2024 ~ 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 1 2 3 4 5 6 7
选项 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 8 9 10
选项 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。
11.（7分）
1 C 1 2 6.70 2 3 ( ) （ ） （ 分）（ ） （ 分）（ ） （2分）（4） （2分）
12. (9分)
1 A 1 2 1 3 0.820 2 4 + （ ） （ 分）（ ）错误（ 分）（ ） （ 分）（ ） （3分）（5）直线（2分）
四、计算题：本题包含 3 小题，共 38 分。
13.（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
G =m ···································································（1分）( )
=
π(R-d)3ρ··································································· （1分）
1
极地地表处
= m ········································································（1分）
M = πR3ρ
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
= + ····································································· （1分）
由地表至井底深处,由动能定理
·d = mv2 - 0··································································（2分）
v = ( ) ··································································（1分）
14.（12分）
（1）物体 A、B整体静止
(mA+mB)gsinθ =kx······························································（2分）
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
( ) = a··············································（2分）
a =2.5m/s ·······································································（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
= [k(x-x1)+k(x-x 1-x2)]······················································（1分）
由动能定理
2
-mBgx2sinθ + x2=0 1 mv2 ··················································（2分）
2
x2=（0.1+0.1 3）m ························································ （1分）
物体 B在斜面上运动的最大位移
x3=x1+x2=（0.2+0.1 3）m ················································（1分）
15.（18分）
(1)设 P球滑到 B点的速度为 v0，碰撞后的速度为 v1，Q碰撞后的速度为 v2，
P由 A到 B，由动能定理
3mg 1 2R 3mv 20 - 0 ………………………………………………………（1分）2
在 B处，P、Q发生弹性碰撞
3mv 0 3mv1 mv 2 ····························································（1分）
1 3mv 2 10 3mv
2 1 mv 2………………………………………………（1分）
2 2 1 2 2
1= ···········································································（1分）
2= ··········································································（1分）
P恰好未从 C点飞出，P由 B到 C，由动能定理
1
3mgR (1 cos ) 0 3mv 2 ···········································（1分）
2 1
cos 1 …………………………………………………………（1分）
2
600 ·········································································· （1分）
(2)Q由 B到 C
mgR(1 cos ) 1 1 mv2 2c mv2 ··········································· （1分）2 2
Q从 C点飞出到最高点，由动能定理
mgh 12 m（vc cos ）
2 1 mv 2c ········································（2分）2 2
h2 3R ·······································································（1分）
h R 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 v
0
c cos15 ································································（1分）
E 1 2k min mvmin ································································ （1分）2
EK min (2 3)mgR ·························································（1分）
4
