Bồi dưỡng HSG VL 10

8/4/2019 Bi dng HSG VL 10

1/23

http://vatlylamdong.com/ L Quc Thnh - GV THPT Tnh Gia 2

Dy bi dng-Nm hc 2011 - 2012 1

Phn mt: ng hc cht im

A.tm tt l thuytI.Vn tc trung bnh

tsvtb

II.Vn tc tc thi

t

Svt

III.Gia tc

0

0

tt

vv

t

va t

. a cng hng vi v

-Trong chuyn ng bin i u th gia tc l mt hng s.

- ln: a=0

0

ttvvt

-Vn tc tc thi: vt=v0+at

-ng i S=v0t+2

1 at2.

-To x=x0+S=x0+v0t+2

1 at2.

-Lin h vn tc, gia tc v ng i:vt

2+v02=2aS

-Ri t do:Chn trc ox thng ng, chiu dng t trn xung. Gc to l v tr vtbt u ri.

vt=gt

x=h=2

1 gt2

vt2=2gh

B.Mt s bi ton th d v phng php giiBi ton 1.1Hai t chuyn ng u cng mt lc t A n B, AB=S. t th nht ina qung ng u vi vn tc v1, na qung ng sau vi vn tc v2. tth hai i vi vn tc v1trong na thi gian u v vi vn tc v2trong nathi gian cn li.a)Tnh vtbca mi t trn c qung ng.

b) Hi t no n B trc v n trc bao nhiu?c) Khi mt trong hai t n B th t cn li cch B mt khong baonhiu?Giia)

+ t 1:2

S =v1.t1t1=1

2vS .


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Dy bi dng-Nm hc 2011 - 2012 2

2

S =v2.t2 t2=2

2v

S

Thi gian i c qung ng l: t=t1+t2=21

21

2

)(

vv

vvS .

vtb1=21

212vvvv

tS

.

+ t 2:

vtb2=2

22 2121 vv

t

vt

vt

t

S

b)

+ t 1 i ht AB trong khong thi gian l: tA=21

21

2

)(

vv

vvS .

+ t 2 i ht AB trong khong thi gian l: tB=21

2vvS

.

tB-tA=)(2

)(

2121

2

21

vvvv

vvS


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Dy bi dng-Nm hc 2011 - 2012 3

Gii:Qung ng S c s o bng s o din tch ca hnh a gic gii hn bing biu din v, trc Ot, ng tung Ov v ng honh t=16. m cc trn th thdin tch a gic l 25 . Vy S=25.4=100m.

Hnh 1Bi ton 1.4:Mt ht c vn tc 18m/s v sau 2,4 s n c vn tc 30m/s theo chiu ngcli.a)Gia tc trung bnh ca ht trong khong thi gian 2,4s l bao nhiu?

b) V th v theo t v ch ra cch tm tc trung bnh trn th.Gii:a)

4,21830

12

12

ttvva =-20m/s

b)Biu thc v theo t c dng nh hnh 2.

v=v0+at=18-20t.v=0 lc t=0,9s.

Trn th biu din v theo t th qung ng S1vt i dc t 0 n 0,9s cgi tr bng din tch hnh tam gic OAB v qung ng S 2vt i c t0,9s n 2,4s-bng din tch hnh tam gic BCD.

S1= 21

(OAxOB)=0,5(18.0,9)=8,1mS2=0,5(DCxBD)=0,5[30(2,4-0,9)]=22,5m.Qung ng i c t 0 n 2,4s lS=S1+S2=8,1+22,5=30,6m.

Tc trung bnh l: vtb=4,2

6,30

t

S=12,75m/s.

v(m/s)

8

4

t

0 2 4 6 8 10 12 14 16

v(m/s)

18 A

0.9 2,40 B D t(s)

-30 C

Hnh 2


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Dy bi dng-Nm hc 2011 - 2012 4

Bi ton 1.5:Mt vt c gia tc khng i l +3,2m/s2. Ti mt thi im no vn tcca n l +9,6m/s. Hi vn tc ca n ti thi im:a)Sm hn thi im trn l 2,5s.

b)Mun hn thi im trn 2,5sl bao nhiu?Gii:

a) v=v0+at=v0+3,2t9,6 =v0+3,2t (1)v- =v0+ 3,2(t-2,5) (2)Tr v vi v ca (2) cho (1) ta c: v-=9,6-3,2.2.5=1,6m /s.

b) v+=v0+3,2(t+2,5) (3).Tr v vi vca (3) cho (1) ta c: v+=9,6+3,2.2,5=17,6m/s.

Bi ton 1.6:Mt ngi ng sn ga nhn on tu chuyn bnh nhanh dn u. Toa (1)i qua trc mt ngi y trong t(s). Hi toa th n i qua trc mt ngi ytrong bao lu?p dng bng s:t=6, n=7.

Gii:Gi chiu di mi toa tu l l. Theo bi ra ta c:

l =2

1 at2 (1)

nl =2

1 at2 (2) vi tl thi gian on tu i ht qua trc mt ngi y.

T (1) v (2) suy ra t=t n . (3)

Tng t: (n-1)l=2

1 at2 (4) vi tl thi gian (n-1) toa tu i ht qua trc

mt ngi y.Do , thi gian toa th n i qua l: t=t-t=( )1 nn t.Bi ton 1.7.( thi chuyn LS)

Cu 1(2,5 im): Mt ngi ng ti im M cch mt con ng thng mtkhong h=50m ch t; khi thy t cn cch mnh mt khong a= 200mth ngi y bt uchy ra ng gp t (hnh 1). Bit t chy vi vntc v1= 36km/gi. Hi:

a)Ngi y phi chy theo hng no gp ng t? Bit rng ngichy vi vn tc v2=10,8 km/gi.

b)Ngi phi chy vi vn tc nh nht bng bao nhiu c th gpc t?

M

h

H

a

Hnh 1M

h

H

a

Hnh 1

A B


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Dy bi dng-Nm hc 2011 - 2012 5

Cu1 (2,0 im)

a)

Mun gp ng t ti B th thi gianngi chy t M ti B phi bng thi gian t

chy t A ti B:12 v

AB

v

MB . (1)...................................................................(0,5im)

Trong tam gic AMB c: sinsin

ABMB . (2)

Vi sina

h . T (1) v (2) ta rt ra

2

1.sinv

v

a

h =0,833 =56030 hoc

=123030.....................................................................................................(0,5im)

b) c th gp c t th phi c12 v

AB

v

MB .........................................(0,5im)

v2min= .a

hv1=2,5m/s...................................................................................(0,5im)

II - Cc bi ton v chuyn ng tng i

Bi 2.1 (4.16*-GTVL10T1)Hai chic tu chuyn ng vi cng vn tc u v hng n O theo qu o

l nhng ng thng hp vi nhau gc =600. Xc nh khong cch nhnht gia cc tu. Cho bit ban u chng cch O nhng khong l1=20km vl2=30 km.GiiGi s khong cch nh nht gia 2 tu khi chng i c thi gian l t.Vy AO=20-vt, BO = 30vt, y2= AO2+BO2-2AO.BO.cos60Hm y2t cc tiu ti (-b/a ; - /a). Vy (y2)Min=75 hay yMin=5 3 (km)Bi 2.2 (4.20*GTVL10T1)Hai tu A v B ban u cch nhau mt khong l.

Chng chuyn ng thng u cng mt lc vi ccvn tc c ln ln lt l v1 v v2.Tu A chuyn ng theo hng AC to vi AB mtgc nh hnh v.a)Hi tu B phi i theo hng no c th gpc tu A. Sau bao lu k t lc chng cc v trA v B th 2 tu gp nhau?

b)Mun 2 tu gp nhau H (xem hnh)th cc lnvn tc v1 v v2phi tho mn iu kin g?

Gii

A

1v

l

H B

C


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Dy bi dng-Nm hc 2011 - 2012 6

a) gp c tu A th tu B phi i theo hnghp vi AB mt gc nh hnh v: =(

2v ,

AB

).Gi s 2 tu gp nhau C. Gi t l thi gian 2 tu

i gp nhau.Theo nh l hm s sin ta c:

sinsinsinsin 2

112

v

vtvtv

Theo nh l hm s cos ta c:AC2=BC2+AB2-2BC.AB.cos

v BC2=AC2+AB2-2AC.AB.cos Tc l v1

2t2=v22t2+l2-2.v2.t.l.cos (1)

v v22t2=v1

2t2+l2-2.v1.t.l.cos (2)

T (1) v (2) ta c t= coscos 21 vv

l

.

b) 2 tu gp nhau ti H tc l tan=1

2

v

v

HA

HB

III- Cng thc cng vn tc

Bi 3.1Mt ngi mun cho thuyn qua sng c dng

nc chy. Nu ngi y cho thuyn theo hng tv tr A sang v tr B (AB vi dng sng, hnh3.1)th sau thi gian t1=10min thuyn s ti v tr C cchB mt khong s=120m. Nu ngi y cho thuynv hng ngc dng th sau thi gian t2=12,5 minthuyn s ti ng v tr B. Coi vn tc ca thuyni vi dng nc khng i. Tnh:a) B rng l ca con sng.

b) Vn tc v ca thuyn i vi dng nc.

c) Vn tc u ca dng nc i vi b.d) Gc Gii:

- Thuyn tham giang thi 2 chuyn ng: chuyn ng cng vi dng ncc vi vn tc u v

A

1v

l

H B

2v

C

B C

M

A

Hnh 3.1

B s C

v

V

u

A

Hnh 3.1.a

B

V

v

u

AHnh 3.1.b


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Dy bi dng-Nm hc 2011 - 2012 8

2) Khi xe chuyn hng m gi khng chuyn hng th 'xdv

gdv , vi '

xdv l

vn tc mi ca xe i vi t. Ta cng c 'dxv

gdv

. Theo bi ra gxv ' gi

nguyn hng c, ngha l gxv ' hp vi gdv

mt gc 450nh hnh trn y.

Theo hnh ny ta c:gx

v ' =

gd

v + '

dx

v ; t suy ra v

gx=v

gd2 =80km/h v

vdx=vxd=vgd=40 2 km/h: xe chy vi tc 40 2 km/h v ngi li xe cmthy gi coa vn tc 80km/h.

IV - Chuyn ng ri t doIV.I-Tnh thi gian ri, qung ng ri v vn tc riPhng php

- Thng chn chiu dng hng xung

-

p dng cc cng thc:s=2

1gt2 ; v=gt ; v2=2gs

Bi tp 4.1.1. Mt vt c bung ri t do ti ni c g=9,8m/s2.a) Tnh qung ng vt ri c trong 3 s v trong giy th 3.b)Lp biu thc qung ng vt ri trong n giy v trong giy th n.

Gii:a)

b)Qung ng vt ri trong n giy v trong giy th n:

sn= 21

gn2

= 2

2n

g; sn-1= 21

g(n-1)2

Suy ra sn=sn-sn-1=

2

g [n2-(n-1)2]=2

)12( n g.

Bi tp 4.1.2Mt vt ri t do ti ni c g=10m/s2. Thi gian ri l 10s. Hytnh:

a) Thi gian ri mt mt u tin.b)Thi gian ri mt mt cui cng

Gii:

a) Qung ng ri trong thi gian t: s= 21

gt

2

. Suy ra s1=1m th t1= g2

=0,45s.b) Thi gian ri (s-1) mt cui cng l:

s=s-1=2

1 gt2g

st

)1(2'

Thi gian ri mt cui cng:

t=t-t=10-5

1102 =0,01s.

Bi tp 4.1.3:Vt A t trn mt phng nghing ca mt ci nm nh hnhv. Hi phi truyn cho nm mt gia tc bao nhiu theo phng nm ngang vt A ri xung di theo phng thng ng?


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Dy bi dng-Nm hc 2011 - 2012 9

Bi tp 4.1.4. Mt bn cu c bn knh R trt u theo mt ng nmngang. Mt qu cu nh cch mt phng ngang mt khong bng R. Ngay khinh bn cu i qua qu cu nh th n c bung ri t do.Tm vn tc nh nht ca bn cu n khng cn tr chuyn ng ri t do

ca qu cu nh. Cho R=40cm.GiiBi 4.1.3.Trong khong thi gian t

nm di: s=2

1 at2.

Khong trng to ra pha di vt: h=s.tan .Qung ng ri ca vttrong khong thi gian t

l: s=2

1 gt2.

Ta phi c: h > s suy ratan

ga

Bi 4.1.4Gi v l vn tc trt ca bn cuQung dng dch chuyn ca bn cu trong thi gian t l : s1= vt.

Trong thi gian , vt ri dc l: s2=2

1 gt2.

qu cu khng b vng vo bn cu th: s1> s2hay s1> 22 OBOA

s21>OA2-OB2 (1)

Vi OA=R, OB=OA-AB=(R-s2)(1) s21> R

2-(R-s2)2

s21> 2Rs2-s22

s12+s2

2-2Rs2>0 (s1

2-2Rs2)+s12> 0 (2)

(2) lun ng ta phi c (s12

-2Rs2)> 0 s1

2> 2Rs2

v2t2 > 2R2

1gt2

v Rg .Vy, vt ri t do m khng b cn tr bi bn cu th vn tc nh nht ca

bn cu l vmin= Rg

IV.2.Lin h gia qung ng,thi gian, vn tc ca 2 vt ri t do

Phng php-p dng cc cng thc v s ri t do cho mi vt v suy ra s lin h v ilng cn xc nh.

ha

AS2

B C

R

O


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Dy bi dng-Nm hc 2011 - 2012 10

Nu gc thi gian khng trng vi lc bung vt, phng trnh qung ng

ri l: s=2

1 (t-t0)2

-C th coi mt vt l h quy chiu v nghin cu cu chuyn ng tng i

ca vt kia.Ta lun c: 021 gga .

Hai vt ri t do lun chuyn ng thng u i vi nhau.Bi tp 4.2.1Hai git nc ri t cng mt v tr, git np sau git kia o,5s.a)Tnh khong cch gia 2 git nc sau khi git trc ri c0,5s, 1s, 1,5s.Hai git nc ri ti t cch nhau mt khong thi gian bao nhiu?(g=10m/s2)GiiChn gc thi gian lc git th nht ri.

Cc qung ng ri: s1= 21

gt2

; s2= 21

g(t-0,5)2

.

a) Khong cch d=s1-s2=4

g (2t-0,5).

b)Thi gian ri bng nhau nn thi dim chm t cch nhau 0,5s.IV.3Chuyn ng ca vtc nm thng ng hng xungPhng php- Chuyn ng c: *gia tc: ga

*vn tcu:0v cng hng vi a

Chuyn ng nhanh dn u.Phng trnh:

s =2

1gt2 + v0t

( Chiu dng hng xung )Ni dung bi ton c gii quyt bng cch*Thit lp cc phng trnh v thc hin tnh ton theo bi.* Xt chuyn ng tng i nu c nhiu vt chuyn ng4.3.1.mt tngthp cch mt t 45m, mt ngi th ri mt vt. Mt giysau, ngi nm vt th hai xung theo hng thng ng. Hai vt chm

t cng lc. Tnh vn tc nm vt th hai (g = 10m/s2).GiiTa c cc phng trnh chuyn ng:

S1=2

1 gt2 =5t2 (1)

S2=2

1g(t-1)2+v02(t-1) (2)

Vi S1=45m suy ra t=g

S1

2=3s.

V S1=S2nn ta dc v02=12,5m/s.Bi tp 4.3.2


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Dy bi dng-Nm hc 2011 - 2012 11

Phi nm mt vt theo phng thng ng t cao h=40m vi vn tc v 0bng bao nhiu n ri ti mt t:

a) Trc 1s so vi trnghp ri t do.b) Sau 1s so vi trng hp rt t do.

Ly g=10m/s

2

.GiiChn trc to Ox hng xung diCc phng trnh ng i:

S=2

1 gt2(ri t do) (1)

S=2

1 gt2 +v0t (2)

a) Theo bi ra S=S=h suyra t0: phi nm hng xung.

Khi chm t t= gh2

= 8 . Vi t-t=1, Thay vo (2) ta c v0=12,7m.c) t>t nn v0


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Dy bi dng-Nm hc 2011 - 2012 12

p dng phng trnh ng i ca chuyn ng bin i u ta suy ra thi

gian ri ca mi vt u bng t=g

R4 .

V - Cc nh lut Niu-tn v cc lc c hc

V.1 Lc n hi, nh lut I Niu tnV.1.1 Tnh cng ca l xoMt l xo khi lng nh khng ng k, c treo vo im c nh O c di t nhin OA =l0. Treo mt vt khi lng m1=100g vo l xo th dnca n l l1=31cm. Treo thm mt vt khi lng m2=100g th di ca nl l2=32cm. Tnh cng K v di t nhin l0ca l xo. Ly g=10m/s

2.

V.1.2 Hai l xo mc song songI-Hai l xo khi lng khng ng k, cng ln lt l k1= 100N/m vk2=150N/m c cng di t nhin l0=20cm, c treo thng ng nh hnhv. u di 2 l xo ni vi mt vt khi lng m=1kg. Ly g=10m/s 2. Tnhchiu di cc l xo khi vt cn bng.II- Hai l xo L1 v L2c cng di t nhin. Khi treo vt nng m vo l xol1 th n dn ra l1=1cm v treo vt nng y vo L2 th n dn ra l2=2cm.

Ni 2 l xo bng c 2 u chng lun c cng di ri treo vt nng m

ni trn vo th 2 l xo cng dn ra

l bng bao nhiu?V.1.3 Vt nm gia 2 l xoHai l xo L1 v L2 cng ln lt l k1 v k2 cmc vo mt qu cu khi lng m=50g (xem hnh).

Cho bit t s2

3

2

1

k

k v 2 l xo u trng thi t

nhin. Nu dng mt lc 5N thf c th y qu cu theo phng ngang i mton 1cm. Tnh cng k1 v k2ca 2 l xo.

V.1.4 Tm cng ca l xotng ng h l xo ghp.I - H 2 l xo c ghp theomt trong 2 cch sau. Tm cng ca l xo tng ng.II - Mt h thng gm 2 l xo c gn vovt, c c nh mt u nh hnh v.Tm cng ca l xo tng ng, t suy ra trng hp tng qut cho h l xo

mc ni tip v ghp song song.V.1.5 Mt l xo nh c treo thng ng,

L1 L2

k1 k2k1 k2

k1 k1m

k2 k2m


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Dy bi dng-Nm hc 2011 - 2012 13

cng k v di t nhin l0.1)Tnh cng k ca na l xo y (l0/2).2)Treo 2 vt nng cng khi lng m vo im cui B v im chnh gia Cca l xo th chiu di l ca l xo l bao nhiu?

V.1.6Chng minh rng cng ca l xo t l nghch vi chiu di ca nL xo c cu to ng u, c di t nhin l0v h s n hi k0. Khi chutc dng ca mt lc F thdn ra mt on l0. Mi n v chiu di ca n

dn ra mt on0

0

l

l . Ta c: k0 l0=F

Mt on l xo y c chiu di l1th khi y b dn mt dn l1=l10

0

l

l .

Do ta c k1 l1 =k1l10

0

l

l =F.

Tng t cho on l xo c chiu di l2:

k2 l2=k2l20

0

l

l = F.

So snh cc ng thc trn ta c:

k0 l0= k1l10

0

l

l =k2l20

0

l

l

Suy ra k0l0=k1l1=k2l2.

Nh vy ta chng minh c l1

2

2

1

l

l

k

k

V.2 - Lc ma st. nh lut II Niu tnV.2.1 Chuyn ng ca vt trn mt phng nghingMt vt c t trn mt mt phng nghing hp vi mt phng nm ngangmt gc =40. Hi:

a) Gii hn ca h s ma st gia vt v mt phng nghing vt c thtrt xung c trn mt phng nghing .

b)Nu h s ma st bng 0,03 th gia tc ca vt bng bao nhiu? Khi mun vt trt ht qung ng s=100m, vt phi mt thi gian bao

lu?c) Trong iu kin cu hi (b), vn tc ca vt cui qung ng 100ml bao nhiu?

Gii tm tt:a) F=mgsin-kPcos>0 hay k


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Dy bi dng-Nm hc 2011 - 2012 14

V.2.2- Chuyn ng ca vt khi phngca lc khng trng phng chuyn ngI - Vt khi lng m=1kg c ko chuynng ngang bi lc F

hp gc =300 vi

phng ngang, ln F=2N. Bit sau khichuyn ng c 2s, vt i c qungng 1,66m. Cho g=10m/s2.a)Tnh h s ma st trt k gia vt v sn.

b)Tnh li k nu vi lc F

ni trn, vt chuyn ng thng u.Lc gii

a) Fcos-k(mg-Fsin)=ma (1)

Trong a=2

2

t

s =0,83 (m/s2).

T (1) suy ra k=0,1b) Gia tc chuyn ng a=0 ta c k=0,19.

II - Cho h nh hnh v: m1=1kg, m2=2kg,k1=k2=0,1, F=6N, =30

0, g=10m/s2.Tnh gia tc chuyn ng v lc cng ca dy.

VI - CHUYN NG TRN U- L XO

VI.1(H C 99-00)

1)Mt l xo R c di t nhin l0=24,3 cm v cng k=100N/m, c u Ogn vi mt thanh cng, nm ngang T(Xem hnh v), u kia c gn mt vtnh A, khi lngm=100g. Thanh Txuyn qua tm vt A, vA c th trt khng mast theo T. Cho bit giatc ri t do lg=10m/s2. Cho thanh T

quay u quanh trcthng ng Oy, vi vntc gc =10rad/s. Tnh di ca R. Xc nh phng, chiu v cng ca lc do R tc dng vo im O.2)Gn thm vo A mt l xo R ging ht R, v cng mang vt B, ging nhvt A. Cho h quay quanh Oy cng vi vn tc =10rad/s. Tnh di ca lxo R, R v lc tc dng vo O.VI.2 (H KTQD 99-00)Mt a phng trn c bn knh R=10cm, nm ngang, quay u quanh mt

trc thng ng i qua tm ca a.1)Nu c mi giy a quay c 1,5 vng th vn tc di ca mt im mp a l bao nhiu?

y

R AO T

y

R A R BO T

y

N

F

msF

x

P

m2 m1 F


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Dy bi dng-Nm hc 2011 - 2012 15

2)Trn mt a c t mt vt kch thc nh, h s ma st gia vt v a lk=0,1. Hi vi nhng gi tr no ca vn tc gc ca a th vt t trn ad v tr no cng khng b trt ra pha ngoi a. Cho gia tc trng trngl g=10m/s2.

3)Treo mt con lc n vo u thanh AB cm thng ng trn mt a, uB cm vo a ti im cch tm quay R/2. ChoAB=2R.a) CMR khi a quay u th phng dy treo hpvi phng thng ng mt gc nm trong mt

phng cha AB v trc quay.b)Bit chiu di con lc l l=R, tm vn tc gc ca a quay =300.VI.3 (H Dc HN-99-00)1)Mt qu cu khi lng m c gn vo u camt si dy, m u kia ca dy c buc vo umt thanh thng ng t c nh trn mt bn quaynm ngang.Bn s quay vi vn tc gcbng baonhiu, nu dy to vi phng vung gc ca mt

bn mt gc=450 Bit rng dy l = 6cm vkhong cch t thanh thng ng n trc quay lr=10cm.2)Mt qu cu khi lng m, treo trn mt dy di l.Qu cu quay u trn mt vng trn nm ngang (con

lc cnic). Dy to mt gc vi phng thngng. Hy tnh thi gian qu cu quay c mt vng.VI.4 (H HH-HP 99-00)V tinh a tnh dng trong thng tin lin lc l v tinh ng yn so vi mtt v trong mt phng xch o. Bit bn knh Qu t R=6370 km, khilng qu t M=6.1024kg, hng s hp dn G=6,67.10-11(N.m2/kg2)a)Tnh cao ca v tinh so vi mt t.

b)Tnh vn tc di ca vtinh trn qu o ca n i vi h quy chiu l tmQu t.

c)Gi s ng thng ni v tinh v tm Qu t i qua kinh s 0. Hinhng vng no nm trn xch o trong khong kinh no nhn c tnhiu ca v tinh nu v tinh pht sng cc ngn (Cho cos810200,15055).

Gii:Bi 81)Goi l l di l xo th lc hng tm tc dng vo vt A trong chuynng trn l: fht=maht=m

2lLc ny chnh bng lc n hi ca l xo: fh=k l=k(l-l0)Do , m2l=k(l-l0). Thay s vo tm c l=27 cm.Lc f do l xo R tc dng vo imO cng chnh l lc hng tm fht(hoclc fh) trn v hng t O n vt A.

A

mB

h l

m


16/23


Dy bi dng-Nm hc 2011 - 2012 16

2)K hiu l1l di ca l xo R, l2l c di ca l xo R, lp lun tng tnh cu 1 ta c 2 phng trnh:Xt vt B: m(l1+l2)

2=k(l2-l0) (1)i vi vt A: ml1

2=k(l1-l0)-k(l2-l0)=k(l1-l2) (2)

Thay s v gii (1) v (2) ta c l1=34,2 cm, v l2= 30,8 cm.Lc tc dng vo O chnh l lc dn hi ca l xo R:f= k(l1-l0) =9,9 (N).

Bi 91) =2 n 3rad/s

v=R=30 cm/s2)Lc ma st ngh c tr s ln nht fms=kmg

Lc qun tnh li tm tc dng ln vt t trnmt a c gi tr ln nht khi vt mp a: Fltmax=m2R

vt khng trt ra khi a phi c Flt max fms m

2R kmg

2 10

R

kg

3)a)Vt m chu tc dng ca trnh lc P, lc cng T,lc li tm (c gi i qua trc quay). Mun m nm cn bng (xt trong h quychiu gn vi a) th cc lc P, Flt, T phi ng phng, ngha l M nm trong

mt phng cha trc quay v thanh AB, khi dy treo AM hp vi phngthng ng AB mt gc .

b)T hnh v ta c: Flt=Ptg , suy ra m2 (OB+lsin )=mgtg

3R

g 7,6rad/s.

Bi 10

1)p dng nh lut II Newton, amPT

Hoc da vo hnh v cc vect lc: m2R=P.tg=mgtg , vi R=lsin+r

T ,

sinlr

gtg

=8,3 rad/s2)Lp lun v tnh ton nh cu 1, ta c

m2(lsin)=P.tg=mgtg , 2=cosl

g

Thi gian qu cu quay c mt vng l T=g

hl

g

2cos

22

Cng thc ny tng ng vi cng thc biu din chu k ca con lc nc cng chiu di h.

Bi 11

A

T

ltF

B P


17/23


Dy bi dng-Nm hc 2011 - 2012 17

Mun mt v tinh trong mt phng xch o v ng yn so vi mt t, nphi chuyn ng trnxung quang Qu t cng chiu v cng vn tc gc nh Tri t quay xung quanh trc ca n vi chu k T=24h.Gi vn tc di ca v tinh trn qu o l v, cao ca n so vi mt t l

h. V chuyn ng trn nn v tinh c gia tc hng tm bng:Fht=

)(

2

Rh

mv

, lc ny chnh l lc hp dn ca Tri t i vi v tinh

Fhd= 2)( RhGmM

. T hai biu thc trn suy ra

)(

2

Rh

mv

=

2)( Rh

GmM

V v=(h+R)22

22

)()(

)(

Rh

GM

Rh

Rh

. Ch rng =T

2 , vi T=24h ta c

h+R= 32

2

32 4

.

TGMGM =42322.103(m)=42322km

Vy, cao ca vtinh so vi mt t l h=42322-6370=35952 kmb)Ta c:

v=(h+R)=T

Rh )(2 =3,1.103m/s hay v=3,1

km/sc)i vi sng cc ngn, ta c th xem nh sngtruyn thng t v tinh xung mt t. T hnhv ta thy vng nm gia kinh tuyn i qua A vB s nhn c tn hiu t v tinh. Ta thy ngay:

cos=hR

R

=0,1505. T =81020.

Nh vy, vng nhn c tn hiu t v tinh nmtrong khong gia 2 kinh 81020 v 278040Bi tp 12(H NT 98-99)1)Mt xe c khi lng m=1600kg chyn ng trn mt ng trn phng,c bn knh R=200m vi mt tc khng i 72 km/h. Hi gi tr ca h sma st gia lp xe v mt ng t nht phi bng bao nhiu xe khngtrt?

2)Nu mt ng nghing gc (so vi ng nm ngang v mt nghinghng v tm ng cong), xe vn i vi tc trn m khng cn ti lcma st th gc bng bao nhiu? Cho bit g=9,8 m/s2.Bi tp 13(H TSNT 98-99)Mt vt nh A bt u trt t nh ca mt mt cu bn knh R=90 cmxung di; khng vn tc ban u, khng ma st. Tnh vn tc ca vt A tiv tr vt bt u tch khi mt cu. Cho gia tc trng trng g=10m/s2.

C NNGA-Tm tt mt s khi nim v cng thc1- Cng

V tinh

h

00A B

R O


18/23


Dy bi dng-Nm hc 2011 - 2012 18

a- Cng c hcb- Cng ca trng lcc- Cng ca lc n hi

2- Cng sut

3- nh lut bo ton cng4-Nng lng5- Th nng6- ng nng7- nh lut bo ton v chuyn ho nng lngB-Cc bi tp v cng v cng sutBi tp 14 ct cnh, mt my bay cn c vn tc l 360 km/h v phi chy trn mton ng bng di 600m. Tm cng sut ti thiu ca ng c my bay ct cnh c. Cho bit khi lng ca my bay l 2 tn, lc cn chuynng t l vi lc nn vung gc ca my bay ln mt ng bng, h s cnny bng 0,2 v chuyn ng ca my bay trn ng bng l nhanh dn u.Gii:Theo nh lut II N ta c: amFF c

F-kmg=ma

V l chuyn ng nhanh dn u nn a=S

v

2

2

T ta c F=m( kgS

v

2

2

)

Cng sut ti thiu ca ng c lN=F.v=m( kg

S

v

2

2

).v=2.106W=2000kW.

Bi tp 15Mt xe t chuyn ng ln dc vi mt vn tc khng i v1=3/s, xung dcvi vn tc 7m/s ri i trn mt ng nm ngang vi vn tc v 0. Tm v0bitrng trong c 3 trng hp, cng sut ca t l nh nhau v lc ko khng

ph thuc vo vn tc ca n. Coi dc l thoai thoi.Gii:Gi F1, F2, F0l lc lm t chuyn ng vi vn tc v1, v2, v3 th theo nhngha v cng sut ta c: F1v1=F2v2=F3v3.Chn h quy chiu gn vi mt ng, chiu dng l chiu chuyn ng. tac:

F1=kmgcos +mg sin F2=kmgcos-mgsin F0=kmg

V cng sut trong 3 trng hp u nh nhau nn:(kmgcos +mg sin)v1=kmgv0 (1)(kmgcos -mg sin) v2=kmgv0 (2)


19/23


Dy bi dng-Nm hc 2011 - 2012 19

Rt gn v gii h phng trnh (1) v (2) ta c v0=2cos21

21

vv

vv

. Dc

thoi, coi cos=1. v0=4,2 m/sBi tp 16

Mt tn la mang ng c bay thng ng t mt t ln ti cao h=40kmv t vn tc v=1,4 km/s. Cho bit khi lng ca tn la l m=500kg vsc cn ca khng kh l khng ng k. Hy tm:

b-Cng do ng c tn la sn ra.c- ng nng v th nng ca tn la cao ny.

Coi chuyn ng ca tn la l nhanh dn u v chuyn ng khng c vntc ban u.Gii:a) Gi F l lc y ng c, ta c A=Fh

Nhng F-mg =ma v a= hv2

2

(chn HT trng qu o chuyn ng,chiudng hng ln) F=mg+ma

Vy A=(mg+mh

v

2

2

)h=mgh+2

1 mv2=686.106 (J)=686kJ

b)Ta thy cng ca lc y tn la bng tng ca ng nng v th nng.Bi tp 17

Ngi ta ko mt kin hng c khi lng l m=100kg trt trn mt onng di S=49,6 m trn mt phng nm ngang. Bit lc ko hp vi mt

phng nm ngang mt gc =310v h s ma st k=0,33, hy tm cng calc ko.Gii:Ta c A=F1.S.Chn chiu dng l chiu chuyn ng ta cF1=kN = k(P-F2) (V vt coi l chuyn ngthng u), nhng F2=F1tg .

Vy F1=kP-kF1tg hay F1=ktg

kP

1v

A= ktgkP

1 S=13500 (J).

NH LUT BO TON C NNGBi tp 18Mt thanh nh, di l1+l2c th quay t doquanh mt trc nm ngang O. Ti cc uca thanh c gn cc vt nng, khi lngtng ng l m1 v m2. Tnh vn tc ca vtm2ti v tr thp nht khi thanh quay t do

t vtr nm ngang n v tr thng ng.Li gii

2F

F

O 1

F

P

1v

l1 l2m1 O m2

2v


20/23


Dy bi dng-Nm hc 2011 - 2012 20

mh

np

p

v

R

Chn cc v tr 1 v 2 l cc v tr tng ng vi thi im thanh nm ngangv thng ng.Chn gc th nng ti v tr thp nht ca m2(khi thanh thng ng)Theo nh lut bo ton c nng ta c W1=W2 hay:

gl2(m1+m2)=m1g(l1+l2)+ 22

2

221

2

1 vmvm (1)

ng thi =2

2

1

1

l

v

l

v (2)

T (1) v (2) ta c v2= 211

2

22

1122 )(2

lmlm

lmlmg

Bi tp 19

Mt vt khi lng m trt khng ma st t nh mt mt cu xung di.Hi t khong cch h no (tnh t nh mt cu) vt bt u ri khi mtcu? Cho bn knh mt cu R=90 cm.Gii

Vt bt u ri khi mt bn cu khi lcnn vt ln mt cu (hay phn lc m mt cutc dng ln vt) bng khng.

Ta c -N+Pn=R

mv2

N=mgcos-R

mv2

=0

v2

=Rgcos (1)Mt khc theo nh lut bo ton c nng ta

c: mg h =2

2mv (2)

T (1) vg (2) ta c32

)(

2

cos Rh

R

hRRRh

=30 cm

Bi tp 20Mt l xo c cng k=100N/m v vt nng khi lng m=100 g c nivi nhau nh hnh v. Lc vt O l xo cha bin dng. Ko l xo sao cho

vt n A vi OA=10cm ri truyn cho vt vn tc v0=2m/s. Tnh vn tc sau vt qua O.GiiHng dn: WO=WAChn gc th nng n hi ti v tr cn

bng ta c 20222

1

2

1

2

1kxmvmv . T

v=3,74 m/s.Bi tp 21Mt qu cu khi lng m=100g treo vol xo c cng k=100N/m. Ly g=10m/s2.a)Tnh dn ca l xo khi vo tr cn bng.

km

O A

k

m


21/23


Dy bi dng-Nm hc 2011 - 2012 21

b)Ko vt theo phng thng ng xung di mt khong x=2cm ri thkhng vn tc ban u. Tnh vn tc ca vt khi qua v tr cn bng.Gii

a) dnk

mgl =1cm.

b) Ta c/m c rng, nu chn gc th nng n hi v th nng trngtrng ti v tr cn bng th th nng ca h vt-l xo ti v tr bt k ch l

th nng ca l xo W=2

1kx

2, x l bin dng so vi mc m ta chn lm

gc th nng.-Ti VTCB: mg=kx0Th nng ca l xo ti A v ti O:

Wt1=2

1k(x0+x)

2, Wt0=

2

1kx0

2 Wt1-Wt0=

2

1kx

2+kxx0.

Vi gc th nng ti O nn Wto=0. Vy Wt1=2

1kx2+kxx0.

-Th nng trng lc ca qu cu ti A l Wt2=mg(-x)

Th nng ca h (vt +l xo) ti A l Wt1+Wt2=2

1kx

2+kxx0-mgx=

2

1kx

2.

Vy,2

1 kx2=2

1 mv2v= xm

k =0,63m/s.

Bi tp 22 o vn tc ca vin n, ta dng con lc th n. l

mt bao ct c khi lng M treo u mt si dy di l.Vin n c khi lng m v vn tc v0 chui vo bao ct vs nm yn. Sau bao ct v vin n s lch khi v trcn bng v dy treo s lch vi phng thng ng mtgc . Tnh vn tc ca vin n. p dng bng s:M=10kg, m=100g, l=1m, =600.GiiHng dn: chn gc th nng ti v tr cn bng ca ti ct.

Theo LBTCN ta c:2

1 (M+m)V2=(M+m)gl(1-cos) )cos1(2 glV .

p dng nh lut bo ton ng lng ta c: mv0=(M+m)V. T ,

v0= )cos1(2

glm

mM =320m/s.

Bi tp 23Mt l xo c th b nn 2cm bi lc 270N. Mt vt c khi lng m=12 kgc th ngh t nh mt mt phng nghing khng ma st c gc nghing=300. Khi dng li nht thi khi n nn l xo mt on x=5,5 cm. Hi:a)Ti thi im l xo b nn cc i ln th nht, khi i theo mt phngnghing mt on bng bao nhiu?

l

m0v

M


22/23


Dy bi dng-Nm hc 2011 - 2012 22

b)Vn tc ca vt l bao nhiu khi n va va chm vo l xo?Giia)Lc n hi ca l xo tnh theo

nh lut Hc: k=l

F

=1.35.104N/m

Gi A l im ngh nh dc v B lim dng nht thi khi i t A n B.Chn gc th nng n hi ti v l xocha bin dng, th nng trng lc tiB, theo nh lut BTCN ta c:

2

1 kx2 =mglsin vi l=AB.

Suy ra AB = l=sin2

2

mg

kx =0,347m=34,7 cm.

b)Khi cha chm vo l xo vt i c mt on AO=l=AB -l= 0,292cm.Gia tc ca vt a=gsin , vn tc ca vt o l v= 'sin2'2 lgal Bi tp 24Mt vt nng khi lng m1=2kg trtkhng ma st dc theo mt mt bn vivn tc v1= 10m/s. Ngay trc mt n,v chuyn ng cng phng vi nc mt vt nng khi lng m2=5kg, chuyn ng vi vn tc v2= 3m/s. Mtl xo khng khi lng, cng 1120N/m c gn vo cnh gn ca m2.Tm nn cc i ca l xo.Li giiTheo LBTL ta c: m1v1+m2v2=(m1+m2)v suy ra v=5m/s.Theo LBT ng nng (c nng bo ton) ta c:

2

1 m1v12 +

2

1 m2v22 =

2

1 (m1+m2)v2 +

2

1 kx2

Trong x l nn cc i ca l xo. T x=0,25m = 25cm.Bi tp 24Mt hn khi lng 8kg nm trn mt ci l xo. Lxo b nn 10 cm. a) Hng s ca l xo l bao nhiu?

Ngi ta y hn xung l xo nn thm 30cm rith ra. b) Tm th nng ca l xo ngay trc khi th.c) Hn ln cao c bao nhiu so vi v tr m nc th? Ly g=10m/s2Li gii:

a)Hng s l xo k=x

mg

x

P =784N/m

b)Th nng ca l xo Wt=2

1 kX2=2

1 (x+30.10-2)2=62,7 (J).

c)Khi th th th nng ca l xo bin thnh th nng trng trng mgh Max.. Tac hMax=

mg

Wt =0,8 m=80 cm.

m A

k o H

B K

D

m1 k m2

m

k


23/23


Bi tp 25 (07.26.NT)Hai vt m1=1kg v m2=4,1 kg ni vi nhau qua l xokhng khi lng, cng k=625N/m, g=9,8 m/s2. KoA khi VTCB 1 on a=1,6 cm ri th cho m1dao ng.

Tm:a)Chu k ca m1.b)Vn tc cc i ca m1.c)Tm lc cc i v cc tiu tc dng ln bn.Li gii

a) ng yn, m1dao ng, m1v l xo l mt con lc n hi

Vy,1m

k =25rad/s , T=0,25s.

b) vMax=A=a=0,4m/s.

c) Tm p lc cc i, cc tiu tc dng ln bn: VTCB O, l xo b nn mt on x0. Ta c:m1g=kx0 (1)v tr bt k no ca m1 th m2u chu tc dng ca3 lc. V m2ng yn nn NFP

2 =0 (2)+Khi l xo b nn ti a:

m2g-NMax+FMax=0,Vi FMax=k(x0+a). Vy,

NMax=m2g+FMax=m2g+m1g+ka (3)

Theo nh lut III Newton, p lc cc i tc dng lnbn l PMax= NMax=59,98N.+Khi l xo b dn ti a:

m2g-Nmin-Fmin=0,Vi Fmin=k(a-x0). Vy, Nmin=m2g-Fmin=m2g+k(x0a)

=m2g+kx0-ka=m2g+m1g-ka. Theo nh lut II Newton,Pmin=Nmin=39,98N.

m1

m2

Oa

m1

N

m2

FP

2

m1a

O

F

N

m2

2P

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