Huong Dan Giai Toan Vat Ly Bang May Tinh

8/3/2019 Huong Dan Giai Toan Vat Ly Bang May Tinh

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NGUYN TRNG SU (Ch bin) - NGUYN VN PHN

HNG DN THC HNH VT L BNGMY TNH CM TAY

CC DNG CU HI V BI TP(ti liu dnh cho lp tp hun GV)

THNG 10/2010


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Phn mt

GII THIU MY TNH CM TAY

Trc khi tnh ton, bn phi chnng Mode theo bng diy:


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PHP TNH N VO MODE

Tnh thng thng COMP

Gii phng trnh EQN

Ch : trli ci t ban u, ta n

Khi y: Tnh ton: COMPn vo gc: DegDng a +10n: Norm 1Dng phn s: ab/cDu cch phn l: chm (Dot)

1. Giai Tha:Tnh X! (X 0)V d: Tnh 12!

Nhp12n Kt qu: 479001600.

2. Cn bc hai, cn bc ba:

V d: 3 12549 Ta ghi vo m hnh ht nh v n =

49 125 Kt qu: 12.

3. Logarit thp Logarit t nhin:My k hiu:

Log: Logarit thpLn: Logarit Nepe

V d: Tnh log10100, Ln e4/7

n 100 Kt qu: 2n 4 7 KQ: 4/7

4. Gii phng trnh m:V d: 6x + 8x = 10x

n 6 8 10

n My hi X? nhp 3 nThao tc thnh cng my hinProcessingKt qu:2Ghi ch:

Chc nng SOLVE gii gn ng theo phng php Newton. Vi biu thc hay gi tri ban u khngcho ra kt qu.

Khi angtm nghim mn hnh hin Processing

5. H phng trnh bc nht hai nTm nghim h phng trnh bc nht hai n

222

111

cybxa

cybxa

V d:

1035

24512

yx

yx

SHIFT X!

SHIFT X! =

+ 3 =SHIFT

log =Ln ALPHA e ^ ( a b/c ) = a b/c

^ ALPHA x + ^ ALPHA x = ^ ALPHA x

SHIFT CALC = SHIFT CALC

Mode 1

Mode Mode Mode 1

SHIFT CLR 2 =


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Ta vo chng trinh gii h phng trnh bc nht hai n bng cch sau:n phm 3 ln bm1 vo chc nng EQN my hiUnKnowns? bm2 thchin gii phng trnh bc nht 2 n

My hi a1n 12

My hi b1n -5

My hi c1n -24

My hi a2n -5

My hi b2n -3My hi c2 n 10

Kt qu: X = -2, n cho kt qu Y = 0.

6. H phng trnh bc nht ba nTm nghim h phng trnh bc nht ba n

3333

2222

1111

dzcybxa

dzcybxa

dzcybxa

V d:

962

7352954

zy

zyx

zyx


My hi a1n 1

My hi b1n -4

My hi c1n 5

My hi d1n 9

My hi a2n 2

My hi b2n 5

My hi c2 n -3

My hi d2n -7

My hi a3n 0

My hi b3n -2

My hi c3 n 6

My hi d3n -9

Kt qu: X = 4.5192,n cho kt qu Y = -5.1346n cho kt qu Z = -3.2115

7. H phng trnh bc nht bnnMun tm nghim ca h phng trnh

MODE

=

=

=

=

==

=

MODE

=

=

=

==

=

=

=

=

=

=

=

=

=


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44444

33333

22222

11111

etdzcybxa

etdzcybxa

etdzcybxa

etdzcybxa

V d 1. Gii h phng trnh

5472372

80384 34825

30024

tzyx

tzyxtzyx

tzyx


Nhp vo cc h s ca h phng trnh:1 4 1 2 300

1 5 2 1 3484 8 1 3 802 7 3 2 547

Kt qu: x = 77, ny = 20, nz = 209, nt = 47

8. Phng trnh bc hai mtn

Tm nghim ca phng trnh bc haiax2 + bx + c = 0V d: Gii phng trnh

x2 + 9x + 8 = 0Ta vo chng trinh gii h phng trnh bc hai mtn bng cch sau:n phm 3 ln bm1 vo chc nng EQN my hinUnKnowns? bmmn hnh xut hinDegree?bm 2 thc hin gii phng trnh bc hai

Nhp vo cc h s ca phng trnh trn:1 9 8

Kt qu: x1 = -1 nx2 = -8.

9. Phng trnh bc ba mtnTm nghim ca phng trnh bc hai

ax3 + bx2 + cx + d = 0V d: Gii phng trnh

2x3 + x2 8x - 4 = 0Ta vo chng trinh gii h phng trnh bc nht hai n bng cch sau:n phm 3 ln bm1 vo chc nng EQN my hinUnKnowns? bmmn hnh xut hinDegree?bm 3 thc hin gii phng trnh bcba

Nhp vo cc h s ca phng trnh trn:2 1 8 4

MODE

= (-) = = =

= (-) = = (-)

= (-) = (-) =

== (-) = (-) = =

=

=

=

=

MODE

= = =

=

MODE

= = (-) = (-) =

=

==


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Kt qu: x1 = 2 nx2 = -2 nx3 = -0.5

10.FIX, SCI, RND ( Chn s ch s l, dng chun a+10n, tnh trn)n Mode 5 ln vo cc chc nng F IX, SCI, NORMa) Fix:nnh ch s l

V d 1. Tnh 200 : 7 = 28.57142857142857 mn hnh ch hin th 4 s sau du phy thp phn ta lm nh saun Mode 5 ln chn1 vo ch Fix mn hnh xut hinFix 0~9 ta nhp s 4 voKt qu: 200 : 7 = 28.5714

b) Sci: nnh s ch s caaV d 1. Tnh 200 : 7 = 28.57142857142857 mn hnh ch hin th kt qu vi 5 s ta lm nh saun Mode 5 ln chn2 vo ch Sci mn hnh xut hinSci 0~9 ta nhp s 5 voKt qu: 200 : 7 = 28.571c) Norm

n Mode 5 ln chn3 vo ch Norm mn hnh xut hinNorm 1~2 xa ci tFix v Sci ta chnNorm1 hocNorm 1

==


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Phn hai

CC DNG CU HI V BI TP

1. BN PHP TNH C BN, LU THA V KHAI CN.

Nhng im cn lu

Trong vic gii cc bi ton Vt l sau khi vn dng cc kin thc cbn v Vt l, mun tnh ran kt qu cui cng chng ta rt hay dng ti cc php tnh cng, tr, nhn, chia, lu tha v khaicn. Cc php tnh cng, tr, nhn, chia, lu tha v khai cn l cc php tnh c bn chng takhng th b qua c. Khi thc hin thnh tho cc php tnh cbn ny s gip ta tm c ktqu ca bi ton mt cch mau l v chnh xc.

Trong vic thc hin cc php tnh cbn ni trn cn phn bit php tr v du tr (-);Exp v 10 ^ , i khi chng cho kt qu nh nhau, nhng ni chung l khc nhau. Mun tnh chnhxc chng ta khng nn ghi cc kt qu trung gian ra giy ri nhp li vo my m nn nhcc ktqu vo nhc lp (Shift Sto) hoc nhmcnh Ans, m ch ghi kt qu cui cng.

Cc v d minh ho

Bi 1: Mt ngi bi dc theo chiu di 50m ca mt b bi ht 20,18s ri quay li v ch xutpht trong 21,34s. Hy xc nh tc trung bnh ca ngi trong cc trng hp sau:

a) Trong khong thi gian bii.

b) Trong khong thi gian bi v.

c) Trong sut c thi gian bii v bi v.

Cch gii Hng dn bm my v kt qu

Theo nh ngha v tc trung bnhtSvTB

a) Trong khong thi gian bii:s/m,vTB 47772 .

b) Trong khong thi gian bi v:s/m,vTB 34302 .

c) Trong sut c thi gian bii v bi v:s/m,vTB 40852

50 20.18 =

KQ: 2.477700694

50 21.34 =

KQ: 2.343017807

100 ( 20.18 + 21.34 ) =

KQ: 2.408477842

Bi 2: Lc 7h mt t chy t Hi Phng v H Ni vi tc khng i 45km/h. Cng lc mt

t chy t H Nii Hi Phng vi tc khng i 65km/h. Bit khong cch H Ni - HiPhng l 105km.

a) Hy lpphng trnh chuyn ng ca hai xe trn cng mt trc to, ly gc to ti HNi, chiu dng hng t H Ni ti Hi Phng, gc thi gian l lc 7h.

b) Xc nh thiim v v tr hai xe gp nhau.


a) Lpphng trnh chuyn ng ca mi xe:

- Xe t Hi Phng v H ni c hng chuyn


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ng ngc vi trc to, v tr ban u ti HiPhng nn phng trnh chuyn ng l:

)h,km(tx 451052 .

- Xe t H nii Hi Phng c hng chuyn ngcng chiu trc to , v tr ban u ti H Ninnphng trnh chuyn ng l: )h,km(tx 652 .

b) Khi hai xe gp nhau th chng phi c cng to

, tc l 21 xx 105 45t= 65t 110t= 105 h,t 95450 .

Thiim hai xe gp nhau l 7,9545 h.

Hai xe gp nhau ti v tr cch H Nikm,x 04545622

105 110 =

KQ: 0.954545454

Ans + 7 =

KQ: 7.954545455

= Ans x 65 =

KQ: 62.04545455

Bi tp vn dng

1.1. Mt ngi chy trn mtngonngt di 200m ht thi gian 30s. Sau ngi nychy thm trn mton ng nha di 150m ht thi gian 20s. Hy xc nh tc trung bnhca ngi trong cc trng hp sau:

a) Trong khong thi gian chy trn ngt.

b) Trong khong thi gian chy trn ng nha.

c) Trong con ngt v ng nha.

p s: a)6,6667 m/s.

b) 7,5m/s.

c) 7m/s.

1.2. Lc 10h mt t chy t Hi Phng v H Ni vi tc khng i 55km/h. Cng lc mtt chy t H Nii Hi Phng vi tc khng i 40km/h. Bit khong cch H Ni - HiPhng l 105km.

a) Hy lpphng trnh chuyn ng ca hai xe trn cng mt trc to, ly gc to ti HiPhng, chiu dng hng t H Ni ti Hi Phng, gc thi gian l lc 10h.

b) Xc nh thiim v v tr hai xe gp nhau.

p s: a) x1 = - 55t km, x2 = -105 + 40t (km).

b) t = 11h6pht19s; x1 = x2 = 60,7895km.

1.3. Trong na thi gian u, mt xe t chuyn ng thng vi tc trung bnh l h/kmv 351 ,trong na thi gian cn li xe chuyn ng thng vi tc trung bnh l h/kmv 452 . Hy tnhtc trung bnh ca xe trn ton thi gian chuyn ng.

p s: 40 km/h.

1.4. Mt xe ln khi lng m = 2kg c ko chuyn ng trn mt phng ngang bi lc ko F =8N hng theo phng ngang. Sau thi gian 3s k t lc btu chuyn ng xe i c 10m. Hytnh h s ma st gia xe ln v mtt. Ly 289 s/m,g .

p s: 0,1814.

2. GII H PHNG TRNH BC NHT HAI N, BA N.


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Nhng im cn lu

Cc h phng trnh bc nht hai n, ba n, bnn vi sphng trnh bng sn th my tnhcm tay VN 570MS c th giic mt cch d dng. c bit vi cc hphng trnh bc nhtc cc h s khng nguyn dn n vic tnh ton thng thng gp nhiu kh khn th my tnhcm tay li thc hin d dng.

Mun gii cc h phng trnh bc nht hai n, ba n, bn n ta a my v ch gii hphng trnh bc nht bng cch bm nh sau:

- Gii hphng trnh bc nht 2 n: Mode (3 ln) 1 2



Nhp cc h s cho h phng trnh, trong khi nhp cc h s c th thc hin cc php tnhthng thng, n khi bm = th gi tr ca h sc gn. Trong khi nhp cc h s ta phi nhp tt c cc h s, cn c bit ch n cc h s c gi tr bng 0 v nhm th t cc h s.Mun trnh nhm ln tt nht ta lp mt ma trn gm m hng v (m + 1) ct (vi m l s phngtrnh).

Cc v d minh ho

Bi 1: Treo ln lt cc vt khi lng m1

= 100g v m2

= 150g vo u di ca mt l xo (utrn ca l xo cnh), th chiu di ca l xo khi vt cn bng ln lt l l1 = 35cm v l2 = 37cm.Hy tnh cng v chiu di t nhin ca l xo. Ly g = 9,8067 m/s2.

Cch gii Hng dn bm my v kt quKhi vt cn bng, lcn hi ca l xo cn bngvi trng lc ca vt. T ta c hphng trnh

gmklkl

gmklkl

gm)ll(k

gm)ll(k

202

101

202

101

Gii hphng trnh ta c

)m(,l

)m/N(,k

,kl

,k

330

033549

181116

033549

00

Mode (3 ln) 1 20.35 =(-) 1 =0.1 x 9.8067 =0.37 =(-) 1 =0.15 x 9.8067 =

KQ: 49.0335=KQ: 16.181055Mode 116.181055 49.0335 =KQ: 0.33

Bi 2: Hai t chuynng thngu trn cng mtng thng, xut pht t hai im A, B cchnhau mt khong S = 100km vi vn tc v1 = 36km/h, v2 = 72km/h ngc chiu nhau. Xc nhthiim v v tr hai xe gp nhau chn A lm gc to, thi im ban u l lc hai xe xut pht.

Cch gii Hng dn bm my v kt quChn chiu dng l chiu chuyn ng ca xe

mt xut pht t A.Phng trnh chuynng ca xe xut pht t A lx1 = v1.t = 36t

Phng trnh chuynng ca xe xut pht t B lx2 = S - v2.t = 100 - 72t

Thiim v v tr hai xe gp nhau khi x1 = x2 = xl nghim ca h phng trnh

10072

036

tx

tx

Mode (3 ln) 1 2

1 =(-) 36 =0 =1 =72 =100 =KQ: 33.33333333=KQ: 0.9259259259


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Gii hphng trnh ta c

)h(,t

)km(,x

92590

333333

Bi tp vn dng

3.1. Mt vt ri t do khng vn tc ban u t cao h = 50m. Hy tnh thi gian t lc th vtn lc vt chuyn ng qua cao h = 13m. Ly g = 9,81m/s2.

p s: 2,7465 (s).

3.2. Ti hai bn xe A, B (AB = 80km) c hai xe cng khi hnh chuyn ng ngc chiu nhauhng v pha nhau. Xe xut pht t A chuyn ng u vi tc 40km/h, xe xut pht t Bchuyn ng nhanh dn u vi tc ban u 20km/h v gia tc 0,5km/h2. Hy xc nh thi imv v tr hai xe gp nhau.

p s: 1,3260 (h), cch A 53,0403 (km).

3.3. Vt khi lng m = 5kg chu tc dng ca mt lc khng i F = 50N, btu chuyn ngth trng thi ng yn. Hy xc nh khong thi gian cn thit vt chuyn ng c qungng 400m k t khi vt c tc 5m/s.

p s: 8,4582 (s).3.4. Mt t ang chuyn ng th t ngt hm phanh, lc hm khng i v bng 25% trng lcca xe. Hy tnh thi gian t lc btu hm phanh n lc xe dng hn. Bit rng ngay sau khihm phanh xe cn i c onng 32m mi dng li. Ly g = 9,81m/s2.

p s: 2,1159 (s).

4. HM M V LGARIT.

Nhng im cn luMy tnh cm tay gip rt ngn thi gian tnh ton ni chung v c bit n thay th hon

ton cc bng tra gi tr lgart thp phn. Gip chng ta gii cc bi ton c lin quan ti hm sm v hm s lgart. My tnh b ti c th tnh ton c gi tr ca hm s m vi cc cs cngha; tnh c lgart ca mt s dng vi cs 10, cs e (cs t nhin) v c th tnh cvi cs bt k (c ngha) m khng cni cs. Vi cc my tnh khng tnh c vi cs bt

k th ta cn dng cng thc i csalog

blog

aln

blnbloga .

Vic tnh ton vi cc hm s v hm s lgart ta my tnh ch Mode 1. Vi hm m vlgart c th tnh ton trong cc ch gii phng trnh, hphng trnh, .... tng t nh bn

php tnh cbn.

Cc v d minh ho

Bi 1: Mt ngun m S (ngunim) pht ra mt m, tiim M cch ngun m mt khong SM =2m c cng m IM = 2.10

-5 (W/m2).

a. Hy tnh mc cng m ti M bit ngng nghe ca m l I0 = 10-9 (W/m2).

b. Tnh cng m v mccng m tiim N cch ngun m mt khong SN = 5,5m.B qua s hp th m ca mi trng.

Cch gii Hng dn bm my v kt qua. Mc cng m ti im M c tnh theocng thc


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0I

IlgL MM = 4,3010 (B)

b. V ngun m S l ngunim v ng hng, bqua s hp th m ca mi trng nn cng m t l nghch vi bnh phng khong cch ti

ngun:2

2

SN

SM

I

I

M

N .

Cng m ti N l2

2

SNSMII MN =2,6446.10

-6

(W/m2).

Mc cng m ti N l0I

IlgL NN = 3,4224(B).

log ( 2 Exp (-) 5 1 Exp (-) 9 ) =KQ: 4.301029996

2 Exp (-) 5 x 2 x2

5.5 x2

=KQ: 2.644628099x10-6

log ( Ans 1 Exp (-) 9 ) =KQ: 3.422364608

Bi 2: Tnh tui ca mt ci tng c bng g, bit rng phng x - ca C146 trong n bng

0,707 ln phng x ca mt khc g va mi cht. Chu k bn r ca C146 l T = 5600 nm.


phng x

-

ca C14

6 c tnh theo cng thc

2200

20

ln

H

)t(Hln

.T)t(H

Hlog.Tt

H)t(H

T/t

t = 2801,2201 (nm)Vy tui ca tng g khong 2800 nm.

(-) 5600 x ln 0.707 ln 2 =KQ: 2801.220127

Bi tp vn dng

4.1. Mt ngun m S (ngunim) pht ra mt m, tiim M cch ngun m mt khong SM =3m c cng m IM = 1,2.10

-5 (W/m2).

a. Hy tnh mc cng m ti M bit ngng nghe ca m l I0 = 10-9 (W/m2).b. Tnh cng m v mccng m tiim N cch ngun m mt khong SN = 6,5m.

B qua s hp th m ca mi trng.

p s: a. LM = 4,0792B.

b. IN = 2,5562.10-6W/m2; LN =3,4076B.

4.2. Tnh tui ca mt ci tng c bng g, b it rng phng x - ca C146 trong n bng 0,57

ln phng x ca mt khc g va mi cht. Chu k bn r ca C146 l T = 5600 nm.

p s: 4541,4106 nm 4500 nm.

5. HM S LNG GIC.


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Nhng im cn luMy tnh cm tay gip vic tnh cc hm s lng gic (sin, cos, tan v ctan) cng cc hm

ngc ca chng trln d dng, vic khng cn phi dng thc tnh gi tr ca hm lng gichoc bn s tra cc gi tr ca hm lng gic. Vi my tnh cm tay c th tnh gi tr ca mthm s lng gic vin v ca bin s l radian (rad) hoc (0). Vi hm ngc acrsinx vacrcosx, gi tr cabin s x phi thucon [ - 1 ; +1]; gi tr ca hm ngcc tnh ra n vrad hoc.

t ch cho my tnh chn v rad hoc vi my tnh VN 570MS ta lm nh sau:

Mode (4 ln) 1 (khi dng n v); hoc Mode (4 ln) 2 (khi dng n v rad). Trong qu trnhtnh ton ta c thin v nhShift DRG. V d my ang ch dng n v (Dec) muntnh sin300 ta bm sin 30 =, my cho kt qu 0,5. Nhng ta khng mun chuyn chn v mtnh ngay sin(30rad) th ta bm sin 30 Shift DRG 2 =, my cho ta kt qu - 0,988031624.

Trong cc bi ton v d minh ho sau y, cc n v gc (0 ; rad) coi nh my tnh c tch ph hp.

Cc v d minh ho

Bi 1: Mt vt c khi lng m = 500g t trn nh mt mt phng nghing hp vi phngngang mt gc = 600. Cho chiu di mt phng nghing l= 2m, g = 9,8m/s2. Tm gia tc ca vttrong cc trng hp sau.

a. B qua mi ma st.b. H s ma st trt gia vt v mt phng nghing l t = 0,2.

Cch gii Hng dn bm my v kt quGia tc ca vt c tnh theo cng thc tng qut

a = k msF F

m

, vi Fms = t.P.cos

Fk= Psin a = g.sin - t.g.cos

a. Thay t = 0 ta tnh c gia tc ca vt la = 4,9m/s2.

b. Gia tc l a 3,2026 (m//s2)

9.8 x cos 60 =

KQ: 4.9

9.8 x cos 60 0.2 x 9.8 x sin 60 =

KQ: 3.202590209

Bi 2: Mt vt m t trn nh mt mt phng nghing hp vi phng ngang mt gc = 300. Mtphng nghing c h s ma st trt vi vt l t = 0,5 cho chiu di mt phng nghing l= 2m, g =9,8m/s2. Cho vt trt t trng thi ngh. Hy tm vn tc ca vt chn mt phng nghing.

Cch gii Hng dn bm my v kt qu bin thin cnng W2 W1 = Asm.

mglsin - 21

2mv = Fms.l= l.t mgcos

Suy ra )cos(singlv t 2 1,6305 m/s

( 2 x 2 x 9.8 x (sin 30 0.5 x cos 30 ) )=

KQ: 1.62046354

Bi tp vn dng5.1. Mt vt c khi lng m = 500g t trn nh mt mt phng nghing hp vi phng ngangmt gc . Cho chiu di mt phng nghing l = 1,2m, g = 9,8m/s2. Th gia tc ca vt trongtrng hp sau.Mt phng nghing khng ma st l a = 2,5m/s2. Xc nh gc .

p s: 1404646,67.

5.2. Mt vt c khi lng m = 500g t trn nh mt mt phng nghing hp vi phng ngangmt gc = 250. Mt phng nghing c h s ma st trt vi vt l t = 0,25 cho chiu di mt

phng nghing l= 2,5m, g = 9,8m/s2. Cho vt trt vi vn tc ban u 05m/s t nh mt phngnghing. Hy tm vn tc ca vt chn mt phng nghing.


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p s: 5,8827 m/s.

5.3. Mt vt c khi lng m = 250g c treo vo mt si dy nm cn bng trn mt phngnghing mt gc so vi phng ngang l =250 v c h s ma st gia vt v mt phng nghingl = 0,025. Cho g = 9,87m/s2. Tnh lc cng cc tiu ca si dy.

p s: 0,9869 N.

6. O HM, VI PHN V TCH PHN

Nhng im cn luCc php tnh o hm bc nht, vi phn bc nht v tch phn mt lp c th dng my tnh

cm tay tnh ton mt cch d dng. Vic dng my tnh cm tay sa chng ta n kt qubng s cui cng ch khng a ra cng thc tng qut, nn cc bi ton cn lyo hm t bchai trln, cc bi ton c s dng tch phn nhiu lp ta vn phi dng cc cng thc ton hca ra cng thc tng qut ri sau thay s mic kt qu.

Dng tng qut ca cchbm my khi tnh o hm v tch phn nh sau (VN 570MS):

- o hm: Shift d/dx , = . c vit di dng

mt bin X, ta c th dng cc php tnh c thtrong my v phm Anpha X lp hm s.Cc v d minh ho

Bi 1: Mt chtim chuyn ng theophng trnh x = 4t2 7t + 5 (x o bng m, t o bng s).Hy tnh vn tc ca chtim ti thiim t = 12s.

Cch gii Hng dn bm my v kt quVn tc l o hm bc nht ca to theo thigian: v = xv = ( 4t2 7t + 5 ) = 89m/s

SHIFT dx ( 4 ALPHA X x2 7 ALPHA

X + 5 ) , 12 ) =KQ: 89

Bi 2: Mt vt chuyn ng vi gia tcph thuc vo thi gian theo cng thc a = 0,2t +1 (m/s2).

Hy tnh vn tc ca vt ti thiim t = 16s.Cch gii Hng dn bm my v kt quGia tc bng o hm bc nht ca vn tc theo

thi gian nn v =16

0

(0,2 1)t dt = 41,6 m/sdx ( 0.2 ALPHA X +1)

, 0 , 16 ) =

KQ: 41.6

Bi tp vn dng

6.1. Mt chtim chuyn ng theophng trnh x = 3t2 5t + 2 (x o bng m, t o bng s). Hytnh vn tc ca chtim ti thiim t = 2s.

p s: 7m/s.6.2. Mt vt chuyn ng vi gia tcph thuc vo thi gian theo cng thc a = 2t +1 (m/s2). Hytnh vn tc ca vt ti thiim t = 5s.

p s: 30m/s.

6.3. Vt m = 2kg chu tc dng ca mt lc biniu theo cng thc F = 5 + 2t (F o bng N, to bng s). Hy tnh vn tc ca vt sau khi tc dng lc 10s.

p s: 75m/s.

6.4. Mt vt c nm ngang vi tc ban u v0 = 20m/s. Hy tnh qung ng m vt chuynng c sau khi nm 3s. Ly g = 9,81m/s2.


14/2614

p s: 77,408 m.

7. HNG S VT L - I N V VT L.

Nhng im cn lu

Vi my tnh cm tay, ngoi cc tin ch nh tnh ton thun li, thc hin cc php tnh nhanh,n gin v chnh xc th phi k ti tin ch tra cu mt s hng s vt l v i mt sn v trong vt l. Cc hng s vt l c ghi trong b nhca my vin v trong hn v SI.

Mun ly cc hng s vt l ta ch cn bm (VN 570MS): Const . C th cc hng sthng dng l:

Hng s vt l M s Cc bm my Gi tr

Khi lng prton (mp) 01 Const 01 1,67262158.10-27 (kg)

Khi lng ntron (mn) 02 Const 02 1,67492716.10-27 (kg)

Khi lng lectron (me) 03 Const 03 9,310938188.10-31 (kg)Bn knh Bo (a0) 05 Const 05 5,291772083.10

-11 (m)

Hng s Plng (h) 06 Const 06 6,62606876.10-34 (Js)

Khi lng 1u (u) 17 Const 17 1,66053873.10-27 (kg)

Hng s Faray (F) 22 Const 22 96485,3415 (mol/C)

in tch lectron (e) 23 Const 23 1,602176462.10-19 (C)

S Avgar (NA) 24 Const 24 6,02214199.1023 (mol-1)

Hng s Bnzman (k) 25 Const 25 1,3806503.10-23 (SI)

Th tch mol kh iukin tiu chun (Vm)

26 Const 26 0,022413996 (m3)

Hng s kh l tng (R) 27 Const 27 8,314472 (J/mol.K)

Tc nh sng trongchn khng (C0)

28 Const 28 299792458 (m/s)

Hng s ini ca chnkhng (0)

32 Const 32 8,854187817.10-12 (SI)

Hng s t mi ca chnkhng (0)

33 Const 33 1,256637061.10-6 (SI)

Gia tc trng trng timtt (g)

35 Const 35 9,80665 (m/s2)

Nhit tuyt i (T) 38 Const 38 273,15 (K)

Hng s hp dn (G) 39 Const 39 6,673.10-11 (Nm2/kg2)

in v vt l ta bm Shift Conv . Vi cc m s c th tra trong bng nm trong npsau ca my.

Cc v d minh ho


15/2615

Bi 1: Mt vt ri t do khng vn tcban u ti ni c gia tc trng trng g. Hy xc nh vntc v qung ng vt ric sau thi gian t = 2,5s.

Cch gii Hng dn bm my v kt quSau 2,5s vttc vn tc

v = gt 24,5166 m/sQung ngi c l

S = 21

2gt 30,6458 m

Const 35 x 2.5 =KQ: 24.516625

Const 35 x 2.5 x2 2 =

KQ: 30.64578125Bi 2: Mt vtc nm thngng ln trn vi vn tc ban u v0 = 15m/s ti ni c gia tctrng trng g. Hy xc nh

a. Thi gian t lc nm n lc vt quay trli ch nm.

b. Vn tc, cao ca vt so vi v tr nm sau thi gian 2s.

Cch gii Hng dn bm my v kt qua. Thi gian t lc nm n lc vt trli l

t =2 ov

g 3,0591s

b. Vn tc ca vt sau 2s l

v = vo gt - 4,6122 m/s cao ca vt l

h = vot - 212

gt 10,3867 m

2 x 15 Const 35 =KQ: 3.059148639

15 - Const 35 x 2 =KQ: - 4.6122

15 x 2 2 x2 Const 35 2 =KQ: 10.3867

Bi tp vn dng

7.1. Mt vt ri t do khng vn tcban u ti ni c gia tc trng trng g. Hy xc nh vn tcv qung ng vt ric sau thi gian t = 3,5s.

p s: s 60,0657 m.

7.2. Mt vtc nm thngng ln trn vi vn tcban u v0 = 20m/s ti ni c gia tc trng

trng g. Hy xc nha. Thi gian t lc nm n lc vt quay trli ch nm.

b. Vn tc, cao ca vt so vi v tr nm sau thi gian 1,2s.

p s: a. t 4,0789s

b. v 8,2320 m/s; h 16,9392 m.

7.3. Mt vtc nm theo phng ngang vi vn tc v0 = 5m/s cao h = 15m so vi mtt.Hy tnh tm xa ca vt.

p s: s 8,7452 m/s.

7.4. Tri t v Mt Tri c khi lng ln lt l kg., 2410965 v 1,97.1030kg. Khong cch trung

bnh gia Tri t v Mt Tri l 150 triu km. Tnh lc ht gia chng.p s: F 3,4822.1016 N.

8. CC BI TON TNG HP.


16/2616

Nhng im cn lu

Cc bi ton vt l cng c khi ch dng mt s t php tnh, nhng i khi cng phi dng nhiuphp tnh, nhiu ch tnh ton. Mt bi ton tng hp phi dng nhiu bc tnh ton, mun tnhton chnh xc th cc kt qu trung gian ta khng c lm trn. khc ph iu ny, chng tann nhcc kt qu trung gian vo cc nhc lp; cng c trng hp khng th nhc ktqu trung gian vo cc nh th bt buc chng ta phi ghi ht cc s m my tnh hin ln mnhnh ra giy, sau chuyn ch tnh ton v nhpy cc s ghi vo my tnh. Vi cch

lm nh vy ta c th hn ch tia sai s.

Cc v d minh ho

Bi tp: Cho c h nh hnh 8.1. Nm c th trt t do trn mt phng ngang. Khi lng canm v trng vt l M = 2 kg v m = 500 g, ca cc rng rc khng ng k. B qua ma st mttip xc. Bit = 300. Tnh:

a. Gia tc ca nm M.

b. Gia tc ca trng vt m.


Ta thy rng nm M chuyn ng sang phi vi giatc a c hng t tri sang phi. Nu M i sang

phi mton x th m cng chuyn ng trn mtphng nghing ca nm mton x. Suy ra gia tcca m so vi M cngc ln l a, hng songsong vi mt phng nghing t trn xung di.

Phng trnh ng lc hc vit cho M v m ln

lt l:- T.cos + N.sin + T = M.a

- T + mg.sin + ma.cos = m.a

N + ma.sin mg.cos = 0.

Suy ra a =)cos1(m2M

sinmg

Gia tc ca m so vi mtt l am = 2asin(/2) =

)cos1(m2M

2sinsinmg2

Thay s a = 1,1489 m/s2; am = 0,5947 m/s

2.

Mode (4 ln) 1

0.5 x CONST 35 x sin 30 ( 2 + 2 x 0.5x ( 1 cos 30 ) ) =KQ: 1.1489

2 x Ans x sin ( 30 2 ) =

KQ: 0.5947

Bi tp vn dng

Mm

Hnh 8.1


17/2617

Mt cht im thc hindao ng iu ho dc theo trc Ox xung quanh v tr cn bng O vi chuk T = 2 s. Ti thiim t1 chtim c to x1 = 2 cm v vn tc v1 = 4 cm/s. Hy xc nh to

v vn tc ca chtim ti thiim t2 = t1 +3

1s.

Cch gii Hng dn bm my v kt quGi s phng trnh dao ng ca vt l x =A.sin(t) (chn pha ban u bng khng). Ti thiim t1 ta c x1 = A.sin(t1) = 2 cm v v1 =

A..cos(t1) = 4 cm/s. Ti thiim t2 = t1 + 1/3 sta c x2 =

)

3tsin(.A 1 A. sin (t1).cos 3

+ A.

cos (t1).sin3

= 2,1027 cm.

v v2 = .

)3

tcos(.A 1 .A.cos(t1).cos 3

-

.A. sin (t1).sin3

= -3,4414 cm/s.

KQ: 2.1027

KQ: -3.4414


18/2618

Phn baGII THIU THI NM 2008

B GIO DC V O TO K THI KHU VC GII TON VT L TRNMY TNH CM TAY NM 2008

Lp 12Thi gian: 150 pht Khng k thi gian giao

Ch :- thi ny gm 3 trang, 10 bi, mi bi 5 im.- Th sinh lm bi trc tip vo bn thi ny.

im ton bi thi Cc gim kho(H tn v ch k)

S phch(Do Ch tch Hi ng thi

ghi)Gim kho 1:Bng s Bng ch

Gim kho 2:

Quy nh:Hc sinh trnh by vn tt cch gii, cng thc p dng, kt qu tnh ton vo trnglin k bi ton. Cc kt qu tnh gn ng, nu khng c ch nh c th, c ngm nh chnh

xc ti 4 ch s phn thp phn sau du phy.

Bi 1: Mt cht im dao ng iu ho dc theo trc Ox theo phng trnh x = 2,5sin(4t + 0,21)+ 1,2cos(4t - 0,62) cm. Hy xc nh:

1. Chu k, bin , pha ban u dao ng ca vt.2. To , vn tc ca vt ti thi im t = 1,25 s.

Cch gii Kt qu


19/2619

Bi 2: T mt im A, mt vin c nm ln cao theo phng thng ng vi vn tc v = 15m/s. Sau mt khong thi gian t0, t mt im B cng cao vi A v cch A mt khong l = 4 m,mt vin th hai c nm xin mt gc = 500 so vi phng ngang, vn tc c ln 15 m/s,sao cho hai vin gp nhau. Hi vin th hai c nm sau vin th nht mt khong thi giant0 l bao nhiu?

Cch gii Kt qu

Bi 3: Cho c h nh hnh 3, cc vt c khi lng m1 =150 g, m2 = 100 g, m3 = 500 g, gc = 70

0, b qua mima st, dy khng dn, khi lng ca dy v rng rckhng ng k.

1. H trng thi cn bng. Hy xc nh gc .2. Hy xc nh gia tc ca mi vt sau khi t dy ni

gia m1 v m2.

Cch gii Kt qu

m2m3

m1

Hnh 3


20/2620

Bi 4: Hnh 4 l th chu trnh ca 1,5mol kh l tng trong mt phng to

p, T. Bit p1 = 1,5 atm, T1 = 320K, T2 =600K. Hy tnh cng m kh thc hintrong chu trnh.

Cch gii Kt qu

Bi 5: Cho mch in c s nh hnh 5, b qua in tr cacc ngun in v cc dy ni. Hy xc nh cng dngin qua cc in tr. Bit E1 = 12 V, E2 = 6 V, E3 = 9 V, R1 =15 , R2 = 33 , R3 = 47 .

Cch gii Kt qu

pp2 (2)

p1 (1) (3)T

T1 T2Hnh 4

E1 E2 E3

R1 R2 R3

Hnh 5


21/2621

Bi 6: Cho mch in xoay chiu RLC mc ni tip c R = 100 , cun thun cm L = 0,5284 Hv t in c in dung C = 100 F. t vo hai u on mch mt hiu in th xoay chiu u =220 2 sin100t V. B qua in tr ca cc dy ni. Hy xc nh:

1. Cng sut tiu th ca on mch.

2. Vit biu thc cng dng in trong mch v biu thc hiu in th tc thi gia hai ut in.

Cch gii Kt qu

Bi 7: Mt ng dy c t cm L = 2,00 H v in tr R0 =

1,00 c ni vi mt ngun in mt chiu c sut inng E = 3,00 V (hnh 7). Mt in tr R = 2,70 c mcsong song vi ng dy. Sau khi dng in trong ng t gi trn nh, ngun in b ct khi mch bng mt kho K. Tnhnhit lng Q to ra trn in tr R sau khi ngt mch. B quain tr ca ngun in v cc dy ni.

Cch gii Kt qu

L, R0

R

E K

Hnh 7


22/2622

Bi 8: Mt tia sng i tim S (S nm trong mi trngc chit sut n1 = 1) qua mt phng phn cch gia haimi trng n im S (Snm trong mi trng trongsut c chit sut n2 = 2 ). Bit v tr ca S v S cxc nh nh trn hnh 8, SH = 4 cm, HK = 2 3 cm, SK= 6 cm. Hy xc nh v tr ca im ti I ca tia sng it S ti mt phn cch gia hai mi trng.

Cch gii Kt qu

Bi 9: Mt thu knh c tiu c f = 25,0 cm, c ca i theo mt phng cha quang trc chnh vvung gc vi tit din ca thu knh, ri mi bt mi na i mt lp c b dy a = 1,00 mm. Sau dn li thnh lng thu knh. Mt khe sng S c t trn trc i xng ca lng thu knh,cch lng thu knh mt khong 12,5 cm, pht ra nh sng n sc c bc sng = 0,60 m. Sauv cch lng thu knh mt khong b = 175 cm ngi ta t mt mn nh vung gc vi trc ixng ca lng thu knh. Xc nh khong vn v s vn quan st c trn mn.

Cch gii Kt qu

SHnh 8

H

K S


23/2623

Bi 10: Ht nhn Po21084 phng x v bin i thnh ht nhn X. Cho khi lng cc ht nhn:

m(Po) = 209,9828u; m() = 4,0015u; m(X) = 205,9744u. Chu k bn r ca plni l 138 ngym.

1. Xc nh ht nhn X v tm nng lng to ra ca mt phn r (tnh ra n v J).

2. Tm khi lng ban u ca khi cht phng x bit phng x ban u ca n l 2 Ci. Tmkhi lng ca cht X to ra trong khong thi gian 30 ngy k t thi im ban u.

Cch gii Kt qu

Ghi ch: Cn b coi thi khng cn gii thch g thm.---------Ht---------

HNG DN CHO IM LP 12 NM 2008Bi Cch gii Kt qu im

1

1. T phng trnh dao ng x = 2,5sin(4t + 0,21) +1,2cos(4t - 0,62) cm, suy ra tn s gc trong dao ng ca vt

l = 4 rad/s chu k dao ng l T =2

= 0,5000 s.

Bin dao ng ca vt l:

)cos(AA2AAA 122122

21 = 3,4810 cm.

Pha ban u trong dao ng ca vt l c

AsinAsinA

sin

cosAcosAsinAsinA

tan

2211

2211

2211

t y ta tnh c = 0,4448 rad.

2. To ca vt ti thi im t = 1,25 s: Thay t = 1,25 s vophng trnh x = 2,5sin(4t + 0,21) + 1,2cos(4t - 0,62) cm tac x = -1,4978 cm.

Vn tc ca vt l v = x = 10cos(4t + 0,21) - 4,8sin(4t -

1. Chu k:T = 0,5000 s.

Bin :A = 3,4810 cm.

Pha ban u: = 0,4448 rad.

2. To :x = -1,4978 cm.

Vn tc:v = - 39,4876

1,0

1,0

1,0

1,0


24/2624

0,62) cm/s (*). Thay t = 1,25 s vo phng trnh (*) ta c v =- 39,4876 cm/s.

cm/s. 1,0

2

Chn h trc to 0x c gc 0 B, 0y hng thng ng lntrn, 0x nm ngang hng t B n A.Phng trnh chuyn ng ca cc vin trong h to trnl :

- Vin th nht: x1 = l. y1 = vt gt2/2.

- Vin th hai: x2 = v.cos.(t t0). y2 = v.sin.(t t0) g(t t0)2/2.

hai vin gp nhau th t v t0 phi tho mn h phng

trnh:

21

21

yy

xx

2t.g

t.v2

)tt.(gsin).tt.(v

lcos).tt.(v22

00

0

0

)cos.v.(2

g.l

cos

sin.lt.v

2

t.g

cos.v

l)tt(

2

22

0

Gii h phng trnh ta c t = 2,7703 s v t0 = 2,3555 s hoct = 0,2888 s v t0 = - 0,1260 s < 0 (loi). Vy vin th haic nm ln sau vin th nht mt khong t0 = 2,3555 s.

Gii h phngtrnh:

)s(1260,0t

)s(2888,0t

)s(3555,2t

)s(7703,2t

0

0

KL: t0 = 2,3555 s.

4,0

1,0

3

1. Khi h cn bng ta c (m1 + m2).g.sin = m3.g.sin suy ra

sin =3

21

m

sin)mm( = 280127,55.

2. Khi t dy ni gia m1 v m2 th h mt cn bng, m3 v m1

cng i xung, m2 i ln.Gia tc ca m1 l a1 = g.sin = 9,2152 m/s.

Gia tc ca m2 v m3 l a2 = a3 =32

23

mm

g)sinmsinm(

=

32

1

mm

sin.g.m

= 2,3038 m/s.

= 280127,55.

a1 = 9,2152 m/s.

a2=a3=2,3038m/s.

2,0

1,0

2,0

4

th biu din chu trnh ca 1,5 mol kh l tng cho trongh trc to p, V nh sau:

Cng m kh thc hin trong c chu trnh l A = A1 + A2 + A3trong :

+ A1 l cng m kh thc hintrong qu trnh ng tch (1)(2):

A1 = 0 J.

+ A2 l cng m kh thc hintrong qu trnh ng nhit (2)(3):

A2 = 3

1

V

V

dV.p vi1

11 p

T.R.nV ,

A1 = 0 J. 1,0pp2 (2)

p1 (1) (3)V

V1 V3Hnh 4 (A)


25/2625

1

23 p

T.R.nV ,

V

T.R.np 2

Tnh tch phn ta c A2 =4703,8942 J.

+ A3 l cng m kh thc hintrong qu trnh ng p (3) (1):

A3 = p1(V1 V3) = n.R.(T1 T2) =

- 3492,0782 J.Cng m kh thc hin trong tonchu trnh l A = 1211,8159 J.

A2 = 4703,8942 J.

A3 = - 3492,0782J.

A = 1211,8159 J.

2,0

1,0

1,0

5

Gi s chiu dng in i nh hnh v. p dng nh lut mcho cc on mch cha ngun v cha my thu ta c h

phng trnh:

321

3

3AB3

2

2AB2

1

1AB1

III

RUI

R

UI

RU

I

E

E

E

0IIIUI.R

UI.R

UI.R

321

3AB33

2AB22

1AB11

E

E

E

Gii h phng trnh bc nht 4 n ta c I1 = 0,1385 A; I2 =0,1189 A; I3 = 0,0196 A; UAB = 9,9226 V.

I1 = 0,1385 A.I2 = 0,1189 A.I3 = 0,0196 A.UAB = 9,9226 V.

2,0

3,0

6

1. Cng sut tiu th trong mch l P = U.I.cos =2

2

ZR.U

=

22

2

)C

1L(R

R.U

= 172,8461 W.

2. Cng dng in trong mch c biu thc: i =1,8593.sin(100t 0,9303) A.Hiu in th gia hai cc ca t in c biu thc: uC =59,1827.sin(100t 2,5011) V.

1. P = 172,8461W

2. i =1,8593.sin(100t

0,9303) A.uC =59,1827.sin(100t

2,5011) V.

1,0

2,0

2,0

7

- Khi dng in trong mch n nh, cng dng in qua

cun dy l IL =0R

E. Cun dy d tr mt nng lng t

trng Wtt =2I.L 2L =

0

2

R2

.LE.

- Khi ngt kho K th nng lng t trng chuyn thnh nhitnng to ra trn hai in tr R v R0, khi ngt mch th cng dng in chy qua R0 v R l nh nhau, suy ra nhit lng

to ra trn R l Q =RR

RW

0tt

=00

2

R)RR(2

.L.R

E

= 6,5676 J.Q = 6,5676 J.

2,0

3,0

8Gi khong cch HI = x. p dng nh lut khc x nh sng ta

c22 )x6(12

x6.2

16x

x

x = 4 cm. 5

BE1 E2 E3

R1 R2 R3

AHnh 5 (A)


26/26

Phng trnh trn tr thnh x4 - 12x3 + 56x2 - 384x + 1152 = 0.Giiphng trnh ta c x = 4 cm. Vyim ti I cch imH v pha S mton x = 4 cm.

9

- Lng thu knh cho hai nh S1 v S2 nm cch lng thuknh 25 cm (trc lng thu knh). Khong cch S1S2 = 2,00mm. Khong cch t hai khe S1S2 ti mn quan st l D = 200cm khong vn i = 0,6000 mm.

- rng trng giao thoa MN = 7.2.a = 14 mm. Trn mnquan st c 23 vn sng.

S1S2 = 2,00 mm.i = 0,6000 mm.

MN =14 mm.23 vn sng.

1,51,0

1,01,5

10

1. p dng nh lut bo ton s khi v bo ton in tch tatm c X l ht nhn Pb (ch). Nng lng to ra t mt phnr l E = (m(Po) m(X) m()).c2 = 1,0298.10-12 J.2. phng x ban u ca mu Po l H0 = 2 Ci suy ra khilng ban u l

m0 = )Po(m.2ln.N

T.H)Po(m.

.N

H

A

0

A

0

= 4,4385.10-4 g.

Khi lng ch to thnh sau 30 ngy k t thi im ban u l

m = )Pb(m..Ne.HH

A

t

00

= )Pb(m.2ln.NT).e1(H

A

tT

2ln

0

= 6,0900.10-

5g.

E=1,0298.10-12J.

m0=4,4385.10-4 g.

m = 6,0900.10-5g.

1,5

1,5

2,0

Ghi ch: Nu HS khng ghi cc kt qu trung gian m ghi kt qu cui cng chnh xc th vn choim ti a.

-----------Ht---------------------

TI LIU THAM KHO1. Vt l lp 10 (Nng cao) NXB GD - Nguyn Th Khi (Tng ch bin)2. Bi tp vt l lp 10 (Nng cao) NXB GD - Nguyn Th Khi (Tng ch bin)

3. Vt l lp 11 (Nng cao) NXGD - Nguyn Th Khi (Tng ch bin)4. Bi tp vt l lp 11 (Nng cao) NXB GD - Nguyn Th Khi (Tng ch bin)5. Vt l lp 12 (Nng cao) NXGD - Nguyn Th Khi (Tng ch bin)6. Bi tp vt l lp 12 (Nng cao) NXB GD - Nguyn Th Khi (Tng ch bin)7. Gii ton Vt l lp 12 (tp 1, tp 2) Bi Quang Hn.8. Gii ton vt l lp 11 (tp 1, tp 2) Bi Quang Hn.9. Gii ton vt l lp 12 (tp 1, tp 2, tp 3) Bi Quang Hn.

10. Hng dn thc hnh Ton L Ho Sinh trn my tnh cm tay - Nguyn Hi Chu(Ch bin)

Cc tc gi chn thnh cm n cc thy c gio v cc bn c ng gp kin cho ti liu ny.

Ch bin: Nguyn Trng Su

a ch: V Gio dc Trung hc, B Gio dc v o to, 49 i C Vit, H Ni

in thoi: 0912.013.739

Email: [email protected]

Blog: http://vn.myblog.yahoo.com/ntsuubgd.

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