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2023～2024 学年第一学期高三年级期中学业诊断
物理参考答案及评分建议
一、单项选择题：本题包含8小题，每小题4分，共32分。
题号 1 2 3 4 5 6 7 8
选项 C C D A D C C C
二、多项选择题：本题包含4小题，每小题4分，共16分。
题号 9 10 11 12
选项 CD CD AC BD
三、实验题：共16分。
13.（8分）
（1）4.80 （2分）
（2） （2分） （2分）
（3）减小遮光片宽度（其它合理建议也可得分）（2分）
14.（8分）
（3）小于 （2分）
（4） （2分）
（2分）
= −
（2分）
= +
四、计算题：共36分。
15. （7分）
（1）赤道处
·····································································（1分）
两极处
·········································································（1分）·············································································（1分）
（2）两极处
·········································································（1分）
········································································（1分）
···················································································（1分）
·········································································（1分）
16. （8分）
（1）根据题意，由平衡条件有
·········································································· （2分）
+ =
由玻意耳定律有
pV pV
0 0 1 1···················································································（1分）
联立解得
·············································································（1分）
= +
带入数据得
V =432cm3·················································································（1分）
1
（2）根据题意，由理想气体状态方程有
pV pV
0 0 0 x
T T
0 2 ·················································································（1分）
解得
T
V 2V
x T 0
0 ····················································································（1分）
则漏出气体与剩余气体质量之比
V V 1
x 0
V 30
0 ············································································（1分）17. （10分）
（1）小球B质量为m ，下落h时速度为v,以下为正方向
1
v 2gh
················································································（1分）
，
与小球A碰后反弹速度大小为 ，小球A碰后瞬间速度为v ，在碰撞的前
2
=
后瞬间
mv m (v)mv
1 1 1 2·····································································（2分）
1 1 1
mv2 m (v)2 mv 2
2 1 2 1 1 2 2 ·························································（2分）
1
v v
2 2 ····················································································· （1分）
····················································································（1分）
=
（2）碰后的小球A再次压缩弹簧x 的过程，设弹簧弹性势能的增加量为△E
0 p
1
mv 2 mgx E
2 2 0 p ···································································（2分）
1
E mghmgx
p 4 0 ····································································（1分）
18. (11分)
（1）以A为研究对象，由动量定理
···············································（1分）
− = = /
物块A以10m/s滑入传送带，物块在传送带上的加速度为
a=μg=2m/s2··················································································（1分）
匀减速阶段所用时间t
1
··········································································（1分）
= −
1s
=
此时A速度 =8m/s ，A以v =8m/s 撞向B
2
−
= ····················································（1分）
+ = +
·············································（1分）
+ = +
解得 v =-1m/s ,v =5m/s·····························································（1分）
4 5
物块A以1m/s返回进入传送带，经过t 时间速度减为零，返回时从传送带离开
2
仍以v =1m/s离开,之后从平台飞出
6
=0.5s
−
=
在传送带上总时间 t=t +2t =2s ·················································· （1分）
1 2
(2)AB下落时间为 t ， =0.6s················································（1分）
3
=
=v t =0.6m············································································（1分）
6 3
=v t =3m ············································································· （1分）
5 3
x -x =2.4m ···············································································（1分）
B A
