Huong Dan Giai Toan Vat Li (TT)

8/4/2019 Huong Dan Giai Toan Vat Li (TT)

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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 mtGII THIU MY TNH CM TAY

Trc khi tnh ton, bn phi chn ng Mode theo bng di y:

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PHP TNH N VO MODE

Tnh thng thng COMP

Gii phng trnh EQN

Ch : tr li ci t ban u, ta nKhi y: Tnh ton: COMPn v o gc: DegDng a+ 10n: Norm 1Dng phn s: a b/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 e4/7n 100 Kt qu: 2n 4 7 KQ: 4/7

4. Gii phng trnh m:V d: 6x + 8x = 10xn 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 khncho ra kt qu.

Khi ang tm nghim mn hnh hin Processing

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

=+=+

222

111

c yb xa

c yb xa

V d:

3

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 =


4/27

==1 035

251 2

y x

y x

Ta vo chng trinh gii h phng trnh bc nht hai n bng cch sau:n phm 3 lnbm1 vo chc nngEQN my hiUnKnowns?bm2 thc

hin gii phng trnh bc nht 2 nMy hi a1 n 12My hi b1 n -5My hi c1 n -24My hi a2 n -5My hi b2 n -3My hi c2n 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

d zc yb xa

d zc yb xa

d zc yb xa

V d:

=+=+ =+

962

7352954

z y

z y x z y x

Ta vo chng trinh gii h phng trnh bc nht hai n bng cch sau:n phm 3 lnbm1 vo chc nngEQN my hiUnKnowns?bm3 thchin gii phng trnh bc nht 3 n

My hi a1 n 1My hi b1 n -4My hi c1 n 5My hi d1 n 9My hi a2 n 2My hi b2 n 5

My hi c2n -3My hi d2 n -7My hi a3 n 0My hi b3 n -2

4

MODE

======

=

MODE

===

==

=

==

===


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My hi c3n 6My hi d3 n -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 bn nMun tm nghim ca h phng trnh

=+++=+++=+++

=+++

44444

33333

22222

11111

et d zc yb xa

et d zc yb xaet d zc yb xa

et d zc yb xa

V d 1. Gii h phng trnh

=+=+=++

=++

5 42372

8 0384

3 4 825

3 0 024

t z y x

t z y x

t z y x

t z y x

Ta vo chng trinh gii h phng trnh bc nht hai n bng cch sau:n phm 3 lnbm1 vo chc nngEQN my hiUnKnowns?bm4 thchin gii phng trnh bc nht 4 n

Nhp vo cc h s ca h phng trnh:1 4 1 2 3001 5 2 1 3484 8 1 3 802 7 3 2 547

Kt qu: x = 77, n

y = 20, nz = 209, nt = 47

8. Phng trnh bc hai mt nTm nghim ca phng trnh bc haiax2 + bx + c = 0

V d: Gii phng trnhx2 + 9x + 8 = 0

Ta vo chng trinh gii h phng trnh bc hai mt n bng cch sau:n phm 3 lnbm1 vo chc nngEQN 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 85

=

==

MODE

= (-) = = == (-) = = (-)

= (-) = (-) =

== (-) = (-) = =

=

=

==

MODE

= = =

===


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

9. Phng trnh bc ba mt nTm nghim ca phng trnh bc haiax3 + bx2 + cx + d = 0

V d: Gii phng trnh2x3 + x2 8x - 4 = 0

Ta vo chng trinh gii h phng trnh bc nht hai n bng cch sau:n phm 3 lnbm1 vo chc nngEQN my hinUnKnowns?bmmn hnh xut hinDegree? bm3 thc hin gii phng trnh bc ba Nhp vo cc h s ca phng trnh trn:

2 1 8 4

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

10.FIX, SCI, RND (Chn s ch s l, dng chun a+ 10n, tnh trn)nMode5 ln vo cc chc nngF IX, SCI, NORMa) Fix:n nh 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 saunMode5 ln chn1 vo ch Fixmn hnh xut hinFix 0~9ta nhp s 4 voKt qu: 200 : 7 = 28.5714

b) Sci: n nh s ch s caaV d 1. Tnh 200 : 7 = 28.57142857142857 mn hnh ch hin th k t qu vi 5 s ta lm nh sau

nMode5 ln chn2 vo ch Scimn hnh xut hinSci 0~9ta nhp s 5 voKt qu: 200 : 7 = 28.571c) NormnMode5 ln chn3 vo ch Normmn hnh xut hinNorm 1~2

xa ci tFixvScita chnNorm 1 hocNorm1

6

=

MODE

= = (-)

=

= (-) =

=


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Phn haiCC 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 c bn v Vt l, mun

n kt qu cui cng chng ta rt hay dng ti cc php tnh cng, tr, nhn, chia, lu tha vcn. Cc php tnh cng, tr, nhn, chia, lu tha v khai cn l cc php tnh c bn chnkhng th b qua c. Khi thc hin thnh tho cc php tnh c bn ny s gip ta tm qu ca bi ton mt cch mau l v chnh xc.

Trong vic thc hin cc php tnh c bn ni trn cn phn bit php tr v du trExp v 10 ^ , i khi chng cho kt qu nh nhau, nhng ni chung l khc nhau. Mun tnh chxc chng ta khng nn ghi cc kt qu trung gian ra giy ri nhp li vo my m nn nh cqu vo nh c lp (Shift Sto) hoc nh mc nh Ans, m ch ghi kt qu cui cng.

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

a) Trong khong thi gian bi i. b) Trong khong thi gian bi v.c) Trong sut c thi gian bi i v bi v.

Cch gii Hng dn bm my v kt qu

Theo nh ngha v tc trung bnht S

vTB =

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

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

50 20.18 =KQ: 2.47770069450 21.34 =KQ: 2.343017807100 ( 20.18 + 21.34 ) =

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s / m ,vTB 40852 KQ: 2.408477842

Bi 2:Lc 7h mt t chy t Hi Phng v H Ni vi tc khng i 45km/h. Cng lc t chy t H Ni i Hi Phng vi tc khng i 65km/h. Bit khong cch H Ni Phng l 105km.

a) Hy lp phng trnh chuyn ng ca hai xe trn cng mt trc to , ly gc to

Ni, chiu dng hng t H Ni ti Hi Phng, gc thi gian l lc 7h. b) Xc nh thi im v v tr hai xe gp nhau.

Cch gii Hng dn bm my v kt qu

a) Lp phng trnh chuyn ng ca mi xe:- Xe t Hi Phng v H ni c hng chuynng ngc vi trc to , v tr ban u ti HiPhng nn phng trnh chuyn ng l:

)h ,km( t x 451052 = .

- Xe t H ni i Hi Phng c hng chuyn ngcng chiu trc to , v tr ban u ti H Ninn phng trnh chuyn ng l:

)h ,km( t x 652 = . b) Khi hai xe gp nhau th chng phi c cng to, tc l 21 x x = 105 45t = 65t 110t = 105 h ,t 95450 .Thi im hai xe gp nhau l 7,9545h.Hai xe gp nhau ti v tr cch H Ni

km , x 04545622

105 110 =KQ: 0.954545454Ans + 7 =KQ: 7.954545455

= Ans x 65 =KQ: 62.04545455

Bi tp vn dng 1.1.Mt ngi chy trn mt ng on ng t di 200m ht thi gian 30s. Sau ngchy thm trn mt on ng nha di 150m ht thi gian 20s. Hy xc nh tc trungca ngi trong cc trng hp sau:

a) Trong khong thi gian chy trn ng t. b) Trong khong thi gian chy trn ng nha.

c) Trong c on ng t 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 t chy t H Ni i Hi Phng vi tc khng i 40km/h. Bit khong cch H Ni Phng l 105km.

a) Hy lp phng trnh chuyn ng ca hai xe trn cng mt trc to , ly gc to

Phng, chiu dng hng t H Ni ti Hi Phng, gc thi gian l lc 10h. b) Xc nh thi im v v tr hai xe gp nhau.p s:a) x1 = - 55t km, x2 = -105 + 40t (km).

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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 k8N hng theo phng ngang. Sau thi gian 3s k t lc bt u chuyn ng xe i c 10mtnh h s ma st gia xe ln v mt t. Ly 289 s / m , g = .

p s:0,1814.

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

Nhng im cn lu Cc h phng trnh bc nht hai n, ba n, bn n vi s phng trnh bng s n th m

cm tay VN 570MS c th gii c mt cch d dng. c bit vi cc h phng trnh bc cc h s khng nguyn dn n vic tnh ton thng thng gp nhiu kh khn th mcm 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 phng trnh bc nht bng cch bm nh sau:

- Gii h phng trnh bc nht 2 n: Mode (3 ln) 1 2- Gii h phng trnh bc nht 3 n: Mode (3 ln) 1 3- Gii h phng trnh bc nht 4 n: Mode (3 ln) 1 4 Nhp cc h s cho h phng trnh, trong khi nhp cc h s c th thc hin cc ph

thng thng, n khi bm = th gi tr ca h s c gn. Trong khi nhp cc h s ta ph

tt c cc h s, cn c bit ch n cc h s c gi tr bng 0 v nhm th t ccMun trnh nhm ln tt nht ta lp mt ma trn gm m hng v (m + 1) ct (vi m l s ptrnh).

Cc v d minh hoBi 1:Treo ln lt cc vt khi lng m1 = 100g v m2 = 150g vo u di ca mt l xo (utrn ca l xo c nh), 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, lc n hi ca l xo cn bngvi trng lc ca vt. T ta c h phng trnh

==

==

gmk lk l

gmk lk l

gm )ll( k

gm )ll( k

202

101

202

101

Gii h phng trnh ta c

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

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=

=

=

=

)m( ,l

)m / N ( ,k

,k l

,k

3 30

0 34 9

1 8 1 11 6

0 3 3 54 9

00

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

Cch gii Hng dn bm my v kt quChn chiu dng l chiu chuyn ng ca xemt xut pht t A.Phng trnh chuyn ng ca xe xut pht t A l

x1 = v1.t = 36tPhng trnh chuyn ng ca xe xut pht t B l

x2 = S - v2.t = 100 - 72tThi im v v tr hai xe gp nhau khi x1 = x2 = xl nghim ca h phng trnh

=+=

1 07 2

03 6

t x

t x

Gii h phng trnh ta c==

)h( ,t

)k m( , x9 2 5 90

3 3 3 33 3

Mode (3 ln) 1 21 =(-) 36 =0 =1 =72 =100 =KQ: 33.33333333=KQ: 0.9259259259

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 thn 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 nhng v pha nhau. Xe xut pht t A chuyn ng u vi tc 40km/h, xe xut phchuyn 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, bt u chuynth trng thi ng yn. Hy xc nh khong thi gian cn thit vt chuyn ng cng 400m k t khi vt c tc 5m/s.

p s:8,4582 (s).

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3.4.Mt t ang chuyn ng th t ngt hm phanh, lc hm khng i v bng 25% trngca xe. Hy tnh thi gian t lc bt u hm phanh n lc xe dng hn. Bit rng ngay sahm phanh xe cn i c on ng 32m mi dng li. Ly g = 9,81m/s2.

p s:2,1159 (s).

4. HM M V LGARIT.

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

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

k th ta cn dng cng thc i c s alog blog

aln

blnblog a ==

.Vic tnh ton vi cc hm s v hm s lgart ta my tnh ch Mode 1. Vi hm m

lgart c th tnh ton trong cc ch gii phng trnh, h phng trnh, .... tng t nh php tnh c bn.

Cc v d minh hoBi 1:Mt ngun m S (ngun im) pht ra mt m, ti im M cch ngun m mt khong 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 mc cng m ti im N cch ngun m mt khong SN =

B qua s hp th m ca mi trng.Cch gii Hng dn bm my v kt qu

a. Mc cng m ti im M c tnh theocng thc

0 I I

lg L M M = = 4,3010 (B)

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

ngun: 22

SN SM

I I

M

N =

.

Cng m ti N l 22

SN SM

I I M N = =2,6446.10-6

(W/m2).

Mc cng m ti N l0 I

I lg L N N = =

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 C 146 trong n bng0,707 ln phng x ca mt khc g va mi cht. Chu k bn r caC 146 l T = 5600 nm.

Cch gii Hng dn bm my v kt qu phng x - ca C 146 c tnh theo cng thc

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2200

20

ln H

)t ( H ln

.T )t ( H

H log .T t

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 (ngun im) pht ra mt m, ti im M cch ngun m mt khong 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 mc cng m ti im N cch ngun m mt khong SN =

B qua s hp th m ca mi trng.p s:a. LM = 4,0792B.

b. I N = 2,5562.10-6W/m2; L N =3,4076B.

4.2.Tnh tui ca mt ci tng c bng g, bit rng phng x -

ca C 14

6 trong n bng 0,57ln phng x ca mt khc g va mi cht. Chu k bn r caC 146 l T = 5600 nm.p s:4541,4106 nm 4500 nm.

5. HM S LNG GIC.

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

ngc ca chng tr ln d dng, vic khng cn phi dng thc tnh gi tr ca hm lnghoc bn s tra cc gi tr ca hm lng gic. Vi my tnh cm tay c th tnh gi tr chm s lng gic vi n v ca bin s l radian (rad) hoc (0). Vi hm ngc acrsinx vacrcosx, gi tr ca bin s x phi thuc on [ - 1 ; +1]; gi tr ca hm ngc c tnh ra

rad hoc .t ch cho my tnh ch n v rad hoc vi my tnh VN 570MS ta lm nhMode (4 ln) 1 (khi dng n v ); hoc Mode (4 ln) 2 (khi dng n v rad). Trong qu ttnh ton ta c th i n v nh Shift DRG. V d my ang ch dng n v (Dec) tnh sin300 ta bm sin 30 =, my cho kt qu 0,5. Nhng ta khng mun chuyn ch n vtnh 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 t ch ph hp.

Cc v d minh hoBi 1: Mt vt c khi lng m = 500g t trn nh mt mt phng nghing hp vi ph

ngang mt gc = 600

. Cho chiu di mt phng nghingl = 2m, g = 9,8m/s2

. Tm gia tc ca vttrong cc trng hp sau.a. B qua mi ma st.

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b. H s ma st trt gia vt v mt phng nghing l t = 0,2.Cch gii Hng dn bm my v kt qu

Gia tc ca vt c tnh theo cng thc tng qut

a = k ms F 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: 4.99.8 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. Mt phng nghing c h s ma st trt vi vt l t = 0,5 cho chiu di mt phng nghingl = 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 c nng W2 W1 = Asm.

mgl sin - 21

2mv = Fms.l = l .t mgcos

Suy ra )cos(sin gl v 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 dng 5.1.Mt vt c khi lng m = 500g t trn nh mt mt phng nghing hp vi phng mt gc . Cho chiu di mt phng nghingl = 1,2m, g = 9,8m/s2. Th gia tc ca vt trong trnghp 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 mt gc = 250. Mt phng nghing c h s ma st trt vi vt l t = 0,25 cho chiu di mt phng nghingl = 2,5m, g = 9,8m/s2. Cho vt trt vi vn tc ban u 05m/s t nh mt phnnghing. Hy tm vn tc ca vt chn mt phng nghing.

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 pnghing 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 lu Cc php tnh o hm bc nht, vi phn bc nht v tch phn mt lp c th dng my

cm tay tnh ton mt cch d dng. Vic dng my tnh cm tay s a chng ta n k bng s cui cng ch khng a ra cng thc tng qut, nn cc bi ton cn ly o hm thai tr ln, cc bi ton c s dng tch phn nhiu lp ta vn phi dng cc cng thc ton ha ra cng thc tng qut ri sau thay s mi c kt qu.

Dng tng qut ca cch bm my khi tnh o hm v tch phn nh sau (VN 570MS):- o hm: Shift d/dx , = . c vit di

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

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Bi 1:Mt cht im chuyn ng theo phng trnh x = 4t2 7t + 5 (x o bng m, t o bng s).Hy tnh vn tc ca cht im ti thi im 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 d ( 4 ALPHA X x2 7 ALPHAX + 5 ), 12 ) =KQ: 89

Bi 2:Mt vt chuyn ng vi gia tc ph thuc vo thi gian theo cng thc a = 0,2t +1 (m2).Hy tnh vn tc ca vt ti thi im 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/s d ( 0.2 ALPHA X +1), 0 , 16 ) =KQ: 41.6

Bi tp vn dng

6.1.Mt cht im chuyn ng theo phng trnh x = 3t2 5t + 2 (x o bng m, t o bng s). Hytnh vn tc ca cht im ti thi im t = 2s.

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

p s: 30m/s.6.3.Vt m = 2kg chu tc dng ca mt lc bin i u theo cng thc F = 5 + 2t (F o bngo 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.

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 nh

n gin v chnh xc th phi k ti tin ch tra cu mt s hng s vt l v i mt strong vt l. Cc hng s vt l c ghi trong b nh ca my vi n v trong h n v S

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

Hng s vt l M s Cc bm my Gi trKhi lng prton (m p) 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)

14


15/27

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 timt t (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)

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

Cc v d minh hoBi 1:Mt vt ri t do khng vn tc ban u ti ni c gia tc trng trng g. Hy xc ntc v qung ng vt ri c sau thi gian t = 2,5s.

Cch gii Hng dn bm my v kt quSau 2,5s vt t c vn tc

v = gt 24,5166 m/sQung ng i c l

S = 212 gt 30,6458 m

Const 35 x 2.5 =KQ: 24.516625

Const 35 x 2.5 x2 2 =KQ: 30.64578125

Bi 2: Mt vt c nm thng ng ln trn vi vn tc ban u v0 = 15m/s ti ni c gia tctrng trng g. Hy xc nha. Thi gian t lc nm n lc vt quay tr li 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 tr li 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

15


16/27

Bi tp vn dng 7.1.Mt vt ri t do khng vn tc ban u ti ni c gia tc trng trng g. Hy xc nh vv qung ng vt ri c sau thi gian t = 3,5s.

p s:s 60,0657 m.7.2.Mt vt c nm thng ng ln trn vi vn tc ban u v0 = 20m/s ti ni c gia tc trngtrng g. Hy xc nh

a. Thi gian t lc nm n lc vt quay tr li 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; h16,9392 m.

7.3.Mt vt c nm theo phng ngang vi vn tc v0 = 5m/s cao h = 15m so vi mt t.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 cchtrung 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.

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 n

php tnh, nhiu ch tnh ton. Mt bi ton tng hp phi dng nhiu bc tnh ton, muton chnh xc th cc kt qu trung gian ta khng c lm trn. khc ph iu ny, chnn nh cc kt qu trung gian vo cc nh c lp; cng c trng hp khng th nh qu trung gian vo cc nh th bt buc chng ta phi ghi ht cc s m my tnh hin lnhnh ra giy, sau chuyn ch tnh ton v nhp y cc s ghi vo my tnh. Vi

lm nh vy ta c th hn ch ti a 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 lngnm v trng vt l M = 2 kg v m = 500 g, ca cc rng rc khng ng k. B qua ma st tip xc. Bit = 300. Tnh:

a. Gia tc ca nm M. b. Gia tc ca trng vt m.

16

M

mHnh 8.1


17/27

Cch gii Hng dn bm my v kt quTa thy rng nm M chuyn ng sang phi vi giatc a c hng t tri sang phi. Nu M i sang phi mt on x th m cng chuyn ng trn mt phng nghing ca nm mt on x. Suy ra gia tcca m so vi M cng c ln l a, hng songsong vi mt phng nghing t trn xung di.Phng trnh ng lc hc vit cho M v m lnlt 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 mt t l am = 2asin(/2) =

)cos1(m2M2

sinsinmg2

+

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

Mode (4 ln) 1

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

2 x Ans x sin ( 30 2 ) =

KQ: 0.5947

Bi tp vn dng Mt cht im thc hin dao ng iu ho dc theo trc Ox xung quanh v tr cn bng O vk T = 2 s. Ti thi im t1 cht im c to x1 = 2 cm v vn tc v1 = 4 cm/s. Hy xc nh to v vn tc ca cht im ti thi im 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 thi im t2 = t1 + 1/3 sta c x2 = =+ )

3tsin(.A 1 A. sin (t1).cos3

+

A. cos (t1).sin 3

= 2,1027 cm.

v v2 = . =+ )3

tcos(.A 1 .A.cos(t1).cos3

-

.A. sin (t1).sin3

= -3,4414 cm/s.

KQ: 2.1027

KQ: -3.4414

17


18/27

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)Bng s Bng ch Gim kho 1:

Gim kho 2:

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

18


19/27

Bi 2:T mt im A, mt vin c nm ln cao theo phng thng ng vi vn tc vm/s. Sau mt khong thi gian t0, t mt im B cng cao vi A v cch A mt khong l = 4 mmt 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 tht0 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 = 700, b qua mi mast, dy khng dn, khi lng ca dy v rng rc khngng 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

19

m2 m3m

1

Hnh 3


20/27

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, T 1 =320K, T2 = 600K. Hy tnh cngm kh thc hin trong chutrnh.

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. BitE1 = 12 V, E2 = 6 V,E3 = 9 V, R 1 =15 , R 2 = 33 , R 3 = 47 .

Cch gii Kt qu

20

p p2 (2)

p1 (1) (3)T

T1 T2Hnh 4

E1 E2 E3

R 1 R 2 R 3

Hnh 5


21/27

Bi 6:Cho mch in xoay chiu RLC mc ni tip c R = 100 , cun thun cm L = 0,52v t in c in dung C = 100 F. t vo hai u on mch mt hiu in th xoay chi220 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

t in.

Cch gii Kt qu

Bi 7: Mt ng dy c t cm L = 2,00 H v in tr R 0 =1,00 c ni vi mt ngun in mt chiu c sut inngE = 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

21

L, R 0

R

E K

Hnh 7


22/27

Bi 8:Mt tia sng i t im 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 = 23 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 ch

vung gc vi tit din ca thu knh, ri mi bt mi na i mt lp c b dy a = 1,00 mm dn li thnh lng thu knh. Mt khe sng S c t trn trc i xng ca lng thcch lng thu knh mt khong 12,5 cm, pht ra nh sng n sc c bc sng = 0,60 mv cch lng thu knh mt khong b = 175 cm ngi ta t mt mn nh vung gc vi trxng ca lng thu knh. Xc nh khong vn v s vn quan st c trn mn.

Cch gii Kt qu

22

SHnh 8

H

K S


23/27

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

khi 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 im1 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 vtl = 4 rad/s chu k dao ng l T =

2 = 0,5000 s.

Bin dao ng ca vt l:)cos(AA2AAA 1221

22

21 ++= = 3,4810 cm.

Pha ban u trong dao ng ca vt l c

+=

++=

A

s inAs inAs in

c o sAc o sAs inAs inA

ta n

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 vo

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

23


24/27

phng 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 -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 ln

trn, 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. (gs i n) .tt. (v

lc o s) .tt. (v22

00

0

=

+

=0

)c o s.v. (2g.l

c o ss in.l

t.v2t.g

c o s.vl

)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 phng

trnh:

==

==

)s(1 2 6,0t

)s(2 8 8 8,0t

)s(3 5 5,2t

)s(7 7 0 3,2t

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

msin)mm( + = 280127,55.

2. Khi t dy ni gia m1 v m2 th h mt cn bng, m3 v m1cng 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

mmg)sinmsinm(

+

=

32

1

mmsin.g.m

+

= 2,3038 m/s.

= 280

127,55.

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 + A3

24


25/27

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

A1 = 0 J.+ A2 l cng m kh thc hin

trong qu trnh ng nhit (2)(3):

A2 = 3

1

V

V

dV. p vi1

11 p

T.R .nV =

,1

23 p

T.R .nV = ,

VT.R .n

p 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.

A1 = 0 J.

A2 = 4703,8942 J.

A3 = - 3492,0782J.

A = 1211,8159 J.

1,0

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

3A B3

2

2A B2

1

1A B1

III

R

UI

R

UI

R

UI

E

E

E

===

=+

0III

UI.R

UI.R

UI.R

321

3A B33

2A B22

1A B11

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 =22

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 qua2,0

25

p p2 (2)

p1 (1) (3)V

V1 V3Hnh 4 (A)

BE1 E2 E3

R 1 R 2 R 3

AHnh 5 (A)


26/27

cun dy l IL =0R

E. Cun dy d tr mt nng lng t

trng Wtt =2I.L 2L =

0

2

R 2.L E .

- Khi ngt kho K th nng lng t trng chuyn thnh nhitnng to ra trn hai in tr R v R 0, khi ngt mch th cng dng in chy qua R 0 v R l nh nhau, suy ra nhit lngto ra trn R l Q = R R

R W

0tt + =

00

2

R )R R (2.L.R

+

E= 6,5676 J.

Q = 6,5676 J. 3,0

8

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

c 22 )x6(12x6

.216x

x

+

=

+

Phng trnh trn tr thnh x4 - 12x3 + 56x2 - 384x + 1152 = 0.Gii phng trnh ta c x = 4 cm. Vy im ti I cch imH v pha S mt on x = 4 cm.

x = 4 cm. 5

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. NT.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.. N

e.HH

A

t00

= )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 c

im 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.

26


27/27

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 Niin thoi: 0912.013.739Email:[email protected]: http://vn.myblog.yahoo.com/ntsuubgd.

-----------------------------------------
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Huong Dan Giai Toan Vat Li (TT)

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