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2023－2024学年度第一学期期中学业水平检测
高三物理答案及评分标准
一、单项选择题：本大题共8小题，每小题3分，共24分。
1．B 2．A 3．D 4．C 5．C 6．B 7．C 8．D
二、多项选择题：本大题共4小题，每小题4分，共16分，选不全得2分，有选错得0分。
9．ACD 10．BC 11．BD 12．BC
三、非选择题
13．（6分）
F
（1）C（2分）；（2）0（2分）、C（2分）。
g
14．（8分）
（1）天平（1分）；（2）C（1分）；
（3）k s ＝k s + s （2分）；ks ＝ks +s （2分）；
1 2 3 1 2 3
（4）小于（2分）。
15．(8分)
（1）由已知条件得
OA 3T
只有A点振动时，t ＝ + ·····················································（1分）
1
v 4
OB T
只有B点振动时，t ＝ + ·······················································（1分）
2
v 4
解得周期T=8s··········································································（2分）
（2）由（1）得v＝10m/s,···························································（1分）
可得波长λ＝vT＝80m·································································（1分）
1
OA-OB＝40m＝ λ······································································（1分）
2
由已知条件得O点为振动加强点··················································（1分）
评分标准：第1问， 4分；第2问， 4分。共 8 分。
16．(10分)
（1）在BC段，餐盘收到的向心力是静摩擦力提供
v2
由牛顿第二定律：μmg＝m ·····················································（2分）
R
解得：μ＝0.1·········································································（2分）
L +L
（2）AC段速率不变，t ＝ 1 2＝4s·············································· （1分）
1
v
物理答案 第1 页 共3页
{#{QQABBQKAggiAAhBAABgCEwXCCkKQkAGACKoGBEAMsAAAARFABCA=}#}由μmg＝ma···········································································（1分）
解得机器人减速的最大加速度：a＝1m/s2
由v2＝2ax
解得机器人减速的位移为：x＝2m·············································（1分）
CD段匀速运动的位移为：L ＝12－x＝10m
3
L
时间为：t ＝ 3＝5s·································································（1分）
2
v
匀减速阶段由v＝at
3
得：t ＝2s············································································ （1分）
3
配送最短时间为：t＝t +t +t ＝11s··············································（1分）
1 2 3
评分标准：第1问，4分；第2问， 6分。共10 分。
17．（12分）
（1）研究Δt时间内出射的水柱
1
由动能定理得水枪对出射水柱做的功W＝ mv 2－0························（1分）
0
2
Δt时间内出射的水柱质量为m＝ρv SΔt········································（1分）
0
W
由P＝ ··············································································· （1分）
Δt
解得：P＝0.22W·····································································（1分）
（2）水柱在空中做斜上抛运动，
水平方向v ＝v cos37°；L＝v t····················································（1分）
x 0 x
1
竖直方向v ＝v sin37°；y＝v t－ gt2·············································（1分）
y 0 y
2
解得y＝－0.35m······································································（1分）
由已知得H＝h+y＝0.15m··························································（1分）
（3）v ＝vcos37°；L＝v t··························································（1分）
x x
v ＝vsin37°；2gy＝v 2·······························································（2分）
y y
解得：v＝5m/s········································································（1分）
评分标准：第1问，4分；第2问，4分；第3问，4分。共12分。
18．(16分)
(1) 滑块A在传送带上运动时，重力和摩擦力做功
由动能定理可得
1
mgsin30°d－μ mgdcos30°d＝ mv 2················································（2分）
1 0
2
物理答案 第2 页 共3页
{#{QQABBQKAggiAAhBAABgCEwXCCkKQkAGACKoGBEAMsAAAARFABCA=}#}16
解得d＝ m＝2.29m································································（1分）
7
(2)A在B板上滑动的过程中，A、B系统动量守恒
由mv ＝(m+m)v ······································································（1分）
0 1
得v ＝2m/s············································································ （1分）
1
由能量守恒定律
1 1
μ mgs＝ mv 2- 2mv 2·································································（1分）
2 0 1
2 2
解得s＝4m·············································································（1分）
A给B的滑动摩擦力等于B受到的合外力
由动能定理
1
μ mgL＝ mv 2－0·····································································（1分）
2 1
2
L＝2m···················································································（1分）
(3)滑块C与挡板碰撞前，A、C系统动量守恒，
有mv ＝ mv +Mv ···································································（1分）
1 A B
滑块C与挡板碰撞后到A、C达到共速的过程中，A、C系统动量守恒，
有mv +M(-v )＝(m+M)v ···························································（1分）
A B 共
2
解得A、C共速时v ＝ (v －1)
共 3 A
全过程A、C与弹簧组成的系统机械能守恒，A、C共速时弹簧弹性势能最大，则有
1 1 1
E ＝ mv 2－ (m+M)v 2＝1－ (v －1)2·········································· （1分）
p 2 1 2 共 3 A
在C与挡板碰撞前，若从A开始压缩弹簧至弹簧再次恢复原长的过程中C与挡板未碰撞，
1 1 1
根据机械能守恒有 mv 2＝ mv 2+ Mv 2······································（1分）
1 A B
2 2 2
又根据mv ＝ mv +Mv
1 A B
2
解得v ＝－ m/s
A
3
2
故C与挡板碰撞前A的速度v 介于－ m/s~2m/s之间
A
3
1
根据E ＝1－ (v －1)2
p A
3
v ＝1m/s时，E ＝1J································································ （1分）
A p
2 2
v ＝－ m/s时，E ＝ J····························································（1分）
A p
3 27
2
可知弹簧弹性势能最大值取值范围为 J≤E ≤1J··························（1分）
p
27
评分标准：第1问，3分；第2问，6分；第3问，7分。共16分。
物理答案 第3 页 共3页
{#{QQABBQKAggiAAhBAABgCEwXCCkKQkAGACKoGBEAMsAAAARFABCA=}#}
