kkkk4444con免费观看

运用动量和能量观点解题的思路
  [2020-04-14 07:57]  浏览次数:420
【字体:放大 正常 缩小】 【打印页面】【关闭窗口

运用动量和能量观点解题的思路

来源:人民教育出版社

  dongliangshouhengdinglv、jixienengshouhengdinglv、nengliangshouhengdinglvbiniudunyundongdinglvdeshiyongfanweigengguangfan,shiziranjiezhongpubianshiyongdejibenguilv,yincishigaozhongwulidezhongdian,yeshigaokaokaochadezhongdianzhiyi。shitichangchangshizongheti,dongliangyunengliangdezonghe,huozhedongliang、nengliangyupingpaoyundong、yuanzhouyundong、rexue、diancixue、yuanziwulidengzhishidezonghe。shitideqingjingchangchangshiwuliguochengjiaofuzade,huozheshizuoyongshijianhenduande,rubianjiasuyundong、pengzhuang、baozha、daji、danhuangxingbiandeng。 

  chongliangshiliduishijiandejilei,qizuoyongxiaoguoshigaibianwutidedongliang;gongshiliduikongjiandejilei,qizuoyongxiaoguoshigaibianwutidenengliang;chonglianghedongliangdebianhua、gonghenengliangdebianhuadoushiyuanyinhejieguodeguanxi,zaicijichushang,haihenrongyilijieshouhengdinglvdetiaojian,yaoshouheng,jiuyingbucunzaiyinqigaibiandeyuanyin。nenglianghaishiguanchuanzhenggewulixuedeyitiaozhuxian,congnengliangjiaodufenxisikaowentishiyanjiuwuliwentideyigezhongyaoerpubiandesilu。 

  yingyongdongliangdinglihedongnengdinglishi,yanjiuduixiangyibanshidangewuti,eryingyongdongliangshouhengdinglvhejixienengshouhengdinglvshi,yanjiuduixiangbidingshixitong;ciwai,zheixieguilvdoushiyunyongyuwuliguocheng,erbushiduiyumouyizhuangtai(huoshike)。yinci,zaiyongtamenjietishi,shouxianyingxuanhaoyanjiuduixiangheyanjiuguocheng。duixiangheguochengdexuanquzhijieguanxidaowentinengfoujiejueyijijiejueqilaishifoujianbian。xuanqushiyingzhuyiyixiajidian: 

kkkk4444con免费观看  1.xuanquyanjiuduixiangheyanjiuguocheng,yaojianlizaifenxiwuliguochengdejichushang。linjiezhuangtaiwangwangyingzuoweiyanjiuguochengdekaishihuojieshuzhuangtai。 

  2.yaonengshiqingkuangduiyanjiuguochengjinxingqiadangdelixianghuachuli。 

kkkk4444con免费观看  3.keyibayixiekansifensande、xianghudulidewutiquanzaiyiqizuoweiyigexitonglaiyanjiu,youshizheiyangzuo,keshiwentidadajianhua。 

kkkk4444con免费观看  4.youdewenti,keyixuanzheibufenwutizuoyanjiuduixiang,yekeyixuanquneibufenwutizuoyanjiuduixiang;keyixuanzheigeguochengzuoyanjiuguocheng,yekeyixuanneigeguochengzuoyanjiuguocheng;zheishi,shouxuandaduixiang、zhangguocheng。 

kkkk4444con免费观看  quedingduixiangheguochenghou,jiuyingzaifenxidejichushangxuanyongwuliguilvlaijieti,guilvxuanyongdeyibanyuanzeshi: 

kkkk4444con免费观看  1.duidangewuti,yixuanyongdongliangdinglihedongnengdingli,qizhongshejishijiandewenti,yingxuanyongdongliangdingli,ershejiweiyideyingxuanyongdongnengdingli。 

kkkk4444con免费观看  2.ruoshiduogewutizuchengdexitong,youxiankaolvlianggeshouhengdinglv。 

  3.ruoshejixitongneiwutidexiangduiweiyi(lucheng)bingshejimocalide,yaokaolvyingyongnengliangshouhengdinglv。 

  例1 图1中轻弹簧的一端固定,另一端与滑块B相连,B静止在水平直导轨上,弹簧处于原长状态。另一质量与B相同的滑块A,从导轨上的P点以某一初速度向B滑行。当A滑过距离时,与B相碰,碰撞时间极短,碰后A、B紧贴在一起运动,但互不粘连。已知最后A恰好回到出发点P并停止。滑块A和B与导轨的摩擦因数都为,运动过程中弹簧最大形变量为,重力加速度为。求A从P点出发时的初速度

  解析:首先要将整个物理过程分析清楚,弄清不同阶段相互作用的物体和运动性质,从而为正确划分成若干阶段进行研究铺平道路。即A先从P点向左滑行过程,受摩擦力作用做匀减速运动。设A刚接触B时的速度为,对A根据动能定理,有

 

  接着A、B发生碰撞,动量守恒。设碰后瞬间A、B的共同速度为,对A、B系统根据动量守恒定律,有 

  随后A、B向左压缩弹簧至阶段,设弹簧获得的势能为,对A、B和弹簧组成的系统,根据功能关系,有: 

  A、B又被弹簧弹回至弹簧恢复到原长阶段,设A、B的速度为,对A、B和弹簧组成的系统,根据功能关系,有 

  最后A、B分离,A滑至P点停下,对A应用动能定理,有 

  由以上各式解得。 

kkkk4444con免费观看  pingxi:donglianghenengliangdezonghewenti,tongchangdoujuyouduogewuliguocheng,fenxishixuyaogenjuzhenggeguochengzaibutongjieduandeshoulitedianheyundongqingkuang,jiangqihuafenweijiaojiandandejigeziguocheng,congerweiyunyongdonglianghenengliangguanxijiejuewentidiandingjichu。

   例2 在地面上方,一小圆环A套在一条均匀直杆B上,A和B的质量均为m,它们之间的滑动摩擦力。开始时A处于B的下端,B竖直放置。在A的下方米处,存在一个“相互作用”区域C,区域C的高度米,固定在空中如图2中划有虚线的部分。当A进入区域C时,A受到方向竖直向上的恒力F作用,。区域C对杆不产生作用力。A和B一起由静止开始下落,已知杆B落地时A和B的速度相同。不计空气阻力,重力加速度。求杆B的长度至少为多少?

 

kkkk4444con免费观看   jiexi:tongguoshenti,jiangwuliguocheng、zhuangtaixifenweirutu2-1、2-2、2-3suoshi。tu2-3zhuangtaiweia、bgangdadaogongtongsudu,cishia、bxiangduiweiyidezhangduweigandezuixiaozhangdu。 

  在物体A、B由图2所示状态变为图2-1所示状态过程:对A、B系统,机械能守恒,有(式中为图2-1所示状态A、B的速度),解得

  zaiwuliqingxingyoutu2-1zhuangtaibiandaotu2-2suoshizhuangtaiguochengzhong: 

  对A物体,由动能定理得 

  (为物体A在图2-2所示状态的速度),解得 

  对A、B系统,由于所受合外力为零,由动量守恒得 

  (为物体B在图2-2所示状态的速度),解得。 

  对B物体,由动能定理得(式中kkkk4444con免费观看为该过程物体B下落的高度) 

  解得。 

kkkk4444con免费观看  zaiwuliqingxingyoutu2-2zhuangtaibiandaotu2-3suoshizhuangtaiguochengzhong: 

  对A由动量定理得 

  由动能定理得(式中为该过程物体A下落的高度;kkkk4444con免费观看为图2-3状态时A、B具有的相同速度。) 

  对B由动量定理得 

  由动能定理得(式中为该过程B下落的高度。) 

  由上式解得m,

  杆的长度至少为m。 

  pingxi:youhenduowuliwentidoushejilinjiezhuangtai,jiejuecileiwentishi,yaoshenqingtiyi,tongguocaohuatuxing,nongqingwuliguocheng,zhaochuzhuanzhedian,zhuazhuchengqianqihoudewuliangliang,quedinglinjietiaojian。yifuhaodeshiyitujiushiyizhongwushengdeqifa,jiezhushiyitukeyibangzhuwomenshenti,kefengfuduiwuliqingjingdexiangxiangli,weizhengquejietikoukaidamen。

   li3 zaihefanyingduili,yongshimozuojiansuji,shiyouheliebiansuochanshengdekuaizhongzitongguoyutanhebuduandepengzhuangerbeijiansu。jiashezhongziyutanhefashengdeshidanxingzhengpeng,qiepengzhuangqiantanheshijingzhide。yizhitanhedezhiliangjinsiweizhongzizhiliangde12bei,zhongziyuanlaidedongnengweie0,shiqiu: 

kkkk4444con免费观看  (1)jingguoyicipengzhuanghouzhongzidenengliangbianweiduoshao? 

kkkk4444con免费观看  (2)ruoe0=1.76mev,zejingguoduoshaocihou,zhongzidenengliangcaikejianshaodao0.025ev。 

kkkk4444con免费观看  jiexi:andanxingzhengpengdeguilvkeqiuchumeicipengzhuanghouzhongzidesudubianweiduoshao,duiyingdedongnengyejiukeyiqiujie;zaigenjumeicipengzhuangqianhoudedongnengzhibiyuxuyaojianshaodao0.025evyuyuandongnenge0debizhiguanxi,quduishuqiuchupengzhuangcishu(bixujinweiquzheng)。 

  (1)danxingzhengpengzunshoudongliangshouhenghenengliangshouhenglianggedinglv。shezhongzidezhiliangm,tanhedezhiliangm。you: 

   

   

  由上述两式整理得 

  则经过一次碰撞后中子的动能 

  (2)同理可得    

  …… 

                   

kkkk4444con免费观看  shejingguoncipengzhuang,zhongzidedongnengcaihuijianshaozhi0。025ev,jien=0.025ev,e0=1.75mev 

  jieshangshide   n≈54 

  pingxi:dongliangyunengliangwenti,yibanyushijiwentijiehejinmi,nengfoujiangyigeshijidewentizhuanhuaweidianxingdewulimoxingheshuxideguocheng,shijiejuezheileiwentideguanjiansuozai。 

  例4 如图3所示,水平金属导轨M、N宽为,足够长金属导轨M’、N’宽为,它们用金属棒EF连接且处在竖直向上的磁感应强度的匀强磁场中,磁场右边界为gh处,cd金属棒垂直M、N静止在M、N导轨上,ab金属棒在光滑水平高台上受到水平向左的外力F=5N的作用,作用时间后撤去力F,ab棒随后离开高台落至cd右侧的M、N导轨上,M、N导轨使ab棒竖直分速度变为零,但不影响ab棒水平分速度。ab、cd棒始终平行且没有相碰,当cd、cd棒先后到达EF时,ab、cd棒均已达到稳定速度,已知,不计一切摩擦阻力。求

 

  (1)ab、cdbangzuizhongsududaxiao。 

kkkk4444con免费观看  (2)zhenggezhuangzhizhongdianliuchanshengdezongreliang。 

  解析:(1)ab棒做平抛运动的初速度为,根据动量定理有,解得 

  由题意可知,ab棒在M、N导轨上水平向左的初速度为,对ab棒、cd棒系统动量守恒,且cd棒到EF前它们已达共同速度,则有,解得 

  当cd棒在M’、N’轨而ab棒在M、N导轨上运动的过程中,每一时刻ab棒所受安培力是cd棒所受安培力的2倍,在相同的时间内两棒所受安培力的冲量大小关系为 

  ab棒到达EF前,ab棒、cd棒已达稳定速度,设分别为,对ab棒有,对cd棒有 

  ab棒、cd棒具有稳定速度时,有 

  解得。 

  当ab棒到达EF后,对ab棒、cd棒系统动量守恒,最终达共同速度,则有 

  解得。 

  (2)ab棒下落到M、N轨后,对整个系统能量守恒,电流产生的总热量等于系统机械能的损失,则有 

  pingxi:zaidianciganyingdewentizhong,jinshubangwangwangzuofeiyunbiansuyundong,youyudaotibangdesudubianhuayinqiledaotibangdeshoulifashengbianhua,yinciduiyufeiyunbiansuyundongdedingliangjisuan,bukeyizhijieyunyongyunbiansuyundongguilvhuoyunyonghenglidechonglianglaijiejue,zheishiwangwangkeyijiezhudongliangdinglilaijiejue。zaishuangjinshubangzhong,wangwangyoufenbieyiliangbangweiyanjiuduixiangyunyongdongliangdinglilaijiejue,dangranyoushiyekeyibashuangjinshubangdangzuoyigexitongzhijieliyongdongliangshouhenglaijiejue。

六安市新安中学 XJL编辑
新安中学办公室  发布


上一篇:没有了
下一篇:没有了
总访问量:21481873 人次 总浏览量:23299912 人次 日均访问:6383 人次 今日访问: 5748 人次 在线人数: 357 人
六安市新安中学版权所有 未经书面允许不得转载信息内容、建立镜像 网站备案号:
 Copyright© 2003 www.schwarzforcongress.com All Rights Reserved 技术支持:皖西电脑有限公司
邮编编码:237151 办公室:0564-2311115 教务处:0564-2315138