bai tap dong luc hoc chat diem co giai chi tiet.6968(2)
TRANSCRIPT
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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.
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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
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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
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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
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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.
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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 :
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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
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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
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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
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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.
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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
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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
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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:
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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
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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:
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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
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!
!
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
-
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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
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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)
-
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!
!
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
-
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