A.图甲磁电式电表铝框骨架能起到电磁驱动的作用
2023~2024学年第二学期高二年级期中学业诊断
B.图乙电流表运输时，把正负接线柱用导线连在一起，利用的是电磁阻尼原理
C.图丙阻尼摆片摆动中产生感应电流，很快停下来
物理试卷
D.图丁变压器硅钢片相互绝缘叠加能减小涡流
4.如图所示，绝缘圆环a与金属圆环b同心共面放置，α带负电并绕中心0点顺时针转动，下列
(考试时间：下午4：30—6：00)
说明：本试卷为闭卷笔答，答题时间90分钟，满分100分。
说法正确的是
A.当α顺时针加速转动时，b产生顺时针方向电流并有扩张趋势
题号
二
三
四
总分
0
B.当a顺时针减速转动时，b产生逆时针方向电流并有扩张趋势
得分
C.当a顺时针匀加速转动时，b产生逆时针方向电流并有扩张趋势
D.当a顺时针匀速转动时，b不产生电流但有收缩趋势
一、单项选择题：本题共10小题，每小题3分，共30分。请将正确选项前字母标号填入下表内
5.如图所示电路中，线圈L电阻不计，D为理想二极管，下列说法正确的是
相应位置。
A.S闭合时，A不亮，B立即变亮
题号
3
4
5
6
9
10
B.S闭合时，A立即变亮，B逐渐变亮
答案
C.S断开时，A不亮，B立即熄灭
1.关于交变电流，下列说法正确的是
D.S断开时，A立即变亮，然后逐渐变暗至熄灭
A.手机电池提供的电流是交变电流
6.如图所示为甲、乙弹簧振子的振动图像，下列说法正确的是
B.交变电流可能是按正余弦规律变化的
个x/cm
C.交变电流的大小一定随时间做周期性变化
A.两弹簧振子的初相位不同
D.家用照明电压的大小随时间做周期性变化，方向不变
B.甲系统的机械能比乙系统的大
2.关于简谐运动，下列说法正确的是
C.甲、乙两弹簧振子加速度最大值之比为2：1
-2
A.弹簧振子的回复力在半个周期内的功一定为0
D.1=2s时，甲具有负方向最大速度，乙具有正方向最大位移
B.弹簧振子的回复力在半个周期内的冲量一定为0
C.若物体做简谐运动，合外力与位移的关系符合F=-:
7,男子蹦床比赛中，运动员某次下落从最高点到着网点用时0.8s,从着网点到最低点用时
D.物体所受合力大小与时间满足一次函数关系，物体一定做简谐运动
0.4s。运动员的质量为60kg,该过程中网对运动员的平均作用力约为
3.关于电磁感应，下列说法不正确的是
A.600N
B.1200N
C.1800N
D.2400N
摆片
mA
8.质量为m的物体静止在光滑水平面上，在水平恒力F的作用下，经时间：走过位移1，动量变
为p、动能变为E。若上述过程中F不变，物体的质量变为2m,下列说法正确的是
A.经过时间1，物体动量变为p
B.经过时间1，物体动能变为E
C.经过位移1，物体动量变为2印
D.经过位移1，物体动能变为2E
高二物理第1页（共8页）
高二物理第2页（共8页）2023～2024学年第二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：本题包含 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 B A A A D D C A D C
二、多项选择题：本题包含 5小题，每小题 3分，共 15分。
题号 11 12 13 14 15
选项 AC AC BC AB ABC
三、实验题：共 14分。
16.（6分）
（1） C （1分）
（2）19.80（2分）805.0（1分）
（3） （2分）
17.（8分）
（1）天平（2分）
（2）D （2分）
（3） 1 1 0 = 1 2 0 + 2 3 0（2分）
1( 1 0) = 1( 2 0) + 2( 3 0) 其他正确答案均可得分（2分）
四、计算题：共 41分。
18.（8分）
（1）甲车向右做减速运动，由动能定理得
f L 1 m v 2 11 m v
2
0 ····························································（1分）2 2
f 0 . 1m g ············································································· （1分）
1 = 2 m/s················································································ （1分）
（2）甲、乙车相碰时，组成系统动量守恒，以右为正方向
m v 1 2 m v 2 ·······································································（2分）
两车碰撞后嵌套在一起，向右做减速运动
2 f x 1 0 2 m v 22 ·····························································（2分）2
= 0.25m···················································································（1分）
19.（9分）
（1）设磁感应强度为 B，导线框边长为 l，导线框电阻为 R，质量为 m，
导线框进入磁场时
2 2 = ···················································································（2分）
2 2
= ······················································································（1分）
安培力方向与运动方向相反，v减小，加速度 a减小····························（1分）
（2）以右为正，导线框全部穿进磁场时速度为 v
由动量定理，导线框穿入磁场，以右为正方向
= 1································································（1分）
导线框穿出磁场
= 2 ································································（1分）
2 3
= ············································································（1分）
1 = 2 ·································································· （1分）
= 1+ 2 ····················································································（1分）
2
20.（11分）
（1）由右手定则 ，R中电流由 P→M······························································(2分)
（2）杆达到最大速度
mg sinθ = ILB··········································································································(1分)
I= = ······································································································· (2分)
+ +
Vm=
sin （ + ）
2 2 ·································································································· (1分)
（3）由动量定理，以沿斜面向下为正方向
mgsinθ· t + （ ) = mV 0
+ m ··········································································································
(2分)
x为杆沿导轨达到最大速度时下滑距离
2 2 = ········································································· (1分)
+
由动能定理可知
mgsinθ· x+（ = 1） 2 0·················································· (1分)安 2
2 2 2
Q = = ( + )sin 4 4 [ 2 2
3 ( + )]······································ (1分)
安 2
21.（13分）
（1）甲棒沿圆弧下落进入磁场时的速度为 v0 ，由动能定理
mgr 1 mv 20 0········································································ (1分)2
甲棒从进入磁场到与乙棒共速，甲、乙组成的系统动量守恒，以右为正方向
mv0 2mv1 ················································································ （2分）
由能量守恒，甲、乙两棒上产生的焦耳热设为 Q
Q 1 mv 2 1 2mv 20 1 ···································································（1分）2 2
甲棒上产生的焦耳热为Q1
Q 1 11 Q mgr ········································································(1分)2 4
（2）从甲棒进入磁场到甲、乙共速，甲棒相对乙棒运动的距离设为 x，
以乙棒为研究对象
B2L2 v
t mv1 0································································ (2分)2R
B2L2 x
mv1 ···········································································（1分）2R
乙棒初始位置 EE'离磁场边界 CC'的距离为 x
mR 2gr
x x ······································································（1分）
B2L2
（3）开关闭合以后，甲、乙棒一起做加速度减小的减速运动，直到静止，
根据能量守恒，回路中产生的总焦耳热设为 Q'
Q 1 2mv 21 ···············································································（1分）2
R上产生的焦耳热为 Q2
Q 22 Q ···················································································（2分）3
Q 12 mgr ················································································ （1分）3
