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