PHN TH NHTBAI TAP ONG LC HOC CHAT IEMBAI 1 :Hai l xo: l xo mt di thm 2 cm khi treo vt m1 = 2kg, l xo 2 di thm 3 cm khi treo vt m2 = 1,5kg. Tm t s k1/k2.Bi gii: Khi gn vt l xo di thm on l. v tr cn bng

mg l K P F0 Vi l xo 1: k1l1 = m1g(1)Vi l xo 1: k2l2 = m2g(2)Lp t s (1), (2) ta c 2235 , 12ll.mmKK122121 BAI 2 :Mt xe ti ko mt t bng dy cp. T trng thi ng yn sau 100s t t vn tc V = 36km/h. Khi lng t l m = 1000 kg. Lc ma st bng 0,01 trng lc t. Tnh lc ko ca xe ti trong thi gian trn.Bi gii:

Chn hng v chiu nh hnh v Ta c gia tc ca xe l: ) s / m ( 1 , 01000 10tV Va2 0Theo nh lut II Newtn :

+ a m f FmsF fms = maF = fms + ma= 0,01P + ma= 0,01(1000.10 + 1000.0,1) = 200 N BAI 3 :Hai l xo khi lng khng ng k, cng ln lt l k1 = 100 N/m, k2 = 150 N/m, c cng di t nhin L0 = 20 cm c treo thng ng nh hnh v. u di 2 l xo ni vi mt vt khi lng m = 1kg. Ly g = 10m/s2. Tnh chiu di l xo khi vt cn bng.Bi gii:

Khi cn bng: F1 + F2 = Vi F1 =K1l; F2 = K21 nn(K1 + K2) l = P) m ( 04 , 025010 . 1K KPl2 1 + Vy chiu di ca l xo l:L = l0 + l = 20 + 4 = 24 (cm) BAI 4 :Tm cng ca l xo ghp theo cch sau:Bi gii: Hng v chiu nh hnh v:Khi ko vt ra khi v tr cn bng mt on x th : dn l xo 1 l x, nn l xo 2 l x Tc dng vo vt gm 2 lc n hi 1F;2 F, + F F F 2 1Chiu ln trc Ox ta c : F = F1 F2 = (K1 + K2)xVy cng ca h ghp l xo theo cch trn l: K =K1 + K2 BAI 5 :Hai vt A v B c th trt trn mt bn nm ngang v c ni vi nhau bng dy khng dn, khi lng khng ng k. Khi lng 2 vt l mA = 2kg, mB = 1kg, ta tc dng vo vt A mt lc F = 9N theo phng song song vi mt bn. H s ma st gia hai vt vi mt bn l m = 0,2. Ly g = 10m/s2.Hy tnh gia tc chuyn ng.Bi gii: i vi vt A ta c:

+ + + +1 1 ms 1 1 1 1a m F T F N PChiu xung Ox ta c: F T1 F1ms = m1a1Chiu xung Oy ta c: m1g + N1 = 0 Vi F1ms = kN1 = km1gF T1 k m1g = m1a1 (1)* i vi vt B:

+ + + +2 2 ms 2 2 2 2a m F T F N PChiu xung Ox ta c:T2 F2ms = m2a2Chiu xung Oy ta c: m2g + N2 = 0 Vi F2ms = k N2 = k m2g T2 k m2g = m2a2 (2)V T1 = T2 = T v a1 = a2 = a nn:F - T k m1g = m1a(3) T k m2g = m2a(4)Cng (3) v (4) ta cF k(m1 + m2)g = (m1+ m2)a 22 12 1s / m 11 210 ). 1 2 ( 2 , 0 9m mg ). m m ( Fa ++ ++ BAI 6 :Hai vt cng khi lng m = 1kg c ni vi nhau bng si dy khng dn v khi lng khng ng k. Mt trong 2 vt chu tc ng ca lc ko F hp vi phng ngang gc a = 300 . Hai vt c th trt trn mt bn nm ngang gc a = 300H s ma st gia vt v bn l 0,268. Bit rng dy ch chu c lc cng ln nht l 10 N. Tnh lc ko ln nht dy khng t. Ly 3 = 1,732.Bi gii: Vt 1 c : + + + +1 1 ms 1 1 1 1a m F T F N PChiu xung Ox ta c: F.cos 300 T1 F1ms = m1a1Chiu xung Oy : Fsin 300 P1 + N1 = 0V F1ms = k N1 = k(mg Fsin 300)F.cos 300 T1k(mg Fsin 300) = m1a1 (1)Vt 2: + + + +2 2 ms 2 2 2 2a m F T F N PChiu xung Ox ta c: T F2ms = m2a2Chiu xung Oy : P2 + N2 = 0M F2ms = k N2 = km2gT2 k m2g = m2a2Hn na v m1 = m2 = m; T1 = T2 = T ; a1 = a2 = a F.cos 300 T k(mg Fsin 300) = ma(3) T kmg = ma(4)T (3) v (4) m0 0t2) 30 sin 30 (cos TT + 2021268 , 02310 . 230 sin 30 cosT 2F0 0 m+ +Vy Fmax = 20 NBi 7:Hai vt A v B c khi lng ln lt l mA = 600g, mB = 400g c ni vi nhau bng si dy nh khng dn v vt qua rng rc c nh nh hnh v. B qua khi lng ca rng rc v lc ma st gia dy vi rng rc. Ly g = 10m/s2. Tnh gia tc chuyn ng ca mi vt. Bi gii: Khi th vt A s i xung v B s i ln do mA > mB vTA = TB = TaA = aB = ai vi vt A: mAg T = mA.ai vi vt B: mBg + T = mB.a* (mA mB).g = (mA + mB).a2B AB As / m 2 10 .400 600400 600g .m mm ma * ++Bi 8: Ba vt c cng khi lng m = 200g c ni vi nhau bng dy ni khng dn nh hnh v. H s ma st trt gja vt v mt bn l = 0,2. Ly g = 10m/s2. Tnh gia tc khi h chuyn ng. Bi gii: Chn chiu nh hnh v. Ta c:

+ + + + + + + + + + a M P T T N P F T T N P F1 1 2 2 2 ms 2 3 4 3 3 3Do vy khi chiu ln cc h trc ta c:' 3 ms 42 ms 3 21 1ma F Tma F T Tma T mg

V a a a a' T T TT T T3 2 14 32 1 ' ma F Tma F T Tma T mgms'ms'

' ma 3 mg 2 mgma 3 F 2 mgms2s / m 2 10 .32 , 0 . 2 1g .32 1a Bi 9: Mt xe trt khng vn tc u t nh mt phng nghing gc = 300. H s ma st trt l = 0,3464. Chiu di mt phng nghing l l = 1m. ly g = 10m/s2 v 3 = 1,732 Tnh gia tc chuyn ng ca vt.

Bi gii: Cc lc tc dng vo vt:1) Trng lc P2) Lc ma st msF3) Phn lc N ca mt phng nghing 4) Hp lc + + a m F N P FmsChiu ln trc Oy: Pcox + N = 0 N = mg cox(1)Chiu ln trc Ox : Psin Fms = maxmgsin N = max

(2)t (1) v (2) mgsin mg cox = maxax = g(sin cox)= 10(1/2 0,3464.3/2) = 2 m/s2 BAI 10 :Cn tc dng ln vt m trn mt phng nghing gc mt lc F bng bao nhiu vt nm yn, h s ma st gia vt v mt phng nghing l k , khi bit vt c xu hng trt xung. Bi gii: Chn h trc Oxy nh hnh v.p dng nh lut II Newtn ta c : 0 F N P Fms + + + Chiu phng trnh ln trc Oy: N Pcox Fsin = 0 N = Pcox + F sinFms = kN = k(mgcox + F sin)Chiu phng trnh ln trc Ox : Psin F cox Fms = 0 F cox = Psin Fms = mg sin kmg cox kF sin + + ktg 1) k tg ( mgsin k cos) kcox (sin mgF BAI 11 :Xem h c lin kt nh hnh vm1 = 3kg; m2 = 1kg; h s ma st gia vt v mt phng nghing l = 0,1 ; = 300; g = 10 m/s2Tnh sc cng ca dy? Bi gii: Gi thit m1 trt xung mt phng nghing v m2 i ln, lc h lc c chiu nh hnh v. Vt chuyn ng nhanh dn u nn vi chiu dng chn, nu ta tnh c a > 0 th chiu chuyn ng gi thit l ng.i vi vt 1: + + +1 1 ms 1 1a m F T N PChiu h xOy ta c: m1gsin T N = mam1g cox + N = 0*m1gsin T m1g cox = ma(1)i vi vt 2: +2 2 2 2a m T Pm2g + T = m2a (2)Cng (1) v (2) m1gsin m1g cox = (m1 + m2)a) s / m ( 6 , 0410 . 1233 . 1 , 021. 10 . 3m mg m cos m sin g ma22 12 1 1 + V a > 0, vy chiu chuyn ng chn l ng* T = m2 (g + a) = 1(10 + 0,6) = 10,6 N BAI 12 :Sn i c th coi l mt phng nghing, gc nghing a = 300 so vi trc Ox nm ngang. T im O trn sn i ngi ta nm mt vt nng vi vn tc ban u V0 theo phng Ox. Tnh khong cch d = OA t ch nm n im ri A ca vt nng trn sn i, Bit V0 = 10m/s, g = 10m/s2. Bi gii: Chn h trc nh hnh v.Phng trnh chuyn ng v phng trnh qu o l:

'20gt21yt V xPhng trnh qu o

) 1 ( xVg21y220Ta c:' sin d OK ycos d OH xAA V A nm trn qu o ca vt nng nn xA v yA nghim ng (1). Do : 220) cos d (Vg21sin d m 33 , 130 cos30 sin.1010 . 2cossin.gV 2d00 2 20 BAI 13 :Mt hn c nm t cao 2,1 m so vi mt t vi gc nm a = 450 so vi mt phng nm ngang. Hn ri n t cnh ch nm theo phng ngang mt khong 42 m. Tm vn tc ca hn khi nm ?GIAIChn gc O ti mt t. Trc Ox nm ngang, trc Oy thng ng hng ln (qua im nm). Gc thi gian lc nm hn .Cc phng trnh ca hn x = V0 cos450t (1)y = H + V0sin 450t 1/2 gt2 (2)Vx = V0cos450(3)Vy = V0sin450 gt (4)T (1) 0045 cos Vxt Th vo (2) ta c : ) 5 (45 cos Vx. g21x . 45 tg 4 y0 2 2020 + Vn tc hn khi nm Khi hn ri xung t y = 0, theo bi ra x = 42 m. Do vy) s / m ( 2042 1 .229 . 4 42H x . 45 tg 45 cos2g. xV045 cos Vxg21x 45 tg H0 000 2 2020++ + BAI 14 :Mt my bay ang bay ngang vi vn tc V1 cao h so vi mt t mun th bom trng mt on xe tng ang chuyn ng vi vn tc V2 trong cng 2 mt phng thng ng vi my bay. Hi cn cch xe tng bao xa th ct bom ( l khong cch t ng thng ng qua my bay n xe tng) khi my bay v xe tng chuyn ng cng chiu.Bi gii:

Chn gc to O l im ct bom, t = 0 l lc ct bom. Phng trnh chuyn ng l:x = V1t (1)y = 1/2gt2(2)Phng trnh qu o:220 xVg21yBom s ri theo nhnh Parabol v gp mt ng ti B. Bom s trng xe khi bom v xe cng lc n B vgh 2gy 2t gh 2V x1 B Lc t = 0 cn xe Agh 2V tV AB 2 2 * Khong cch khi ct bom l : ) 2V (Vgh 2) V V ( AB HB HA1 2 1BAI 15 :T nh mt mt phng nghing c gc nghing so vi phng ngang, ngi ta nm mt vt vi vn tc ban u V0 hp vi phng ngang gc . Tm khong cch l dc theo mt phng nghing t im nm ti im ri. Bi gii; Cc phng thnh to ca vt:) 2 (gt 21t sin V H y) 1 ( t cos V x200' + T (1) cos Vxt0 Th vo (2) ta c:(3)cos Vxg21x tg H y2 202 + Ta c to ca im M: ' sin l H ycos l xMMTh xM, yM vo (3) ta c: + 2 202 2cos V 2cos glcos l tg H sin l H + + + 22022022 20cos g) sin(cos V 2cos gsin cos cos sincos V 2cos gsin cos tg. cos V 2 lBAI 16 : mt i cao h0 = 100m ngi ta t 1 sng ci nm ngang v mun bn sao cho qu n ri v pha bn kia ca to nh v gn bc tng AB nht. Bit to nh cao h = 20 m v tng AB cch ng thng ng qua ch bn l l = 100m. Ly g = 10m/s2. Tm khong cch t ch vin n chm t n chn tng AB. Bi gii: Chn gc to l ch t sng, t = 0 l lc bn.Phng trnh qu o 220 xVg21y n chm t gn chn tng nht th qu o ca n i st nh A ca tng nn 2A20AxVg21y s / m 25 100 .80 . 210 . 1x .yg21VAA0 Nh vy v tr chm t l C m ) m ( 8 , 1110100 . 225gh 2Vgy . 2V x0C0 C Vy khong cch l: BC = xC l = 11,8 (m)BAI 17 :Mt vt c nm ln t mt t theo phng xin gc ti im cao nht ca qu o vt c vn tc bng mt na, vn tc ban u v cao h0 =15m. Ly g = 10m/s2.Tnh ln vn tcBi gii: Chn: Gc O l ch nm* H trc to xOy* T = 0 l lc nmVn tc ti 1 imy xV V V + Ti S: Vy = 0 cos V V Vo x sM o os6021cos2VV V ( )s / m 202315 x 10 x 2singy 2Vg 2sin Vyso2ox BAI 18 :Em b ngi di sn nh nm 1 vin bi ln bn cao h = 1m vi vn tc V0 = 10 2 m/s. vin bi c th ri xung mt bn B xa mp bn A nht th vn tc oV phi nghing vi phng ngang 1 gc bng bao nhiu?Ly g = 10m/s2. Bi gii: vin bi c th ri xa mp bn A nht th qu o ca vin bi phi i st A. Gi 1V l vn tc ti A v hp vi AB gc 1 m:g2 sin VAB12(coi nh c nm t A vi AB l tm AB ln nht th 41 2 sin1 1 V thnh phn ngang ca cc vn tcu bng nhau V0cos = V.cos11ocos .VVcos Vi ' 21cosgh 2 V V12oNn ( )2110 21 x 1021Vgh2121.Vgh 2 Vcos2 2oo2o o60 BAI 19 :Mt bn nm ngang quay trn u vi chu k T = 2s. Trn bn t mt vt cch trc quay R = 2,4cm. H s ma st gia vt v bn ti thiu bng bao nhiu vt khng trt trn mt bn. Ly g = 10 m/s2 v 2 = 10Bi gii: Khi vt khng trt th vt chu tc dng ca 3 lc: ngh F ; N , PmsTrong :0 N P +Lc vt chuyn ng trn u nnmsF l lc hng tm:' ) 2 ( mg . F) 1 ( R mw Fms2msgR wg . R w22 Vi w = 2/T = .rad/s 25 , 01025 , 0 .2 Vy min = 0,25BAI 20 :Mt l xo c cng K, chiu di t nhin l0, 1 u gi c nh A, u kia gn vo qu cu khi lng m c th trt khng ma st trn thanh () nm ngang. Thanh () quay u vi vn tc gc w xung quanh trc (A) thng ng. Tnh dn ca l xo khi l0 = 20 cm; w = 20rad/s; m = 10 g ; k = 200 N/m

Bi gii: Cc lc tc dng vo qu cudhF ; N ; P( )( )2o2o2 2o2mw Kl mwll mw mw K ll l mw l K + vi k > mw2( )( )m 05 , 020 . 01 , 0 2002 , 0 . 20 . 01 , 0l22 BAI 21 :Vng xic l mt vnh trn bn knh R = 8m, nm trong mt phng thng ng. Mt ngi i xe p trn vng xic ny, khi lng c xe v ngi l 80 kg. Ly g = 9,8m/s2 tnh lc p ca xe ln vng xic ti im cao nht vi vn tc ti im ny l v = 10 m/s.Bi gii: Cc lc tc dng ln xe im cao nht l N ; PKhi chiu ln trc hng tm ta cN 216 8 , 981080 gRvm NRmvN P2 22

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+BAI 22 :Mt qu cu nh c khi lng m = 100g c buc vo u 1 si dy di l = 1m khng co dn v khi lng khng ng k. u kia ca dy c gi c nh im A trn tr quay (A) thng ng. Cho trc quay vi vn tc gc w = 3,76 rad/s. Khi chuyn ng n nh hy tnh bn knh qu o trn ca vt. Ly g = 10m/s2.Bi gii: Cc lc tc dng vo vt P ; TKhi () quay u th qu cu s chuyn ng trn u trong mt phng nm ngang, nn hp lc tc dng vo qu cu s l lc hng tm.T P F + vi

'R mw FP F2gR wmgFtg v 2 R = lsin cossingsin l wtg2V o2 245 707 , 01 . 76 , 3 10l wgcos 0 Vy bn knh qu o R = lsin = 0,707 (m)BAI 23 :Chu k quay ca mt bng quanh tri t l T = 27 ngy m. Bn knh tri t l R0 = 6400km v Tri t c vn tc v tr cp I l v0 = 7,9 km/s. Tm bn knh qu o ca mt trng.Bi gii:Mt trng cng tun theo quy lut chuyn ng ca v tinh nhn to.Vn tc ca mt trngRGMvoTrong M0 l khi lng Tri t v R l bn knh qu o ca mt trng.Vn tc v tr cp I ca Tri t( ) ( )( )km 10 . 38 R14 , 3 . 49 , 7 x 24 . 3600 . 27 . 64004v . T RRRRTvR 2R .T2v ;RRvvRGMv522 222o o 3 ooooooo BAI 24 :Qu cu m = 50g treo u A ca dy OA di l = 90cm. Quay cho qu cu chuyn ng trn trong mt phng thng ng quanh tm O. Tm lc cng ca dy khi A v tr thp hn O. OA hp vi phng thng ng gc = 60o v vn tc qu cu l 3m/s, g = 10m/s2.Bi gii: Ta c dng: a m P ; TChiu ln trc hng tm ta cN 75 , 09 321x 10 05 , 0Rv60 cos g m TRvm maht 60 cos P T2 202o

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+ PHN TH HAIMT S BI TP VT L VN DNG SNG TO PHNG PHP TA Phng php ta l phng php c bn trong vic gii cc bi tp vt l phn ng lc hc. Mun nghin cu chuyn ng ca mt cht im, trc ht ta cn chn mt vt mc, gn vo mt h ta xc nh v tr ca n v chn mt gc thi gian cng vi mt ng h hp thnh mt h quy chiu.Vt l THPT ch nghin cu cc chuyn ng trn mt ng thng hay chuyn ng trong mt mt phng, nn h ta ch gm mt trc hoc mt h hai trc vung gc tng ng.Phng php+ Chn h quy chiu thch hp.+ Xc nh ta ban u, vn tc ban u, gia tc ca cht im theo cc trc ta : x0, y0; v0x, v0y; ax, ay. ( y ch kho st cc chuyn ng thng u, bin i u v chuyn ng ca cht im c nm ngang, nm xin).+ Vit phng trnh chuyn ng ca cht im'+ + + + 0 0y2y0 0x2xy t v t a21yx t v t a21x + Vit phng trnh qu o (nu cn thit) y = f(x) bng cch kh t trong cc phng trnh chuyn ng.+ T phng trnh chuyn ng hoc phng trnh qu o, kho st chuyn ng ca cht im:- Xc nh v tr ca cht im ti mt thi im cho.- nh thi im, v tr khi hai cht im gp nhau theo iu kin'2 12 1y yx x- Kho st khong cch gia hai cht im 22 122 1) y (y ) x (x d + Hc sinh thng ch vn dng phng php ta gii cc bi ton quen thuc i loi nh, hai xe chuyn ng ngc chiu gp nhau, chuyn ng cng chiu ui kp nhau,trong cc cht im cn kho st chuyn ng tng minh, ch cn lm theo mt s bi tp mu mt cch my mc v rt d nhm chn. Trong khi , c rt nhiu bi ton tng chng nh phc tp, nhng nu vn dng mt cch kho lo phng php ta th chng tr nn n gin v rt th v.Xin a ra mt s v d:Bi ton 1Mt vt m = 10kg treo vo trn mt bung thang my c khi lng M = 200kg. Vt cch sn 2m. Mt lc F ko bung thang my i ln vi gia tc a = 1m/s2. Trong lc bung i ln, dy treo b t, lc ko F vn khng i. Tnh gia tc ngay sau ca bung v thi gian vt ri xung sn bung. Ly g = 10m/s2.Nhn xtc xong bi, ta thng nhn nhn hin tng xy ra trong thang my (chn h quy chiu gn vi thang my), rt kh m t chuyn ng ca vt sau khi dy treo b t. Hy ng ngoi thang my quan st (chn h quy chiu gn vi t) hai cht im vt v sn thang ang chuyn ng trn cng mt ng thng. D dng vn dng phng php ta xc nh c thi im hai cht im gp nhau, l lc vt ri chm sn thang.GiiChn trc Oy gn vi t, thng ng hng ln, gc O ti v tr sn lc dy t, gc thi gian t = 0 lc dy t.Khi dy treo cha t, lc ko F v trng lc P = (M + m)g gy ra gia tc a cho h M + m, ta cF - P = (M + m)a 2310N g) m)(a (M F + + + Gia tc ca bung khi dy treo tLc F ch tc dng ln bung, ta cF Mg = Ma1, suy ra211,55m/sMMg Fa + Thi gian vt ri xung sn bungVt v sn thang cng chuyn ng vi vn tc ban u v0.Phng trnh chuyn ng ca sn thang v vt ln lt lt v t a21y021 1+ ; 02 022 2y t v t a21y + + Vi a1 = 1,55m/s2, y02 = 2m, vt ch cn chu tc dng ca trng lc nn c gia tc a2 = -gVyt v 0,775t y021+ v2 t v 5t y022+ + Vt chm sn khi Vt chm sn khi y1 = y2, suy ra t = 0,6s.Bi ton 2Mt toa xe nh di 4m khi lng m2 = 100kg ang chuyn ng trn ng ray vi vn tc v0 = 7,2km/h th mt chic vali kch thc nh khi lng m1 = 5kg c t nh vo mp trc ca sn xe. Sau khi trt trn sn, vali c th nm yn trn sn chuyn ng khng? Nu c th nm u? Tnh vn tc mi ca toa xe v vali. Cho yOFTP0v0vy02bit h s ma st gia va li v sn l k = 0,1. B qua ma st gia toa xe v ng ray. Ly g = 10m/s2.Nhn xty l bi ton v h hai vt chuyn ng trt ln nhau. Nu ng trn ng ray qua st ta cng d dng nhn ra s chuyn ng ca hai cht im vali v mp sau ca sn xe trn cng mt phng. Vali ch trt khi sn xe sau khi ti mp sau sn xe, tc l hai cht im gp nhau. Ta a bi ton v dng quen thuc.GiiChn trc Ox hng theo chuyn ng ca xe, gn vi ng ray, gc O ti v tr mp cui xe khi th vali, gc thi gian lc th vali.+ Cc lc tc dng lnVali: Trng lc P1 = m1g, phn lc N1 v lc ma st vi sn xe Fms, ta c1 1 ms 1 1a m F N P + +Chiu ln Ox v phng thng ng ta c:Fms = m1a1 v N1 = P1 = m1g, suy ra 2111ms11m/s kgmkNmFa Xe: Trng lc P2 = m2g, trng lng ca valig m P1,1 , phn lc N2 v lc ma st vi vali Fms. Ta c2 2 ms 2 2'1a m ' F N P P + + +Chiu ln trc Oxta c-Fms = m2a2 2212ms2ms20,05m/smg kmmFmF'a Phng trnh chuyn ng ca vali v xe ln lt2t 0,025t t v t a21x4 0,5t x t a21x2022 220121 1+ + + + Vali n c mp sau xe khi x1 = x2, hay 0,5t2 + 4 = -0,025t2 + 2t0v1NmsF1P'2N1P2PmsF'xOPhng trnh ny v nghim, chng t vali nm yn i vi sn trc khi n mp sau ca xe.Khi vali nm yn trn sn, v1 = v2Vi v1 = a1t + v01 = t , v2 = a2t + v0 = -0,05t + 2, suy rat = - 0,05t + 2 suy ra t = 1,9sKhi vali cch mp sau xe mt khong2t 0,025t 4 0,5t x x d2 22 1 + + Vi t = 1,9s ta c d = 2,1mVn tc ca xe v vali lc v1 = v2 = 1,9m/s.Bi ton 3Mt b vc mt ct ng c dng mt phn parabol (hnh v). TimAtrnsn b vc, cao h = 20m so vi y vc v cch imBi din trnbbnkia(cngcao, cng nm trong mt phng ct) mt khongl=50m, bnmt qunphoxinlnvi vn tc v0= 20m/s, theo hng hp vi phng nm ngang gc = 600. B qua lc cn ca khng kh v ly g = 10m/s2. Hy xc nh khong cch t im ri ca vt n v tr nm vt. Nhn xtNu ta v phc ha qu o chuyn ng ca vt sau khi nm th thy im nm vt v im vt ri l hai giao im ca hai parabol. V tr cc giao im c xc nh khi bit phng trnh ca cc parabol. GiiChn h ta xOy t trong mt phng qu o ca vt, gn vi t, gc O ti y vc, Ox nm ngang cng chiu chuyn ng ca vt, Oy thng ng hng ln. Gc thi gian l lc nm vt.Hnh ct ca b vc c xem nh mt phn parabol (P1) y = ax2 i qua im A c ta (x = -) h y ;2lSuy ra 20 = a(- 25)2 a = 1254Phng trnh ca (P1): 2x1254y Phng trnh chuyn ng ca vt:hl0vA Bh0vA BCx(m) Oy(m)'+ + + + 20 t 3 10 5t h sin v gt21y25 10t2cos v x2020tltKh t i ta c phng trnh qu o (P2):9) 3 (2045x25 3 2x201y2 ++ im ri C ca vt c ta l nghim ca phng trnh:' ++ 9) 3 (2045x25 3 2x201yx20001y22 vi 20m y 25m, x Suy ra ta im ri: xC = 15,63m v yC = 7,82mKhong cch gia im ri C v im nm A l42,37m2)ByA(y2)CxA(x AC + Mt s bi ton vn dngBi 1T nh dc nghinggcso vi phngngang, mt vt c phng i vi vn tc v0 c hng hp vi phng ngang gc.Hy tnh tm xa ca vt trn mt dc. S: gcos) ( sin . cos 2vs220+Bi 2Trn mt nghing gc so vi phng ngang, ngi ta gi mt lng tr khi lng m. Mt trn ca lng tr nm ngang, c chiu di l,c t mt vt kch thc khng ng k, khi lng 3m, mp ngoi M lng tr (hnh v). B qua ma st gia vt v lng tr, h s ma st gia lng tr v mt phng nghing l k. Th lng tr v n bt u trt trn mt phng nghing. Xc nh thi gian t lc th lng tr n khi vt nm mp trong M lng tr.0vm3mlMMS: cos ) cos sin ( 2 k gltBi 3Hai xe chuyn ng thng u vi cc vn tc v1, v2 (v1