大一下礦物理習(xí)題上

advdvdxvdv dx 2xdx

x21(v2v2 1616 w

ddt

anRw2144RtatR當(dāng)t2san230.4m/s2at4.8m/s2144Rt'424Rt得t'316(t')24t'3sv0勻速收繩時(shí)，試求船的速率和加速度大小（假設(shè)繩子不可xaektdybkekt

b

d2

ake,

ak2ekt

dt

,

d2 bk2ekt,dt2

2kt

2kt(2)dy

aak

ibk ydy t ybbekt0

bektybekt箭在t=50s時(shí)用完那一瞬間所能達(dá)到的高度及該時(shí)刻火箭的速度。a1tdv1tdv1 積分得vdvt1tdtv1t 0 當(dāng)t20sv100ms。由vdx1t2 積分有xdxt1t 0x1t3t20sx2000 adv1t dv(1t20 t 積分dv6t

3得v1t220t dxt(1t220t

3

20 x1t310t2200t 當(dāng)t50sv475msh8916.7m10m/s5m/s的速度向南。問(wèn)：從 vv甲地+vvv10i5 11.2m x (2)v10i5j,方向與v第二章運(yùn)動(dòng)定RRKv，式中負(fù)號(hào)表示阻力，K0t0將小球浸沒(méi)在水中，小球初速度v0t 1.AB分別重GA200N和GB400NAPNma

d2xmhw2cos

dt NPAmAhw2coswtmAw2x Pmw2hP(1w2h) Pmw2hP(1w2h) RmaxPBNmaxRminPB

Fmgxgg(l 定律建立方程：mdvg(lmdvdxg(lx)mdvv,dtdx dxlvdvg(lx)d 0lvdv0(glg'x lv22gl2gl2v 垂線上，離桿的距離為l。求：細(xì)桿間受到質(zhì)點(diǎn)的萬(wàn)有引力。2FNman Fma μv2 tdtRv μv0v Rv0tRstvdtt dt 0Rv0sRμ某彈 定律,若施力F,則相應(yīng)伸長(zhǎng)x,力與伸長(zhǎng)的關(guān)系F=2x+3x2(SI),將彈簧橫放在水平光滑桌面上,一端固定,0.2kg的物體,然x2=2m,再將物體由靜止釋放,x1=1m時(shí),物體的解：(1)由功的定義，力F所作的功為 ????d????=?(2????+3????2)d????= ????=????(x)=2x+ ?2????+A ?2????+

d????=?d????=?

v=?2????/????=?≡ ?????

?d????=????????????? d

1=????????????

?????+?

?=????(????+)1

????

=????????2?2?????=?2????????(????+能零點(diǎn)，設(shè)物體脫離球面時(shí)距離頂點(diǎn)為h0，由圖可知?0=????(1?1????????2??????????= mgcos????0=?????0=3，????????????????0=CBBAd是多少？????????=mg=m????2

故????2

????????2=????????2+ R=d=L?R=mF將物體自平衡位置開始向右拉動(dòng)，試求此系統(tǒng)所具有的最大勢(shì)能。簧拉升最長(zhǎng)處，物體達(dá)到最遠(yuǎn)距離，這時(shí)物體的速度為0，因而動(dòng)能為0。????=?????????????????????

?

????????????=2

=2(????????? =2(????????? ????????=

左側(cè)下滑離開圓柱面，試求地面系下看，過(guò)程中圓柱面支持力N對(duì)小球所作功?！鋗gcos????= 1????????′2=???????cos

cos????= ????′=?2 ????????=????0?′cos???? ????????=??sin22W=

2???????????????(1?

+????????02可算得：W2????????2 0

假設(shè)一個(gè)質(zhì)量為0.5kg的球被木棒，木棒對(duì)球的擊打力滿足拋物線規(guī)律，

0.25a+0.5b+c=4a+2b+c I=?????(????)????????=4400?????????????=4.4?????又mvmv0=????，其中v0=可解得v=m

=始下滑，當(dāng)它滑到光滑水平面C點(diǎn)時(shí)，有一個(gè)質(zhì)量為m=0.02kg的射入木塊中，已知θ=30°)）

1????????2=????????????,

=???

????????0?????????1=(????+????????0????????2 ????+ （????+????）????2 （????+????）????2+(????+??????+ ????2=????2?????+????????????????)=????2?

(M+m)gsin????=(????+????)????2=????????sin???? 代入????2

7 ???? 2

?2????+

+

????????=314.32在m1內(nèi)。試求：m????0=(????+故

????+m+m1，m2和彈簧k為系統(tǒng)，系統(tǒng)在運(yùn)動(dòng)過(guò)程中水平方向動(dòng)量守恒，且等于質(zhì)心的動(dòng)量，設(shè)質(zhì)心的速度為vc，則有m????0=(????+????1+????????=????????0/(????+????1+????′=

????+????1????+????1+????′20=0

=????+????1+ E=(????+??????′ 10+2 E2=2

=????????

0????(????+????1)(????+????1+m+m1的速度為????′1，m2的速度為????′2，取此為終態(tài)，由機(jī)械能守恒定律，有1(????+??????′ ′

1+2????2????2=2(m+1??′+????2????′2=聯(lián)立以上兩式，并將????????????′2=

????+????1+????2=????′2+????????= ????+????1+m????0=(????+ (????+????1)????1=(????+????1+ (????+????)????2=(????+????+????)????

2+2

=????????

0????(????+????1)(????+????1+(m+m1)速度，用????′2表示此時(shí)m2速度，則有 (????+????)????2

(????+????)????′

1+2????′2+2??′?2????????0]=0，??′= ，????+????1+=因微小擾動(dòng)而下滑，試問(wèn)????達(dá)到何值時(shí)A球離墻？AAθB在θ角時(shí)設(shè)A向下的速度為????????,B向右的速度為????????A、B1???????? 2=????????????(1?cos ????+2????????cos????=????????sin

????????=??????cos????)?cos所以系統(tǒng)的水平方向的動(dòng)量為（令????=cos????????=0+????????????=????2????(?????)??得???????????1 2√1??????????+???? ?????????)=0?3

一個(gè)箭體質(zhì)量為M0（不含部分載有m0的火箭在太空中由靜止開始點(diǎn)火，燃燒后的熾熱氣體相對(duì)于火箭以u(píng)的速度向后噴出。求(2)若火箭在均勻重力場(chǎng)g中垂直起飛，T時(shí)間后耗盡，則耗盡瞬間火箭由題意知在dt時(shí)間后，噴出的質(zhì)量為dm，則火箭的剩余質(zhì)量為M–dm，速度為v+dv，噴出的為dm，速度為v-u，Md????????d????=0，Md????????d????=0，（dm=d????=

兩邊積分，得?d????= 0即，耗盡時(shí)，火箭速度????=????ln

?????????d????=(M?dm)(????+d????)+dm(?????????)?

d????d????，得?????????d????=????d?????同樣的，因d????=?????????d????=????d????+d????=?????? d????=??????d????? 得

????=?gT+????ln????0+1????? g? ?3??? \?? ?\??? Y??v0R\?????:?u-E?-\?-\BlOCBlOCl2lA O?:?- ??J0 ? J0=l2-E?v=vB=2vC=-NX =lmvA+lmvB+l=2ml2ω9

l4 4- NX??=Jω=29 ????k ????ǒ??? ??????u??????? ?ǒb=a??? a?? :??: ?????????vR?vvbv

2×1mv2+1k(b?a)2=2×12??5?=2a(

rv ??l ?v0???M ?1? ?ǒ?Ol??I>????? (???b>?? ?lb b0?A;ATT(L????Rmv=1mv2?GMm=1 b 1+0

M00RMvl???R?! ? ?′R? ?? 2 ?? ′?K ′2 M=

′=1 ???{ǒ ?m .

IM

MR22

23

R m=2(?m)

=?24 ?u: R m=I+(?m)· ?

=?24

–122

=?8Im=IM+m=24??ǒm?!??H.RI7g???.HHRhhr

V=3

ρ=d dV=2πr(H?5?A?'Xh= .

dV=2πrH1?rR r2dm=ρdV r Z 6mZ=

r3

r4R 6m1R4?1R4R24 3X????X????Y??1w?-f ?\?m1g?T1=TR?TR= I=1 2T2=a=?a 2(m1+m2)+Dx? ?????! 7?g ?=???=??1??^o???????O2????O1,m2 ???,ω2?3 ? ?ω′, " ?????ω′R1=ω′ ??.???T1?T2? ??)Z1R1(T1?T2)dt=I1(ω′?1Z2R (T1?

I1(ω′?= = I1R2+ R1(m1+ω1′=R2(I1ω1R2?I2ωI1R2+ R1(m1+ ω′=R1(I2ω2R1?I1ω1R2)=m2R2ω2? I1R2+ R2(m1+ ??1 =.?X??3FG ?/w???? ??7????-?-??!Nm1,2?.?I?N\???-?_???ǒm1g?T1=m1a1T2?m2g=T1R?T2r=?

a1= a2=a=(m1R?m2r)R a2=(m1R?m2r)r I+m1R2+ I+m1R2+α= m1R?m2r I+m1R2+m2r2T=I+m2r(R+r)m1 T2=I+m2R(R+r)m2 I+m1R2+ I+m1R2+??![\????? ?\ ??ω07????=?? ?L???\?=?l?(1 ?η=m?? ?dm=l ?\?r??d dM=r×μgdm= l

Z lZl=20 14

2μmgrdr(2)d?

?Mdt=0? ?t=?ǒl??ǒm!?\?7?L???Y?=?,????^? ?-? .? πe???\u ?.? θθlmm\? !1×1ml2ω2=1mgl1?cos ω ?\5-???lmv0=lmv+131mv2=1mv2+1×1ml2× ?

v0

3mgl(1?cosθ)=1 )cosθ=3于平衡位置且向x軸的負(fù)方向運(yùn)動(dòng)。求：解：(1)vm=ωA(2)am=ω2A=vmω=4.5×10-2

12，x=

2m(1)用k,M,R,x給出系統(tǒng)總能量的表達(dá)式（因?yàn)檩棗l質(zhì)量可忽略，的轉(zhuǎn)動(dòng)慣量為MR2)；(2)根據(jù)能量守恒，dE/dt0.們一起在光滑的水平面上作簡(jiǎn)諧運(yùn)動(dòng)，彈簧的勁度系數(shù)為k，砝碼與滑塊在振動(dòng)過(guò)程中不發(fā)生相對(duì)運(yùn)動(dòng)，已知t=0時(shí)刻，滑塊恰經(jīng)過(guò)平衡位置且速度為v.求

各物理量為國(guó)際單位制(SI).振動(dòng)振幅最大?又????3應(yīng)為何值，才能使x1+x2+x3的合振動(dòng)振幅最小?已知一平面簡(jiǎn)諧波的表達(dá)式為????=0.02cos(4????????+求該波的波長(zhǎng)????，頻率????和波速????的值寫出????0.6s時(shí)刻各波峰位置的坐標(biāo)表達(dá)式，并求出此時(shí)離坐標(biāo)原點(diǎn)最近的求????=0.6s時(shí)離坐標(biāo)原點(diǎn)最近的那個(gè)波峰通過(guò)坐標(biāo)原點(diǎn)的時(shí)刻t。解：(1)對(duì)比機(jī)械波????=????cos[????(?????????)+????]，得到????=0.02????,????= 4????????????????/????,????0=0????????????,????=?2波速大小為2m/s，方向朝x負(fù)方向。所以????=????=2又根據(jù)????=????????,????????=2????(????2?????1)/????

????

=1????=0.6s時(shí)，波峰的位置滿足4?????0.6+2????????????=2????????(????=0,±1,±2,? ????????=????1.2(????=01,±2,? k=1|????????|最小，離原點(diǎn)最近,所以????????=?0.2波峰從原點(diǎn)到????=?0.2m所需要的時(shí)間為????????=????????=0.2=0.1????.波沿著 負(fù)方向的，波峰先經(jīng)過(guò)原點(diǎn)，所以經(jīng)過(guò)坐標(biāo)原點(diǎn)的時(shí)刻????=0.6?0.1=0.5平面簡(jiǎn)諧波沿x軸正方向，振幅為2cm，頻率為50Hz，波速為200m/s。表達(dá)式及該點(diǎn)在t=2s時(shí)的振動(dòng)速度。解：x0y0Acos(t∴y0Acos(2t

122102cos(100t1 由波 2102cos(100t11y0Acos(2t2x/y2102cos(100t12

v2102100πsin(200π1π)=6.28m2一平面簡(jiǎn)諧波沿x軸正向，其振幅為A，頻率為 yuuOt=t′解：(1)x0yAcos(2t由圖可知，ttyAcos(2tdy/dt2Asin(2t)

2t/2

12t

yAcos[2(tt)12yAcos[2(ttx/u)12S1S2的振動(dòng)相位分別為????1和????2，在????1 即

?? =(2????+? (?????????)?2?????????2????1=(2????+ (?????????)?2?????????2????2=(2????+ 由(2)-(1)得：4????????2?????1=2????????=2(????????)=6

?

2?????????2????12????1)????=(2????5)????,當(dāng)????=?2,3位差最?。????????=????2?????1=如圖，一列平面簡(jiǎn)諧橫波沿著x軸負(fù)方向，波長(zhǎng)為????，角頻率為????。P點(diǎn)距離原點(diǎn)為????，已知P點(diǎn)的振幅為A，????0=0P點(diǎn)的位移????/2,速度方向指向平衡位置。試求P點(diǎn)的振動(dòng)方程，????????若此平面簡(jiǎn)諧波在距離原點(diǎn)一個(gè)波長(zhǎng)處的Q點(diǎn)發(fā)生反射，????????=?????,Q點(diǎn)為解：(1)設(shè)????????=????cos(????????+????????(????=0)=????/2=????cos(????????)得到????????=又因?yàn)樗俣确较蛑赶蚱胶馕恢?，所????????=振動(dòng)方程為????????=????cos(????????+可設(shè)波函數(shù)為????=????cos(????????+2????????+???? 當(dāng)????????????=????cos(????????2????????????????cos(????????????0=????/3

所以波函數(shù)為：????=????cos(????????+2????????+?????

=????cos(?????????2????????+當(dāng)????=?????時(shí)，入射波相位為????????=?????????2????+????/3?????????時(shí)，反射波相位為????????????????2????Q為固定端，反射波有半波損失，????????=????????+(2????+1)????.即????????+2????+????=?????????2????+????/3?2????????/????+(2????+1)????.取????=2????4????/32????????/????.

=????cos(?????????2????????+4????/3?兩輛汽車????,????分別以40????/????,20????/????的速率在馬相向而駛,此時(shí)汽車A按下????????

????0=340?40?400≈453.3????=????+????????

=340+20?400=480????????????? 340?飛船的固有長(zhǎng)度是Lct由題意知，飛船的固有長(zhǎng)度為L(zhǎng)ct1905年提出了狹義相對(duì)論，狹義相對(duì)論是以兩條基本假設(shè)為前提的， S系中平面上一個(gè)靜止的圓的面積為12cm2，已知S系在tt0時(shí)與S系坐標(biāo)軸重合以-0.8c的速度沿公共軸xx運(yùn)動(dòng)則在S系測(cè)得該圓面積為7.2cm2 2R，所以測(cè)得的面積為（1v1vc

RR

1v1vc一艘宇宙飛船的船身固有長(zhǎng)度為L(zhǎng)090m，相對(duì)于地面以v00.8c（c為真11c0L

t1

2.25 11c0

3.75107SS系中記錄到兩 空間間隔x600m時(shí)間間隔t8107s而s系中記錄t0，則s系相對(duì)s系的速度為0.4C。 的時(shí)空坐標(biāo)分別(x',t')為及(x',t') 1 2 t't t't't'(tt)v(xx)tvx

1

根據(jù)題意得：t0x600m,t8101 0tc2 vxt1.210m/s 111和2m/V。式中m、v為靜止質(zhì)量和體積，v/c1 方向以速度v相對(duì)于觀察者測(cè)得立方體的長(zhǎng)、寬、高分別為：1v1vcVxyz

yy0zz0y 1 v00 m m V

2C7.設(shè)某微觀粒子的總能量是它靜止能量的k倍，則其運(yùn)動(dòng)速度的大小是Cv K21Kmmc2km0 km0c所以：vK

K28.中觀察到宇宙射線一介子的是它的固有的8倍，則介子的動(dòng)能是7m0c2。（已知該介子的靜止質(zhì)量為m01V2解：由tt0 8t 得 1V1V2 x mc2－mc2 C2mc28mc2mc2 xECT3RT38.3150623(J AQA623(J

ECT3RT38.3150623(J AV2pdVT2RdTRT8.3150416(J QAE1039(J

QMC(TT)p1V1C(TT)pVCV(T2 1 1.01050.0825(4001)683(J2QMC(TT)p1V1C(TT)pVCp(T2 1.01050.0827(4001)956(J2

pVAV2pdVVV

dVC(11pV

1

MAC(11)(pVpV) R(TT)M

2 1

p ECT5RT58.31601246.5(J ARTln

8.31(27380)ln22033.3(JQAE3279.8(J

ARTln

8.31(27320)ln21687.7(JECT5RT58.31601246.5(J QAE2934.2(JT2(V2)12V1TV 解：由絕熱方 2，得

2，得

2 Cp

332

2222

C雙雙

2552

2222 解：

1

1300

RTln

8.31400ln0.0055350(JAQ10.2553501338(JQ2(1)Q1(10.25)53504013(J

1

1

p

1

V 72.7%70%

1

1

80%70%V1V12P21MQ 1M

CT

(p

pV1 2 Q CTCp(pVpV M M

2 2V1

(p

pV 1 2

221 (p1PabOcPabOcTPaVa4.151052.0

V 5TPbVb4.151053.0

5 nC(TT)55R(TT)558.31(300200)10375 nC(TT)53R(TT)6225 ca，等溫過(guò)程，VbVcTc

ln

P

ln2

a A

QQ

QabQbc

103756225336536g水蒸汽(視為剛性分子理想氣體)abcda循環(huán)過(guò)程如圖所示．其中a－b、c－d為等體過(guò)程，b－c為等溫過(guò)程，d－a為等壓過(guò)程．試求：p62 VT 914b

(M/

mol (4) 1 1

1T21

解 1

A1 2T 4002533(K1 2 QQ

QQ2 2 1 1

2 1 1 1T2

T

400

500(K 11

1 2T2

1 2400源（T2=300K）放出熱量Q2=2.09×104J，11Q212.091

3.3401T213000 (2)AQ103.3410450%1.67104(J0.76m0.12m。管0.76m水銀柱壓強(qiáng)為1.013×105Pa）M

pVM

0.004 8.31(273的氦氣和們的溫度為30℃，試求容器中的混合氣體的壓強(qiáng)。

M p1

V，

M2molMpp1

HM

H H

)VHe 4.0 8.31(272(4.01