2023～2024学年第一学期高一年级期末学业诊断
物理参考答案及评分建议
一、单项选择题：本题包含 10小题，每小题 3分，共 30分。
题号 1 2 3 4 5 6 7 8 9 10
选项 A C D A C C D D B B
二、多项选择题：本题包含 5小题，每小题 3分，共 15分。
题号 11 12 13 14 15
选项 CD BC BD BD BC
三、实验题：共 14分。
16.（7分）
（1）A（2分） （2）C（2分） （3）F '（1分） （4）B（2分）
17.（7分）
（1）D（2分） （2）甲（2分） （3）C（1分） B（1分） A（1分）
四、计算题：共 41分。
18.（7分）
v v
（1）由 a ······································································ （1分）
t t
a=3m/s2················································································ （1分）
x 1 at 2由 ············································································（1分）
2
x = 6m··················································································（1分）
（2）由牛顿第二定律
F mg ma ········································································（2分）
μ=0.3····················································································（1分）
{#{QQABKQSEogggAhBAAAhCQwl4CkKQkAEACCoGREAEsAIAiQNABAA=}#}
19.（7分）
（1）人在救生滑梯上做匀加速直线运动
mgsin37。- μmgcos37。= ma···················································································（1分）
a=2m/s2·······················································································（1分）
x=10m························································································（1分）
x 1 at 2 ····················································································· （1分）
2
t = 10s = 3.16s············································································（1分）
（2）vt= a t·················································································（1分）
vt= 7.32 m/s················································································· （1分）
20.（8分）
（1）小球 1与 2组成的整体，由平衡条件
° = ········································································· (1分)
° = ··········································································· (1分)
= mg················································································ (1分)
= mg
细线b对小球 2的拉力大小
= +
= mg················································································· (1分)
（2）剪断细线b的瞬间，a绳弹力发生突变，小球 1所受合力
= ° = ·····························································合 (1分)
= ····················································································(1分)
弹簧c 弹力不变，小球 2
{#{QQABKQSEogggAhBAAAhCQwl4CkKQkAEACCoGREAEsAIAiQNABAA=}#}
'合 =
'合 = 2m ··············································································· (1分)
= ···················································································· (1分)
21.（9分）
（1）物块的加速度大小为 a
μ1mg＝ma···············································································································(2分)
a＝2m/s2
行李箱在传送带上做匀加速直线运动，加速至与传送带共速，位移为 x1
2ax1＝v 2 2传 - v 物 ··········································································· (1分)
x1＝2m
行李箱在传送带上先做匀加速直线运动，加速运动时间为t1；
与传送带共速后做匀速直线运动，匀速运动时间为t2
v 传＝v 物+a t1·················································································(1分)
t1＝1s
x2＝L - x1＝3m ，x2＝ v 传 t2 ，t2＝1s
t＝t1＋t2＝2s·················································································(1分)
（2）行李箱在水平接物台上做匀减速直线运动
μ2mg＝ma'··············································································································(2分)
2（- a'）x1＝0 - v 2传 ···································································· (1分)
μ2＝0.75······················································································ (1分)
22.（10分）
（1）滑块和木板发生相对运动，滑块加速度为
F1 F2 mg ma1 ········································································(1分)
{#{QQABKQSEogggAhBAAAhCQwl4CkKQkAEACCoGREAEsAIAiQNABAA=}#}
1= 2m/s2 ·················································································· (1分)
木板加速度为
F2 Mg Ma2 ············································································(1分)
2= 1m/s2 ·················································································· (1分)
a1 a2
（2）设经过时间 t 滑块到达木板的中点
=
······································································· (1分)
t = 0.8s ······················································································ (1分)
（3）滑块与木板不发生相对滑动，滑块与木板间静摩擦力增大为最大静摩擦力
木板： F2 Mg Ma ·································································· (1分)
= 2m/s2
滑块： F1 F2 mg ma ····························································· (1分)
F1=18N
整体向上加速 F1>（M+m）g························································· (1分)
F1满足条件为 15N < F1 ≤ 18N························································ (1分)
{#{QQABKQSEogggAhBAAAhCQwl4CkKQkAEACCoGREAEsAIAiQNABAA=}#}4.如图所示，质量为m的木块在质量为3m长木板上滑行，长木板一直处于静止状态。若木块
2023~2024学年第一学期高一年级期末学业诊断
与木板之间的动摩擦因数为，重力加速度为g。长木板受到地面的摩擦力为
物理试卷
豪
(考试时间：上午10：30一12：00)
说明：本试卷为闭卷笔答，答题时间90分钟，满分100分。
A.mg,向左
B.mg,向右
题号
二
三
四
总分
C.4umg,向左
D.4mg,向右
得分
5.一名工人师傅用两侧内壁均粗糙的钳子把5块相同的砖块整齐的竖
一、单项选择题：本题包含10小题，每小题3分，共30分。请将正确选项前的字母填入表内相
直提起，沿直线匀速向前走去，每块砖的质量为3kg,g取10m/s2。左
应位置。
侧钳子与砖块的摩擦力为
题号
1
4
5
o
A.15N
B.30N
答案
C.75N
D.150N
1.下列说法正确的是
6.一轻弹簧原长为10cm,在弹性限度内悬挂10N的物体并静止时，弹簧长度为12cm,下列选
A.由于地球的吸引而使物体受到的力叫作重力
项正确的是
B.物体重心的位置与物体的形状无关
A.弹簧劲度系数k=5N/m
C.物体受到的重力一定指向地心
B.悬挂6N的物体，静止时其伸长量为11.2cm
D.物体悬挂在弹簧测力计下加速向上运动，测力计的示数等于物体重力的大小
C.悬挂8N的物体，静止时弹簧长度为11.6cm
2.玩具汽车停在模型桥面上，下列说法正确的是
2.5
D.悬挂8N的物体，静止时弹簧劲度系数发生改变
A.桥梁受到向下的弹力，是因为桥梁发生了弹性形变
7.如图所示，阻力伞共有7根等长的伞绳和1根中心轴线绳，每根伞绳与中心轴线绳的夹角均
B.桥梁受到向上的弹力，是因为汽车发生了弹性形变
为37°，每根伞绳能承受的最大拉力均为30N,忽略伞的重力，c0s37°=0.8，若中心轴线绳
C.汽车受到向上的弹力，是因为桥梁发生了弹性形变
D.汽车受到向下的弹力，是因为汽车发生了弹性形变
始终未被拉断，阻力伞可提供的最大阻力为
中心轴线绳
3.如图所示，F、F、F恰好构成封闭的直角三角形，这三个力合力最大的是
A.24N
B.108N
C.126N
D.168N
料
高一物理第1页（共8页）
高一物理第2页（共8页）
8.蹦极是一项极限体育项目，运动员身拴弹性绳从高处跳下，在弹性绳被拉直前做自由落体
12.关于牛顿第二定律，下列说法正确的是
运动。当弹性绳被拉直后，运动员继续下降，到达最低点后再向上运动。不计空气阻力，下
A.牛顿第二定律的表达式中，a与F的方向可能不同
列说法正确的是
B.物体某一瞬间的加速度，由这一瞬间的合外力决定
A.运动员下降过程中，在弹性绳被拉直后，一直处于失重状态
B.运动员下降过程中，在弹性绳被拉直后，先处于超重状态后处于失重状态
C.公式中若F为合力，a等于每一个分力单独作用在该物体上产生加速度的矢量和
C.运动员上升过程中，在弹性绳恢复原长前，一直处于超重状态
D.物体的运动方向一定与物体所受合力的方向相同
D.运动员上升过程中，在弹性绳恢复原长前，先处于超重状态后处于失重状态
13.学校教室里竖直平面内的磁性黑板上通常贴挂一些“笑脸”形状的小磁铁。“笑脸”被吸在
9.如图所示为某新型电动汽车试车时的-t图像，下列选项正确的是
↑/(ms)
黑板上用于辅助教学。下列说法正确的是
A.在4~14s内，电动汽车做减速运动
B.在5s末，电动汽车的加速度为-1m/s2
A.“笑脸”受到黑板的磁吸引力大于黑板对其的弹力才能被吸在黑板上
C.在4s末，电动汽车的运动方向改变
B.“笑脸”与黑板间存在三对作用力与反作用力
D.在4~6s内的加速度比10~14s内的加速度大
02468101214tWs
C.“笑脸”受到的支持力与黑板受到的压力是一对平衡力
10.水平面上有A、B两物体，同时同地以某一初速度同向运动，运动规律如图所示，下列选项
D.若黑板光滑，无论“笑脸”磁性多强，它都不能够静止贴在黑板上
正确的是
42/(m2s2）
A.A物体的加速度为-10m/s2
14.下列有关单位制的说法正确的是
100
B.两物体在x=10m位置处相遇
64
A.国际单位制中，力学的基本单位是牛顿、米、秒
C.两物体在图像交点时刻相遇
B.实验表明，物体在空气中高速行进时，空气阻力f与受力面积S及速度v的关系为
1013
D.当物体A运动到x=5m的位置时，两物体相距最远
x/m
fS2,则在国际单位制中，比例系数k的单位是kgm
二、多项选择题：本题包含5小题，每小题3分，共15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3分，选不全的得2分，有错者或不答的得0分。请将正确答案
C.1N是指使质量为1kg的物体产生9.8m/s2的加速度所需要的力
填入下表内。
D.在单位制中，“N/g”与“m/s2”是等价的
题号
11
12
13
14
15
15.如图所示，一根细线系着一个小球，细线上端固定。现给小球施加力F,小球平衡后细线跟
答案
竖直方向的夹角为30°，F与细线的夹角为α，a心90°。要使小球保持平衡，下列判断正确的是
11.从第46届世乒赛开始，乒乓球改用“大球”，球的直径由38mm改为40mm,球的质量随之变
A.若小球位置不动，改变F的方向，夹角α的变化范围是0≤a≤180°
大，下列说法正确的是
B.若小球位置不动，F由图示位置逆时针缓慢转动一定角度，力F可
30
A.乒乓球被削回飞行途中不受力并做匀速直线运动
能先减小后增大，细线的弹力一定减小
B.乒乓球改用“大球”后惯性减小
C.在研究乒乓球的旋转对轨迹的影响时，不可将乒乓球视为质点
C.若保持夹角α不变，细线逆时针缓慢转至水平，F一直增大
D.被削回的乒乓球，速度的方向发生变化，惯性不变
D.若保持夹角α不变，细线逆时针缓慢转至水平，细线的弹力先增大后减小
高一物理第3页（共8页）
高一物理第4页（共8页）
