2025届山西省太原市高三11月期中诊断考-物理试题（含答案）.docx

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024 ~ 2025 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 7 小题，每小题 4 分，共 28 分。
题号
选项
1
2
3
4
5
6
7
C
B
B
A
D
C
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
选项
8
9
10
BC
CD
ACD
三、实验题：本题包含 2 小题，共 16 分。请将正确答案填在题中横线上或按要求作答。
1.（7 分）
1
ꢂ
ꢃ
ꢄ
ꢃꢀꢁ− ꢂ
（
1） C （1 分）（2） 6.70（2 分）（3） (ꢀ − )
（2 分）（4）
（2 分）
ꢁ
ꢃꢅꢆ
ꢇ
1
2. (9 分)
ꢈ+ꢉ
（1）A（1 分）（2）错误（1 分）（3）0.820（2 分）（4）
（3 分）（5）直线（2 分）
ꢈ
四、计算题：本题包含 3 小题，共 38 分。
1
3.（8 分）
（1）由题意可知在深度为 d 的深井处，半径为 R-d 的球体吸引物体
ꢈꢊꢉ
(ꢋ−ꢂ)ꢃ
G
=mꢄꢊ ··································································· （1 分）
ꢇ
ꢈ = π(R-d)3ρ··································································· （1 分）
ꢊ
ꢆ
1

极地地表处
ꢌꢈꢉ
ꢋꢃ
=
mꢄ········································································（1 分）
ꢇ
M = πR3ρ
ꢆ
ꢋ−ꢂ
ꢋ
g1 =
g········································································· （1 分）
（2）地球内部的重力加速度大小与到地心的距离成正比
ꢉꢄ+ꢉꢄ
ꢍ
ꢌ =
ꢊ····································································· （1 分）
ꢃ
由地表至井底深处,由动能定理
ꢌꢍ ·d = mv2 - 0·································································· （2 分）
ꢊ
ꢃ
(ꢃꢋ−ꢂ)
ꢋ
v =
ꢄꢂ ··································································（1 分）
1
4.（12 分）
1）物体 A、B 整体静止
(m +m )gsinθ =kx······························································（2 分）
（
A
B
x=0.4m············································································（1 分）
（2）物体 B 由静止向上做匀加速直线运动，位移为 x1=0.1m 时，
物体 A、B 间弹力消失，此位置物体 B
ꢎ(ꢀ − ꢀ ) − ꢉ ꢄꢐꢑꢒꢓ = ꢉ a··············································（2 分）
ꢊ
ꢏ
ꢏ
a =2.5m/s² ·······································································（1 分）
（3）物体 B 由静止向上做匀加速直线运动，位移 x1=0.1m 时速度为 v
v2 - 0=2ax1 ·······································································（1 分）
物体 B 继续向上运动的最大位移为 x2，物体 B 在运动过程中受到弹簧弹
力的平均值为ꢔꢍ
ꢊ
ꢍ
ꢔ= [k(x-x )+k(x-x -x )]······················································ （1 分）
1
1
2
ꢃ
由动能定理
2

1
2
ꢍ
-
m gx sinθ +ꢔx =0
-
mv2 ·················································· （2 分）
B
2
2
x2=（0.1+0.1 3 ）m ························································ （1 分）
物体 B 在斜面上运动的最大位移
x =x +x =（0.2+0.1 3 ）m ················································ （1 分）
3
1
2
1
5.（18 分
(1)设 P 球滑到 B 点的速度为 v ，碰撞后的速度为 v ，Q 碰撞后的速度为 v ，
0
1
2
P 由 A 到 B，由动能定理
1
3
mg ´ 2R = 3mv 02 - 0 ………………………………………………………（1 分）
2
在 B 处，P、Q 发生弹性碰撞
3
mv = 3mv + mv ····························································（1 分）
0
1
2
1
2
1
1
3
mv02 = 3mv12 + mv22 ………………………………………………（1 分）
2
2
ꢕꢁ
ꢃ
ꢕ1=
···········································································（1 分）
ꢆ
ꢕ2= ꢕꢁ··········································································（1 分）
ꢃ
P 恰好未从 C 点飞出，P 由 B 到 C，由动能定理
1
-
3mgR (1 - cos q ) = 0 - 3mv12 ···········································（1 分）
2
1
2
cosq =
…………………………………………………………（1 分）
q
= 600 ·········································································· （1 分）
(2)Q 由 B 到 C
1
2
1
2
-
mgR(1-cosq) = mv2 - mv2 ··········································· （1 分）
c
2
Q 从 C 点飞出到最高点，由动能定理
1
1
-
mgh2 = m（vc cosq）2 - mvc2 ········································（2 分）
2
2
h2 = 3R ·······································································（1 分）
R
h = + 3R = 3.5R ·························································· （1 分）
2
3

（3）当重力与风力的合力与速度垂直时，Q 动能最小··············（1 分）
vc = 2 2gR ···································································· （1 分）
vmin = vc cos150 ································································（1
分）
1
2
Ek min
=
mvm2in ································································ （1 分）
EK min = (2 + 3)mgR ·························································（1 分）
4