青岛市2023年高三年级第一次适应性检测物理答案及评分标准一、单项选择题：本大题共8小题，每小题3分，共24分。1．D2．C3．B4．B5．A6．D7．D8．B二、多项选择题：本大题共4小题，每小题4分，共16分，选不全得2分，有选错得0分。9．BC10．AC11．ACD12．CD三、非选择题13．（6分）222mg()xxmgxxx21（1）水平（2分）；（2）10122（2分）；（3）（1分）、偏小（1分）4hx014（8分）（1）如图（2分）（2）C（1分）、H（1分）（3）2.9（1分）、2.0（1分）（4）多次测量减小了偶然误差；（1分）避免了电压表分流作用或电流表的分压作用，消除了系统误差（1分）。15．(8分)pp（1）对容器A中的气体：1＝2·····················································（2分）T1T2解得：T2＝298.3K··········································································（1分）（2）环境温度为360K时：pA′＝pB+1.16pVp′V对于总气体：AA＝A································································（2分）T1T3对于进入容器B的气体：pA′(V－VA)＝pBVB··········································（2分）解得：pB＝0.2atm···········································································（1分）评分标准：第1问，3分；第2问，5分。共8分。16．（8分）（1）只要气嘴灯位于最高点时ab接触即可保证全程灯亮，mg弹簧原长时ab的距离为：+L=2L··················································（1分）kmv2气嘴灯位于最高点时的向心力为：＝mg+2kL＝3mg··························（2分）R可解得满足要求的最小速度为v＝3gR·············································（1分）2πRπ（2）速度为2gR时轮子滚动的周期为：T＝＝2gR····················（2分）2gRg物理答案第1页共3页(2gR)2此速度下气嘴灯所需的向心力为:m＝2mg·································（1分）R此力恰好等于ab接触时弹簧的弹力，即无重力参与向心力，对应与圆心等高的点，Tπ故当气嘴灯位于下半圆周时灯亮，即t＝＝2gR·····························（1分）22g评分标准：第1问，4分；第2问，4分。共8分。17．(14分)mv2（1）粒子所受的洛伦兹力充当向心力：0＝qvB·····························（2分）R00mv可解得R＝0·············································································（1分）qB0πm（2）在第1、4个的时间内，粒子向x轴正方向移动了2R的距离·····（2分）2qB0πm在第2、3个的时间内，粒子向y轴负方向移动了2R的距离············（2分）2qB02mv2mv2mv故荧光屏在x轴上的位置应为0，粒子打在屏幕上的坐标为(0,l－0,0)（1分）qB0qB0qB0（3）由粒子的轨迹易知：当粒子的速度与z轴正方向的夹角为60°时：Rx＝+2nR····················································································（2分）2mvqBl同时满足：l≥2n0，得n≤0，且取整数····································（1分）qB02mv0当粒子的速度与z轴负方向的夹角为60°时：3Rx＝+2nR··················································································（2分）2mvqBl同时满足：l≥2(n+1)0，得n≤0-1，且取整数···························（1分）qB02mv0评分标准：第1问，3分；第2问，5分；第3问，6分。共14分。18．(16分)（1）物块在凹坑中滑动时，其运动情况与摆长为R的单摆相同，若该单摆的周期为Tl，质量为m的弹簧振子的周期为Tk1，则：RmT＝2π＝2π＝T·································································（2分）lgkk1R故物块第一次回到出发点所用时间为一个单摆周期，即：t＝2π·········（1分）1g（2）由简谐振动的运动特点可知，振子在同一位置有等大的速度，由动量守恒定律有：mv－mv＝0····················································································（1分）即碰撞产生的组合体C的初速度为零。设组合体C与弹簧组成的弹簧振子的周期为Tk2，：2mT＝2π＝2T······································································（1分）k2klT碰撞之前两个物块运动时间相同，均为l···········································（1分）8物理答案第2页共3页1121(5+42)πR故所求时间为：t＝(+++)T＝································（2分）28424l4g（3）设碰撞位置的球面半径与竖直方向夹角为φ，则此处向心力为：2mv22mgtanφ＝c············································································（1分）πrsin4πrsin4其中，tanφ≈··········································································（1分）R由水平切向的弹性碰撞可列出动量与机械能守恒方程，即：mv0＝mvt+2mvc···············································································（1分）111mv2＝mv2+2mv2····································································（1分）202t2C32rg解之，得v·····································································（1分）04R（4）若组合体在被推至最低点时的动量为p，并与来自Q点的物块碰撞时，组合体会获得来自Q的物块的动量pQ，由于单摆周期与振子质量无关，当组合体从振幅处返回时必然再次与下一个来自Q的物块碰撞，由于之前在碰撞中获得的动量pQ已经反向，在第二次碰撞中该部分动量会被再次获得的动量pQ抵消，仅留下与第一次碰撞前等大反向的动量－p。故碰后组合体将重新沿弹簧轴向运动并开始压缩弹簧。……………………………………………………（2分）（或者，也可以如下方式分析：显然，在任何情况下，当组合体在被推至最低点并与来自Q点的物块碰撞时都满足：p+pQ＝p′·····················································································（1分）由于单摆周期与振子质量无关，故当组合体回到最低点时必与下一个来自Q点的物块碰撞：－p′+pQ＝－p················································································（1分）易知碰后组合体将重新沿弹簧轴向运动并开始压缩弹簧。）由以上结论可知：第1块既然与组合体碰撞，则第2块必定也能碰撞；此时组合体质量为4m，周期为2Tl；第3块到达最低点时，组合体在弹簧压缩最大处，无碰撞；第4块到达最低点时，组合体恰好到达最低点，物块与组合体碰撞，则第5块必定也能碰撞；第6块到达最低点时，组合体正在压缩弹簧，无碰撞；故前6个物块中，第1、2、4、5个物块可与组合体发生碰撞，碰撞物块个数为4。（1分）评分标准：第1问，3分；第2问，5分；第3问，5分；第4问，3分。共16分。物理答案第3页共3页