物理答案_2024-2025高二（7-7月题库）_2024年11月试卷_1120山西省太原市2024-2025学年高二上学期11月期中考试_山西省太原市2024-2025学年高二上学期11月期中考试物理试题（含答案

2024 ~ 2025 学年第一学期高二年级期中学业诊断
物理参考答案及评分建议
一、单项选择题（本题包括10小题，每小题3分，共30分）
题号 1 2 3 4 5 6 7 8 9 10
选项 D C B A C C A C A A
二、多项选择题（本题包括5小题，每小题3分，共15分）
题号 11 12 13 14 15
选项 AC CD AD AB BCD
三、实验题（本题包括2小题，共14分）
16.（6分）
（ ）
（1）AB（1分） （2）BC（2分） （3） （3分）
+    

17.（8分）
（2）22.30（2分） （3）①B（1分）D（1分）②丙（1分） （4） （3分）



四、计算题（本题包括4小题，共41分）
18.（9分）
（1）小球下摆过程，由机械能守恒定律
·······················································································（1分）

=     
小球与A发生弹性碰撞
m v =m v ＋m v ····················································································（2分）
B B B 1 A 2
·········································································（2分）

=      +     
碰后瞬间，物块A速度的大小
v = m/s·································································································（1分）
2


1
{#{QQABDYSUgggAAAAAAAhCAQEQCEOQkhCACSgGAEAEoAIBiQNABAA=}#}（2）碰后瞬间，小球B速度的大小
v = m/s···································································································（1分）
1


·······················································································（1分）

=     
小球B上摆的最大高度
m·································································································（1分）

=  
19.（9分）
（1）W qU ······················································································（1分）
AB AB
= ····························································································· （1分）
−   
电荷从B点移动到C点，电势能增加6.0×10-8J，电场力做负功
····················································································（1分）

=  =−   
········································································（1分）
=    +    =−    
（2） ·················································································（1分）
=   −  
A点的电势
····························································································（1分）
=−    
，B点的电势
=   −  
····························································································（1分）
=−   
······························································································ （1分）
=   
··················································································（1分）
− 
=−  .  ×   
20.（10分）
（1）平行金属板A、B间的电场强度
U
E  ····································································································（2分）
d
qE=ma·································································································· （2分）
2
{#{QQABDYSUgggAAAAAAAhCAQEQCEOQkhCACSgGAEAEoAIBiQNABAA=}#}······················································································（1分）

=  ×    / 
（2）粒子在 T到 时间内，向右做匀加速运动，在 T到 时间内

=   =    =   =  
粒子向A板做匀减速运动···········································································（1分）
粒子向右运动的最大位移
2
1 T  1
x2 a   aT2·············································································（2分）
2  4 16
粒子恰好不能到达A板
xd······································································································（1分）
······················································································（1分）
− 
=   ×   
21.（13分）
（1）根据粒子甲的运动情况可知，场强方向由F指向O··································（2分）
方法一：对于粒子甲， 沿初速度方向
Rcos45°=v t ··························································································（1分）
01
沿加速度方向
1
Rsin45°= a t 2 ·······················································································（1分）
11
2
q E=m a ······························································································ （1分）
1 1 1
2 2m v 2
E= 1 0 ··························································································（2分）
q R
1
方法二：对于粒子乙， 沿垂直场强方向
Rsin45°=v cos45°t ··················································································（1分）
0 2
沿加速度方向
1
Rcos45°=v sin45°t + a t 2 ······································································（1分）
0 2 22
2
q E=m a ·····························································································（1分）
2 2 2
3
{#{QQABDYSUgggAAAAAAAhCAQEQCEOQkhCACSgGAEAEoAIBiQNABAA=}#}2 2m v 2
E= 2 0 ··························································································（2分）
q R
2
2 2m v 2
（2）方法一：若第一问已算得E= 1 0
q R
1
1
由Rcos45°=v t Rsin45°= a t 2 q E=m a
01 11 1 1 1
2
1
Rsin45°=v cos45°t Rcos45°=v sin45°t + a t 2 q E=m a
0 2 0 2 22 2 2 2
2
可得：q m =q m ······················································································ （2分）
1 2 2 1
对于粒子乙，从O到D，由动能定理
1 1
q ERcos45°= m v2- m v 2··········································································（2分）
2 2 2 0
2 2
v= 5v ···································································································（2分）
0
2 2m v 2
方法二：若第二问已算得E= 2 0
q R
2
对于粒子乙，从O到D，由动能定理
1 1
q ERcos45°= m v2- m v 2 ········································································（4分）
2 2 2 0
2 2
v= 5v ···································································································（2分）
0
4
{#{QQABDYSUgggAAAAAAAhCAQEQCEOQkhCACSgGAEAEoAIBiQNABAA=}#}