河海大学物理习题答案

1一1E2D二12M/S5M/S02200HTVLXXTVLVV000变加速力学一）2/5/910SMJSMJMJRRR3/24044964/324045123163/12/642325,254122233SMJIJITVASMJIJJITRVSMJTJDTDVIDTDVDTVDASMJTIJDTYDIDTDXDTRDVMXYMJTITRYXRRRRRRRRRRRRRRRRRRRRRRRRRRRRDDDD三123002231,0KTVVDTKTDVDTKTDVTVVTXDTKTVDXKTVDTDXV03003031311240KTTVX三23力学二）一1C2B二160M/S54M/S2221COSVVQQSIN1V325M/S866M/S25M/S220M三、1、1/8,/182SRADDTDSRADTDTDWAQW（2）SMTRV/8132024324W222/24,/081SMRASMRVATNATDTDSV1020222/80,2201020SMASTTRVANN当2/10SMDTDVAT2、4一1C2C二12G02056M/S2588N力学三）三1解0NRYXNRGMRFRQ0COSSIN0200MGNYRMVNXQQ方向方向RGVTG20Q以设计标准速度行驶时，无侧向摩擦力RMVFNX2COSSINQQ方向以V行驶时，有侧向摩擦力2340M22022220VVGRRGVVMGMRQNFM0SINCOSMGFNYQQ方向5三2解分别以M1和M2为研究对象，分析各自受力。NRTR设M1相对升降机的加速度A,方向向左；则M2相对升降机的加速度A,方向向下。AR1AMT3GR311GMGMN322AGMGMT2/78SMA/337821SMJIAMRRR对地/4522SMJAMRR对地GMR1GMR23GRTRAR1M2MYXO6JSFDSFRDFA29422301PTTRR一1C2C二1A/2CA/2C285N85W3力学四）JDYFDXFRDFARYRX229002RRFRFFAAA0212MMMKXMGXFXM2MGFKXMM22221MGFKKXEMPM此时三1解（1）F一直保持与圆弧相切，作功（2）F与X、Y轴方向不变，作功2解运动中受摩擦力FR，弹性力F，外力F。设最大伸长量XM，此时速度为0。过程中作功KMGKMGGH22KGMKMGH2MG227力学五）一1C2B二1KMGQSIN2MKLGL2SIN2QRMMGRMMG6,33SMJ6,182三1解摩擦力MGNFMM设所需距离为S由功能原理2210MVSF221MVMGSMGVSM228（1）小环下滑过程中，小环、弹簧、地球系统机械能守恒，设B处重力势能为0222121LKMVMGHD222122121SIN2QQCOSRKMVRMG222COS14SIN4QQKRMGRVNAR（2）分析B处受力COS12QDRLKFRVMMGFN2COS12COS14SIN422QQQRKKRMGMGNGMRFRNR三2解9力学六）一1B2C二1230222,2,BVABABAQV2与X轴1350MMVMMMV2,2三1解VR（1）M、M系统在子弹穿透过程中，水平方向动量守恒，设M获得初速为V0VVMMVM获得初速V后瞬时受力为T、MG，有加速度LVMMGT25262022NVVLMMGMTMVMVMV0（2）子弹受到的冲量740SNIVMVMIRRRRYXOGMRTR10222121AMVMVMGHMMGHMV2M从A底部匀速滑到B底部，速度不变。FIM上升最大高度H时，M和M具有共同速度VB，水平向右。由水平方向上动量守恒和上滑过程中系统机械能守恒得BVMMMVMGHVMMMVB222121HMMMH2三2解由动量守恒，下滑到底部时AMVMV0由机械能守恒有11三1解10VMMMV20212121MVVMMMGHGMMMVH220离开小车的过程中，则从入射到小车的速度为离开轨道时速度为设VVM,MVVMMV02220212121MVVMMV0VMMMMV一1A2B3A二1守恒不守恒2力学七）36400M/S42222222,2MRLMRLMRL222MMVMMMMGL2512AB121212020LKXVMB设A、B达到共同速度为V，则20LLVMMVMBAB由式1、2可得032XMKVB（2）A、B速度相同时，弹簧的伸长量为最大MAX2220212121KXVMMKXBA0MAX33XX三2解（1）弹簧为原长时，B的速度为V013力学八）OOKJJKWW2,92LGMGL53,21一1B2B二1202120TTBWQST420BWSRADTT80BWWSMTV/69W三1解2解AARGMRGMR2BA0T1T2TARAAJRTRT10MAMGT1轮AAB02JRTRTMATMG222RAAMGT811轮BM2M解得14一1C2A二1628RAD133J133J94M/S47M/S2590M/S214NSMKG/1017233OW41力学九）23三解碰撞过程角动量守恒VRM43LMMLWW31434322LMLMJJLMVMMSRAD/373W上摆过程机械能守恒MH0610HMGHMGJJMM2143212WH43H15一1C2B3C二1三解振动（一）2/150,/30,800,261,6SMSMSMKMKM22,22PPSM750,10244223/2500MNLGMK110SMKWMX2010463SMV/200MA0406PJ振动方程610COS040SITXP16振动（二）一1B2C3B二12212141三解0,34,6,12,100VSMPPCMSS179,571,41324COS800PTX24SIN203PTVXTA1624COS812PSMVSMVTM/23,/232SIN23,010P2/4616,402SMXAMXNMAF28117一1A2C3B二1410COS1060PTXSRAD/423402P410COS1060SITXP振动（三）42,452,0850PPPPPKK23三解（1）AMMAMGF2MAXMAXWM5002GAWMXFRARMXNRARMGMRHZAG23221PN（若已知W，问物体在何处分离）（2）如图MANMG时不分离，即0N0MAMGCOS2JWWTAAGAA2MAXWNPW2即必须18210050COS10PPXTY波动（一）一1C2C3C二1322COS10PPTY52322COS10XTYPPPSI05HZ,05M/S,1M,005M23AYTSRADT00/21PPW2COS01022SIXTYMSUPPMUT43LMYMX010135三1解COS010SITYPPPJ2解SRADU/8/2PLPW248COS05000SITXTYAYTPPPJMYXT0500,2501258310SMAVMYXT0503,2502258310SMAV19一1B2B3B210COS010PPTY310COS010PPTY183M2122RR22COS00SIXXTAVAYPLP三1解二12PP228COS0101SITYPP2208COS0102PPXTY波动（二）3MYMXA0OX2000,000PJXVDTDYYTXX2COS000MAX0PWWTAVAYAVVADTDYX2解2000,00PJXDTDYYTXSCMUCM/20402秒内波形移动了在8223PLPWLUM20波动（三）一1A2A3A3同频率、同振动方向、同振幅、相差恒定为P12K二12三1解04,0IMU10NLPPLP542DFBPAPPMMAAAAAP832COS2212221DF21222221LLLLKBBAXXA2解X处两波振动位相差为222222212212XABBAXPLPLPLPLPDFPK2时干涉加强。解得K为整数21一1D2C3C0UYX0YX25020COS080XTPPSMMHZ/40410122MKXMKXTXYPP20COS50COS1602COSSIXTTAYPLP22COS2SIN221SITTXAYYYPPLP二12M50HZ01MHZM1720250HZSM2468/330波动（四）2,1,0K34三解（1）反射端固定，X0处反射波的振动方程为/2COSPPTTAY（2）22一1C2D二1同频率,同振动方向,相位差恒定2间距不变,条纹明暗对比度变小39R104MMMX4632条共12三1解由干涉相消条件OOAA4615400065LLOOAA6667545587LL光学一212LDDKX122KDXDL得在可见光范围取8,7,6,5K分别对应2解由图示几何关系2DDLD2X1X0X22221LDDLDX得到MMLDLDX1611类似地MMDDX1902DL31212DXXXN干涉条纹范围干涉条纹数MMX463161条纹间距23二1光学二一1D2B221111DNDNOA1217OOOAAA508067744064RAD41001234三1解干涉条纹上移。01S2SD2RE1RPX012NEERRDENRR112XDDRR12Q5211EDXDN即2解LAODDL2/MLAOD510942DLLNAODDL2/MNDAOL410682LD光程差24光学（三）二10823M；0627MM2341M3109MM4一1C2AMM4758只能看到4条明纹；FL既不最亮也不最暗。三1解2222NRKRKL11N1121NRKRKL331104734NN222221222112KKKKDDRR3312N2解321NNN5Q10000E7500E5000E250004030201AAAAE；FI各明纹对应的宽度；25光学（四）23SINSINLQFA340二161明；一1C2B2212KASINLJQFXJSINFKAX122L00A7000A7600LQ又3K72860AL三解26光学（五）二118CM21083向上124；4116CM52,1,0K一1D2B三解LJKBSINAM60104230SIN2BAL3ABAM610803BAA第三级缺级FI1SINMAXJ41MAXLBAK,2,13LKK级缺级屏上只能看到五条，对应4K在屏上无穷远，看不到。27ANII20COS329,800II光学六613二1一1B2C075401020TGITGIARCTGI2345I0II0I28光学七一1A2D二1椭圆偏振,线偏振2M475M3偏振；1/4波片；平行；P/4。1一1E2D二12M/S5M/S02200HTVLXXTVLVV000变加速力学一）2/5/910SMJSMJMJRRR3/24044964/324045123163/12/642325,254122233SMJIJITVASMJIJJITRVSMJTJDTDVIDTDVDTVDASMJTIJDTYDIDTDXDTRDVMXYMJTITRYXRRRRRRRRRRRRRRRRRRRRRRRRRRRRDDDD三123002231,0KTVVDTKTDVDTKTDVTVVTXDTKTVDXKTVDTDXV03003031311240KTTVX三23力学二）一1C2B二160M/S54M/S2221COSVVQQSIN1V325M/S866M/S25M/S220M三、1、1/8,/182SRADDTDSRADTDTDWAQW（2）SMTRV/8132024324W222/24,/081SMRASMRVATNATDTDSV1020222/80,2201020SMASTTRVANN当2/10SMDTDVAT2、4一1C2C二12G02056M/S2588N力学三）三1解0NRYXNRGMRFRQ0COSSIN0200MGNYRMVNXQQ方向方向RGVTG20Q以设计标准速度行驶时，无侧向摩擦力RMVFNX2COSSINQQ方向以V行驶时，有侧向摩擦力2340M22022220VVGRRGVVMGMRQNFM0SINCOSMGFNYQQ方向5三2解分别以M1和M2为研究对象，分析各自受力。NRTR设M1相对升降机的加速度A,方向向左；则M2相对升降机的加速度A,方向向下。AR1AMT3GR311GMGMN322AGMGMT2/78SMA/337821SMJIAMRRR对地/4522SMJAMRR对地GMR1GMR23GRTRAR1M2MYXO6JSFDSFRDFA29422301PTTRR一1C2C二1A/2CA/2C285N85W3力学四）JDYFDXFRDFARYRX229002RRFRFFAAA0212MMMKXMGXFXM2MGFKXMM22221MGFKKXEMPM此时三1解（1）F一直保持与圆弧相切，作功（2）F与X、Y轴方向不变，作功2解运动中受摩擦力FR，弹性力F，外力F。设最大伸长量XM，此时速度为0。过程中作功KMGKMGGH22KGMKMGH2MG227力学五）一1C2B二1KMGQSIN2MKLGL2SIN2QRMMGRMMG6,33SMJ6,182三1解摩擦力MGNFMM设所需距离为S由功能原理2210MVSF221MVMGSMGVSM228（1）小环下滑过程中，小环、弹簧、地球系统机械能守恒，设B处重力势能为0222121LKMVMGHD222122121SIN2QQCOSRKMVRMG222COS14SIN4QQKRMGRVNAR（2）分析B处受力COS12QDRLKFRVMMGFN2COS12COS14SIN422QQQRKKRMGMGNGMRFRNR三2解9力学六）一1B2C二1230222,2,BVABABAQV2与X轴1350MMVMMMV2,2三1解VR（1）M、M系统在子弹穿透过程中，水平方向动量守恒，设M获得初速为V0VVMMVM获得初速V后瞬时受力为T、MG，有加速度LVMMGT25262022NVVLMMGMTMVMVMV0（2）子弹受到的冲量740SNIVMVMIRRRRYXOGMRTR10222121AMVMVMGHMMGHMV2M从A底部匀速滑到B底部，速度不变。FIM上升最大高度H时，M和M具有共同速度VB，水平向右。由水平方向上动量守恒和上滑过程中系统机械能守恒得BVMMMVMGHVMMMVB222121HMMMH2三2解由动量守恒，下滑到底部时AMVMV0由机械能守恒有11三1解10VMMMV20212121MVVMMMGHGMMMVH220离开小车的过程中，则从入射到小车的速度为离开轨道时速度为设VVM,MVVMMV02220212121MVVMMV0VMMMMV一1A2B3A二1守恒不守恒2力学七）36400M/S42222222,2MRLMRLMRL222MMVMMMMGL2512AB121212020LKXVMB设A、B达到共同速度为V，则20LLVMMVMBAB由式1、2可得032XMKVB（2）A、B速度相同时，弹簧的伸长量为最大MAX2220212121KXVMMKXBA0MAX33XX三2解（1）弹簧为原长时，B的速度为V013力学八）OOKJJKWW2,92LGMGL53,21一1B2B二1202120TTBWQST420BWSRADTT80BWWSMTV/69W三1解2解AARGMRGMR2BA0T1T2TARAAJRTRT10MAMGT1轮AAB02JRTRTMATMG222RAAMGT811轮BM2M解得14一1C2A二1628RAD133J133J94M/S47M/S2590M/S214NSMKG/1017233OW41力学九）23三解碰撞过程角动量守恒VRM43LMMLWW31434322LMLMJJLMVMMSRAD/373W上摆过程机械能守恒MH0610HMGHMGJJMM2143212WH43H15一1C2B3C二1三解振动（一）2/150,/30,800,261,6SMSMSMKMKM22,22PPSM750,10244223/2500MNLGMK110SMKWMX2010463SMV/200MA0406PJ振动方程610COS040SITXP16振动（二）一1B2C3B二12212141三解0,34,6,12,100VSMPPCMSS179,571,41324COS800PTX24SIN203PTVXTA1624COS812PSMVSMVTM/23,/232SIN23,010P2/4616,402SMXAMXNMAF28117一1A2C3B二1410COS1060PTXSRAD/423402P410COS1060SITXP振动（三）42,452,0850PPPPPKK23三解（1）AMMAMGF2MAXMAXWM5002GAWMXFRARMXNRARMGMRHZAG23221PN（若已知W，问物体在何处分离）（2）如图MANMG时不分离，即0N0MAMGCOS2JWWTAAGAA2MAXWNPW2即必须18210050COS10PPXTY波动（一）一1C2C3C二1322COS10PPTY52322COS10XTYPPPSI05HZ,05M/S,1M,005M23AYTSRADT00/21PPW2COS01022SIXTYMSUPPMUT43LMYMX010135三1解COS010SITYPPPJ2解SRADU/8/2PLPW248COS05000SITXTYAYTPPPJMYXT0500,2501258310SMAVMYXT0503,2502258310SMAV19一1B2B3B210COS010PPTY310COS010PPTY183M2122RR22COS00SIXXTAVAYPLP三1解二12PP228COS0101SITYPP2208COS0102PPXTY波动（二）3MYMXA0OX2000,000PJXVDTDYYTXX2COS000MAX0PWWTAVAYAVVADTDYX2解2000,00PJXDTDYYTXSCMUCM/20402秒内波形移动了在8223PLPWLUM20波动（三）一1A2A3A3同频率、同振动方向、同振幅、相差恒定为P12K二12三1解04,0IMU10NLPPLP542DFBPAPPMMAAAAAP832COS2212221DF21222221LLLLKBBAXXA2解X处两波振动位相差为222222212212XABBAXPLPLPLPLPDFPK2时干涉加强。解得K为整数21一1D2C3C0UYX0YX25020COS080XTPPSMMHZ/40410122MKXMKXTXYPP20COS50COS1602COSSIXTTAYPLP22COS2SIN221SITTXAYYYPPLP二12M50HZ01MHZM1720250HZSM2468/330波动（四）2,1,0K34三解（1）反射端固定，X0处反射波的振动方程为/2COSPPTTAY（2）22一1C2D二1同频率,同振动方向,相位差恒定2间距不变,条纹明暗对比