t- is },j.s ve-Io ci W' (.. +! (v,)? c 0/7 f1 fA B- I eJ~ e. etgvo./.s M f\ 0 'V r1 ()../ fho.f h" ~ h ::;.Dl Y po ten r;0..1 c. e.re; y h is f i X ,. n e -hi c e J1 erB Y : f\'\ ~ rV\ v, ~ 50 j in +-he. 1"10 VJ :::. m enhJm fii;h. c..OllS u'"l/",-t10l1 ~~ vat ion is no &01': II 0 cS 0.1 I.l, ~ J ). ~ h :. J b0 I~ ~ V;t r h. e. SfJ e,fd... wn; c,n -rh L l' -!5~ r~'jh s Ii r).." is: V;l'::' = 5 - ~8 ""'Is ~~::e~: If the coefficient of kinetic friction of their bodies with the floor is Jlr.: --. 0245 , h ow fiar d 0 they . Ihe... forc.e, f::: ~ 0 J of fr; ~t'io~ ,'s CA.c-til'ltj "n +nt.. ""e}1: YY\}L ~ =- m 0... Wl the.. ole. c. e.,lIyo.:t ,.an (){- f h e... I,~r" e. n is: 0... =:. -.P- j use.. f o..,m; e.~u 0.+i 0 " 1/, 1. : = Vi ':l. ~ +- ~ ()... AX ljA-~ AX v,. = ~.J-A-J ( b. c2g~h):l =- l:i X == 5.8(j m Problem 8.78 A 5.00-g bullet is shot through a 1.00-kg wood block suspended on a string 2.000 m long. The center of mass of the block rises a distance of 0.45 cm. Find the speed of the bullet as it emerges from the block if its initial speed is 450 mls. W)O liII\e." t-U-W1 W~ ~.s0lJ""'e.. n'\ 0..('" the.. j.,itio..l e.. S y s te e.1 llncA. /1''''0..1 t); the. lJo...(, , o fi b p,,; yY'\ Vb=- .. , .. , tJ V bloc."" t:J P ~ - m b1od" mbVb -I- Ynb Vba b;:' . (J 0 5 I~ 3 Vh'oc..K i.s '-I 5 D m/s \lb~ r, e ht' is the. sp~etA of f-ne. bloc~ o.Jte.,- +h e. bv IIe.t emerfj es. +he., e.1Y1er6in~ bu 1I~+-. tke.. spee..oA- of \j./ h ot . ij v' b 10 c../A. ~ ,he. """",imum heil()hr of the. blocI'. ~v ~ (F'r=.) is. flrofJorriol'lq..1 tD t"he. it'\itio..l ve,lo ciry pf~h ~ '/.7v::. J;t~h ~ I!bIOc:.l",::.~OO'f5mJ V b/ocK::-- o. 1q 7 ""Is W('i+i~ 0 ut the- W\,omt.t\tum c..Ol1ser VQ.+-iotl e.~u of(. (A.+iOl1: (. 00 5 I~~) ('f 50 '%)::." (I /~5)(o. :t '! 7 ""Is) 005 ""5) Vb .1. Sol v t -Eo r Vb" : [Vb~~3"/0. b ~ Problem 8.82 A blue puck with a mass of 3.90xlO-2 kg, sliding with a speed of 0.250 m/s on africtionless, horizontal air table, makes a perfectly elastic, head-on collision with a red puck with mass mfilll, initially at rest. After the collision, the velocity of the blue puck is 8.00xlO-2 lUis in the same direction as its initial velocity. Part A: Find the velocity of the red puck after the collision. We... I'\(lOW ... -e. h re.lo..tive. o..f I--e.r f'h \/elociri~s ho..Vt.. 1') +he sCAftI e.- m 1k3 n ; hI cJ..e,... be. fa re, Qnt! e c.o U;sio ViS I VB, ~ VA, .= - = :::=I tJ. ~5 trt/s 0 (\18.;1. - V~/) v~ V ::r. "i fj (;l.1 8 ;a. =- 8 x It) -~ IY'/s \AI h~"" 1.5 VA ~ 7 . O. J 6 "'Is ;;. - ( 8 .x I ()-2 "'/S - VA,.) VII) ::.~ (0. + mA VA:t 3 3~ !!! mome."+V"": Part B: Find the mass mredofthe red puck. USe. tAe. e'llU",tioh for consl2.rVo-TilJ'IJ DI Pi ::.P .f YYlb VOl:;;:' Tnb Vb~ \II e. fI e..e. do- -I- <> sol IIe.. fd r rnb Vbl - ,."R (/- It e Vl\~ .o..S s 0 -f +h... re"l. I' vc.-I'o) mb V&;L =- ~ .2 5 VY1" ( Vb - , VO.1) _.=. yV1" 0... VA~ If X 10 -~ K,J (. 'r.s -." 8 "'/sl ::;.YI1 A O. 3 :3 ""'/5 l~ o. (J~K37 Problem 8.94 A 45.0-kg woman stands up in a 60.0-kg canoe of length 5.00 m. She walks from a point 1.00 m from one end to a point 1.00 m from the other end. If you ignore resistance to motion of the canoe in the water, how far does the canoe move during this process? ~ Slal1 1.00m Finish 3.00 III ) ( 1.00 1~1 'W h e,re, is t~ c e.. 0 e. rtre.y- 0 f m GlS j .[ ? C.0-'" 0 .r'" t.. c en lot..I" f ... sS 0 -fh e. e.. 1s i!-s c e h I-e,r 0 1...e.t + he. j,,; h 0..1 Ce.nl-e..r ({z. t fA e. t:J- --:;::/ \./vJL-v COd''\ 0 e... b e- -rh~ t.ro ,\ c.oore)..; I\~fe. 'm 1 I.SM 0 \.J "e..- e.- i.s t-h e. c t.." I- t.. r 0 f- ,.,. lAS S ? ?(,,,,,:: (~O 1>\":) (~ .,.. ('is ~:'\) { (, 0 ~~) I.~ r ;:;; (L{ 5 K~) x(",:::.:!.?!'~ I 5 0 I"'" I . 5 h'l O. b '1 3 I'Y1 IV 0 IN') t hi" o..re. a-.c1-j n, O. K a.bo 0" ho.pplMi"'j' No e-x fe r 1\<>.1 f~rc.e.5 f he. S yS I-eth So fh e. c e l'll-e."- "f WI o..s..s <1 -I- "",1,0.+ ~ i'..s s r 0,...i0 f\ o..r y. ''13", d 0 .. ,jm I U Whe..n the. \AI' CA.I W 0 yY\ ()... Y\ h ('A, S 0 I" er).. ro +- h e 0 +A e r S icJ.. e/ ~ / -11. t '?::=:~""""------~ , I~ f"h e... c. f..,,+er l-S J'\ Xcm; : I:;; -f f'h.e. c..~n ()! o . ~'1 3 m 1-0 /-h e0 fi"q..l fositl'~ I of CQ.Y\ () e P os i ~io n of -I-h e. c. C\ 0 e.. \ n a'i~1I Y 0 f- J t,f. t f fA e, eM. fA e. ClA.n 0 e YY\O ves 0... +-0 +-~ I [/.:H M I Problem 8.96 A projectile of mass 19.5 kg is fired at an angle of 65.0 above the horizontal and with a speed of 0 80.0 . At the highest point of its trajectory the projectile explodes into two fragments with equal mass, one of which falls vertically with zero initial speed. You can ignore air resistance. Part A: How far from the point offiring does the other fragment strike ijthe terrain is level? , m/s .. .... \ \ \ at fAe.. rop of -f J..-e, fro.. in )ee...+-or y./ I-hehorizo..nf-t'olo.,Xl"s: 1'1S (21" 11Q().... P,' =fJF fXi I I =- f~f < ) mom e.n of u r1'\ is c.0 ,-.n X be. c..a..u..S e.. 2.-F =- 0 ..,.. ~ 1'\ I rIO. 'L- /1 y.. Pv ,.; - m V0 C 05 e +- Y.:~JY\ V}t' f ~\I" co.s .J e ::./.:1. '/. V,x of J f tJow) %~ =:. i rn (v ~O) Vx f =fo f..(J 6 7" 6;J, "3 we.. use. ''''-tifleYJ1a.hc. fe.chrb'tues j-h~ ~roll~ol. fjflrJ h;t-s t-he.. E>ro"~,,~ : /'r"fY\ where. lLfJl!...x: f::. ~VJ<.flt rAe. ~ ~Y'(;1..(jMe~d-- l{,.sj)'\8::::.[/.'tosj So the.. -tol-.I ai.staY\ce.. (05: AX ::{v6cose Part B: How much energy is released during the explosion? l ~X:=. 750 ~ W~~t i-s +Jt e., In" t '-0.1 1"\ E ot- +h e- site II? XE J"\ == Y;l m (vo Y;t (~ C o.:s e) ~ A Y\ r1. t-he of; "'0..1 e.. V1er~ 2.. y .. + E - ) (0) M Vo i: (':. ) (:J. v"r) J., l~Et: =- ~co s ~e e: '" 1l'I Vo W he>-- /5 + .b.""E.::::. fA~ C.h.o..I\3 1'\ IEI' -1<' EI "-cos >'e - h 1Y\lIo "-CO S "e ~ I~E.'::' y;>. m Vo ~cos~e [b K IE =- I "1'15 J~/t!..SJ ...
View Full Document

This note was uploaded on 05/22/2008 for the course PHYS 2A taught by Professor Hicks during the Winter '07 term at UCSD.

Ask a homework question - tutors are online