济宁市第一中学2024届高三4月份定时检测物理答案和评分标准

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济宁市第一中学 2024 届高三 4月份定时检测
物理参考答案及评分标准 2024 年4月12 日
一、选择题：本题共 8小题，每小题 3分，共 24 分。每小题只有一个选项符合题目要求。
1.B 2.A 3.C 4.D 5.B 6.D 7.C 8.B
二、多项选择题：本题共 4小题，每小题 4分，共 16 分。每小题有多个选项符合题目要求，全部选
对得 4分，选对但不全的得 2分，有选错的得 0分。
9.BD 10.ABC 11.BC 12.BD
三、非选择题：本题共 5小题，共 60 分。
13. (6 分)(1)
4
3
(2)
35
24
d
c
(3)能 (每空 2分)
14. (8 分) (1)红 (2)36 (3)×1k (4)
100
k
(每空 2分)
15. (8 分) (1) 1500kg (2) 580W
【解析】(1)设轿厢和配重的加速度大小为 a，有
2
1
2
h at=
······································ (1 分)
解得 a=2m/s2 ······························································································ (1 分)
以轿厢和配重为研究对象，
()mg Mg m M a−=+
················································· (1 分)
解得配重 B的质量 m=1500kg ·········································································· (1 分)
(2)对配重 B列平衡方程，
0
()F mg M m g
+=+
，解得 F=3000N ······························ (1 分)
电动机的输出功率
PF=
出v
············································································· (1 分)
电动机内阻消耗的功率
P UI P= −
出
内
·································································· (1 分)
解得
=580W ····························································································· (1 分)
16. (8 分) (1)0.417s (2)6mg
【解析】(1) 图乙可知半个周期为 0.5s，故一个周期为 T=1.0s ································ (1 分)
质点由平衡位置运动到 5cm 所用的时间为 Δt，
Δ
6
2
π
t
πT
=
，解得
Δ
12
T
t=
···················· (1 分)
时间
1
Δ
2
T
tt= −
，解得
1
5
12
tT=
=0.417s ······························································· (1 分)
(2) 物体 A在最高点时，物体 B与水平面间的作用力刚好为零，对B分析，
此时弹簧的拉力为 F1=2mg ············································································ (1 分)
对A分析
1
F mg ma+=
，解得 A在最高点的加速度 a=3g ····································· (1 分)
{#{QQABbQCEggAIAJJAARgCAQ2ACEIQkACCAAoGxEAIoAABSAFABAA=}#}
2024 届高三 4月考答案 第2页（共 4页）
当物体 A运动到最低点时，物体 B对水平面的压力最大，
由简谐运动的对称性可知，物体 A在最低点时加速度向上，且大小等于 3g， ··········· (1 分)
对A，有
2
F mg ma−=
，解得
2
4F mg=
····························································· (1 分)
由物体 B的受力可知，物体 B受水平面的最大支持力
N2
2F F mg= +
由牛顿第三定律得 B对地面的最大压力
N6F F mg= =
压
········································· (1 分)
17. (14 分) (1)
2qEd
m
，与 x轴正向夹角 60°；(2)
2
4( 3 )
3
md
qE
π
+
；(3)
2
48d
【解析】(1)设粒子第一次通过坐标原点 O时速度大小为
v
，方向与 x轴正向夹角为 θ，粒子从
Q到O时间为 t0，由类平抛运动规律，水平方向有
0
3 cosdt
θ
=⋅⋅v
························ (1 分)
竖直方向有
2
0
1
1.5 2
d at=
·················································································· (1 分)
0
sin at
θ
=v
··································································································· (1 分)
根据牛顿第二定律有
qE ma=
联立解得
2qEd
m
=v
，θ=60° ·········································································· (2 分)
(2)由于题目中没有告诉磁场的具体位置，所以首先要确定磁场的具体位置，
设粒子在匀强磁场中做匀速圆周运动的半径为 r，则
2
ar
=v
··································· (1 分)
又
qE ma=
，解得 r=4d ············································ (1 分)
根据粒子进入磁场和粒子飞出磁场的速度方向可确定磁场的位
置，如图所示，假设 OP 间的距离为 s，
由几何关系得
2 cos30 4 3sr d= °=
····························· (1 分)
粒子从 O点到刚进入磁场所用的时间
1
43sd
t= =
vv
······· (1 分)
粒子的匀强磁场中运动时间
2
2 2 16
33
rd
t
ππ
=×=
vv
·················································· (1 分)
从粒子第一次经过 O点开始计时，到粒子再次回到 O点的时间
12
28
2 (3 )
3
d
t tt
π
= += + v
将
2qEd
m
=v
代入上式，解得
2
4( 3 )
3
md
tqE
π
= +
··············································· (1 分)
(3)满足条件的磁场为图中的矩形 abcd，最小面积为
( )
2
2 1 cos 3S rr r
θ
=⋅+ =
·············· (2 分)
代入 r=4d，解得
2
48Sd=
··············································································· (1 分)
{#{QQABbQCEggAIAJJAARgCAQ2ACEIQkACCAAoGxEAIoAABSAFABAA=}#}
济宁市第一中学2024届高三4月份定时检测物理答案和评分标准.pdf

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