一模物理答案_2025年3月_250328太原市2025年高三年级模拟考试（一）（全科）_太原市2025年高三年级模拟考试（一）物理

太原市 2025 年高三年级模拟考试（一） 物理参考答案与评分建议 一、选择题：本题共7小题，每小题4分，共28分。在每小题给出的四个选项中，只有一 项是符合题目要求的。 题号 1 2 3 4 5 6 7 答案 D B D A B D A 二、多项选择题：本题包含3小题，每小题6分，共18分。在每小题给出的四个选项中， 至少有两个选项正确，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 8 9 10 答案 BC AC ABD 三、实验题：本题包含2小题，共15分。请将正确答案填在题中横线上或按要求作答。 11.(6分） k br （1）A（2分） （2） （2分） （2分） r kg 12.（9分） （1）分压式（1分） （2）截止（或极限）（1分） （3）Oa （1分） （4）Ob（1分） 3.91015（3.6~ 4.31015）（2分） ke（1分） （1分） （5）右（1分）四、计算题：本题包含3小题，共39分。解答应写出必要的文字说明、方程式和重要演算 步骤，只写出最后答案的不能得分。有数值计算的题，答案中必须明确写出数值和单位。 13.（1）M到达bb时速度为v 1 v2 F mg m 1 ···················································································1分 N R F  F 2mg N 压 v  gR ····························································································1分 1 方法一： M从aa到bb，依能量守恒 1 mgR  mv2 Q·················································································1分 2 1 Q r 1 N   ·······················································································1分 Q 2r 2 1 Q  mgR························································································1分 N 4 方法二： M从aa到bb，依动能定理   1 mgR W  mv2 0······································································1分 安 2 1 W Q 安 Q Q Q 2Q ·············································································1分 M N N 1 Q  mgR························································································1分 N 4 （2）M到达bb时，N解除锁定，M减速，N加速，最后分别做匀速运动 N切割磁感线的有效长度为L, v 、v 水平向右 M N B2Lv  BLv ····················································································1分 M N M减速过程，依动量定理，以右为正 I2LBt mv mv ······································································1分 M 1 N加速过程，依动量定理，以右为正 ILBt mv 0··············································································1分 N1 v  gR ························································································1分 M 5 2 v  gR ························································································1分 N 5 14.（1）电子在间距为r 的轨道上运动时 n e2 v2 k  m ························································································3分 r2 r n n 2 nh r mv  n 2π ke2 v 2mr n n2h2 r  ······················································································1分 n 2π2ke2m （2）方法一：电子偶素中电子的总动能、势能分别为 1 E 2 mv2 ·····················································································1分 k 2 1 e2 E  k ·························································································1分 k 2 r n e2 E k ··························································································1分 p r n E  E E n k p π2k2e4m E  ···················································································1分 n n2h2 方法二：电子偶素中电子或正电子的动能、势能分别为 1 E  mv2 ··························································································1分 k 2 1 e2 E  k ·························································································1分 k 4 r n e2 E k ··························································································1分 p r n E 2E E n k pπ2k2e4m E  ···················································································1分 n n2h2 （3）电子偶素由第一激发态跃迁到基态时，放出光子的频率为 h E E ························································································1分 2 1 π2k2e4m E  ···················································································1分 2 4h2 π2k2e4m E  ··················································································· 1分 1 h2 3π2k2e4m  ······················································································1分 4h3 15.（1）方法一： 火箭在点火预热阶段喷出气体，火箭对气体的平均作用力为F，依动量定理，以下为正 Ft mv 0······················································································3分 0 0 火箭获得的平均推力F F  F································································································1分 mv F  0 ·····························································································1分 t 0 方法二： 火箭在点火预热阶段喷出气体，火箭对气体的平均作用力为F， Fm a·····························································································2分 0 v a  0 ································································································1分 t 0 火箭获得的平均推力F F  F································································································1分 mv F  0 ·····························································································1分 t 0 （2）运载物、燃料在喷出气体的过程中动量守恒，以上为正  Mv 3m v 0················································································3分 2 1 3mv v  1 ···························································································1分 2 M （3）方法一： 以地面为参考系，以上为正，第一级燃料燃烧后，运载物获得的速度大小为v ， 运1 依动量守恒定律 (M 2m)v m（-vv )0····························································1分 运1 运1 第二级燃料燃烧后，运载物获得的速度大小为v 运2 (M m)v m（-vv )(M 2m)v ·············································1分 运2 运2 运1 第三级燃料燃烧后，运载物获得的速度大小为v 3 Mv m(vv )(M m)v ······························································1分 3 3 运2 mv 解得：v  运1 M 3m mv mv v   运2 M 3m M 2m mv mv mv v    ···················································2分 3 M 3m M 2m M m 方法二： 以火箭为参考系，以上为正，第一级燃料燃烧后，运载物获得的速度增量为v ，第二级燃 1 料燃烧后，运载物获得的速度增量为v ，第三级燃料燃烧后，运载物获得的速度增量为v ， 2 3 依动量守恒定律 (M 2m)v m（-vv )0····························································1分 1 1 (M m)v m（-vv )0·····························································1分 2 2 Mv m（-vv )0······································································1分 3 3 mv v  1 M 3m mv v  2 M 2m mv v  3 M mmv mv mv v v v v    ···································2分 3 1 2 3 M 3m M 2m M m 同理：若把质量为3m的燃料分成n份，运载物最终获得的速度为v n 3m 3m 3m 3m 3m v v v v v v  n  n  n  n  n ·····2分 n M n 3m M  n1 3m M  n2 3m M 2 3m M 1 3m n n n n n 3mv 3mv 3mv 3mv 3mv v      n nMn3m nM(n1)3m nM(n2)3m nM23m nM13m 依题目已知条件，当n时 M 3m v vln ··················································································1分 n M