高三一模物理答案

太原市2023年高三年级模拟考试（一）物理参考答案及评分建议二、选择题:本题共8小题，每小题6分，共48分。在每小题给出的四个选项中，第14～18题只有一项符合题目要求，第19～21题有多项符合题目要求。全部选对的得6分，选对但不全的得3分，有选错的得0分。题号1415161718192021选项CCDBDBDBDBD三、非选择题：共62分。22.（6分）(1)AB(2)1��4�评分标准：每空各2分。23.（12分）（1）正1.1（2）○1○20.96○32.94.8评分标准：每空（或每图）各2分。24.（10分）（1）设“冰壶”被推出时瞬间速度的大小为v，继续滑行时加速度的大小为，位移大小=40m，时间t=10s，由运动学规律有11=t··�························�····························································（1分）��120=v-t························································································（1分）解得�：1=0.8m/，v=8m/s2设“冰壶�”1与冰面�的动摩擦因数为μ，由牛顿运动定律有μmg=m······························································（1分）解得：μ�=10.08························································（1分）运动员推“冰壶”的过程中位移大小=5m，加速度的大小为，水平恒力的大小为F，�2�2=2······························································（1分）F�2-μm�g2=�m2···························································（1分）解得：F=�1244N·······················································（1分）（2）当“冰壶”与冰面的动摩擦因数为μ´=0.1时，设“冰壶”被水平恒力推出时瞬间速度的大小为v´，+=t�12解�得�：v´=29m/s·······················································（1分）“冰壶”被推出后滑行的加速度为´，“冰壶”被水平恒力推出后滑行的距离为´�1μ´mg=m´�1=2´�1´‘2�解得：�1�´1=40.5m····················································（1分）水平恒力�1的大小变为F´，其作用距离为´，加速度的大小为´+=´+´�2�2�解1得�：2�1´=4�.25m=2�´2‘2F�´-μ´m�2g=�m2´´´解得：F´=2�002N······················································（1分）另：用动能定理或动量定理求解，方法更优。25．（14分）v0sin30°（1）粒子在电场中运动时间t1=2····················（1分）aqEa=································································（2分）m2mv0粒子在磁场中运动半径为R，qv0B···················（1分）RαR粒子在磁场中运动时间t2=································（1分）v0π由几何关系可知α=··············································（1分）3粒子从a点运动到b点所用的时间t=t1＋t2··························（1分）mv0πm联立可得t=＋··········································（1分）qE3qB（2）Ob两点间距离l1=2Rsin30°·······························（1分）粒子第二次在a点速度方向与x轴正方向夹角为θv0sinθv0cos30°t1=2v0cosθ···········································（2分）aOc两点间距离l2=2Rsinθ························································（1分）bc两点间距离l=l1-l2········································································（1分）mv0联立得l=(3-1)················································（1分）qB26.（20分）（1）Q下落：设在最低点速度为，0··················�·······························（1分）12�1�∙�0=2�1�0在最低点，（分）2····································1�0�110解得�−��=��···································（1分）（2）�Q0=和4Pm相/s碰：�设�=碰1后8NQ、P速度分别为、，以左为正····························�··1····�··2········（1分）101122��=��+��······················································（1分）121212210211222解�得�=��+��································（2分）碰后：�1P=在−滑1m板/s上滑动�2，=P3与m滑/s板共速时，相对AB的高度最大，设此高度为H，比较H与R即可判断能否从C点飞出。设此时P和滑板的速度大小均为�3······························································（1分）22233��=�+��·····················（1分）12122223322解2�得�=2×�；+��+����+�????R�故3=不1能m飞/s出····�···=···0·.·0··5·m······································（2分）设H<小球碰撞后能摆到的最高点与的连线和竖直方向的夹角为θ，′由机械能守恒定律得�1mv2mgL(1cos)··········································（1分）21110解得=0.9375，<0可知θco>s�，故碰后小co球s�Qc的os运5动不能视为简谐运动········（2分）0（3）计5算P两次经过B时的速度，以左为正···························································（1分）′′222233��=��+��································（1分）22121′1′2�2�2=2�2�2+2×�3�3+��2��解得·················································（1分）′3±6�2=3m/s第一次经过B点速度′3+6�2+=3m/s第二次经过B点速度′3−62−故物体P相对滑板滑动�时，=3m/s>0没有相对地面向右运动的速度····································（1分）在段相对滑板从时间AB′�2−�2+6−6�→��1=��=15s在段相对滑板从相对静止的时间AB′�3−�2−62物体P相对滑板的水�平→部分�=��=15sAB运动的总时间····························（2分）�=�1+�2=0.4s