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7/31/2019 Bai Tap Dong Luc Hoc Vat Ran
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z
ng lc hc
Vt rn
1
7/31/2019 Bai Tap Dong Luc Hoc Vat Ran
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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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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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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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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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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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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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+
=+
=
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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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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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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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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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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( )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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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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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)
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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
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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
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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
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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
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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)
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++=++=
==
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
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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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