giao trinh day them 10

7/24/2019 Giao Trinh Day Them 10

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Cc dng bi tp vt l 10

Bin son: Kiu Quang V 1GV: Tr. THPT NCP_ Qung Ngi

MC LCPHN I: C HC................................................................................................................... 4

Chng I. NG HC CHT IM.............................................................................. 4A. H THNG C S L THUYT............................................................................ 4

Bi 1: CHUYN NG C......................................................................................... 4Bi 2: CHUYN NG THNG U....................................................................... 5Bi 3: CHUYN NG THNG BIN I U..................................................... 5Bi 4: S RI T DO .................................................................................................. 7

Bi 5: CHUYN NG TRN U.......................................................................... 8

Bi 6: TNH TNG I CA CHUYN NG. CNG THC CNG VN TC ............................................................................................................................................. 9

Bi 7: SAI S CA PHP O CC I LNG VT L................................... 10

B. H THNG BI TP.............................................................................................. 11DNG 1: XC NH VN TC TRUNG BNH TRONG CHUYN NG THNGU................................................................................................................................... 11

DNG 2: VIT PHNG TRNH CHUYN NG THNG U...................... 15DNG 3: TH CA CHUYN NG THNG U....................................... 20DNG 4: XC NH VN TC, GIA TC, QUNG DNG I TRONG

CHUYN NG THNG BIN I U................................................................... 21DNG 5: TNH QUNG NG VT I C TRONG GIY TH N V

TRONG N GIY CUI CA CHUYN NG THNG BIN I U................ 23

DNG 6: VN DNG CNG THC TNH QUNG NG, VN TC TRONGRI T DO....................................................................................................................... 25

DNG 7: TNH QUNG NG VT RI TRONG N GIY CUI, V TRONGGIY TH N.................................................................................................................... 26

DNG 8: XC NH V TR 2 VT GP NHAU C TH RI VI CNGTHI IM KHC NHAU............................................................................................. 31

DNG 8: VN DNG CC CNGTHC TRONG CHUYN NG TRN U

........................................................................................................................................... 32

DNG 9: XC NH VN TC TNG I, TUYT I, KO THEO.......... 35BI TP P DNG CHNG I............................................................................... 39

Chng II. NG LC HC CHT IM.................................................................. 44A. H THNG C S L THUYT.......................................................................... 44

Bi 9: TNG HP V PHN TCH LC. IU KIN CN BNG CA CHTIM................................................................................................................................. 44

Bi 10 : BA NH LUT NIUTN........................................................................... 44

Bi 11: LC HP DN. NH LUT VN VT HP DN................................ 46

Bi 12: LC N HI CA L XO. NH LUT HC....................................... 46


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Bi 13: LC MA ST................................................................................................. 47Bi 14: LC HNG TM....................................................................................... 48Bi 15: BI TON V CHUYN NG NM NGANG........................................ 48

B. H THNG BI TP.............................................................................................. 49DNG 2: P DNG 3 NH LUT NIU-TN........................................................ 50

DNG 3: VN DNG CNG THC TNH LC HP DN V GIA TC TRNGTRNG. ......................................................................................................................... 52

DNG 4: VN DNG NH LUT HC............................................................... 53DNG 5: VN DNG CNG THC TNH MA ST V PHNG PHP NG

LC HC. ........................................................................................................................ 54DNG 6: VN DNG CC CNG THC CA LC HNG TM................. 56

DNG 7: VN DNG CNG THC CHUYN NG NM NGANG............... 58

BI TP CHNG II................................................................................................ 61

Chng III. CN BNG V CHUYN NG CA VT RN................................ 66A. H THNG C S L THUYT.......................................................................... 66

Bi 17: CN BNG CA MT VT CHU TC DNG CA HAI LC V CABA LC KHNG SONG SONG..................................................................................... 66

Bi 18: CN BNG CA MT VT C TRC QUAY C NH. M MEN LC ........................................................................................................................................... 66

Bi 19: QUI TC HP LC SONG SONG CNG CHIU..................................... 67Bi 20: CC DNG CN BNG. CN BNG CA MT VT C MT CHN

...................................................................................................................................... 67Bi 21: CHUYN NG TNH TIN CA VT RN. CHUYN NG QUAY

CA VT RN QUANH MT TRC C NH......................................................... 68Bi 22: NGU LC .................................................................................................... 69

B. H THNG BI TP.............................................................................................. 70DNG 1: HP LC CA HAI LC SONG SONG CNG CHIU....................... 70DNG 2: BI TON V M MEN LC................................................................. 72BI TP TNG HP CHNG 2 V CHNG 3............................................... 74

Chng IV. CC NH LUT BO TON................................................................. 75A. H THNG C S L THUYT.......................................................................... 75

Bi 23: NG LNG. NH LUT BO TON NG LNG..................... 75Bi 24: CNG V CNG SUT............................................................................... 76Bi 25: NG NNG................................................................................................. 77Bi 26: TH NNG..................................................................................................... 77Bi 27: C NNG....................................................................................................... 78

B. H THNG BI TP.............................................................................................. 79

DNG 1: TM LN CA NG LNG........................................................ 79DNG 2: BI TP V CNG V CNG SUT.................................................... 82


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DNG 3: BI TP V NG NNG V NH L NG NNG.................... 83DNG 4: BI TP V TH NNG TRNG TRNG V TH NNG N HI

........................................................................................................................................... 85

DNG 5: C NNG CA VT TRONG TRNG TRNG NH LUT BOTON C NNG............................................................................................................. 88

BI TP CHNG IV............................................................................................... 90PHN HAI: NHIT HC .................................................................................................... 96

Chng V. CHT KH.................................................................................................... 96A. H THNG C S L THUYT.......................................................................... 96

Bi 28: CU TO CHT. THUYT NG HC PHN T................................. 96Bi 29: QU TRNH NG NHIT. NH LUT BI-L-MA-RI-T............... 97

Bi 30: QU TRNH NG TCH. NH LUT SC-L..................................... 97

Bi 31: PHNG TRNH TRNG THI CA KH L TNG............................ 97

B. H THNG BI TP.............................................................................................. 99DNG 1: BI TON NH LUT BI - L V MA - RI - T............................ 99

DNG 2: BI TP V QU TRNH NG TCH............................................... 101

DNG 3: PHNG TRNH KH L TNG....................................................... 102

BI TP CHNG V.............................................................................................. 104Chng VI. C S CA NHIT NG LC HC.................................................. 107

A. H THNG C S L THUYT........................................................................ 107Bi 32: NI NNG V S BIN I NI NNG............................................... 107

Bi 33: CC NGUYN L CA NHIT NG LC HC.................................. 107B. H THNG BI TP............................................................................................ 108

DNG 1: BI TP V NHIT LNG TA RA V THU VO TRONG CCQU TRNH NHIT...................................................................................................... 108

DNG 2: BI TP V NGUYN L CA NHIT NG LC HC............... 111Chng VII. CHT RN V CHT LNG. S CHUYN TH............................. 115

A. H THNG C S L THUYT........................................................................ 115Bi 34: CHT RN KT TINH. CHT RN V NH HNH............................ 115

Bi 36: S N V NHIT CA CHT RN.......................................................... 115Bi 37: CC HIN TNG B MT CA CHT LNG................................... 116Bi 38: S CHUYN TH CA CC CHT........................................................ 118Bi 39: M CA KHNG KH......................................................................... 119

B. H THNG BI TP............................................................................................ 120DNG 1: N V NHIT........................................................................................... 120DNG 2: BI TP V LC CNG MT NGOI................................................ 121BI TP CHNG VII ........................................................................................... 123


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PHN I: C HC

Chng I. NG HC CHT IM

Bi 1: CHUYN NG CI. Chuyn ng c Cht im1. Chuyn ng c

Chuyn ng ca mt vt l s thay i v tr ca vt so vi cc vt khc theo thigian.

2. Cht imNhng vt c kch thc rt nh so vi di ng i (hoc vi nhng khong cch m

ta cp n), c coi l cht im.Khi mt vt c coi l cht im th khi lng ca vt coi nh tp trung ti cht im

.3. Qu o

Qu o ca chuyn ng l ng m cht im chuyn ng vch ra trong khng gian.II. Cch xc nh v tr ca vt trong khng gian.1. Vt lm mc v thc o

xc nh chnh xc v tr ca vt ta chn mt vt lm mc v mt chiu dng trn quo ri dng thc o chiu di on ng t vt lm mc n vt.2. H to a) H to 1 trc (s dng khi vt chuyn ng trn mt ng thng): To ca vt v tr M : x =

b) H to 2 trc (s dng khi vt chuyn ng trn mt ng cong trong mt mt phng):To ca vt v tr M :

x = y = III. Cch xc nh thi gian trong chuyn ng .1. Mc thi gian v ng h.

xc nh tng thi im ng vi tng v tr ca vt chuyn ng ta phi chn mc thigian v o thi gian tri i k t mc thi gian bng mt chic ng h.2. Thi im v thi gian.

Vt chuyn ng n tng v tr trn qu o vo nhng thi im nht nh cn vt i t

v tr ny n v tr khc trong nhng khong thi gian nht nh.IV. H qui chiu.

OM


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Mt h qui chiu gm :+ Mt vt lm mc, mt htogn vi vt lm mc.+ Mt mc thi gian v mt ng h

Bi 2: CHUYN NG THNG UI. Chuyn ng thng u

1. Tc trung bnh: vtb= Vi : s = x2x1; t = t2t1

2. Chuyn ng thng u: L chuyn ng c qu o l ng thng v c tc trungbnh nh nhau trn mi qung ng.3. Qung ng i trong chuyn ng thng u: s = vtbt = vt

Trong chuyn ng thng u, qung ng i c s t l thun vi thi gian chuynng t.II. Phng trnh chuyn ng v th to thi gian ca chuyn ng thng u.1. Phng trnh chuyn ng.

x = xo+ s = xo+ vt

Trong :+ s: l qung ng i+ v: l vn tc ca vt hay tc + t: l thi gian chuyn ng+ x0: l ta ban u lc t = 0+ x: l ta thi im t

2. th to thi gian ca chuyn ng thng u.a) Bng

t(h) 0 1 2 3 4 5 6

x(km) 5 15 25 35 45 55 65

b) th

Bi 3: CHUYN NG THNG BIN I UI. Vn tc tc thi. Chuyn ng thng bin i u.1. ln ca vn tc tc thi.

i lng: v =l ln vn tc tc thi ca vt ti M khi t rt nh, s rt ngn.

n v vn tc l m/s2. Vc t vn tc tc thi.


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Vecto vn tc tc thi ti mt im trong chuyn ng thng c:+ Gc nm trn vt chuyn ng khi qua im + Hng trng vi hng chuyn ng

+ di biu din ln vn tc theo mt txch no v c tnh bng:v =

Vi+ s l qung ng i rt nh tnh t im cn tnh vn tc tc thi+ t l khong thi gian rt ngn i on s

3. Chuyn ng thng bin i u- Chuyn ng thng nhanh dn u l chuyn ng c qu o l mt ng thng v c

vn tc tc thi tng u theo thi gian.- Chuyn ng thng chm dn u l chuyn ng c qu o l mt ng thng v c

vn tc tc thi gim u theo thi gian.Ni tm li chuyn ng thng bin i u l chuyn ng thng c vn tc thc thi

bin i u theo thi gian (tng hoc gim u theo thi gian)II. Chuyn ng thng nhanh dn u v thng chm dn u.1. Gia tc trong chuyn ng thng nhanh dn u v thng chm dn u.a) Khi nim gia tc.

a == hng s

Vi: v = vvo; t = tton vgia tc l m/s2.

b) Vc t gia tc

= =

- i vi chuyn ng nhanh dn u: vect gia tc lun cng chiu vi cc vect vntc.

- i vi chuyn ng chm dn u: vect gia tc lun ngc chiu vi cc vect vntc.2. Vn tc, qung ng i, phng trnh chuyn ng ca chuyn ng thng nhanh

dn v thng chm dn u:- Cng thc vn tc: v = v0+ at

- Cng thc tnh qung ng i: s = v0t +

- Phng trnh chuyn ng: x = x0+ v0t +

- Cng thc lin h gia a, v v s ca chuyn ng thng bin i u:v2vo2= 2asTrong :

+ v0l vn tc ban u+ v l vn tc thi im t


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+ a l gia tc ca chuyn ng+ t l thi gian chuyn ng+ x0l ta ban u+ x l ta thi im t* Nu chn chiu dng l chiu chuyn ng th:

+ v0> 0 v a > 0 vi chuyn ng thng nhanh dn u+ v0> 0 va < 0 vi chuyn ng thng chm dn u.

Bi 4: SRI TDOI. S ri trong khng kh v s ri t do.1. S ri ca cc vt trong khng kh.

Cc vt ri trong khng kh xy ra nhanh chm khc nhau l do lc cn ca khng kh tcdng vo chng khc nhau.2. S ri ca cc vt trong chn khng (s ri t do).

-Nu loi b c nh hng ca khng kh th mi vt s ri nhanh nh nhau. S rica cc vt trong trng hp ny gi l s ri t do.nh ngha:

- S ri t do l s ri ch di tc dng catrng lc.II. Nghin cu s ri t do ca cc vt.1. Nhng c im ca chuyn ng ri t do.

+ Phng ca chuyn ng ri t do l phng thng ng (phng ca dy di).+ Chiu ca chuyn ng ri t do l chiu t trn xung di.

+ Chuyn ng rit do l chuyn ng thng nhanh dn u vi gia tc g.2. Cc cng thc ca chuyn ng ri t do khng c vn tc u:

+ Vn tc ri ca vt theo thi gian: v = g.t

+ Qung ng vt ri c sau thi gian t: S=

+ Mi lin h gia qung ng v vn tc ti mt thi im bt k: v2= 2gSTrong :+ v vn tc ca vt ti thi im t

+ S l qung ng vt i c trong t(s)3. Gia tc ri t do.

+ Ti mt ni nht nh trn Tri t v gn mt t, cc vt u ri t do vi cng mtgia tc g.

+ nhng ni khc nhau trn Tri t v nhng cao khc nhau, gia tc ri t do skhc nhau

- a cc g ln nht : g = 9,8324m/s2.- xch o g nh nht : g = 9,7872m/s2

+ Nu khng i hi chnh xc cao, ta c th ly g = 9,8m/s2

hoc g = 10m/s2

.

2

2

1gt


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Bi 5: CHUYN NG TRN UI. nh ngha.1. Chuyn ng trn.

Chuyn ng trn l chuyn ng c qu o l mt ng trn.2. Tc trung bnh trong chuyn ng trn.

Tc trung bnh ca chuyn ng trn l i lng o bng thng s gia di cungtrn m vt i c v thi gian i ht cung trn .

vtb= di ca cung trn

thi gian i ht cung trn3. Chuyn ng trn u.

Chuyn ng trn u l chuyn ng c qu o trn v c tc trung bnh trn micung trn l nh nhau.II. Tc di v tc gc.

1. Tc di.v =

Trong chuyn ng trn u tc di ca vt c ln khng i.2. Vc t vn tc di trong chuyn ng trn u.

=Vct vn tc di trong chuyn ng trn u lun c phng

tip tuyn vi ng trn qu o.Trong chuyn ng trn u vct vn tc c phng lun lun

thay i.3. Tn s gc, chu k, tn s.a) Tc gc.

Tc gc ca chuyn ng trn u l i lng o bng gc m bnknh quay qut c trong mt n v thi gian.

= trong : : gc m bn knh qut c trong thi gian t.

+ Tc gc ca chuyn ng trn u l mt i lng khng i.+ n v tc gc l rad/s.

b) Chu k.Chu k T ca chuyn ng trn u l thi gian vt i c mt vng trn qu o.Lin h gia tc gc v chu k :

T =2

n v chu k l giy (s).c) Tn s.

Tn s f ca chuyn ng trn u l s vng m vt i c trong 1 giy.

1

2


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Lin h gia chu k v tn s : f =1T

n v tn s l vng trn giy (vng/s) hochc (Hz).d) Lin h gia tc di v tc gc.

v = R.

II. Gia tc hng tm.1. Hng ca vct gia tc trong chuyn ng trn u.

Trong chuyn ng trn u, tuy vn tc c ln khng i, nhng c hng lun thayi, nn chuyn ng ny c giatc. Gia tc trong chuyn ng trn u lun hng vo tmca quo nn gi l gia tc hng tm.2. ln ca gia tc hng tm.

aht = =

2.R

trong R l bn knh qu o

Bi 6: TNH TNG I CA CHUYN NG. CNG THC CNG VN TCI. Tnh tng i ca chuyn ng.1. Tnh tng i ca qu o.

Hnh dng qu o ca chuyn ng trong cc h qui chiu khc nhau th khc nhau. Quo c tnh tng i2. Tnh tng i ca vn tc.

Vn tc ca vt chuyn ng i vi cc h qui chiu khc nhau th khc nhau. Vn tcc tnh tng i

II. Cng thc cng vn tc.1. H qui chiu ng yn v h qui chiu chuyn ng.

+ H qui chiu gn vi vt ng yn gi l h qui chiu ng yn.+ H qui chiu gn vi vt vt chuynng gi l hqui chiu chuyn ng.

2. Cng thc cng vn tc.+ Cng thc cng vn tc: = Trong :

*

vn tc tuyt i ( vn tc ca vt i vi hquy chiu ng yn)

* vn tc tng i ( vn tc ca vt i vi hquy chiu chuyn ng)* vn tc ko theo ( vn tc ca hquy chiu chuyn ng i vi hquy chiu ng

yn)* Trng hp cng phng, cng chiu

- Vln: v13= v12+ v23- Vhng: cng hng vi v

* Trng hp cng phng, ngc chiu - Vln: v13= |v12- v23|- Vhng:


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+ cng hng vi khi 12 23v v + cng hng vi khi 23 12v v

Bi 7: SAI SCA PHP O CC I LNG VT LI. Php o cc i lng vt l H n v SI.1. Php o cc i lng vt l.

Php o mt i lng vt l l php so snh n vi i lng cng loi c qui c lmn v.

+ Cng c so snh gil dng c o.+ o trc tip: So snh trc tip qua dng c.+ o gin tip: o mt s i lng trc tip ri suy ra i lng cn o thng qua cng

thc.2. n v o.H SI qui nh 7 n v c bn:

+ di: mt (m)+ Thi gian: giy (s).+ Khi lng: kilgam (kg)+ Nhit tuyt i: kenvin (K)+ Cng dng in: ampe (A)+ Cng sng: canla (Cd)+ Lng cht: mol (mol).

II. Sai s ca php o.1. Sai s h thng.

L s sai lch do phn l khng c c chnh xc trn dng c (gi l sai s dng c

A) hoc im 0 ban u b lch.

Sai s dng c A thng ly bng na hoc mt chia trn dng c.2. Sai s ngu nhin.

L s sai lch do hn ch v kh nng gic quan ca con ngi do chu tc ng ca ccyu t ngu nhin bn ngoi.3. Gi tr trung bnh.

A= A1A2Ann 4. Cch xc nh sai s ca php o.

+ Sai s tuyt i ca ln o th i: Ai= |A - Ai|

+ Sai s tuyt i trung bnh ca n ln o:1

n

i

i

AA

n

+ Sai s ca dng c A' bng 1/2 khong chia nhnht i vi dng co sdng vchchia. i vi dng co hin sbng gi trsai scho php c ghi trn dng co.

+ Sai stuyt i ca php o l tng sai stuyt i trung bnh v sai sdng c:


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= 5. Cch vit kt qu o.

A = A AHoc:

A - A A A - A

6. Sai sti .

A = 100%

7. Cch xc nh sai s ca php o gin tip.+ Sai s tuyt i ca mt tng hay hiu th bng tng cc sai s tuyt i ca cc s hng.+ Sai s t i ca mt tch hay thng th bng tng cc sai s t i ca cc tha s.+ Nu trong cng thc vt l xc nh cc i lng o gin tip c cha cc hng s th

hng s phi ly n phn thp phn l nh hn ttng cc sai s c mt trong cng cng

thc tnh.+ Nu cng thc xc nh i lng o gin tip tng i phc tp v cc dng c o

trc tip c chnh xc tng i cao th c th b qua sai s dng c.

DNG 1: XC NH VN TC TRUNG BNH TRONG CHUYN NG THNG

U

Hng gii quyt:+ S dng cngthc tnh qung ng S = vt trong qu trnh tm S, v, t.

+ tnh vn tc trung bnh p dng cng thc: vtb== ++++++ .

BI TP C HNG DNBi 1: Mt xe chy trong 5h: 2h u xe chy vi tc trung bnh 60km/h, 3h sau xe chyvi tc trung bnh 40km/h.Tnh tc tc trung bnh ca xe trong sut thi gian chuynng.

GiiQung ng i trong 2h u: S1= v1.t1= 120 kmQung ng i trong 3h sau: S2= v2.t2= 120 km

Vn tc trung bnh: = ++ = 48km/hBi 2: Mt xe i na on ng u tin vi tc trung bnh v1=12km/h v na on

ng sau vi tc trung bnh v2 =20km/h. Tnh tc trung bnh trn c on ng.Gii

10

1


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Thi gian i na on ng u: 11

1 2.12 24

S S St

v

Thi gian i na on ng cui: 22

2 2.20 40

S S St

v

Tc trung bnh: vtb=

+= 15km/h

Bi 3: Mt t i t A n B. u chng t i tng thi gian vi v = 50km/h. Gia chng t i

thi gian vi v = 40km/h. Cui chng t i

tng thi gian vi v = 20km/h. Tnh

vn tc trung bnh ca t?Gii

Qung ng i u chng: S1= v1.= 12,5t

Qung ng chng gia: S2= v2.= 20t

Qung ng i chng cui: S1= v2.= 5tVntc trung bnh: vtb=

++ = 37,5 (km/h)

Bi 4:Mt ngi i xe my t A ti B cch 45km. Trong na thi gian u i vi vn tc

v1, na thi gian sau i vi v2 =v1. Xc nh v1, v2bit sau 1h 30min ngi n B.

GiiQung ng S ngi i xe my vit theo 2 giai on: S = S1+ S2= 45km

Trong : S1= v1,

v S2= v2,

v1, v1, = 45 v1= 10,4 km/h v v2= 6,9 km/h

Bi 5:Mt t i trn con ng bng phng vi v = 60 km/h trong thi gian 5min, sau leo dc 3 min vi v = 40km/h. Coi t chuyn ng thng u. Tnh qung ng t itrong c giai on.

GiiQung ng t i c trong thi gian 5min: S1= v1t1= 5km.Qung ng t i c trong thi gian 3min: S2= v2t2= 2km.

Tngqung ng: S = S1+ S2= 7kmBi 6:Mt t i trn qung ng AB vi v = 54km/h. Nu tng vn tc thm 6km/h tht n B sm hn d nh 30 pht. Tnh qung ng AB v thi gian d nh i qungng .

Gii+ Biu din qung ng t i theo vn tc v1= 54km/h: S1= v1.t1= 54t1+ Biu din qung ng t i theo vn tc v2= 60km/h: S2= v2.t2= 60(t10,5)V S1= S2 t1= 5h

Vy AB = S = v1.t1= 270km.


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Bi 7:Mt t i trn qung ng AB vi v = 54km/h. Nu gim vn tc i 9km/h th tn B tr hn d nh 45 pht. Tnh qung ng AB v thi gian d tnh i qung ng.

Gii+ Biu din qung ng t i theo vn tc v1= 54km/h: S1= 54t1

+ Biu din qung ng t i theo vn tc v2= 45km/h: S2= 45( t1+ )V S1= S2t1= 3,75hVy qung ng AB = v1.t1= 202,5km.

Bi 8 :Hai xe cng chuyn ng u trn ng thng. Nu chng i ngc chiu th c 30pht khong cch ca chng gim 40km. Nu chng i cng chiu th c sau 20 pht khongcch gia chng gim 8km. Tnh vn tc mi xe.

GiiChn chiu dng l chiu chuyn ng ca mi xe.

+ Nu i ngc chiu th S1+ S2= 40 1 2 402

v v (1)

+ Nu i cng chiu th S1S2= (v1v2)t = 8 1 2 83

v v (2)

Gii (1), (2) v1= 52km/h ; v2= 28km/hBi 9:Mt ngi i xe my chuyn ng thng u t A lc 5gi sng v ti B lc 7gi 30

pht, AB = 150km.a) Tnh vn tc ca xe.

b) Ti B xe dng li 45 pht ri i v A vi v = 50km/h.Hi xe ti A lc my gi.Gii

a) Thi gian lc i: t = 7h305h = 2,5h

+ Vn tc ca xe: v == 60km/h

+ Thi im ngi lc bt u v: t = 7h30+ 45= 8h15

b) Thi gian xe i tB vA vi vn tc v = 50km/h: t == 3h

+ Xe tiA lc: 8h15+ 3h = 11h15

Bi 10: Mt ngi i xe my t A n B cch nhau 2400m. Na qung ng u, xe i viv1, na qung ng sau i vi v2=

v1. Xc nh v1, v2sao cho sau 10 pht xe ti B.

Gii

+ Qung ng xe i vi vn tc v1: S1= v1.t1t1==

+ Qung ng xe i vi vn tc v2: S2= v2.t2t2==

=

+ M ta c t1+ t2= 600

= 600v1= 6m/s v v2= 3m/s


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Bi 11: Mt t chuyn ng trn onng MN. Trongqung ng u i vi v =

40km/h. Trongqung ng cn li i trong

thi gian u vi v = 75km/h v trong

thi

gian cui i vi v = 45km/h. Tnh vn tc trung bnh trn on MN.Gii

Qung ng i vi vn tc v1 = 40km/h: S1= v1.t1= 40t1t1= Qung ng i vi vn tc v2= 75km/h: S2= v2.

=

7()

Qung ng i vi vn tc v3= 45km/h: S3= v3.

=()

S3+ S2=

() 7() = 60 60 = 60 = = 48

Vy vn tc trung bnh trn on ng MN l: vtb== 48km/h

Bi 12: Mt t chy trn on ng thng t A n B phi mt khong thi gian t. Tc

ca t trong na u ca khong thi gian ny l 60km/h. Trong na khong thi gian cuil 40km/h. Tnh tc trung bnh trn con AB.

GiiTrong na thi gian u: S1= v1.t1= 30tTrong na thi gian cui: S2= v2.t2= 20t

Vn tc trung bnh ca oto trn cqung ng l: vtb==

++= 50km/h

Bi 13: Mt ngi ua xe p i trnqung ng u vi 25km/h. Tnh vn tc ca ngi

i trn on ng cn li. Bit rng vtb= 20km/h.Gii

Trongqung ng u: S1= v1.t1t1=

7

Trongqung ng u: S2= v2.t2t1=

Vn tc trung bnh trn ton bqung ng l: vtb=

+=

+ = 20 km/h

=

km/h

Bi 14: Mt ngi i xe p trn mt on ng thng AB. Trn on ng u i viv = 12km/h,

on ng tip theo vi v = 8km/h v

on ng cui cng i vi v =

6km/h. Tnh vtbtrn con AB.Gii

Trong 1/3 on ng u: S1= v1.t1t1==

Thc hin tng t c: t2==

; t =

=

Tng thi gian vt i t = t1+ t2+ t3= vtb= = 8 km/h


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Bi 15:Mt ngi i xe my chuyn ng theo 3 giai on: Giai on 1 chuyn ng thngu vi v1= 12km/h trong 2km u tin; giai on 2 chuyn ng vi v2= 20km/h trong 30

pht; giai on 3 chuyn ng trn 4km trong 10 pht. Tnh vn tc trung bnh trn conng.

Gii

+ Giai on 1: t1= = (h)+ Giai on 2: S2= v2. t2= 10km;+ Tng qung ng xe my i: S = S1+ S2+ S3= 16km+ Thi gian xe my i ht qung ng S: t= t1+ t2+ t3= 5/6 gi.

Vn tc trung bnh trn ton bqung ng: vtb== 19,2 (km/h)

DNG 2: VIT PHNG TRNH CHUYN NG THNG U1. Bi ton lp phng trnh chuyn ng:

- Chn trc chiu dng ca h trc, gc ta , gc thi gian v v hnh biu din ccvecto vn tc, v tr ban u ca cht im.- Vit phng trnh chuyn ng:+ Nu t0= 0 th x = x0+ v.t.+Nu t00 th x = x0+ v.(t - t0)Trong :+ x0l v tr ban u ca cht im c th m hoc dng ty thuc vo v tr tng i

so vi gc ta v chiu dng. x0> 0 khi nm phn dng ca trc ta , x0nm phn

m ca trc ta , x0= 0 ti gc ta .+ v vn tc ca cht im. Vecto vn tc c gi trv > 0 khi cng hng vi chiu dng,vecto vn tc c gi trv < 0 khi ngc hng vi chiu dng.2. Bi ton xc nh v tr v thi gian gp nhau:-p dng iu kin gp nhau:x1= x2.- Gii phng trnh x1= x2tm gi tr t.- Thay t vo x1hoc x2xc nh v tr gp nhau.

BI TP C HNG DNBi 1:Trn ng thng AB, cng mt lc xe 1 khi hnh t A n B vi v = 40km/h. Xeth 2 t B i cng chiu vi v = 30km/h. Bit AB cch nhau 20km. Lp phng trnh chuynng ca mi xe vi cng h quy chiu.

GiiChn gc to ti A, gc thi gian lc 2 xe xut pht.Chiu dng cng chiu vi chiu chuyn ng vi hai xe.+ Phng trnh chuyn ng xe A: xA= x0+ vA.t = 40t+ Phng trnh chuyn ng xe B: xB= x0+ vB.t = 20 + 30t.


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Bi 2:Lc 7 gi, mt ngi A chuyn ng thng u vi v = 36km/h ui theo ngi B ang chuyn ng vi v = 5m/s. Bit AB = 18km. Vit phng trnh chuyn ng ca 2ngi. Lc my gi v u 2 ngi gp nhau nhau.

GiiChn gc to ti A, gc thi gian lc 7 gi.

Phng trnh chuyn ca xe A: xA= 36tPhng trnh chuyn ca xe A: xB= x0+ vB.t = 18 + 18tKhi hai xe gp nhau: x1= x2 t = 1h. xA= xB= 36km

Vy hai xe gp nhau cch gc to 36km v vo lc 8 giBi 3:Lc 6 gi sng, mt ngi i xe my khi hnh t A chuyn ng vi vn tc khngi 36km/h ui theo mt ngi i xe p chuyn ng vi v = 5m/s i c 12km kt A. Hai ngi gp nhau lc my gi.

GiiChn gc to ti v tr A, gc thi gian lc xe my chuyn ng.Phng trnh chuyn ng ca xe my: xm= 36tPhng trnh chuyn ng ca xe p:x= 12 + 18tKhi hai xe gp nhau: xm= x t = 2/3 phtVy hai xe gp nhau lc 6 gi 40 pht

Bi 4: Hai t xut pht cng mt lc, xe 1 xut pht t A chy v B, xe 2 xut pht t B

cng chiu xe 1, AB = 20km. Vn tc xe 1 l 50km/h, xe B l 30km/h. Hi sau bao lu xe 1gp xe 2.

GiiChngc to ti v tr ti A, gc thi gian lc 2 xe xut pht.Phng trnh chuyn ng ca xe mt: x1= 50tPhng trnh chuyn ng ca xe mt: x2= 20 + 30tKhi hai xe gp nhau nhau: x1= x2 t = 1h

Bi 5: Lc 6 gi sng, mt ngi i xe my khi hnh t A chuyn ng vi v = 36km/h iv B. Cng lc mt ngi i xe p chuyn ng vi vkxut pht t B n A. Khong cchAB = 108km. Hai ngi gp nhau lc 8 gi. Tm vn tc ca xe p.

GiiGc thi gian lc 2 xe xut pht, gc to ti A.Hai xe xut pht t lc 6gi v gp nhau lc 8 gi t = 2hPhng trnh chuyn ng ca xe my: xm= 36t = 72Phng trnh chuyn ng ca xe p: x= 108 - 2v2

Khi hai xe gp nhau nhau: xm= x v2= 18km/h


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Bi 6:Lc 7 gi sng mt t khi hnh t A chuyn ng vi v k= 54km/h ui theomt ngi i xe p chuyn ng vi vk= 5,5 m/s i c cch 18km. Hi 2 xe gp nhaunhau lc my gi.

GiiChn gc to v tr A, gc thigian lc t xut pht.

Chn gc thi gian lc 7 gi.Phng trnh chuyn ng ca xe A: x1= 54tPhng trnh chuyn ng ca xe B: x2= 18 + 19,8.tKhi 2 xe gp nhau: x1= x2 54t = 18 + 19,8.t

t = 0,52 h = 31phtVy hai xe gp nhau lc 7 gi 31 pht.

Bi 7:Lc 5 gi hai xe t xut pht ng thi t 2 a im A v B cch nhau 240km v

chuyn ng ngc chiu nhau. Hai xe gp nhau lc 7 gi. Bit vn tc xe xut pht t A l15m/s. Chn trc Ox trngvi AB, gc to ti A.a) Tnh vn tc ca xe B.

b) Lp phng trnh chuyn ng ca 2 xe.c) Xc nh to lc 2 xe gp nhau.

Giia) Qung ng xe A i: S1= v1.t=108kmDo hai xe chuyn ng ngc chiu S2= 132 km l qung ng xe B i.

v2= = 66km/hb) Phng trnh chuyn ng xe th nht: x1= 54tPhng trnh chuyn ng xe th hai: x2= 24066tc) Khi hai xe gp nhau: x1= 54.4 = 108km

Bi 8:Lc 8 gi sng, xe 1 khi hnh t A chuyn ng thng uv B vi v = 10m/s. Nagi sau, xe 2 chuyn ng thng u t B n A v gp nhau lc 9 gi 30 pht. Bit AB =72km.

a) Tm vn tc ca xe 2.b) Lc 2 xe cch nhau 13,5km l my gi.

Giia) Chn gc to ti A, gc thi gian lc xe 1 khi hnh.Phngtrnh chuyn ng ca xe A: x1= 36tPhng trnh chuyn ng ca xe B: x2= 72v2( t0,5 )Sau t = 1,5 gi hai xe gp nhau.

iu kin hai xe gp nhau: x1= x236t = 72v2( t0,5 ) v2= 18km/hb) Khi hai xe cch nhau 13,5kmx2x1= 13,5 t = 1,25h tc l lc 9h25


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x1x2= 13,5 t = 1,75h tc l lc 9h45Bi 9:Lc 8 gi sng, mt t khi hnh t A n B vi vk= 40km/h. thi im 1 xep khi hnh t B n A vi v2= 5m/s. Coi AB l thng v di 95km.a) Tm thi im 2 xe gp nhau.

b) Ni gp nhau cch A bao nhiu km.

Giia) Chn gc to ti A, chiu dng t A n B. Gc thi gian lc 8h.Phng trnh chuyn ng ca xe A: x1= 40tPhng trnh chuyn ng ca xe B: x2= 9518t

b) Khi hai xe gp nhau: x1= x2 t = 1,64h = 1h38Thi im gp nhau l 9h38v cch A: x1= 40.1,64 = 65,6km

Bi 10:Mt xe khch chy vi v = 95km/h pha sau mt xe ti ang chy vi v = 75km/h.Nu xe khch cch xe ti 110m th sau bao lu n s bt kp xe ti? Khi xe ti phi chy

mt qung ng bao xa.Gii

Chn gc to ti v tr xe khch chyPhng trnh chuyn ng ca xe 1: x1= 95tPhng trnh chuyn ng ca xe 2: x2= 0,11 + 75tKhi hai xe gp nhau: x1= x2

t = 5,5.10-3

Vy S2= v2.t = 0,4125km

Bi 11:Lc 14h, mt t khi hnh t Hu n Nng vi vk= 50km/h. Cng lc , xeti i t Nng n Hu vi vk= 60km/h, bit khong cch t Hu n Nng l 110km.Hai xe gp nhau lc my gi?

GiiChn gc to ti Hu, gc thi gian lc 14h.Phng trnh chuyn ng ca xe1: x1= 50tPhng trnh chuyn ng ca xe2: x2= 11060tKhi hai xe gp nhau: x1= x2 t = 1h

Vy hai xe gp nhau lc 15 giBi 12: Hai t cng lc khi hnh ngc chiu t 2 im A, B cch nhau 120km. Xe chyt A vi v = 60km/h, xe chy t B vi v = 40km/h.a) Lp phng trnh chuyn ng ca 2 xe, chn gc thi gian lc 2 xe khi hnh, gc to A, chiu dng t A n B.

b) Xc nh thi im v v tr 2 xe gp nhau.c) Tm khong cch gia 2 xe sau khi khi hnh c 1 gi.d) Nu xe i t A khi hnh tr hn xe i t B na gi, th sau bao lu chng gp nhau.

Giia) Phng trnh chuyn ng ca xe A: x1= 60t


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Phng trnh chuyn ng caxe B: x2= 12040t

b) Khi hai xe gp nhau: x1= x2t = 2hTo khi gp nhau: x1= 60. 1,2 = 72kmc) Khi khi hnh c 1 gix1= 60km ; x2= 80km

x = |x1- x2| = 20kmd) Nu xe A xut pht tr hn na gi.Phng trnh chuyn ng ca xe A: x1= 60 (t0,5 )Phng trnh chuyn ng ca xe B:x2= 12040tKhi hai xe gp nhau: x1= x2 t = 1,5h

Bi 13:Mt vt xut pht t A chuyn ng u v B cch A 630m vi v = 13m/s. Cng lc, mt vt khc chuyn ng u t B n A. Sau 35 giy 2 vt gp nhau. Tnh vn tc cavt th 2 v v tr 2 vt gp nhau.

GiiChn gc to ti v tr A, chiu dng l chiu chuyn ng t A n B.Phng trnh chuyn ng ca vt A: x1= 13.t = 455mPhng trnh chuyn ng ca vt B: x2= 63035v2Khi hai vt gp nhau: x1= x2455 = 63035v2 v2= 5m/sV tr hai vt gp nhau cch A 455m

Bi 14: Hai vt xut pht t A v B cch nhau 340m, chuyn ng cng chiu hng t A

n B. Vt t A c v1, vt t B c v2= v1. Bit rng sau 136 giy th 2 vt gp nhau. Tnhvn tc mi vt.

GiiChn gc to ti APhng trnh chuyn ng ca vt A: x1= v1t = 136v1Phng trnh chuyn ng ca vt B: x2= 340 + 68v1Khi hai vt gp nhau: x1= x2 v1= 5m/s; v2= 2,5m/s

Bi 15: Xe my i t A n B mt 4 gi, xe th 2 i t B n A mt 3 gi. Nu 2 xe khihnh cng mt lc t A v B n gn nhau th sau 1,5 gi 2 xe cch nhau 15km. Hi qungng AB di bao nhiu.

Gii

Vn tc xeA: vA=S

Vn tc xe B: vB=S vA =

vB

Chn gc to ti v tr A, gc thi gian lc 2 xe xut pht.

Phng trnh chuyn ng ca vt A: x1=vB.t

Phng trnh chuyn ng ca vt B: x2= 3.vBvB.t


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Sau 1,5 gi:

x = |x1- x2| = 15m vB= 40km/h S = 3.vB= 120km.

DNG 3: THCA CHUYN NG THNG U.Hng gii quyt:* Tnh cht chuyn ng:- th ta theo thi gian xin ln, vt chuyn ng thng u cng chiu dng.- th ta theo thi gian xin xung, vt chuyn ng thng u ngc chiu

dng.-thta theo thi gian nm ngang, vt nm yn.

* Tnh tc chuyn ng theo cng thc: v=| |

BI TP C HNG DN

Bi 1:Mt ngi i xe p t A v mt ngui i b t B cng lc v cng theo hng AB.

Ngi i xe p i vi vn tc v =12km/h, ngi i b i vi v = 5 km/h. AB = 14km.a) H gp nhau khi no, u?b) V th ta theo thi gian theo hai cch chn A lm gc v chn B lm gc

Giia) Chn gc to ti A, chiu dng l chiu chuyn ng ca xe.Phng trnh chuyn ng ca xe p: x1= x0+ v1.t = 12.tPhng trnh chuyn ng ca ngi i b: x2= x0+ v2.t = 14 + 5tKhi hai xe gp nhau: x1= x2

12.t = 14 + 5t t = 2 hTo khi gp nhau: x1= 12. 2 = 24km

b) V th:Bng gi tr

t 0 1 2

x1 0 12 24

x2 14 19 24

th:

Bi 2:Hai t xut pht cng mt lc t 2 a im A v B cch nhau 20km trn mt ngthng i qua B, chuyn ng cng chiu theo hng A n B. Vn tcca t xut pht tA vi v = 60km/h, vn tc ca xe xut pht t B vi v = 40km/h.

a)Vit phng trnh chuyn ng.b) V th to - thi gian ca 2 xe trn cng h trc.c) Da vo th xc nh v tr v thi im m 2 xe ui kp nhau.

Giia) Chn gc to ti A, gc thi gian lc 2 xe xut pht

Phng trnh chuyn ng ca oto A: x1= 60tPhng trnh chuyn ng ca oto B: x2= 20 + 40t

0.5 1.0 1.5 2.0t s

5

10

15

20

x m

x1

x2


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b) Bng ( x, t )t (h) 0 1 2

x1 (km) 0 60 120

x2 (km) 20 60 100

th hnh bn

c) Da vo th ta thy 2 xe gp nhau v tr cchA 60km v thi im m hai xe gp nhau 1h

Bi 3:Cho th nh hnh v. Da vo th.a) Tnh vn tc ca xe.

b) Lp phng trnh chuyn ng ca xec) Xc nh thi im v v tr 2 xe gp nhau.

Gii

a) Vn tc xe1: v1=

=

= 40 km/h

Vn tc xe 2: v2= = =

= 20 km/hb) Phng trnh chuyn ng ca xe 1: x1= 40tPhng trnh chuyn ng ca xe 2: x2= 12020t

c) Khi hai xe gp nhau: x1= x240t = 12020t t = 2hV tr gp nhau cch O: x1= 80km

DNG 4: XC NH VN TC, GIA TC, QUNG DNG I TRONGCHUYN NG THNG BIN I U.

Vn dng cc cng thc sau:- Tnh gia tc: a =

v - v0t

- Cng thc vn tc theo t: v = v0+ at

- Qung ng theot: S = v0.t +at

2

-H thc c lp thi gian: v2v02= 2.a.STrong :+ a > 0 nu CND

+ a < 0 nu CCD

BI TP C HNG DNBi 1: Mt on tu ang chuyn ng vi v0= 72km/h thhm phanh chuyn ng chmdn u, sau 10 giy t v1= 54km/h.a) Sau bao lu k t lc hm phanh th tu t v = 36km/h v sau bao lu th dng hn.

b) Tnh qung ng on tu i c cho n lc dng li.Gii

Chn chiu dng l chiu chuyn ng ca tu, gc thi gian lc bt u hm phanh.

a) Gia tc ca on tu: a = = -0.5m/s2

0.5 1.0 1.5 2.0t h

20

40

60

80

100

120

x km

x1

x2

0.5 1.0 1.5 2.0t h

20

40

60

80

100

120

x km

x1

x2


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Thi gian on tu t vn tc v2= 36km/h: v2= v0+ at t2=

= 20sKhi dng li hn: v3= 0 t2=

= 40s

b) Qung ng tu i c n khi dng hn: = 2. S = = 400mBi 2:Mt xe la dng li hn sau 20s k t lc bt u hm phanh. Trong thi gian xe

chy c 120m. Tnh vn tc ca xe lc bt u hm phanh v gia tc ca xe.Gii

Vn tc ca tu la khi dng li: v = v0+ at = 0 v0= -20a. (1)

on ng tu la i c khi dng hn: S = v0t +at

2(2)

T (1), (2)a = -0,6m/s2, v0= 12m/sBi 3: Mt on tu bt u chuyn ng nhanh dn u khi i ht 1km th nht th v 1=10m/s. Tnh vn tc v sau khi i ht 2km.

GiiTheo h thc c lp thi gian tng ng vi S1: v2v02= 2.a.S1a = 0,05m/s2

Vn tc sau khi i ht on ng S2: v12= 2.a.S2+ v02v1= 102m/sBi 4: Mt chic xe la chuyn ng trn on thng qua im A vi v = 20m/s, a = 2m/s 2.TiB cch A 100m. Tm vn tc ca xe.

Gii

p dng hthc: = 2. . vB= 2. = 28m/sBi 5:Mt chic can chy vi v = 16m/s, a = 2m/s2cho n khi t c v = 24m/s th bt

u gim tc cho n khi dng hn. Bit can bt u tng vn tc cho n khi dng hnl 10s. Hi qung ng can chy.

Gii

Thi gian ca n tng tc n 24km/h: t1=

= 4s.Nn thi gian gim tc : t2= tt1= 6s

Qung ng i c khi tng tc : S1 = v0t1+at1

2= 80m

Qung ng i c t khi bt u gim tc n khi dng hn:

S2= v01t2+ at22 = 72mTng qung ng ca n i chy: S = S1+ S2= 152m

Bi 6:Mt xe chuyn ng nhanh dn u i c S1= 24m, S2= 64m trong 2 khong thigian lin tip bng nhau l 4s. Xc nh vn tc ban u v gia tc.

Giip dng cng thc tnh qung ng trong chuyn ng nhanh dn u:

S1 = v01t1+at1

24.v01+ 8a = 24 (1)

S2= v02t2+ at22 4.v01+ 8a = 64 (2)M v02= v01+ at2(3)


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Gii h 3 phng trnh (1), (2), (3) ta c: v01= 1m/s, a = 2,5m/s2Bi 7:Mt t chuyn ng thng nhanh dn u vi v0= 10,8km/h. Trong giy th 6 xe ic qung ng 14m.a) Tnh gia tc ca xe.

b) Tnh qung ng xe i trong 20s u tin.

Giia) Qung ng i trong 5s u: S1= v0t1+

Qung ng i trong 6s: S2= v0t2+

Qung ng i trong giy th6: S = S2- S1= 14 a = 2m/s2

b) Qung ng xe i trong 20s u tin:S20= v0t20+ = 460m

Bi 8:Mt xe chhng chuyn ng chm dn u vi v0= 25m/s, a = - 2m/s2.a) Tnh vn tc khi n i thm c 100m.

b) Qung ng ln nht m xe c thi c.Gii

a) Theo hthc: v2v02= 2.a.S v = 2. = 15m/sb) Theo hthc: v2v02= 2.a.S (v = 0) S =

= 156.3 m.

Bi 9:Mt xe my ang i vi v = 50,4km/h bng ngi li xe thy c g trc mt cchxe 24,5m. Ngi y phanh gp v xe n g th dng li.a) Tnh gia tc

b) Tnh thi gian hm phanh.Gii

a) Theo hthc: v2v02= 2.s.S a =

= -4m/s2

b) Thi gian hm phanh: t =

= 3.5sBi 10:Mt vin bi ln nhanh dn u t nh mt mng nghing vi v0= 0, a = 0,5m/s2.a) Sau bao lu vin bi t v1= 2,5m/s

b) Bit vn tc khi chm t v2= 3,2m/s. Tnh chiu di mng v thi gian vin bi chm t.

Giia) Thi gian vin bi t vn tc v1= 2.5m/s l: t =

= 5s

b) p dng hthc: v2v02= 2.a.S S =

. = 10,24 mThi gian vin bi chm t: t =

= 6.4s

DNG 5: TNH QUNG NG VT I C TRONG GIY THN VTRONG N GIY CUI CA CHUYN NG THNG BIN I U.

Cch gii:

* Qung ng vt i trong giy th n.


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- Tnh qung ng vt i trong n giy: S1= v0.n +a.n

2

- Tnh qung ng vt i trong (n 1) giy: S2= v0.( n- 1) +a.(n1 )

2

- Tnh qung ng vt i trong giy th n: S = S1S2* Qung ng vt i trong n giy cui.

- Tnh qung ng vt i trong t giy: S1= v0.t + a.t2- Tnh qung ng vt i trong (t n) giy: S2= v0.( t - n) +

a.(tn )

2

- Tnh qung ng vt i trong n giy cui : S = S1S2BI TP C HNG DN

Bi 1:Mt xe chuyn ng nhanh dn u vi v = 18km/h. Trong giy th 5 xe i c5,45m.

a) Tnh gia tc ca xe.

b) Tnh qung ng i c trong giy th 10.Gii

a) Qung ng i trong 5s u: S5= v0t5+at5

2= 25 + 12,5a

Qung ng i trong 4s:S4= v0t4+at4

2= 20 + 8a

Qung ng i trong giy th 5: S = S5- S4= 5,45 a = 0,1 m/s2

b) Qung ng i trong 10s u: S10= v0t10+at10

2= 55m

Qung ng i trong 9s: S9= v0t9+

at9

2= 49,05m

Qung ng i trong giy th 10: S = S10- S9= 5,45Bi 2:Mt vt chuyn ng nhanh dn u trong 10s vi a = 4m/s2. Qung ng vt ic trong 2s cui cng l bao nhiu?

Gii

Qung ng vt i c trong 10s: S10= v0t10+at10

2= 200m

Qung ng vt i c trong 8s u: S8= v0t8+at8

2= 128m

Qung ng vt i trong 2s cui: S = S10S8= 72m

Bi 3: Mt vt chuyn ng thng bin i u khng vn tc u v i c qung ngS mt 3s. Tm thi gian vt i c 8/9 on ng cui.

Gii

Qung ng i c trong 3s: S= v0t +at

2= 4,5a

Thi gian vt i trong 1/9 qung ng u: S= v0t+at

2= 0,5a.t

1/9 S = 0,5a.tt= 1s

Thi gian vt i trong 8/9 qung ng cui: t= tt= 2s


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DNG 6: VN DNG CNG THC TNH QUNG NG, VN TC TRONGRI TDO

- Cng thc tnh qung ng: S =gt

2

- Cng thc vn tc: v = g.t-H thc lin h v v S: v2= 2g.S

BI TP C HNG DNBi 1:Mt vt ri t do t cao 20m xung t, g = 10m/s2.a) Tnh thi gian vt ri n t.

b) Tnh vn tc lc va chm t.Gii

a) Tcng thc S =gt

2t = = 2sb) Vn tc vt khi chm t v = gt = 20 m/s

Bi 2:Mt vt c th ri khng vn tc u khi va chm t c v = 70m/s, g = 10m/s2a) Xc nh qung ng ri ca vt.

b) Tnh thi gian ri ca vt.Gii

a) p dng hthc v2= 2.g.S S =.= 245m

b) Theo cng thc: v = gtt = 7sBi 3:T cao 120m ngi ta th mt vt thng ng xung vi v = 10m/s, g = 10m/s2.

a) Sau bao lu vt chm t.b) Tnh vn tc ca vt lc va chm t.

Gii

a) T cng thc S = v0t +gt

2100 = 20t + t2t = 4s ( nhn ) hoc t = - 6s ( loi )

b) Vn tc ca vt lc va chm t: v = v0+ gt = 50 m/sBi 4:Th mt hn t cao h xung y, hn ri trong 1s. Nu th hn th=4h th thi gian ri l bao nhiu?

Gii

p dng cng thc: S = h =gt

2t = = 1(s)TS' = h=

gt1

2t1= = 2(s)Bi 5:Mt vt ri t do khi chm t th vt t v = 30m/s. Hi vt c th ri t caono? g = 9,8m/s2.

Giip dng cng thc: v = gt t = 3,06s

Qung ng vt ri: h = S = gt2= 45,9m


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Bi 6:Ngi ta th mt vt ri t do, sau 4s vt chm t, g = 10m/s2. Xc nh.a) Tnh cao lc th vt.

b) Vn tc khi chm t.c) cao ca vt sau khi th c 2s.

Gii

a) cao lc th vt: h = S = gt2= 80mb) Vn tc khi chm t: v = v0+ gt = 40 m/s

c) Qung ng vt ri 2s u tin: S1=gt1

2= 20m

cao ca vt sau khi th 2s: h = S2= SS1= 60mBi 7:Mt ngi th vt ri t do, vt chm t c v = 30m/s, g = 10m/s2.a) Tm cao th vt.

b) Vn tc vt khi ri c 20m.

c) cao ca vt sau khi i c 2s.Gii

a) cao lc th vt: h = S =gt

2= 45m

Vn tc ca vt khi chm t: v = gtt = 3s

b) Thi gian vt ri 20m u tin:S=gt

2t= 2s

Vn tc khi vt ri c 20m: v= gt= 20m/sc) cao ca vt khi ri c 2s: h= SS= 25m

DNG 7: TNH QUNG NG VT RI TRONG N GIY CUI, V TRONGGIY THN.

Hng gii quyt:* Qung ng vt i c trong n giy cui.

- Qung ng vt i trong t giy: S1=g.t

2

- Qung ng vt i trong (t n) giy: S2=g.(t - n)

2

- Qung ng vt i trong n giy cui: S = S1S2= g(2.t - n).n* Qung ng vt i c trong giy th n.

- Qung ng vt i trong n giy: S1=g.n

2

- Qung ng vt i trong (n 1) giy: S2=g.(n - 1)

2

- Qung ng vt i c trong giy th n: S = S1S2=g(2.n - 1)

* Thi gian ri c S (m) cui cng:

- Thi gian ri n khi chm t: t =


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- Thi gian ri trong khong S -S l: t1= () - Thi gian ri trong S (m) cui cng: t = t - t1= - ()

BI TP C HNG DN

Bi 1:Mt vt ri khng vn tc u t cao 80m xung t.a) Tm vn tc lc va chm t v thi gian ca vt tlc ri ti lc chm t.b) Tnh qung ng vt ri c trong 0,5s u tin v 0,5s cui cng, g = 10m/s2

Gii

a) Vn tc: v = 2= 40m/sThi gian lc chm t: t =

= 4s

b) Qung ng vt ri trong 0,5s u tin: S1=gt

2= 1.25m/s

Qung ng vt i c trong 0.5s cui: S = g(2t - n).n = 18,75mBi 2:Mt vt ri t do ti mt a im c g = 10m/s2. Tnha) Qung ng vt ri c trong 5s u tin.

b) Qung ng vt ri trong giy th 5.Gii

a) Qung ng vt ri trong 5s u: S5=gt5

2= 125m

b) Qung ng vt ri trong giy th 5: S =g(2.n - 1) = 45m

Bi 3:Trong 3s cui cng trc khi chm t, vt ri t do c qung ng 345m.Tnhthi gian ri v cao ca vt lc th, g = 10 m/s2.

GiiGi t l thi gian vt ri.

p dng cng thc: S =g(2.t - n).n

Trong n = 3

Suy ra: 345 = 53(2t - 3) t = 10

cao lc th vt: S=

gt

2

= 500m


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Bi 4:Mt vt ri t do t cao h. Bit rng trong 2s cui cng vt ri c qung ngbng qung ng i trong 5s u tin, g = 10m/s2.a) Tm cao lc th vt v thi gian vt ri.

b) Tm vn tc cu vt lc va chm t.Gii

a) Chn chiu dng hng xung, gc to ti vtr vt bt u ri, gc thi gian lcvt ri.

Gi t l thi gian vt ri.


Trong n = 2

Ngoi ra S =g= 125

Suy ra: 125 = 52(2t - 2) t = 7.25s

cao lc th vt: S= gt2= 252,81mb) Vn tc lc va chm t: v = gt = 72,5m/s

Bi 5:Mt vt ri t do t cao 50m, g = 10m/s2. Tnha) Thi gian vt ri 1m u tin.

b) Thi gian vt ri c 1m cui cng.Gii

a) Thi gian vt ri 1m u tin: S1=gt1

2t1= 0,45s

b) Thi gian vt ri n mt t: S = gt2

t = 3,16s

Thi gian vt ri 49m u tin: S2=gt2

2t2= 3,13s

Thi gian vt ri 1m cui cng: t= tt2= 0,03sBi 6: Mt vt c th ri t do khng vn tc u, g = 10m/s2.a) Tnh on ng vt i c trong giy th 7.

b) Trong 7s cui cng vt ri c 385m. Xc nh thi gian ri ca vt.c) Tnh thi gian cn thit vt ri 45m cui cng

Gii

a) Qung ng i trong giy th t = 7: S =g(2.t - 1) = 65mb) Gi t l thi gian vt ri chm t.


Trong n = 7

Suy ra: 385 = 57(2t - 7) t = 9s

c) Qung ng vt ri trong 9s: S =gt

2= 405m

Thi gian vt ri trong 45m cui l t = - () = 0,51s


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Bi 7:Mt vt ri t do trong 10 s. Qung ng vt ri trong 2s cui cng l bao nhiu?,ly g = 10m/s2.

GiiGi t l thi gian vt ri chm t.

p dng cng thc: S =

g(2.t - n).n

Trong n = 7

S = 52(210 - 2) = 180mBi 8: Mt vt ri t do khng vn tc u t cao 45m xung t. Ly g = 10m/s.a) Tnh thi gian ri v tc ca vt khi va khi va chm t.

b) Tnh thi gian vt ri 10m u tin v thi gian vt ri 10m cui cng trc khi chm t.Gii

a) Thi gian ri: t =

= 3s

Vn tc ca vt lc chm t: v = g.t = 30m/s

b) p dng cng thc S1=gt

2 = 10m t = =2sThi gian vt ri 10m cui cng trc khi chm t: t = -

() = 0.35s

Bi 9:Mt vt ri t do khng vn tc u t cao 80m xung t. Ly g = 10m/s2. Tnh:a) Thi gian t lc bt u ri n khi vt chm t v tc ca vt khi chm t

b) Qung ng vt ri c trong 2s u tin v qung ng vt ri trong 2s cui cngtrc khi chm t

Gii

a) Thi gia roi ca vt: t = = 4sVn tc ca vt lc chm t: v = g.t = 40m

b) Qung ng ri trong 2s u tin: S2= = 20m

Qung ng vt ri trong 2s cui: S =

g(2t - 2).2 = 60m

Bi 10:Mt vt c th ri t do khng vn tc u t cao h so vi mt t. Cho g=10m/s2. Tc ca vt khi chm t l 30m/s.a) Tnh cao h, thi gian tlc vt bt u ri n khi vt chm t.

b) Tnh qung ng vt ri trong hai giy u v trong giy thhai.Gii

a) p dng hthc: v2= 2gh h == 45 m

Thi gian ri: t =

= 3s

b) Qung ng vt ri trong 2s u: S2= = 20m


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Qung ng vt ri trong giu thhai: S = (2. 1)= 15m vi t = 2s.

Bi 11:Mt vt c th ri t do khng vn tc u t cao h so vi mt t. Cho g=10m/s2. Thi gian vt ri l 4 giy.a) Tnh cao h, vn tc ca vt khi vt chm t.

b) Tnh qung ng vt ri trong giy cui cng trc khi chm t.Gii

a) cao lc th vt: S =g.t

2= 80m

Vn tc ca vt khi chm t: v = g.t = 40m/s

b) Qung ng vt ri trong giu thhai: S = (2. 1)= 35m vi t = 4s.

Bi 12: Mt vt c th ri t do khng vn tc u t cao h so vi mt t. Cho g=10m/s2. Thi gian vt ri 10 m cui cng trc khi chm t l 0,2s. Tnh cao h, thigian ri v tc ca vt khi chm t.

GiiGi S l qung ng vt ri trong t(s).

p dng hthc tnh thi gian trong S (m) cui c: t = - ()

Trong : t = 0.2s, S = 10m, g = 10m/s2

Suy ra: ()

= 0.2S = h = 130 m

Thi gian vt vt chm t: t = = 5,1s

Vn tc khi va chm t: v = g.t = 51m/sBi 13:Mt vt ri t do khng vn tc u ti ni c gia tc trng trng g. Trong giy th3, qung ng ri c l 24,5m v tc ca vt khi va chm t l 39,2m/s. Tnh g v cao ni th vt.

Gii

Qung ng vt ri trong 3 giy: S1=g.t1

2= 4,5.g

Qung ng vt ri trong 2s u: S2= g.t22

= 2.gQung ng vt ri trong giy th 3: S = S1S224,5 = 4,5g - 2.g g = 9,8 m/s2

cao lc thvt: h == 78.4 m

Bi 14:Mt vt c th ri t do khng vn tc u t cao h so vi mt t ti ni cgia tc trng trng g = 10m/s2. Qung ng vt ri trong na thi gian sau di hn qungng vt ri trong na thi gian u 40m. Tnh h, thi gian ri v tc ca vt khi chmt.

GiiQung ng vt ri na thi gian u: S1=

g.(

)

2=


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Qung ng vt ri na thi gian cui: S2= 2 =

M theo gia thit S2= 40 + S140 +

=

t = 4s cao lc th vt: S =

g.t

2= 80m

Vn tc khi chm t: v = g.t = 40m/s

DNG 8: XC NH VTR 2 VT GP NHAU C THRI VI CNGTHI IM KHC NHAU.

Hng gii quyt bi ton:- Chn chiu dng hng xung, gc to ti v tr vt bt u ri, gc thi gian lc

bt u ri ( ca vt ri trc )

-Phng trnh chuyn ng c dng: y = y0+g(tt0)

2

+ Vt 1: y1= y01+

g .t 2

+ Vt 2:y2= y02+ g (tt0)2

-Hai vt gp nhau khi chng c cng to : y1= y2t- Thay t vo y1hoc y2 tm c v tr gp nhau.

BI TP C HNG DNBi 1:T tng 9 ca mt to nh, Nam th ri vin bi A. Sau 1s, Hng th ri vin bi B tngthp hn 10m. Hai vin bi s gp nhau lc no ( Tnh t khi vin bi A ri ), g = 9,8 m/s2.

Gii

Chn trc to thng ng, chiu dng hng xung, gc to ti v tr th, gc thigian lc bi A ri.

Phng trnh chuyn ng cho tng vin bi c dng:

+ Bi A: y1= y01+gt

2=gt

2(vi y01= 0)

+ Bi B: y2= y02+g(t - t0)

2= 10 +g(t - 1)

2(vi y02= 10m, t0= 1s)

Khi 2 vin bi gp nhau: y1= y2gt

2= 10 +g(t - 1)

2t = 1,5s

Bi 2:T 1 nh thp cao 20m, ngi ta bung mt vt. Sau 2s th ngi ta li bung vt th

2 tng thp hn nh thp 5m. Chn trc Oy thng ng, gc O nh thp, chiu (+)hng xung, thi gian lc vt 1 bt u ri, g = 10m/s2a) Lp phng trnh chuyn ng v phng trnh vn tc ca 2 vt.

b) Hai vt c chm t cng lc khng.c) Vn tc lc chm t ca mi vt l bao nhiu?

Giia) Phng trnh chuyn ng cho tng vt c dng:

+ Vt 1: y1=

gt2= 5t2& v1= gt = 10t

+ Vt 2: y2= y0+ g(t- t0)2= 5(t24t +5) & v2= g(t2) = 10(t -2)


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b) Thi im vt 1 chm t: y1= 20m t1= 2sThi im vt 2 chm t: y2= 5 (t24t +5 ) = 20

t2= 3,73s ( nhn ) hoc t2= 0,27s < 2 ( loi)V t1t2: 2 vt khng chm t cng lc.c) Vn tc ca vin bi1: v1= 10t1= 20m/s

Vn tc ca vin bi 2: v2= 10(t22) = 17,3 m/sBi 3: Mt vin bi A c th ri t cao 30m. Cng lc , mt vin bi B c bn theo

phng thng ng t di t ln vi v = 25m/s ti va chm vo bi A. Chn trc Oy thngng,gc O mt t, chiu dng hng ln, gc thi gian lc 2 vin bi bt u chuynng, g = 10m/s2. B qua sc cn khng kh.a) Lp phng trnh chuyn ng ca mi vin bi.

b) Tnh thi im v ta 2 vin bi gp nhau.c) Vn tc mi vin bi khi gp nhau.

Giia) Phng trnh chuyn ng c dng:

+ Vt 1: y1= y01 + v01t+gt

2= 305.t2

+ Vt 2: y2= y02+v02t +at

2= 25t - 5t2vi a = - g

b) Khi hai vin vi gp nhau: y1= y2305t2= 25t - 5t25t = 6 t = 1,2sc) Vn tc ca mi vin bi khi gp nhau:- Vin bi 1: v1= - gt = -12m/s

- Vin bi 2: v2= v0- gt = 13m/sDNG 8: VN DNG CC CNG THC TRONG CHUYN NG TRN U

Hng gii quyt:

+ Cng thc chu k: T =

+ Cng thc tn s: f ==

+ Cng thc gia tc hng tm:aht = 2 = 2r+ Cng thc lin h gia tc di, tc gc: v = r

BI TP C HNG DNBi 1: Xe p ca 1 vn ng vin chuyn ng thng u vi v = 36km/h. Bit bn knh

ca lp bnh xe p l 32,5cm. Tnh tc gc v gia tc hng tm ti mt im trn lpbnh xe.

GiiVn tc xe p cng l tc di ca mt im trn lp xe: v = 10 m/s

Tc gc: == 30.77 (rad/s)

Gia tc hng tm: aht=

= 307,7 (m/s2

)


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Bi 2:Mt vt im chuyn ng trn ng trnbn knh 15cm vi tn skhng i 5vng/s. Tnh chu k, tn sgc, tc di.

GiiTc gc: = 2f = 10 (rad/s)

Chu k: T =

= 0,2 (s)

Tc di: v = r.= 4,71 (m/s)Bi 3:Trong 1 my gia tc e chuyn ng trn qu o trn c R = 1m. Thi gian e quay ht5 vng l 5.10-7s. Hy tnh tc gc, tc di, gia tc hng tm ca e.

Gii

Chu k: T == 10

-7(s)

Tc gc: = = 2.10

7(rad/s)

Tc di: v = .R = 2.107(m/s)

Gia tc hng tm: aht= 2.R = 4.1015(m/s2) vi 2= 10Bi 4:Mt xe ti c bnh xe c ng knh 80cm, chuyn ng u vi tc 10m/s. Tnhchu k, tn s, tc gc ca u van xe.

Gii

Tc gc: == 12.5 (rad/s)

Chu k: T = = 0,5 (s)

Tn s: f =

= 2(Hz)

Bi 5:Mt a quay u quanh trc qua tm O, vi vn tc qua tm l 300vng/ pht.a) Tnh tc gc, chu k.

b) Tnh tc di, gia tc hng tm ca 1 im trn a cch tm 10cm, g = 10m/s2.Gii

Tn s quay ca a: f = 300 vng/ pht = 5 vng/sa) Tc gc: = 2f = 10 (rad/s)

Chu k: T == 0,2 (s)

b) Tc di: v = r. = 3,14 m/s;Gia tc hng tm: aht =

= 98,7 (m/s

2)

Bi 6:Mt a ng cht c dng hnh trn c R = 30cm ang quay trn u quanh trc can. Bit thi gian quay ht 1vng l 2s. Tnh tc di, tc gc ca 2 im A, B nmtrn cng 1 ng knh ca a. Bit im A nm trn vnh a, im B nm trn trung imgia tm O ca vng trn v vnh a.

GiiTa c bn knh: RA= 30cm v RB= 15cmTc gc ca tng im:


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+ A: A==

+ B: Chu k khng ph thuc vo R nn: A= BTc di ca tng im:+ A: vA= rA.A= 0,94 m/s+ B: v

B= r

B.

B= 0,47 m/s

Bi 7:Mt v tinh quay quanh Tri t ti cao 200km so vi mt t. cao g =9,2m/s2. Hi tc di ca vtinh l bao nhiu?

Gii

p dng cng thc aht= g =

+v = ( )= 7785,8 (m/s)Bi 8:Mt v tinh nhn to c qu o l mt ng trn cch mt t 400km, quay quanhTri t 1 vng ht 90 pht. Gia tc hng tmca vtinh l bao nhiu, RT= 6389km.

Gii

Chu kchuyn ng ca vtinh nhn to: T = 90 pht = 5400sTc gc: = Tc gc: =

= 1,16.10

-3(rad/s)

Gia tc hng tm: aht = =

[(+)]+ = 9,13 (m/s)

Bi 9:V tinh A ca Vit Nam c phng ln qu o ngy 19/4/2008. Sau khi n nh, vtinh chuyn ng trn u vi v = 2,21 km/h cao 24000km so vi mt t. Bn knh Tl 6389km. Tnh tc gc, chu k, tn s ca v tinh.

Gii

Tc di: v = 2,21km/h = 0,61m/sBn knh quo: r = R + h = 24689km = 24689.103mTc gc: = v.r = 15060290 rad/s

Chu k: T = = 4,17.10-7s

Tn s: = = 2398135 vng/sBi 10: Gia tc hng tm ca chuyn ng trn u tng hay gim bao nhiu nu vn tcgc gim cn mt na nhng bn knh quo tng 2 ln.

GiiGia tc hng tm ng vi bn kinh r v tc gc : aht=

= r

2

Gia tc hng tm ng vi bn knh r' = 2r v tc gc ' = : = . ()=

Bi 11: Mt ng htreo tng c kim gidi 2,5cm, kim pht di 3cm. So snh tc gc,tc di ca 2 u kim ni trn.

Gii+ i vi kim gi:

Chu kca kim gii: Th= 43200(s)Tc gc: h=


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Tc di: vh= Rh.h+ i vi kim ph:Chu k ca kim gii: Tp= 3600(s)

Tc gc: p=

Tc di: vp= Rp.hVy ts:

+ Tc gc:=

=

p= 12h

+ Tc di:=

.

=

. , = 7vp= 14,4vh

Bi 12:Mt bnh xe p c ng knh l 20cm, khi chuyn ng c vn tc gc l 12,56rad/s. Vn tc di ca mt im trn vnh bnh xe l bao nhiu?.

Gii

Vn tc di ca mt im trn vnh bnh xe: v = R.= 2,512 m/sBi 13: Mt im nm trn vnh ngoi ca lp xe my cch trc bnh xe 30cm. Bnh xequay u vi tc 8vng/s. S vng bnh xe quay s ch trn ng h tc ca xe snhy 1 s ng vi 1km v thi gian quay ht s vng y l bao nhiu?

GiiTa c 1 vng c chiu di l: 2R S = 1000m = N.2.R N = 531 vng.

Thi gian bnh xe quay ht N vng: t = N.T = = 66(s)

DNG 9: XC NH VN TC TNG I, TUYT I, KO THEO.Hng gii quyt bi ton-t:+ v13: vn tc ca vt i vi h qui chiu ng yn.+ v12: vn tc ca vt i vi hquy chiu chuyn ng.+ v23: vn tc ca hqui chiu chuyn ng i vi hquy chiu ng yn-Xc nh cc i lng: v13; v12; v23- Vn dng cng thc cng vn tc: = + -Khi cng chiu: v13= v12+ v23-Khi ngc chiu: v13= v12v23

- Qung ng: = BI TP C HNG DN

Bi 1: Hai xe my ca Nam v An cng chuyn ng trn on ng cao tc, thng vi vntc vN= 45km/h, vA= 65km/h. Xc nh vn tc tng i ( ln v hng ) ca Nam sovi An.a) Hai xe chuyn ng cng chiu.

b) Hai xe chuyn ng ngc chiuGii


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Gi:+ v12 l vn tc ca Nam i vi An+ v13l vn tc ca Nam i vi mt ng+ v23l vn tc ca An i vi mt ng

a) Khi chuyn ng cng chiu: v13= v12+ v23v12= -20km/h

Hng: ngc li vi hng chuyn ng ca 2 xe.ln: l 20km/hb) Khi chuyn ng ngc chiu: v13= v12- v23v12= 110km/hHng: theo hng ca xe Nam ln: l 110km/h

Bi 2: Lc tri khng gi, mt my bayt a im M n N theo 1 ng thng vi v =120km/s mt thi gian 2 gi. Khi bay tr li, gp gi nn bay mt thi gian 2 gi 20 pht.Xc nh vn tc gi i vi mt t.

GiiGi:+ v12 l vn tc ca my bay i vi gi+ v13l vn tc ca my bayi vi mt t+ v23l vn tc ca gi i vi mt t.Khi my bay bay tM n N lc khng gi: v23= 0

v13= 120m/s v12= 120m/s

Khi bay tN n M ngc gi

=

= 102,9m/s

M v13' = v12v23 v23= v12v13= 17,1 m/s.Bi 3:Mt can i xui dng nc t A n B mt 4 gi, cn nu i ngc dng nc t Bn A mt 5 gi. Bit vn tc ca dng nc so vi b sng l 4 km/h. Tnh vn tc ca canso vi dng nc v tnh qung ng AB.

GiiGi v12lvn tc ca can so vi dng nc.

Khi i xui dng: SAB= v13.t1= (v12+ v23).4 (v12+ 4).4 = SAB

Khi i ngc dng: SAB= v13' .t2= ( v12v23).5 (v12- 4).5 = SAB(v12+ 4 ).4 = (v124).5 v12= 36km/h SAB= 160kmBi 4:Mt chic thuyn chuyn ng ngc chiu dng nc vi v = 7,5 km/h i vi dngnc. Vn tc chy ca dng nc i vi b sng l 2,1 km/h. Vn tc ca thuyn i vi

b sng l bao nhiu?Gii

Khi i ngc dng ta c: v13= v12v23= 7,52,1 = 5,4 km/hBi 5:Mt can chuyn ng u v xui dng t A n B mt 1 gi. Khong cch AB l

24km, vn tc ca nc so vi b l 6km/h.a) Tnh vn tc ca can so vi nc.


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b) Tnh thi gian can quay v t B n A.Gii

Gi v12l vn tc ca can so vi nc.

a) Ta c: v13== 24km/h

Khi xui dng: v13= v12+ v23v12= v13v23= 18km/hb) Khi ngc dng: v13= v12v23= 12km/h

Thi gian cano quay v: t =

= 2hBi 6:Mt ngi li xung my d nh m my cho xung chy ngang con sng rng 320m,mi xung lun lun vung gc vi b sng. Nhng do nc chy nn xung sang n b

bn kia ti mt im cch bn d nh 240m v mt 100s. Xc nh vn tc cu xung so vidng sng.

Gii

Khong cch gia 2 b sng l 360m, xung n b cch bn240m:S = = 400 mVn tc ca xung khi xui dng: v =

= 4 (m/s)

Bi 7:Mt tu ho chuyn ng thng u vi v = 10m/s so vi mt t. Mt ngi i utrn sn tu c v = 1m/s so vi tu. Xc nh vn tc ca ngi so vi mt t trong cctrng hp.a) Ngi v tu chuyn ng cng chiu.

b) Ngi v tu chuyn ng ngc chiu.c) Ngi v tu chuyn ng vung gc vi nhau.

GiiGi+ v13l vn tc ca ngi so vi mt t.+ v12l vn tc ca ngi so vi tu.+ v23lvn tc ca tu so vi mt t.a) Khi cng chiu: v13= v12+ v23= 11 (m/s)

b) Khi ngc chiu: v13= v23v12= 9 (m/s)

c) Khi vung gc: v13= = 10,05 (m/s)Bi 8:Mt chic thuyn xui dng t A n B v quay v A. Bit vn tc ca nc so vi

bl 2km/h, AB = 14km. Tnh thi gian tng cng i v vca thuyn.Gii

Vn tc thuyn i vi nc: v12= 12km/hVn tc ca nc i vi b: v23= 14km/h

Khi xui dng: v13= v12+ v23= 14km/h t1=

= 1h

Khi ngc dng: v13= v12v23= 10km/h t2= = 1,4h


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Thi gian tng cng: t = t1+ t2= 2,4hBi 9: Mt xung my i trong nc yn lng vi v = 30km/h. Khi xui dng t A n Bmt 2 gi, ngc dng t B n A mt 3 gi.a) Tnh qung ng AB.

b) Vn tc ca dng nc so vi b sng.

GiiGi+ v12l vn tc ca xung i vi nc: v12= 30km/h+ v13l vn tc ca xung i vi b+ v23l vn tc ca dng nc i vi b sng.a) Khi xui dng: v13= v12+ v23= 30 + v23Khi ngc dng: v13= v12v23= 30v23

v13+ v13=

S +

S = 60 S = 72km

b) Vn tc ca dng nc i vi bsng: v23= 30= 6 (km/h)Bi 10:Mt can chy thng u xui dng t A n B cch nhau 36km mt khong thigian 1,5h. Vn tc ca dng chy l 6km/h.a) Tnh vn tc ca can i vi dng chy.

b) Tnh khong thi gian nhnht can ngc dng tB n A.Gii

a) Vn tc cano i vi b: v13=

= 24 (km/h)

Khi xui dng: v13= v12+ v23v12= 18km/h

b) Khi ngc dng: v13= v12- v23= 12km/h t' =

= 3hBi 11:Mt can i t bn sng P n Q ri t Q n P. Hai bn sng cch nhau 21km trnmt ng thng. Bit vn tc ca can khi nc khng chy l 19,8km/h v vn tc cadng nc so vi b sng l 1,5m/s. Tm thi gian chuyn ng ca can.

Gii

Khi xui dng: v13= v12+ v23= 7m/s t1=

= 3000(s)

Khi ngc dng: v13= v12- v23= 4m/s t2= = 5250sTng thi gian: t = t1+ t = 8250s.

Bi 12:Mt thuyn my chuyn ng xui dng t M n N ri chy ngc dng t N nM vi tng cng thi gian l 4 gi. Bit dng nc chy vi v = 1,25m/s so vi b, vn tcca thuyn so vi dng nc l 20km/h. Tm qung ng MN.

Gii

Khi xui dng: v13= v12+ v23= 6,81m/s t1=

Khi ngc dng: v13= v12- v23= 4,31m/s t2=


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t1+ t2= 4

= 4.3600S = 37894,7 (m) = 37,9km

Bi 13:Mt chic thuyn xui dng sng t A n B ht 2 gi 30 pht. Khi quay ngc dngt B n A mt 3 gi. Vn tc ca nc so vi bsng v vn tc ca thuyn so vi nc lkhng i. Tnh thi gian 1 cnh ci kh ttri tA n B l bao nhiu?.

GiiKhi xui dng: v13= v12+ v23Khi ngc dng: v13= v12- v23

v13- = 2v23 , = 2. v23= ( , )tc= = 30hBI TP P DNG CHNG I

Chuyn ng thng u

1. T B lc 8h, mt ngi i v C, chuyn ng thng u vi vn tc 60km/h.a. Vit phng trnh chuyn ng v xc nh v tr ca ngi ny lc 10h.

b. Bit BC = 270km. dng phng trnh ta xc nh thi im ngi y n C.2. Mt xe t chuyn ng thng u qua A vi tc khng i v = 40km/h. Chn trc

ta Ox trng vi hng chuyn ng, gc ta O trng vi v tr A. Gc thi gian l lcxut pht.

a. Vit phng trnh chuyn ng.b. Dng phng trnh chuyn ng xc nh v tr t sau 1,5hc. Tm thi gian t i n B cch A l 30km.3. Hai t cng mt lc i qua hai a im A v B cch nhau 40km, chuyn ng thng

u cng chiu t A n B vi tc ln lt l 60km/h v 40km/h. Chn trc ta Oxtrng vi ng thng AB, gc ta O trng vi A, chiu dng AB. Gc thi gian llc hai xe xut pht.

a. Vit cng thc tnh qung ng i ca mi xe?b. Vit phng trnh chuyn ng ca mi xe?c. Tm thi gian xe t A ui kp xe t B v v tr hai xe gp nhau?d. V th ta - thi gian chuyn ng ca hai xe.4. Hai ngi cng lc i b t hai im A v B i n im C cch A 7,2km v cch

B 6km, vi vn tc khng i ln lt l 20km/h v 15km/h.a. Lp phng trnh chuyn ng ca hai ngi.

b. Hai ngi c gp nhau trc khi n C hay khng ?5. Lc 6h mt ngi i xe p xut pht t A chuyn ng thng u vi tc 12km/h

ui theo mt ngi i b ang i thng u vi tc 4km/h ti B cch A 12km. Chn trcta Ox trng vi ng thng AB, gc ta O trng vi A, chiu dng t AB. Gcthi gian l lc ngi i xe p xut pht.

a. Vit phng trnh chuyn ng ca mi ngib. Tm thi im ngi i xe p ui kp ngi i b v v tr lc gp nhau.c. Hai ngi cch nhau 4km vo nhng thi im no?


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6. Mt ngi i b xut pht t A chuyn ng thng u vi tc 4km/h, 1gi sau mtngi i xe p cng xut pht t A chuyn ng thng u vi tc 12km/h ui theongi i b. Chn trc ta Ox trng vi hng chuyn ng ca hai ngi, gc ta Otrng vi A, gc thi gian l lc ngi i b xut pht.

a. Vit phng trnh chuyn ng ca hai ngi

b. Tm thi gian chuyn ng ca mi ngi i gp nhau v v tr lc gp nhau?c. V th ta - thi gian ca hai ngi.

7. th ta - thi gian ca mt vt chuyn ng thngu nh hnh v. Da vo th tm vn tc v vit phngtrnh chuyn ng ca vt.

8. th ta - thi gian ca hai vt chuyn ng thngu nh hnh v:

a. Da vo th tm vn tc v lp phng trnh chuynng ca mi vt ?

b. Bng php tnh tm thi gian chuyn ng hai vt gpnhau v v tr lc gp nhau?

9. th ta - thi gian ca hai ng t chuyn ngthng u nh hnh v.a. Da vo th tm vn tc v lp phng trnh chuyn ngca mi ng t?

b. Bng php tnh tm thi gian chuyn ng hai ngt gp nhau v v tr lc gp nhau?

Chuyn ng thng bin i u

10. Mt xe chuyn ng thng trong 5 gi: 2 gi u xe chy vi tc trung bnh 60km/h;3 gi sau xe chy vi tc trung bnh 40km/h. Tnh vn tc trung bnh ca xe trong sutthi gian chuyn ng?

11. Mt xe chuyn ng thng t A n B. Na on ng u xe chuyn ng vi tc khng i 12km/h; na on ng cn li xe chuyn ng vi tc khng i 20km/h.Tnh vn tc ca xe trn c on ng?

12. Mt xe chuyn ng thng, i 1/3 on ng u vi tc 30km/h, i 2/3 on ng cn li vi tc 60km/h. Tnh tc trung bnh ca xe trn ton b qung ng.

13. Mt xe chuyn ng thng nhanh dn u, sau khi khi hnh c 10s th t vn tc54km/h.

0

0

x(km)

t(

h)

t

(h)

2

0

4

0

x(k

m) (I

)

II)

10

40

x(km)

t

(h)

(

I)

II)


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a. Tm gia tc ca xe?b. Tm vn tc v qung ng xe i c sau khikhi hnh c 6s?14. Mt t chuyn ng thng nhanh dn u t A n B trong 1 pht th vn tc tng t

18km/h ln n 72km/h.a. Tm gia tc ca t?

b. Tm qung ng AB?c. Nu t i t A n C vi AC=400m th mt thi gian bao lu?15. Mt on tu ang chy vi vn tc 36km/h th hm phanh chuyn ng thng chm

dn u, sau 10s th dng li.a. Tm gia tc ca on tu?

b. Sau thi gian 4s k t lc hm phanh, th tu chy c mt on ng bao nhiu?Tm vn tc ca tu khi ?

16. Mt on tu chuyn ng thng chm dn u vi vn tc u 0 72 /v km h sau 10s

vn tc ca on tu cn li 15m/s.a. Tm gia tc ca on tu?

b. Sau bao lu th tu dng hn?17. Mt xe ang chuyn ng vi vn tc 36km/h th hm phanh, chuyn ng chm dn

u i c 100m th dng hn.a. Tm gia tc ca xe?

b. Qung ng xe i c v vn tc ca xe sau khi hm phanh 10s?18. Mt ngi i xe p chuyn ng chm dn u ln mt dc di 50m. Vn tc chn

dc l 18km/h, nh dc l 3m/s.a. Tmgia tc v thi gian xe ln ht dc?

b. Nu ln dc c 10s th vn tc ca xe khi l bao nhiu? Cn bao nhiu mt nath ti nh dc?

19. Mt on tu dng hn li sau 20s k t lc bt u hm phanh, trong thi gian tuchy c 120m. Tm vn tc lc tu hm phanh v gia tc ca tu?

20. Mt qu cu chuyn ng thng nhanh dn u ln t nh mt dc di 100m, sau 10sth n chn dc. Sau qu cu chuyn ng thng chm dn u tip tc ln trn mt phng

nm ngang c 50m th dng li.a. Tm gia tc ca qu cu trn dc v trn mt phng ngang?

b. Thi gian qu cu chuyn ng?c. Vn tc trung bnh ca qu cu?21. Mt t ang chuyn ng vi vn tc36km/h th xung dc chuyn ng thng nhanh

dn u vi gia tc 0.1 m/s2n cui dcth t vn tc 72km/h.a. Tm thi gian xe xung ht dc?

b. Tm chiu di ca dc?


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c. Khi xung dc c 625m th vn tc t l bao nhiu? Cn bao lu na th t xunght dc?

22. th vn tc- thi gian ca mt vt chuyn ngthng nh hnh v:

a. Cho bit tnh cht chuyn ng ca tng giai on?

b. Xc nh gia tc ca tng giai on?c. Lp cng thc vn tc ca giai on I?23. Cc cng thc vn tc trong chuyn ng thng l:a. v = 5 - 2t (m/s)

b. v = 2 + 4t (m/s)

c. v = 4 (m/s)

Hy vit cng thc tnh qung ng i tng ngS ri t do

24. Mt vt c th ri t cao 20m so vi mt t. Ly g = 10m/s2.a. Tm thi gian vt ri n t?

b. Tm vn tc ca vt khi chm t?c. Sau khi ri c 1s th vt cn cch mt t bao nhiu?25. Mt vt c th ri t do, khi vt chm t th vn tc ca vt l 20m/s. Ly g =

10m/s2.

a. Tm cao lc th vt?b. Tm thi gian ri n t?

c. Khi vn tc ca vt l 10m/s th vt cn cch mt t bao nhiu? Cn bao lu na thvt ri n t?

26. Mt hn ri t mingmt ci ging cn xung n y ging mt 3s. Ly g =10m/s2.

a. Tnh su ca ging v vn tc hn khi chm y ging?b. Tnh qung ng hn ri trong giy th ba?27. Mt vt ri t do, trong giy cui cng ri c qung ng 45m. Tnh thi gian ri

v cao vt ri?

28. Mt vt ri t do ti ni c g = 10m/s2, thi gian ri n t l 10s. Tm thi gian vtri 10m cui cng?

29. T mt v tr, sau 2s k t lc git nc th hai ri, th khong cch gia git ncth nht vi git nc th hai l 25m. Tnh xem git nc th hai ri tr hn git nc thnht bao lu?

30. T mt nh thp ngi ta bung ri mt vt. Mt giy sau tng thp hn 10m ngita bung ri vt th hai. Hai vt cng ri chm t mt lc. Tnh thi gian ri ca vt thnht ? Ly g = 10m/s2.

Chuyn ng trn u

2

0

6

0

20

40

v(m/

s)

t(s)

8

0


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31. Mt t c bnh xe bn knh 30cm quay u mi giy c 10 vng. Tnh vn tcca t ?

32. Tm tc gc ca mt im trn Tri t i vi trc quay ca Tri t?33. Mt ngi ngi trn gh ca mt chic u quay ang quay vi tn s 5vng/pht.

Khong cch t ch ngi n trc quay ca chic u l 3m. Tm gia tc hng tm ca ngi

?34. Mt da trn bn knh 15cm, quay u quanh mt trc i qua tm da mi vng mt

0,1s. Tnh tc gc, tc di, gia tc hng tm ca da trn.35. Mt bnh xe bn knh 60cm quay u 100 vng trong 2s.Tm chu k, tn s, tc

gc v tc di ca mt im trn vnh bnh xe?36. Mt con tu v tr chuyn ng trn u quanh tri t, mi vng mt 90pht. Con

tu bay cao h = 320m cch mt t. Bit bn knh tri t l 6400km. Tnh tc dica con tu v tr?

37. Vnh ngoi ca mt bnh xe t c bn knh l 25cm. Tnh tc gc v gia tchng tm ca mt im trn vnh ngoi ca bnh xe khi t ang chy vi tc di khngi l 36km/h.

38. Chiu di kim pht ca mt ng h gp 1,5 ln kim gi ca n, chiu di kim giygp 4/3 ln kim pht. Hy so snh tc gc, tc di ca u kim pht vi u kim gi,gia ukim giy vi u kim gi?

Cng thc cng vn tc

39. Trn mt on tu ang chy vi vn tc 10m/s, mt ngi i t u toa xung cui

toa vi vn tc 2m/s. Tnh vn tc ca ngi i vi mt t?40. Mt ca-n chuyn ng thng trn dng nc, vn tcca ca-n vi dng nc l

30km/h. Ca-n xui dng t A n B mt 2 gi v ngc dng t B v A mt 3 gi. Tm:a) Khong cch AB?

b) Vn tc ca dng nc so vi b?41. Hai bn sng A v B cch nhau 6km. Mt thuyn chuyn ng thng xui dng t A

n Bri ngc dng quay tr li A. Vn tc ca thuyn i vi dng nc l 5km/h, vntc ca dng nc i vi b l 1km/h. Tnh thi gian chuyn ng ca thuyn?

42. Mt chic ph xui dng t A n B mt 6 gi. Nu ph tt my tri theo dngnc th thi gian ph tri t A n B l bao nhiu?

43. Mt t chy u trn mt ng thng vi vn tc 40km/h. Mt t B ui theo t A vi vn tc 60km/h. Xc nh vn tc ca t B i vi t A.44. A ngi trn mt toa tu chuyn ng vi vn tc 15km/h ang ri ga. B ngi trn mt toatu khc chuyn ng vi vn tc 10km/h ang vo ga. Hai ng tu song song vi nhau.Tnh vn tc ca B i vi A.

_____________THE END CHAPTER 1______________


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Chng II. NG LC HC CHT IM

Bi 9: TNG HP V PHN TCH LC. IU KIN CN BNG CA CHTIM

I. Lc. Cn bng lc.- Lc l i lng vc t c trng cho tc dng ca vt ny ln vt khc m kt qu l gyra gia tc cho vt hoc lm cho vt bin dng.- Hai lc cn bng l hai lc cng tc dng ln mt vt, cng gi, cng ln v ngc chiu.- n v ca lc l Niutn (N).II. Tng hp lc.1. nh ngha.Tng hp lc l thay th cc lc tc dng ng thi vo cngmt vt bng mt lc c tc dng ging ht cc lc y.Lc thay thny gi l hp lc.2. Qui tc hnh bnh hnh.

Nu hai lc ng qui lm thnh hai cnh ca mt hnh bnhhnh, th ng cho ktim ng qui biu din hp lc ca chng.

= III. iu kin cn bng ca cht im.Mun cho mt cht im ng cn bng th hp lc ca cc lc tc dng ln n phi bngkhng.

= = 0IV. Phn tch lc.1. nh ngha.Phn tch lc l thay th mt lc bng hai hay nhiu lc c tc dng ginght nh lc .

Cc lc thay thgi l cc lc thnh phn.2. Phn tch mt lc thnh hai lc thnh phn trn hai phng chotrc.

Bi 10 : BA NH LUT NIUTNI. nh lut I Newton.1. nh lut I Newton.

Nu mt vt khng chu tc dng ca lc no hoc chu tc dng ca cc lc c hp lc bngkhng. Th vt ang ng yn s tip tc ng yn, ang chuyn ng s tip tc chuyn

ng thng u.2. Qun tnh.


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L tnh cht ca mi vt c xu hng bo ton vn tc ca n c v hng v ln.V d:- ang ngi trn xe chuyn ng thng u, xe r sang tri, tt c cc hnh khch u nghingsang phi theo hng chuyn ng c.- ang ngi trn xe chuyn ng thng u, xe t ngt hm phanh, tt c cc hnh khch

trn xe u b chi v pha trc.II. nh lut II Newton.1. nh lutGia tc ca mt vt cng hng vi lc tc dng ln vt. ln ca gia tc tlthun viln ca lc v tlnghch vi khi lng ca vt.

= hay = Trong trng hp vt chu nhiu lc tc dng , , , th l hp lc ca cc lc :

= 2. Khi lng v mc qun tnh.a) nh ngha.Khi lng l i lng c trng cho mc qun tnh ca vt.b) Tnh cht ca khi lng.+ Khi lng l mt i lng v hng, dng vkhng i i vi mi vt.+ Khi lng c tnh cht cng.3. Trng lc. Trng lng.

a) Trng lc.- Trng lc l lc ca Tri t tc dng vo vt, gy ra cho chng gia tc ri tdo. Trng

lc c k hiu l .- gn tri t trng lc c phngthng ng, chiu t trn xung. im t ca trng lctc dng ln vt gi l trng tm ca vt.b) Trng lng. ln ca trng lc tc dng ln mt vt gi l trng lng ca vt, k hiu l P. Trnglng ca vt c o bng lc k.

c) Cng thc ca trng lc. = III. nh lut III Newton.1. S tng tc gia cc vt.Khi mt vt tc dng ln vt khc mt lc th vt cng b vt kia tc dng ngc tr limt lc. Ta ni gia 2 vt c s tng tc.2. nh lut.Trong mi trng hp, khi vt A tc dng ln vt B mt lc, th vt B cng tc dng li vt

A mt lc. Hai lc ny c cng gi, cng ln nhng ngc chiu.=


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3. Lc v phn lc.Mt trong hai lc tng tc gia hai vt gi l lc tc dng cn lc kia gi l phn lc.c im ca lc v phn lc :+ Lc v phn lc lun lun xut hin (hoc mt i) ng thi.+ Lc v phn lc c cng gi, cng ln nhng ngc chiu. Hai lc c c im nh vy

gi l hai lc trc i.+ Lc v phn lc khng cn bngnhau v chng t vo hai vt khc nhau.

Bi 11: LC HP DN. NH LUT VN VT HP DNI. Lc hp dn.Mi vt trong v tr u ht nhau vi mt lc, gi l lc hp dn.Lc hp dn gia Mt Tri v cc hnh tinh gi cho cc hnh tinh chuyn ng quanh MtTri.Lc hp dn l lc tc dng t xa, qua khong khng gian gia cc vt.

II. nh lut vn vt hp dn.1. nh lut:Lc hp dn gia hai cht im bt k t l thun vi tch hai khi lng ca chng v t lnghch vi bnh phng khong cch gia chng.2. Hthc:

Fhd= . Trong :+ m1v m2 l khi lng ca hai cht im (kg)+ r l khong cch gia hai cht im (m)+ Fhd ln lc hp dn (N)+ G = 6,67.10-11(N.m2/kg2)hng s hp dn3. nh lut c p dng cho cc trng hp:+ Hai vt l hai cht im+ Hai vt ng cht hnh cu vi khong cch gia chng c tnh t tm vt ny n tmvt kia.III. Trng lc l trng hp ring ca lc hp dn.Trng lc tc dng ln mt vt l lc hp dn gia Tri t v vt .Trng lc t vo mt im c bit ca vt, gi l trng tm ca vt.

ln ca trng lc (trng lng): P = . (+)Gia tc ri tdo: g =

Nu gn mt t (h


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+ Lc n hi xut hin hai u ca l xo v tc dng vo vt tip xc (hay gn) vi l xo,lm n bin dng.+ Hng ca mi lc n hi mi u ca l xo ngc vi hng ca ngoi lc gy bindng.II. ln ca lc n hi ca l xo. nh lut Hc.

1. Gii hn n hi ca l xo.Mi l xo hay mi vt n hi c mt gii hn n hi nht nh.2. nh lut Hc (Hookes).Trong gii hn n hi, ln ca lc n hi ca l xo t l thun vi bin dng ca lxo.

Fh= k.|l|

Trong :+ Fdhl ln ca lc n hi (N)

+ l= ll0l bin dng ca l xo (m)+ k l cng hay h s n hi ca l xo (N/m)3. Ch .+ i vi dy cao su hay dy thp, lc n hi ch xut hin khi b ngoi lc ko dn. V thlc n hi trong trng hp ny gi l lc cng.+ i vi mt tip xc b bin dn khi b p vo nhau th lc n hi c phng vung gcvi mt tip xc.

Bi 13: LC MA ST

I. Lc ma st trt.1. Cch xc nh ln ca ma st trt.Mc lc k vo vt ri ko theo phng ngang cho vt trt gn nh thng u. Khi , lck ch ln ca lc ma st trt tc dng vo vt.2. c im ca ln ca ma st trt.+ Khng ph thuc vo din tch tip xc v tc ca vt.+ Tlvi ln ca p lc.+ Phthuc vo vt liu v tnh trng ca hai mt tip xc.

3. Hsma st trt.

t =

H s ma st trt tph thuc vo vt liu v tnh trng ca hai mt tip xc.4. Cng thc calc ma st trt.

Fmst= t.N

Trong :+ Fmst l ln lc ma st trt.

+ N l p lc vt ln mt tip xc.+ tl h s ma st trt, khng c n v.


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Bi 14: LC HNG TMI. Lc hng tm.1. nh ngha.Lc (hay hp lc ca cc lc) tc dng vo mt vt chuyn ng trn u v gy ra cho vtgia tc hng tm gi l lc hng tm.

2. Cng thc.

Fht= maht= . = m2rTrong :+ Fht l lc hng tm (N)+ m l khi lng ca vt (kg)+ aht l gia tc hng tm (m/s2)+ v l tc di ca vt chuyn ng trn u (m/s)+ r l bn knh qu o trn (m)+ l tc gc ca vt chuyn ng trn u (rad/s)3. V d.+ Lc hp dn gia Tri t v v tinh nhn to ng vai tr lc hng tm, gi cho v tinhnhn to chuyn ng trn u quanh Tri t.+ t mt vt trn bn quay, lc ma st ngh ng vai tr lc hng tm gi cho vt chuynng trn.+ ng t v ng st nhng on cong phi lm nghin v pha tm cong hp lcgia trng lc v phn lc ca mt ng to ralc hng tm gi cho xe, tu chuyn ngd dng trn qu o.

Bi 15: BI TON VCHUYN NG NM NGANGI. Kho st chuyn ng ca vt nm ngang.1. Chn h trc to v gc thi gian.+ Chn h trc to -cc xOy, trc Ox hng theo vc t vn tc , trc Oy hng theovc t trng lc + Chn gc thi gian lc bt u nm.

2. Phn tch chuyn ng nm ngang.Chuyn ng ca cc hnh chiu Mxv Mytrn cc trc Ox v Oy gi l cc chuyn ngthnh phn ca vt M.+ Trn trc Ox ta c : ax= 0 ; vx= vo; x = vot

+ Trn trc Oy ta c : ay= g ; vy= gt ; y =12gt

2

II. Xc nh chuyn ng ca vt.1. Dng ca quo v vn tc ca vt.

+ Phng trnh quo: y =g

2v0 .


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+ Vn tc ca vt nm ngang thi im t bt k: v= (gt)2v022. Thi gian chuyn ng.

t = 2hg 3. Tm nm xa.

L = xmax= vot = vo2hg

DNG 1: TNG HP CC LC TC DNG LN VTHng gii quyt:+ Nu 2 lc cng phng, cng chiu th lc tng hp: F = F1+ F2vc chiu cng

chiu vi 2 lc.+ Nu 2 lc cng phng, ngc chiu th lc tng hp: 1 2F F - F v c chiu cng

chiu vi lc c ln ln hn.

Nu 2 lc khng cng phng th lc tng hp: = 2 v cchiu theo quy tc hnh bnh hnh.

BI TP C HNG DNBi 1:Mt vt c trng lng 60N c treo vo vng nhn O

(coi l cht im). Vng nhn c gi yn bng dy OA vOB. Bit OA nm ngang hp vi OB gc 1350. Tm lc cngca dy OA v OB.

GiiBiu din cc lc tc dng vo vt c:Gc l gc gia OA v OB: = 450.Thnh vc: TOA= P.tan = 60 N

Tng t: TOB= P.sin= 602NBi 2:Cho F1= F2= 30 N, = 60

0

. Hp lc ca , l bao nhiu? V hp lc.Giip dngcng thc: = 2 = 303NHnh vhp lc.

= 450


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DNG 2: P DNG 3 NH LUT NIU-TNHng gii quyt:

+ nh lut II Niu-tn: = = + nh lut III Niu-Tn: = -

BITP C HNG DNBi 1: Mt t c khi lng 1 tn ang chuyn ng vi v = 54km/h th hm phanh, chuynng chm dn u. Bit lc hm 3000N.a) Xc nh qung ng xe i c cho n khi dng li.

b) Xc nh thi gian chuyn ng cho n khi dng li.Gii

Chn chiu (+) l chiu chuyn ng, gc thi gian lc bt u hm phanh.

Theo nh lut II Newton: = .Chiu ln phng chuyn ng: - Fh= m.a a = = - 3 (m/s2)p dng hthc: v2- = 2a.S S = 37,5m.

b. Ta c: v = v0+at t = 5sVy thi gian chuyn ng cho n khi dng hn l: t = 5s

Bi 2:Mt qu bng m = 0,4kg ang nm yn trn mt t. Mt cu th d bng vi lc300N. Thi gian chn tc dng vo qu bng l 0,015s. Tnh tc ca qu bng lc bay i.

Gii

Theo nh lut II Newton c: a =

= 750 m/s

2

Vn tc ca qu bng lc bay i: v = at = 11,25 m/s

Bi 3:Cho vin bi A chuyn ng ti va chm vo bi B ang ng yn, v A= 20m/s sau vachm bi A tip tc chuyn ng theo phng c vi v = 10m/s, thi gian xy ra va chm l0,4s. Tnh gia tc ca 2 vin bi, bit mA= 200g, mB= 100g.

Gii

Gia tc ca bi A: aA=

= - 2,5 (m/s2)

Theo nh lut III Niu-tn c: FAB= FBAaB= =

.||

= 5m/s2

Vy gia tc ca vin bi B: aB= 5m/s2

= 600


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Bi 4: Mt vt ang ng yn, c truyn 1 lc F th sau 5s vt ny tng v = 2m/s. Nu ginguyn hng ca lc m tng gp 2 ln ln lc F vo vt th sau 8s, vn tc ca vt l

bao nhiu?Gii

Gia tc ca vt: a1=

= 0,4 (m/s2) F1= ma1= 0,4m (N)

Khi tng F= 2.F1= 0,8m a2= 0,8m/s2Vn tc ca vt sau 8s: v2= 6,4m/s

Bi 5:Lc F1tc dng ln vin bi trong khong t = 0,5s lm thay i vn tc ca vin bi t0 n 5 cm/s. Tip theo tc dng lc F2= 2.F1ln vin bi trong khong t =1,5s th vn tcti thi im cui ca vin bi l? ( bit lc tc dng cng phng chuyn ng).

Gii

Gia tc ca vt khi chu tc dng ca F1: a1=

= 0,1 (m/s2) F1= ma1= 0,1m (N)

Khi tng F

= 2.F1= 0,2m a2= 0,2m/s2

Vn tc ca vt sau 1,5s: v2= 0,3 m/s

Bi 6:Mt t c khi lng 500kg ang chuynng thng u th hm phanh chuyn ngchm dn u trong 2s cui cng i c 1,8 m. Hi lc hm phanh tc dung ln t c ln l bao nhiu?

Giip dng hthc:v2- = 2. = 3,6(1)Ta li c: v = v0+ a.t v0= - a.t = - 2.a (2)

T (1) v (2) ta c: a = - 0,9 m/s2Fh= m.a = - 450N.

Vy lc hm c ln l 450NBi 7:Lc F truyn cho vt khi lng m1th vt c gia tc a1= 2m/s2, truyn cho vt khilng m2th vt c a2= 3m/s2. Hi lc F struyn cho vt c khi lng m3= m1+ m2thvt c gia tc l bao nhiu?

Gii

p dng nh lut II Newton ta c: m1=

, m2=

Gia tcca h hai vt m1+ m2l: a3= +=

+

= .+= 1,2m/s

2


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DNG 3: VN DNG CNG THC TNH LC HP DN V GIA TC TRNGTRNG.

Hng gii quyt:

- Lc hp dn: Fhd= . - Trng lng ca vt khi lng m khi vt trn mt t: P = .

. = mg0

- Trng lng ca vt khi lng m khi vt cao h so vi mt t: P = . .(+)=mg

- Gia tc ri tdo ca vt khi vt mt t: g0= . - Gia tc ri t do ca vt khi vt cao h so vi mt t: g = . (+)= g0

(+)

BI TP C HNG DN

Bi 1:Tnh gia tc ri t do ca mt vt cao h = 5R ( R = 6400km), bit gia tc ri tdo ti mt t l 9,8m/s2.

Gii

Gia tc mt t: g0= . = 9,8 m/s2Gia tc cao h: g = . (+)= 0,27 m/s2

Bi 2:Mt vt c m = 10kg khi t mt y c trng lng l 100N. Khi t ni cchmt t 3R th n c trng lng l bao nhiu?

Gii

mt t: P = F = . . cao h: Ph= . .(+)= = 6,25 N

Bi 3:Nu khi lng ca 2 vt u tng gp i lc hp dn gia chng khng i thkhong cch gia chng phi l bao nhiu?

Gii

Lc hp dn ban u: F1=

..

Lc hp dn khi tng khi lng m1v m2gp i: F2= . . Khi F1= F2

=

r2= 2r1

Bi 4:Tm gia tc ri t do ca mt vt cao bng na bn knh T. Cho bit gia tc rit do trn b mt t l 9,81m/s2.

Gii

Gia tc mt t: g0= .

= 9,81 m/s2

Gia tc cao h: g = . (+)= . = 4,36 m/s2


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Bi 5:Gia tc ri t do trn b mt ca mt trng l 1,6m/s2v RMT= 1740km. Hi caono so vi mt trng th g = 1/9 gMT.

Gii

Gia tc mt trng: gT= . Gia tc cao h: g = .

(+)

=

(+) = 9 h = 3480 km

Vy cao h l: 3480kmBi 6:Mt vt c m = 20kg. Tnhtrng lng ca vt 4R so vi mt t, R = RT. Bit giatc trng trng trn bmt T l 10m/s2.

Gii

p dng hthc: Ph= m.gh= m.g0

(+)= 8N

DNG 4: VN DNG NH LUT HCHng gii quyt:p dng cng thc ca nh lut Hc: Fdh= k.|l|Trong :+l = |l - l0| bin dng ca l xo+ l l chiu di lc sau ca l xo+ l0l chiu di t nhin ( ban u)

+ k cng ca l xo.Khi l xo treo thng ng trng thi cn bng th: Fdh= P

BI TP C HNG DNBi 1:Mt l xo dn ra on 3cm khi treo vt c m = 60g, g = 10m/s2a) Tnh cng ca l xo.

b) Mun l = 5cm th ml bao nhiu?Gii

a) Khi cn bng:Fh= P k|l| = mg k = 20(N/m)b) Khi l = 5cm m =

|| = 0,1 kg

Bi 2:Mt l xo c l0= 40cm c treothng ng. Treo vo u di ca l xo mt qucn 500g th chiu di ca l xo l 45cm. Hi khi treo vt c m = 600g th chiu di lc saul bao nhiu? g = 10m/s2

Giiiu kin cn bng: Fh= P k|l| = mg k = 100(N/m)

Khi tng khi lng ln n m = 600g: Fh

= P k( - 0) = mg = 0,46 (m)


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Bi 3:Mt l xo c chiu di t nhin 20cm c treo thng ng. Treo vo u t do cal xo vt c m = 25g th chiu di ca l xo l 21cm, g = 10m/s 2. Nu treo thm vt c m =75g th chiu di ca l xo l bao nhiu?

GiiKhi treo vt 25g: k( - 0) = m1g k = 25 (N/m)

Khi treo thm 75g: k(' - 0) = mg ' = 0,24 (m)Bi 4:Mt l xo c chiu di t nhin l0, c treo vo im c nh O. Nu treo vo l xovt 100g th chiu di ca l xo l 31cm, treo thm vt m2= 200g th chiu di ca l xo l33cm. Tm cng v di t nhin ca l xo, g = 9,8m/s2, b qua khi lng l xo.

GiiKhi treo vt m1: k(1- 0) = m1g (1)Khi treo thm m2: k(2- 0) = (m1 + m2)g (2)

T (1) v (2) 0= 30cm k = 97 N/m

Bi 5:Treo vt c m = 200g vo mt l xo lm n dn ra 5cm, g = 10m/s2. Tm cng cal xo.

Giiiu kin cn bng: F = P k|l| = mg k = 40 (N/m)

DNG 5: VN DNG CNG THC TNH MA STV PHNG PHP NGLC HC.

Hng gii quyt:-Phn tch cc lc tc dng ln vt.-Biu din cc lc tc dng ln hnh v, chn h trc xOy thch hp+ Ox: trng vi hng chuyn ng.+ Oy: trng vi phng vung gc vi chuyn ng- Cng thc lc ma st: Fmst=t.N. Ch trong trng hp:+ Nm ngang: Fmst=t.m.g+ Nm trn mt phng nghing gc : Fmst=t.m.g.cos

-p dng phng trnh nh lut II: = (1)+ Chiu phng trnh (1) ln trc Ox: F1x+ F2x+ +Fnx= max(2)+ Chiu phng trnh (1) ln Oy: F1y+ F2y+ +Fny= may(3)Thng thng ax= a v ay= 0.T (2) v (3) suy ra i lng cn tm

BI TP C HNG DNBi 1:Mt t con chuyn ng thng u trn mt ng. H s ma st ln 0,023. Bit rngm = 1500kg, g = 10m/s2. Tnh lc ma st ln gia bnh xe v mt ng.

GiiTa c N = P


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Lc ma st lng gia bnh xe v nn ng: Fmst= t.N = 0,023150010 = 345N

Bi 2:Mt vt chuyn ng trt u trn mt phng nghing khi h s ma st l 3, g =10m/s2. Tm gc hp bi mt phng nghing vi phng ngang, m = 0,1kg. F = 10N.

GiiVhnh phn tch cc lc tc dng ln vt.

Hp lc tc dng ln vt: = Chiu ln Ox: - Fmst + Psin = 0 v chuyn ng thng

u nn a =0

sin =

(1)Khi chiu ln trc Oy: N - Pcos= 0 N = Pcos

cos =(2)

T(1) v (2) tan =

= t=

3= 54,70

Bi 3:Mt t c khi lng 3,6 tn bt u chuyn ng trn ng nm ngang vi lc koFk. Sau 20s vn tc ca xe l 12m/s. Bit lc ma st caxe vi mt ng bng 0,25Fk, g = 10m/s2. Tnh lc mast, lc ko.

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