Bai Tap Dong Luc Hoc Vat Ran

7/31/2019 Bai Tap Dong Luc Hoc Vat Ran

1/22

z

ng lc hc

Vt rn

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2/22

PHN THNHTBAI 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 ==

Vi l xo 1: k1l1 = m1g (1)Vi l xo 1: k2l2 = m2g (2)Lp t s (1), (2) ta c

223

5,12

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. Tnh lcko 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

010t

VVa 20 =

=

=

Theo nh lut II Newtn :

=+ 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 xoni vi mt vt khi lng m = 1kg. Ly g = 10m/s2. Tnh chiu di l xo khi vt cn bng.

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3/22

Bi gii:

Khi cn bng: F1 + F2=Vi F1 = K1l; F2 = K21nn (K1 + K2) l = P

)m(04,025010.1

KKP

l21

==+

=

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

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

Bi gii:

Hng v chiu nh hnh v:

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

Tc dng vo vt gm 2 lc n hi

1F ; 2F

,

=+ FFF 21Chiu ln trc Ox ta c :

F = F1F2 = (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 dy

khng dn, khi lng khng ng k. Khi lng 2 vt l mA = 2kg, mB = 1kg, ta tc dng vovt A mt lc F = 9N theo phng song song vi mt bn. H s ma st gia hai vt vi mtbn l m = 0,2. Ly g = 10m/s2. Hy tnh gia tc chuyn ng.

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4/22

Bi gii:

i vi vt A ta c:

=++++11ms1111amFTFNP

Chiu xung Ox ta c: F T1F1ms = m1a1Chiu xung Oy ta c: m1g + N1 = 0Vi F1ms = kN1= km1g

F T1k m1g = m1a1 (1)* i vi vt B:

=++++ 22ms2222 amFTFNP

Chiu xung Ox ta c: T2F2ms = m2a2

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

T2k 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)a2

21

21 s/m112

10).12(2,09mm

g).mm(Fa =

++

=+

+=

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

khi lng khng ng k. Mt trong 2 vt chu tc ng ca lc koF hp vi phng ngang

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

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

Bi gii:

Vt 1 c :

=++++11ms1111 amFTFNP

Chiu xung Ox ta c: F.cos 300T1 F1ms = m1a1Chiu xung Oy : Fsin 300P1 + N1 = 0V F1ms= k N1 = k(mg Fsin 300)F.cos 300 T1k(mg Fsin 300) = m1a1 (1)

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5/22

Vt 2:

=++++ 22ms2222 amFTFNPChiu xung Ox ta c: T F2ms= m2a2Chiu xung Oy : P2 + N2 = 0M F2ms= k N2 =km2g

T2k m2g = m2a2Hn na v m1 = m2= m; T1 = T2 = T ; a1 = a2 = aF.cos 300T k(mg Fsin 300) = ma (3) T kmg = ma (4)T (3) v (4)

m

00

t2

)30sin30(cosTT

+=

20

21

268,023

10.2

30sin30cos

T2F

00m =

+=

+

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 si dynh khng dn v vt qua rng rc c nh nh hnh v. B qua khi lng ca rng rc v lcma st gia dy vi rng rc. Ly g = 10m/s2. Tnh gia tc chuyn ng ca mi vt.

Bi gii:

Khi th vt A s i xung v B s i ln do mA > mB vTA = TB = TaA = aB = ai vi vt A: mAg T = mA.ai vi vt B: mBg + T = mB.a* (mAmB).g = (mA + mB).a

2

BA

BA s/m210.400600400600

g.mm

mma* =

+

=+

=

Bi 8:

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6/22

Ba vt c cng khi lng m = 200g c ni vi nhau bng dy ni khng dn nh hnh v.H s ma st trt gja vt v mt bn l = 0,2. Ly g = 10m/s2. Tnh gia tc khi h chuynng.

Bi gii:

Chn chiu nh hnh v. Ta c:

=++++++++++ aMPTTNPFTTNPF 11222ms234333

Do vy khi chiu ln cc h trc ta c:

==

=

3ms4

2ms32

11

maFT

maFTT

maTmg

V

aaaa

'TTTTTT

321

43

21

===

====

=

=

=

maFT

maFTT

maTmg

ms'

ms'

=

=

m3mg2mg

ma3F2mg ms

2s/m210.3

2,0.21g.321a ===

Bi 9:Mt xe trt khng vn tc u t nh mt phng nghing gc = 300. H s ma st trt l = 0,3464. Chiu di mt phng nghing l l = 1m. ly g = 10m/s2

v

3 = 1,732 Tnh gia tc chuyn ng ca vt.

Bi gii:

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7/22

Cc lc tc dng vo vt:

1) Trng lcP

2) Lc ma st

msF

3) Phn lc

N ca mt phng nghing4) Hp lc

=++= amFNPF ms

Chiu ln trc Oy: Pcox + N = 0N = mg cox (1)Chiu ln trc Ox : PsinFms = maxmgsin N = max (2)t (1) v (2) mgsin mg cox = maxax = g(sin cox)

= 10(1/2 0,3464. 3/2) = 2 m/s2BAI 10 :Cn tc dng ln vt m trn mt phng nghing gc mt lc F bng bao nhiu vt nm yn, h s ma st gia vt v mt phng nghing l k , khi bit vt c xu hng trtxung.

Bi gii:

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

0FNPF ms=+++

Chiu phng trnh ln trc Oy: N PcoxFsin = 0 N = Pcox + F sin

Fms = kN = k(mgcox + F sin)Chiu phng trnh ln trc Ox : PsinF coxFms = 0 F cox = PsinFms = mg sinkmg coxkF sin

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8/22

+

=+

=

ktg1)ktg(mg

sinkcos)kcox(sinmg

F

BAI 11 :Xem h c lin kt nh hnh vm1 = 3kg; m2 = 1kg; h s ma st gia vt v mt phng nghing l = 0,1 ; = 300; g = 10m/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:

=+++ 11ms11 amFTNP

Chiu h xOy ta c: m1gsinT N = ma m1g cox + N = 0* m1gsinT m1g cox = ma (1)i vi vt 2:

=+ 2222 amTP

m2g + T = m2a (2)Cng (1) v (2) m1gsin m1g cox = (m1 + m2)a

)s/m(6,04

10.1233.1,0

21.10.3

mm

gmcosmsingma

2

21

211

=

+

=

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

BAI 12 :Sn i c th coi l mt phng nghing, gc nghing a = 30 0 so vi trc Ox nmngang. T im O trn sn i ngi ta nm mt vt nng vi vn tc ban u V0 theo

phng Ox. Tnh khong cch d = OA t ch nm n im ri A ca vt nng trn sn i,Bit V0 = 10m/s, g = 10m/s2.

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9/22

Bi gii:

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

=

=

2

0

gt21

y

tVx

Phng trnh qu o

)1(x

V

g21

y 220

=

Ta c:

====

sindOKycosdOHx

A

A

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

220

)cosd(V

g21

sind =

m33,130cos

30sin.

1010.2

cossin

.g

V2d

0

0220 ==

=

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 42 m.Tm vn tc ca hn khi nm ?GIAIChn gc O ti mt t. Trc Ox nm ngang, trc Oy thng ng hng ln (qua im nm).Gc thi gian lc nm hn .Cc phng trnh ca hn

x = V0cos450t (1)y = H + V0sin 450t 1/2 gt2 (2)Vx = V0cos450 (3)Vy = V0sin450gt (4)

T (1)

00 45cosV

xt =

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10/22

Th vo (2) ta c :

)5(45cosV

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

421.22

9.442

Hx.45tg45cos

2g

.xV

045cosV

xg21x45tgH

000

0220

20

=+

=+

=

=+

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

ng qua my bay n xe tng) khi my bay v xe tng chuyn ng cng chiu.Bi gii:

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

x = V1t (1)y = 1/2gt2 (2)

Phng trnh qu o:

220

xV

g21

y=

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

v

g

h2

g

y2t ==

g

h2Vx 1B =

Lc t = 0 cn xe A

gh2

VtVAB 22 ==

* Khong cch khi ct bom l :

)=== 2V(Vgh2

)VV(ABHBHA 121

BAI 15 :T nh mt mt phng nghing c gc nghing so vi phng ngang, ngi tanm mt vt vi vn tc ban u V0 hp vi phng ngang gc . Tm khong cch l dc theo

mt phng nghing t im nm ti im ri.

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11/22

Bi gii;Cc phng thnh to ca vt:

)2(

gt2

1tsinVHy

)1(tcosVx

20

0

+=

=

T (1)

=

cosV

xt

0 Th vo (2) ta c:

(3)cosV

xg

21

xtgHy22

0

2

+=

Ta c to ca im M:

=

=

sinlHy

coslx

M

M

Th xM, yM vo (3) ta c:

+= 220

22

cosV2

cosglcosltgHsinlH

+=

+

=

+

=

220

220

222

0

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 cho

qu n ri v pha bn kia ca to nh v gn bc tng AB nht. Bit to nh cao h = 20 mv 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.

Bi gii:

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12/22

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

220

xV

g21

y=

n chm t gn chn tng nht th qu o ca n i st nh A ca tng nn2A2

0A x

V

g

2

1y =

s/m25100.80.210.1

x.yg

21

V AA

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 = xCl = 11,8 (m)BAI 17 :Mt vt c nm ln t mt t theo phng xin gc ti im cao nht ca quo 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

yx VVV +=Ti S: Vy = 0

== cosVVV oxsM

oo

s 602

1

cos2

V

V ===

V

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13/22

( )s/m20

23

15x10x2

sin

gy2V

g2

sinVy so

2o

x ==

=

=

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 bng bao nhiu?Ly g = 10m/s2.

Bi gii:

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

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

g

2sinVAB 1

2 =

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

412sin 11

==

V thnh phn ngang ca cc vn tcu bng nhau V0cos = V.cos1

1o

cos.VV

cos =

Vi

==

21

cosgh2VV

1

2

o

Nn

( ) 21

102

1x1021

V

gh21

21

.V

gh2Vcos

22oo

2o ===

=

o60=

BAI 19 :Mt bn nm ngang quay trn u vi chu k T = 2s. Trn bn t mt vt cch

trc quay R = 2,4cm. H s ma st gia vt v bn ti thiu bng bao nhiu vt khng trttrn mt bn. Ly g = 10 m/s2 v 2 = 10Bi gii:

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14/22

Khi vt khng trt th vt chu tc dng ca 3 lc:

nghF;N,P msTrong :

0NP =+

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

=

=

)2(mg.F

)1(RmwF

ms

2ms

gRw

g.Rw2

2

Vi w = 2/T = .rad/s

25,010

25,0.2=

Vy min = 0,25BAI 20 :Mt l xo c cng K, chiu di t nhin l 0, 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= 20 rad/s; m = 10 g ; k = 200 N/m

Bi gii:

Cc lc tc dng vo qu cu

dhF;N;P

( )

( )2

o2

o

22

o2

mwK

lmwl

lmwmwKl

llmwlK

=

=

+=

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15/22

vi k > mw2( )

( )m05,0

20.01,0200

2,0.20.01,0l

2

2

=

=

BAI 21 :Vng xic l mt vnh trn bn knh R = 8m, nm trong mt phng thng ng.Mt ngi i xe p trn vng xic ny, khi lng c xe v ngi l 80 kg. Ly g = 9,8m/s2

tnh lc p ca xe ln vng xic ti im cao nht vi vn tc ti im ny l v = 10 m/s.Bi gii:

Cc lc tc dng ln xe im cao nht l N;P

Khi chiu ln trc hng tm ta c

N2168,98

1080g

Rv

mN

Rmv

NP

22

2

=

=

=

=+

BAI 22 :Mt qu cu nh c khi lng m = 100g c buc vo u 1 si dy di l = 1m

khng co dn v khi lng khng ng k. u kia ca dy c gi c nh im A trntr quay (A) thng ng. Cho trc quay vi vn tc gc w = 3,76 rad/s. Khi chuyn ng nnh hy tnh bn knh qu o trn ca vt. Ly g = 10m/s2.Bi gii:

Cc lc tc dng vo vt P;T

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

TPF +=vi

=

RmwF

PF2

gRw

mgF

tgv2

==

R = lsin

=

=

cos

sin

g

sinlwtg

2

Vo

2245707,0

1.76,3

10

lw

gcos0 ====

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

15


16/22

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 ca mttrng.Bi gii:Mt trng cng tun theo quy lut chuyn ng ca v tinh nhn to.Vn tc ca mt trng

RGMv 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 9,7x24.3600.27.64004

v.TRRR

R

TvR2

R.T2

v;R

R

vv

R

GMv

5

2

22

2

2

oo3oo

o

o

o

oo

=

===

==

=

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

tr thp hn O. OA hp vi phng thng ng gc = 60o

v vn tc qu cu l 3m/s, g =10m/s2.

Bi gii:

Ta c dng:

= amP;TChiu ln trc hng tm ta c

N75,093

21

x1005,0Rv

60cosgmT

Rv

mmaht60cosPT

220

2o

=

+=

+=

==

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

16


17/22

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 i uv chuyn ng ca cht im c nm ngang, nm xin).

+ Vit phng trnh chuyn ng ca cht im

++=

++=

00y

2

y

00x

2

x

ytvta2

1y

xtvta

2

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

== 21

21

yyxx

- 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 chiuui 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 cch kho lo

phng php ta th chng tr nn n gin v rt th v.

Xin a ra mt s v d:

Bi ton 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/s2. Trong lc

bung i ln, dy treo b t, lc ko F vn khng i. Tnh gia tc ngay sau cabung v thi gian vt ri xung sn bung. Ly g = 10m/s2.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 dy treo

b t. Hy ng ngoi thang my quan st (chn h quy chiu gn vi t) hai chtim vtvsn thangang chuyn ng trn cng mt ng thng. D dng vn dng

17


18/22

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 v tr sn lc dy t,gc thi gian t = 0 lc dy t.Khi dy treo cha t, lc ko F v trng lc P = (M + m)g gy ra gia tc a cho h M +m, ta c

F - P = (M + m)a 2310Ng)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

11,55m/s

M

MgFa =

=

+ Thi gian vt ri xung sn bungVt v sn thang cng chuyn ng vi vn tc ban u v0.Phng trnh chuyn ng ca sn thang v vt ln lt 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 a2 =-gVy

tv0,775ty 02

1 += v 2tv5ty 02

2 ++=Vt chm sn khi

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 ng ray

vi vn tc v0 = 7,2km/h th mt chic vali kch thc nh khi lng m1 = 5kg ct nh vo mp trc ca sn xe. Sau khi trt trn sn, vali c th nm yn trn snchuyn ng khng? Nu c th nm u? Tnh vn tc mi ca toa xe v vali. Cho

18

y

O

F

T

P

0v

0v

y02


19/22

bit h s ma st gia va li v sn l k = 0,1. B qua ma st gia toa xe v ng ray.Ly g = 10m/s2.

Nhn 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 sn xe,tc l hai cht im gp nhau. Ta a bi ton v dng quen thuc.

GiiChn trc Ox hng theo chuyn ng ca xe, gn vi ng ray, gc O ti v

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

11ms11 amFNP =++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'FNPP =+++

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

2t0,025ttvta

2

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

19

1N

F1P'

2N

1P

2P

msF'

xO


20/22

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

Khi vali cch mp sau xe mt khong2t0,025t40,5txxd 22

21 ++==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 ngc dng mt phn parabol (hnhv). T im A trn sn bvc, cao h = 20m so vi yvc v cch im B i din

trn b bn kia (cng cao,cng nm trong mt phng ct)mt khong l = 50m, bn mtqu n pho xin ln vi vntc v0 = 20m/s, theo hng hp vi phng nm ngang gc = 600. B qua lc cnca khng kh v ly g = 10m/s2. Hy xc nh khong cch t im ri ca vt n vtr nm vt.

Nhn xt

Nu ta v phc ha qu o chuyn ng ca vt sau khi nm th thy im nmvt v im vt ri l hai giao im ca hai parabol. V tr cc giao im c xc nhkhi 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 = ax 2 i qua im Ac ta

(x = - )hy;2

=l

Suy ra 20 = a(- 25)2a =1254

Phng trnh ca (P1):2x

125

4y =

Phng trnh chuyn ng ca vt:

20

h

l0

v

A B

h0

v

A B

C

x(m)O

y(m)


21/22

++=++=

==

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:

+

+=

=

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 = 7,82mKhong cch gia im ri C v im nm A l

42,37m2)ByA(y2)CxA(xAC =+=

Mt s bi ton vn dng

Bi 1

T nh dcnghing gc so

vi phng ngang,mt vt c phngi vi vn tc v0 chng hp vi

phng ngang gc . Hy tnh tm xa ca vt trn mt dc.

S:gcos

)(sin.cos2vs

2

2

0 +=

Bi 2

Trn mt nghing gc so vi phng ngang, ngi ta gi mt lng tr khi lng m.

Mt trn ca lng tr nm ngang, c chiu dil, c t mt vt kch thc khng ng k,

khi lng 3m, mp ngoi M lng tr (hnhv). B qua ma st gia vt v lng tr, h sma st gia lng tr v mt phng nghing lk. Th lng tr v n bt u trt trn mt

phng nghing. Xc nh thi gian t lc thlng tr n khi vt nm mp trong M lngtr.

21

0v

m

3m

l

MM


22/22

S: cos)cossin(2

=kg

lt

Bi 3

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

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