Bai Tap Dong Luc Hoc Chat Diem Co Giai Chi Tiet.6968(2)

8/8/2019 Bai Tap Dong Luc Hoc Chat Diem Co Giai Chi Tiet.6968(2)

1/21

PHN TH NHTBAI TAP ONG LC HOC CHAT IEM

BAI 1 :Hai l xo: l xo mt di thm 2 cm khi treo vt m1 = 2kg, l xo 2 di thm 3 cm khitreo vt m2 = 1,5kg. Tm t s k1/k2.Bi gii:

Khi gn vt l xo di thm on (l. v tr cn bng

mglKPF0 !(!pp

Vi l xo 1: k1(l1 = m1g (1)

Vi l xo 1: k2(l2 = m2g (2)

Lp t s (1), (2) ta c2

2

3

5,1

2

l

l.

m

m

K

K

1

2

2

1

2

1 !!(

(!

BAI 2 :Mt xe ti ko mt t bng dy cp. T trng thi ng yn sau 100s t t vntc V = 36km/h. Khi lng t l m = 1000 kg. Lc ma st bng 0,01 trng lc t. Tnhlc ko ca xe ti trong thi gian trn.Bi gii:

Chn hng v chiu nh hnh vTa c gia tc ca xe l:

)s/m(1,0100

010

t

VVa

0 !

!

!

Theo nh lut II Newtn :ppp

! amfF ms F fms = maF = fms + ma

= 0,01P + ma= 0,01(1000.10 + 1000.0,1)= 200 N

BAI 3 :Hai l xo khi lng khng ng k, cng ln lt l k1 = 100 N/m, k2 = 150N/m, c cng di t nhin L0 = 20 cm c treo thng ng nh hnh v. u di 2 l xo

ni vi mt vt khi lng m = 1kg. Ly g = 10m/s 2. Tnh chiu di l xo khi vt cn bng.


2/21

Bi gii:

Khi cn bng: F1 + F2=

Vi F1 = K1(l; F2 = K

2(1

nn (K1 + K2) (l = P

)m(04,00

10.1

KK

Pl

21

!!

!(

Vy chiu di ca l xo l:L = l0 + (l = 20 + 4 = 24 (cm)

BAI 4 :Tm cng ca l xo ghp theo cch sau:

Bi gii:

Hng v chiu nh hnh v:Khi ko vt ra khi v tr cn bng mt on x th : dn l xo 1 l x, nn l xo 2 l x

Tc dng vo vt gm 2 lc n hip

1F ; 2p

,ppp

! FFF 21 Chiu ln trc Ox ta c :

F = F1 F2 = (K1 + K2)xVy cng ca h ghp l xo theo cch trn l:

K = K1 + K2BAI 5 :Hai vt A v B c th trt trn mt bn nm ngang v c ni vi nhau bng dykhng dn, khi lng khng ng k. Khi lng 2 vt l m A = 2kg, mB = 1kg, ta tc dng


3/21

vo vt A mt lc F = 9N theo phng song song vi mt bn. H s ma st gia hai vt vi

mt bn l m = 0,2. Ly g = 10m/s2. Hy tnh gia tc chuyn ng.Bi gii:

i vi vt A ta c:pppppp

! 11ms1111 amFTFNP Chiu xung Ox ta c: F T1 F1ms = m1a1

Chiu xung Oy ta c: m1g + N1 = 0Vi F1ms = kN1= km1g

F T1 k m1g = m1a1 (1)* i vi vt B:

pppppp! 22ms2222 amFTFNP

Chiu xung Ox ta c: T2 F2ms = m2a2Chiu xung Oy ta c: m2g + N2 = 0Vi F2ms= k N2= k m2g

T2 k m2g = m2a2 (2)

V T1= T2 = T v a1 = a2 = a nn:

F - T k m1g = m1a (3)

T k m2g = m2a (4)

Cng (3) v (4) ta c F k(m1 + m2)g = (m1+ m2)a

2

21

21 s/m112

10).12(2,09

mm

g).mm(Fa !

!

Q!

BAI 6 :Hai vt cng khi lng m = 1kg c ni vi nhau bng si dy khng dn v

khi lng khng ng k. Mt trong 2 vt chu tc ng ca lc kop

F hp vi phng

ngang gc a = 300 . Hai vt c th trt trn mt bn nm ngang gc a = 30 0H s ma st gia vt v bn l 0,268. Bit rng dy ch chu c lc cng ln nht l 10 N.

Tnh lc ko ln nht dy khng t. Ly 3 = 1,732.

Bi gii:

Vt 1 c :pppppp

! 11ms1111 amFTFNP


4/21

Chiu xung Ox ta c: F.cos 300 T1 F1ms = m1a1

Chiu xung Oy : Fsin 300 P1 + N1 = 0

V F1ms= k N1 = k(mg Fsin 300)

F.cos 300 T1k(mg Fsin 30

0) = m1a1 (1)Vt 2:

pppppp

! 22ms2222 amFTFNP Chiu xung Ox ta c: T F2ms= m2a2

Chiu xung Oy : P2 + N2 = 0M F2ms= k N2 =km2g T2 k m2g = m2a2Hn na v m1 = m2= m; T1 = T2 = T ; a1 = a2 = a

F.cos 300 T k(mg Fsin 300) = ma (3) T kmg = ma (4)T (3) v (4)

m

00

t2

)30sin30(cosTT e

Q!

20

2

1268,0

2

3

10.2

30si30c s

T2F

00

m !

!Q

e

Vy Fmax = 20 N

Bi 7:Hai vt A v B c khi lng ln lt l mA = 600g, mB = 400g c ni vi nhau bng sidy nh khng dn v vt qua rng rc c nh nh hnh v. B qua khi lng ca rng rc

v lc ma st gia dy vi rng rc. Ly g = 10m/s2

. Tnh gia tc chuyn ng ca mi vt.

Bi gii:

Khi th vt A s i xung v B s i ln do mA > mB vTA = TB = TaA = aB = ai vi vt A: mAg T = mA.a


5/21

i vi vt B: mBg + T = mB.a* (mA mB).g = (mA + mB).a

2

BA

BA s/m210.400600

400600g.

mm

mma* !

!

!

Bi 8:Ba vt c cng khi lng m = 200g c ni vi nhau bng dy ni khng dn nh hnh v.

H s ma st trt gja vt v mt bn l Q = 0,2. Ly g = 10m/s2. Tnh gia tc khi h chuynng.

Bi gii:

Chn chiu nh hnh v. Ta c:pppppppppppp

! aMPTTNPFTTNPF 11222ms23333 Do vy khi chiu ln cc h trc ta c:

!

!

!

3ms4

2ms32

11

maFT

maFTT

maTmg

V

aaaa

'TTT

TTT

321

43

21

!!!

!!

!!

!

!

!

maFT

maFTT

maTmg

ms'

ms'

!Q!

ma3mg2mgma3F2mg ms

2s/m210.

3

2,0.21g.

3

21a !

!

Q!

Bi 9:Mt xe trt khng vn tc u t nh mt phng nghing gc E = 300. H s ma st trt l

Q = 0,3464. Chiu di mt phng nghing l l = 1m. ly g = 10m/s 2v3 = 1,732 Tnh gia tc chuyn ng ca vt.


6/21

Bi gii:

Cc lc tc dng vo vt:

1) Trng lcp

P

2) Lc ma stp

msF

3) Phn lc

p

N ca mt phng nghing4) Hp lc

ppppp!! amFNPF ms

Chiu ln trc Oy: PcoxE + N = 0 N = mg coxE (1)Chiu ln trc Ox : PsinE Fms = max mgsinEQN = max (2)t (1) v (2) mgsinEQ mg coxE = max ax = g(sinEQ coxE)

= 10(1/2 0,3464. 3 /2) = 2 m/s2

BAI 10 :Cn tc dng ln vt m trn mt phng nghing gc E mt lc F bng bao nhiu vt nm yn, h s ma st gia vt v mt phng nghing l k , khi bit vt c xu hngtrt xung.

Bi gii:

Chn h trc Oxy nh hnh v.p dng nh lut II Newtn ta c :


7/21

0FNPF ms !pppp

Chiu phng trnh ln trc Oy: N PcoxE FsinE = 0 N = PcoxE + F sinE

Fms = kN = k(mgcoxE + F sinE)Chiu phng trnh ln trc Ox : PsinE F coxE Fms = 0 F coxE = PsinE Fms = mg sinE kmg coxE kF sinE

EE!

EEEE!

ktg1)ktg(mg

sikc s)kc x(simgF

BAI 11 :Xem h c lin kt nh hnh v

m1 = 3kg; m2 = 1kg; h s ma st gia vt v mt phng nghing l Q = 0,1 ; E = 300; g = 10

m/s2Tnh sc cng ca dy?

Bi gii:

Gi thit m1 trt xung mt phng nghing v m2 i ln, lc h lc c chiu nh hnh v.

Vt chuyn ng nhanh dn u nn vi chiu dng chn, nu ta tnh c a > 0 th chiuchuyn ng gi thit l ng.i vi vt 1:

ppppp

! 11ms11 amFTNP Chiu h xOy ta c: m1gsinE T QN = ma

m1g coxE + N = 0

* m1gsinE T Q m1g coxE = ma (1)i vi vt 2:

ppp

! 2222 amTP

m2g + T = m2a (2)Cng (1) v (2) m1gsinEQ m1g coxE = (m1 + m2)a

)s/m(6,04

10.12

33.1,0

2

1.10.3

mm

gmcosmsingma

2

21

211

}

!

EQE!

V a > 0, vy chiu chuyn ng chn l ng* T = m2 (g + a) = 1(10 + 0,6) = 10,6 N


8/21

BAI 12 :Sn i c th coi l mt phng nghing, gc nghing a = 300 so vi trc Ox nm

ngang. T im O trn sn i ngi ta nm mt vt nng vi vn tc ban u V0 theophng Ox. Tnh khong cch d = OA t ch nm n im ri A ca vt nng trn sn i,

Bit V0 = 10m/s, g = 10m/s2.

Bi gii:

Chn h trc nh hnh v.Phng trnh chuyn ng v phng trnh qu o l:

!

!

2

0

gt2

1y

tVx

Phng trnh qu o

)1(Vg

21y 2

20

!

Ta c:

E!!

E!!

sindOKy

cosdOH

A

A

V A nm trn qu o ca vt nng nn xAv yA nghim ng (1). Do :

2

20

)cosd(V

g

2

1sind E!E

m33,1

30cos

30sin.

10

10.2

cos

sin.

g

V2d

0

0220 !!

E

E

!

BAI 13 :Mt hn c nm t cao 2,1 m so vi mt t vi gc nm a = 450 so vimt phng nm ngang. Hn ri n t cnh ch nm theo phng ngang mt khong 42m. Tm vn tc ca hn khi nm ?GIAIChn gc O ti mt t. Trc Ox nm ngang, trc Oy thng ng hng ln (qua im nm).Gc thi gian lc nm hn .Cc phng trnh ca hn


9/21

x = V0cos450t (1)

y = H + V0sin 450t 1/2 gt2 (2)

Vx = V0cos450 (3)

Vy = V0sin450 gt (4)

T (1)

00 4c sV

xt !

Th vo (2) ta c :

)5(45c sV

x.g

2

1x.45tg4y

0220

20 !

Vn tc hn khi nmKhi hn ri xung t y = 0, theo bi ra x = 42 m. Do vy

)s/m(20

421.2

2

.442

Hx.45tg45c s

2g.x

V

045c sV

xg

2

1x45tgH

000

0220

20

!

!

!

!

BAI 14 :Mt my bay ang bay ngang vi vn tc V1 cao h so vi mt t mun th

bom trng mt on xe tng ang chuyn ng vi vn tc V2 trong cng 2 mt phng thngng vi my bay. Hi cn cch xe tng bao xa th ct bom ( l khong cch t ngthng ng qua my bay n xe tng) khi my bay v xe tng chuyn ng cng chiu.Bi gii:

Chn gc to O l im ct bom, t = 0 l lc ct bom.Phng trnh chuyn ng l:

x = V1t (1)

y = 1/2gt2 (2)

Phng trnh qu o:2

20

xV

g

2

1y !

Bom s ri theo nhnh Parabol v gp mt ng ti B. Bom s trng xe khi bom v xe cnglc n B

vg

h2

g

y2t !!

g

h2Vx 1B !

Lc t = 0 cn xe A


10/21

g

h2VtVAB 22 !!

* Khong cch khi ct bom l :

!!! V(Vg

h2)VV(ABHBHA 121

BAI 15 :T nh mt mt phng nghing c gc nghing F so vi phng ngang, ngi ta

nm mt vt vi vn tc ban u V0 hp vi phng ngang gc E . Tm khong cch l dctheo mt phng nghing t im nm ti im ri.

Bi gii;Cc phng thnh to ca vt:

)2(

gt2

1tsiVHy

)1(tc sVx

20

0

E!

E!

T (1)

E

!cosV

xt

0 Th vo (2) ta c:

(3)cosV

xg2

1xtgHy 22

0

2

EE! Ta c to ca im M:

F!

F!

sinlHy

coslx

M

M

Th xM, yM vo (3) ta c:

E

FFE!F

220

22

c sV2

c sglc sltgHsilH

F

FEE!

F

FEFEE!

F

FFEE!

2

20

2

20

2

220

cosg

)sin(cosV2

cosg

sincoscossincosV2

cosg

sincostg.cosV2l

BAI 16 : mt i cao h0= 100m ngi ta t 1 sng ci nm ngang v mun bn sao choqu n ri v pha bn kia ca to nh v gn bc tng AB nht. Bit to nh cao h = 20 m

v tng AB cch ng thng ng qua ch bn l l = 100m. Ly g = 10m/s 2. Tm khongcch t ch vin n chm t n chn tng AB.


11/21

Bi gii:

Chn gc to l ch t sng, t = 0 l lc bn.Phng trnh qu o

2

20

xV

g

2

1y !

n chm t gn chn tng nht th qu o ca n i st nh A ca tng nn

2A2

0

A xV

g

2

1y !

s/m25100.80.2

10.1x.

y

g

2

1V A

A

0 !!!

Nh vy v tr chm t l C m

)m(8,1110

100.225

g

h2V

g

y.2Vx 0

C0C !!!!

Vy khong cch l: BC = xC l = 11,8 (m)BAI 17 :Mt vt c nm ln t mt t theo phng xin gc ti im cao nht ca qu

o vt c vn tc bng mt na, vn tc ban u v cao h0 =15m. Ly g = 10m/s2.Tnh ln vn tc

Bi gii:

Chn: Gc O l ch nm* H trc to xOy* T = 0 l lc nmVn tc ti 1 im


12/21

yx VVV ! Ti S: Vy = 0

E!! c sVVV oxs M

oos 60

2

1cos

2

VV !E!E!

V

s/m20

23

15x10x2

si

gy2V

g2

siVy

so

2o

x !!E!

E!

BAI 18 :Em b ngi di sn nh nm 1 vin bi ln bn cao h = 1m vi vn tc

V0 = 102 m/s. vin bi c th ri xung mt bn B xa mp bn A nht th vn tc oV phi nghing vi phng ngang 1 gc E bng bao nhiu?

Ly g = 10m/s2.

Bi gii:

vin bi c th ri xa mp bn A nht th qu o ca vin bi phi i st A.

Gi 1V l vn tc ti A v hp vi AB gc E1 m:

g

2siVAB 1

2 E!

(coi nh c nm t A vi AB l tm AB ln nht th

412sin 11

T!E!E

V thnh phn ngang ca cc vn tc

u bng nhau V0cosE = V.cosE1

1o

cos.V

Vcos E!E

Vi

!E

!

2

1cos

gh2VV

1

2o

Nn


13/21

21

102

1x10

2

1

V

gh

2

1

2

1.

V

gh2Vcos

22oo

2o !!!

!E

o60!E

BAI 19 :Mt bn nm ngang quay trn u vi chu k T = 2s. Trn bn t mt vt cchtrc quay R = 2,4cm. H s ma st gia vt v bn ti thiu bng bao nhiu vt khng trt

trn mt bn. Ly g = 10 m/s2 v T2 = 10Bi gii:

Khi vt khng trt th vt chu tc dng ca 3 lc:nghF;N,P ms

Trong :

0NP !

Lc vt chuyn ng trn u nn msF l lc hng tm:

Q!

!

)2(mg.F

)1(RmwF

ms

2ms

g

Rwg.Rw

22 uQQe

Vi w = 2T/T = T.rad/s

25,010

25,0.2!

TuQ

Vy Qmin = 0,25BAI 20 :Mt l xo c cng K, chiu di t nhin l0, 1 u gi c nh A, u kia gnvo qu cu khi lng m c th trt khng ma st trn thanh (() nm ngang. Thanh (()quay u vi vn tc gc w xung quanh trc (A) thng ng. Tnh dn ca l xo khi l 0 =20 cm; w = 20Trad/s; m = 10 g ; k = 200 N/m

Bi gii:


14/21

Cc lc tc dng vo qu cu

dhF;N;P

2

o2

o22

o2

mwK

lmwl

lmwmwKl

llmwlK

!(

!(

(!(

vi k > mw2

m05,0

20.01,02002,0.20.01,0l2

2

!T

T!(

BAI 21 :Vng xic l mt vnh trn bn knh R = 8m, nm trong mt phng thng ng.

Mt ngi i xe p trn vng xic ny, khi lng c xe v ngi l 80 kg. Ly g = 9,8m/s 2tnh lc p ca xe ln vng xic ti im cao nht vi vn tc ti im ny l v = 10 m/s.Bi gii:

Cc lc tc dng ln xe im cao nht l N;P Khi chiu ln trc hng tm ta c

N2168,98

1080g

R

vmN

R

mvNP

22

2

!

!

!

!

BAI 22 :Mt qu cu nh c khi lng m = 100g c buc vo u 1 si dy di l = 1mkhng co dn v khi lng khng ng k. u kia ca dy c gi c nh im A trntr quay (A) thng ng. Cho trc quay vi vn tc gc w = 3,76 rad/s. Khi chuyn ng

n nh hy tnh bn knh qu o trn ca vt. Ly g = 10m/s2.Bi gii:

Cc lc tc dng vo vt P;T


15/21

Khi (() quay u th qu cu s chuyn ng trn u trong mt phng nm ngang, nn hplc tc dng vo qu cu s l lc hng tm.

TPF ! vi

!

B

RmwF

PF

2

g

Rw

mg

Ftgv

2

!!E

R = lsinE

EE

!E

!Ecos

sin

g

sinlwtg

2

V

o

2245707,0

1.76,3

10

lw

gcos0 !E!!!E{E

Vy bn knh qu o R = lsinE = 0,707 (m)

BAI 23 :Chu k quay ca mt bng quanh tri t l T = 27 ngy m. Bn knh tri t lR0 = 6400km v Tri t c vn tc v tr cp I l v0 = 7,9 km/s. Tm bn knh qu o camt trng.Bi gii:Mt trng cng tun theo quy lut chuyn ng ca v tinh nhn to.Vn tc ca mt trng

R

GMv

o!

Trong M0 l khi lng Tri t v R l bn knh qu o ca mt trng.Vn tc v tr cp I ca Tri t

km10.38R

14,3.4

9,7x24.3600.27.6400

4

v.TRR

R

R

Tv

R2

R.T

2v;

R

R

v

v

R

GM

v

5

2

22

2

2oo3o

o

o

o

o

oo

!

!T

!!T

T!!

!

BAI 24 :Qu cu m = 50g treo u A ca dy OA di l = 90cm. Quay cho qu cu chuynng trn trong mt phng thng ng quanh tm O. Tm lc cng ca dy khi A v tr thp

hn O. OA hp vi phng thng ng gc E= 60o v vn tc qu cu l 3m/s, g = 10m/s2.

Bi gii:


16/21

Ta c dng:

p

! amP;T Chiu ln trc hng tm ta c

N75,093

2

1x1005,0

R

v60cosgmT

R

vmmaht60cosPT

220

2o

!

!

!

!!

PHN TH HAIMT S BI TP VT L VN DNG SNG TO PHNG PHP TA

Phng php ta l phng php c bn trong vic gii cc bi tp vt lphn ng lc hc. Mun nghin cu chuyn ng ca mt cht im, trc ht ta cnchn mt vt mc, gn vo mt h ta xc nh v tr ca n v chn mt gcthi gian cng vi mt ng h hp thnh mt h quy chiu.

Vt l THPT ch nghin cu cc chuyn ng trn mt ng thng hay chuynng trong mt mt phng, nn h ta ch gm mt trc hoc mt h hai trc vunggc tng ng.

Phng php + Chn h quy chiu thch hp.+ Xc nh ta ban u, vn tc ban u, gia tc ca cht im theo cc trc

ta : x0, y0; v0x, v0y; ax, ay. ( y ch kho st cc chuyn ng thng u, bin iu v chuyn ng ca cht im c nm ngang, nm xin).

+ Vit phng trnh chuyn ng ca cht im

!

!

00y

2

y

00x

2

x

ytvta21y

xtvta2

1x

+ Vit phng trnh qu o (nu cn thit) y = f(x) bng cch kh t trong ccphng trnh chuyn ng.

+ T phng trnh chuyn ng hoc phng trnh qu o, kho st chuynng ca cht im:

- Xc nh v tr ca cht im ti mt thi im cho.- nh thi im, v tr khi hai cht im gp nhau theo iu kin


17/21

!

!

21

21

yy

xx

- Kho st khong cch gia hai cht im 2212

21 )y(y)x(xd ! Hc sinh thng ch vn dng phng php ta gii cc bi ton quen

thuc i loi nh, hai xe chuyn ng ngc chiu gp nhau, chuyn ng cng chiu

ui kp nhau,trong cc cht im cn kho st chuyn ng tng minh, chcn lm theo mt s bi tp mu mt cch my mc v rt d nhm chn. Trong khi, c rt nhiu bi ton tng chng nh phc tp, nhng nu vn dng mt cchkho lo phng php ta th chng tr nn n gin v rt th v.

Xin a ra mt s v d:

Bi ton 1

Mt vt m = 10kg treo vo trn mt bung thang my c khi lng M =200kg. Vt cch sn 2m. Mt lc F ko bung thang my i ln vi gia tc a = 1m/s 2.Trong lc bung i ln, dy treo b t, lc ko F vn khng i. Tnh gia tc ngay sau ca bung v thi gian vt ri xung sn bung. Ly g = 10m/s 2.Nhn xt

c xong bi, ta thng nhn nhn hin tng xy ra trong thang my (chnh quy chiu gn vi thang my), rt kh m t chuyn ng ca vt sau khi dytreo b t. Hy ng ngoi thang my quan st (chn h quy chiu gn vi t) haicht im vtv sn thang ang chuyn ng trn cng mt ng thng. D dngvn dng phng php ta xc nh c thi im hai cht im gp nhau, l lc vtri chmsn thang.Gii

Chn trc Oy gn vi t, thng ng hng ln, gc O ti vtr sn lc dy t, gc thi gian t = 0 lc dy t.Khi dy treo cha t, lc ko F v trng lc P = (M + m)g gy ra giatc a cho h M + m, ta c

F - P = (M + m)a 2310g)m)(a(MF !! + Gia tc ca bung khi dy treo tLc F ch tc dng ln bung, ta c

F Mg = Ma1, suy ra

2

1 1,55m/sM

MgFa !

!

+ Thi gian vt ri xung sn bungVt v sn thang cng chuyn ng vi vn tc ban u v 0.

Phng trnh chuyn ng ca sn thang v vt ln lt l

tvta2

1y 0

2

11 ! ; 0202

22 ytvta2

1y !

Vi a1 = 1,55m/s2, y02 = 2m, vt ch cn chu tc dng ca trng lc nn c gia tc a 2 =

-gVy

tv0,775ty 02

1 ! v 2tv5ty 02

2 ! Vt chm sn khi

y

O

FT

TT

PT

0vT

0vT

y02


18/21

Vt chm sn khi y1 = y2, suy ra t = 0,6s.

Bi ton 2

Mt toa xe nh di 4m khi lng m2 = 100kg ang chuyn ng trn ngray vi vn tc v0 = 7,2km/h th mt chic vali kch thc nh khi lng m 1 = 5kgc t nh vo mp trc ca sn xe. Sau khi trt trn sn, vali c th nm yn

trn sn chuyn ng khng? Nu c th nm u? Tnh vn tc mi ca toa xe vvali. Cho bit h s ma st gia va li v sn l k = 0,1. B qua ma st gia toa xe vng ray. Ly g = 10m/s 2.Nhn xt

y l bi ton v h hai vt chuyn ng trt ln nhau. Nu ng trn ngray qua st ta cng d dng nhn ra s chuyn ng ca hai cht im vali v mp sauca sn xe trn cng mt phng. Vali ch trt khi sn xe sau khi ti mp sau snxe, tc l hai cht im gp nhau. Ta a bi ton v dng quen thuc.Gii

Chn trc Ox hng theo chuynng ca xe, gn vi ng ray, gc O ti

v tr mp cui xe khi th vali, gc thigian lc th vali.+ Cc lc tc dng lnVali: Trng lc P1 = m1g, phn lc N1 vlc ma st vi sn xe Fms, ta c

11ms11 amFNPTTTT

! Chiu ln Ox v phng thng ng ta c:

Fms = m1a1 v N1 = P1 = m1g, suy ra

2

1

1

1

ms1 1m/skgm

kN

m

Fa !!!!

Xe: Trng lc P2 = m2g, trng lng ca vali gmP 1,1 ! , phn lc N2 v lc ma st vivali Fms. Ta c

22ms22

'

1 am'FNPPTTTTT

! Chiu ln trc Ox ta c

-Fms = m2a2

2

2

1

2

ms

2

ms2 0,05m/sm

gkm

m

F

m

F'a !

!

!

!

Phng trnh chuyn ng ca vali v xe ln lt

2t0,025ttvta2

1x

40,5txta2

1x

2

0

2

22

2

01

2

11

!!

!!

Vali n c mp sau xe khi x1 = x2, hay 0,5t2 + 4 = -0,025t2 + 2t

Phng trnh ny v nghim, chng t vali nm yn i vi sn trc khi n mp sauca xe.Khi vali nm yn trn sn, v1 = v2Vi v1 = a1t + v01 = t , v2 = a2t + v0 = -0,05t + 2, suy ra

t = - 0,05t + 2 suy ra t = 1,9s

0vT

1N

msF 1PT

2N

1P

2P

msF xO


19/21

Khi vali cch mp sau xe mt khong 2t0,025t40,5txxd 2221 !! Vi t = 1,9s ta c d = 2,1mVn tc ca xe v vali lc v1 = v2 = 1,9m/s.

Bi ton 3

Mt b vc mt ct ng c dng mt phn parabol(hnh v). T im A trn sn b vc, cao h = 20mso vi y vc v cch im B i din trn b bn kia(cng cao, cng nm trong mt phng ct) mt khongl= 50m, bn mt qu n pho xin ln vi vn tc v0 =20m/s, theo hng hp vi phng nm ngang gc =600. B qua lc cn ca khng kh v ly g = 10m/s 2. Hyxc nh khong cch t im ri ca vt n v tr nmvt.Nhn xt

Nu ta v phc ha qu o chuyn ng ca vt sau khi nm th thy imnm vt v im vt ri l hai giao im ca hai parabol. V tr cc giao im c xcnh khi bit phng trnh ca cc parabol.Gii

Chn h ta xOy t trong mt phng qu o ca vt, gn vi t, gc O tiy vc, Ox nm ngang cng chiu chuyn ng ca vt, Oy thng ng hng ln.Gc thi gian l lc nm vt.

Hnh ct ca b vc c xem nh mt phn parabol (P1) y = ax2 i qua imA c ta

(x = - )hy;2

!l

Suy ra 20 = a(- 25)

2

a = 125

4

Phng trnh ca (P1): 2x125

4y !

Phng trnh chuyn ng ca vt:

!!

!!

20t3105thsinvgt2

1y

2510t2

cosvx

2

0

2

0

t

lt

Kh t i ta c phng trnh qu o (P2):

9)3(204

5x

2

532x

20

1y 2

!

im ri C ca vt c ta l nghim ca phng trnh:

h

l

0v

T

E

A B

h

0vT

E

A B

C

x(m)O

y(m)


20/21

!

!

9)3(204

5x

2

532x

20

1y

x2000

1y

2

2

vi 20my25m,x {{

Suy ra ta im ri: xC = 15,63m v yC = 7,82m

Khong cch gia im ri C v im nm A l42,37m2)yA(y

2)CxA(xAC !!

Mt s bi ton vn dng

Bi 1

T nh dc nghing gc so vi phng ngang, mt vt c phng ivi vn tc v0 c hng hp vi phng ngang gc . Hy tnh tm xaca vt trn mt dc.

S: gcos

)(sin.cos2v

s 2

2

0

! Bi 2

Trn mt nghing gc so vi phng ngang, ngi ta gi mt lng tr khi lngm. Mt trn ca lng tr nm ngang, cchiu di l, c t mt vt kch thckhng ng k, khi lng 3m, mpngoi M lng tr (hnh v). B qua ma stgia vt v lng tr, h s ma st gia lngtr v mt phng nghing l k. Th lng trv n bt u trt trn mt phng

nghing. Xc nh thi gian t lc th lngtr n khi vt nm mp trong M lng

tr.

S:EEE cos)cossin(2

!kg

lt

Bi 3

Hai xe chuyn ng thng u vi cc vn tc v1, v2 (v1


21/21

Documents

Bai Tap Dong Luc Hoc Chat Diem Co Giai Chi Tiet.6968(2)