2023－2024 学年度第一学期期中学业水平检测
高二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．C 3．A 4．B 5．C 6．D 7．B 8．A
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．AB 10．BC 11．ACD 12．BC
三、非选择题
13．（6分）（1）C（2分）；（2）BC（2分）；（3）OM + ON ＝ OP （2分）。
14．（8分）
（1） （2分）； 100Ω（2分）； （2）2.43V（2分）；1.00Ω（2分）。
15．(8分)(1)玩具汽车匀速行驶时太阳能电池的总功率为
P1 EI ······························································································（1分）
太阳能集光板的接收功率为
P2 420 1.6 10
3W 67.2W ··································································（1分）
P
太阳能集光板把太阳能转化为电能的效率为 1P ·····································（1分）2
联立并代入数据得 62.5% ··································································（1分）
（2）依题意，可得驱动电机的输入电压 U＝E－Ir＝17v······························（1分）
驱动电机输入的电功率
P入 IU 17 2W 34W ······································································（1分）
驱动电机的热功率
P I 2热 RM 2
2 0.4W 1.6W ·································································（1分）
驱动电机输出的机械功率 P P入 P热 32.4W （1分）机 ·········································
评分标准：第 1问， 4分；第 2问， 4分。共 8 分。
物理答案 第 1 页 共 3 页
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16．(10分)
（1）设 B物体自由下落至与 A碰撞前其速度为 v0，
根据 v02＝2gh·························································································（2分）
v0 2gh 2 10 1.8m/s 6m/s ······························································（1分）
设 A、B碰撞结束之后瞬间二者共同速度为 v，根据动量守恒定律
m2v0 (m1 m2 )v ····················································································（2分）
解得 v 4.0m/s ·······················································································（1分）
（2）选择竖直向下为正方向，从二者一起运动到速度变为零的过程中，选择 B作为研究对象，
根据动量定理得
m2g FN t 0 m2v ···············································································（2分）
解得 FN 120N ······················································································ （2分）
评分标准：第 1问，6分；第 2问， 4分。共 10 分。
17．（12分）
（1）2U E ·························································································（1分）
U 6V ························································································（1分）
由 I—U图得，
当U 6V时
I 0.5A ·······················································································（2分）
P IU 3W ·················································································（2分）
（2）设流过每个灯的电流为 I灯两端的电压为 U，则
E U 3IR0 ··················································································
（1分）
得 I 1 U 0.4 （1分）
30 ···········································································
由 I—U图得出···············································································（1分）
工作点的
U 3V ························································································（1分）
I 0.3A ·······················································································（1分）
所以 P IU 0.9W ·········································································（1分）
评分标准：第 1问，6分；第 2问，6分。共 12分。
物理答案 第 2 页 共 3 页
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18．(16分)
（1）物块下滑，第一次碰撞前做匀加速运动，
根据牛顿第二定律：2mgsinθ＝2ma1·························································（1分）
根据速度与位移关系公式： v2－v02＝2a1x
得物块与盒子发生第一次碰撞前的速度为：v＝ 2a1L ＝6m/s·······················（1分）
物块与盒子发生第一次碰撞满足动量守恒和机械能守恒，则有：
2mv＝2mv1 + mv2··················································································（1分）
1 2mv2 1 2mv 2+1× ＝ × 1 mv22 ····································································（1分）
2 2 2
解得： v1＝2m/s，v2＝8m/s··································································· （2分）
（2）第一次碰撞后盒子做匀减速运动，
根据牛顿第二定律有：mgsinθ－3μmgcosθ＝ma2 ········································（1分）
物块与盒子速度相等时有：v1+a2t0＝v2+a1t0
解得：t0＝0.3s·····················································································（1分）
在t0＝0.3s时，盒子与物块的位移差为：
v2t0+1a2t 102－(v1t0+ a1t02)＝0.9m＜3m，故不会与上板相碰····························（1分）
2 2
盒子减速到零所用时间Δt 4＝ s ·······························································（1分）
7
此时盒子位移x 16 1042＝ m ；物块的位移为x1＝ m ；即第二次碰撞时，盒子已经停止，设物块再
7 49
经过时间t与盒子发生第二次碰撞，则有：x2＝v1t+1a1t2································（1分）
2
t 385－7解得：＝ s·············································································· （1分）
21
（3）物块与盒子发生的是弹性碰撞，分析可知物块与盒子只要运动就一定会碰撞，当物块与盒
子均静止时就不在发生碰撞···································································（1分）
设盒子下滑的距离为x，根据能量守恒可得：
3μmgxcosθ＝mgxsinθ+2mgsinθ(x+L) ·······················································（2分）
解得：x=18m······················································································（1分）
评分标准：第 1问，6分；第 2问，6分；第 3问，4分。共 16分。
物理答案 第 3 页 共 3 页
{#{QQABDQCAoggIABIAARhCAwGCCECQkBEAAKoGBEAEoAAAQQNABCA=}#}2023一2024学年度第一学期期中学业水平检测
4.如图，货车在一水平恒力F作用下沿光滑水平面运动，当货车经过一竖直固定的漏斗下方
时，沙子由漏斗连续地落进货车，单位时间内落进货车的沙子质量恒为和。某时刻，货车
高二物理
2023.11
(连同已落入其中的沙子)质量为M,速度为y,此时货车的加速度为
A.F+mov
注意事项：
M
固定漏斗
1.答卷前，考生务必将自已的姓名、考生号等填写在答题卡和试卷指定位置。
B.
F-mov
M
2.回答选择题时，选出每小题答案后，用铅笔把答题卡上对应题日的答案标号涂黑。如需
改动，用橡皮擦干净后，再选涂其他答案标号。回答非选择题时，将答案写在答题卡上。写在
c
777777777777777777
本试卷上无效。
3.考试结束后，只需要上交答题卡。
D.F-mogv
一、单项选择题：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。
5.如图，四根与纸面垂直的长直导线a、b、c、d,位于边长为L的正方形四个顶点上，导线
1.关于电源，下列说法正确的是
中通有大小相同的电流1，导线周围的磁场满足B=,下列关于正方形中心O点的磁场大
A,电源单位时间内向外提供的电能越多，表明电动势越大
B.电源电动势大，电源对外提供电能的本领大
小和方向正确的是
C.电源内部，屯流由负极流向正极，表明内部由负极到正极电势降低
A2向上
4
D.电动势为2V的电源，将1C正电荷从负极经电源内部移动到正极，静电力做功2J
B.22,向下
2.舞中幡是我国传统杂技节目。如图，演员用手顶住中幡，将中幡从
A⑧
⑧6
胸口处竖直向上抛起，2s后在自已的胸口处开始按幡，并下蹲缓冲，
c22向左
经0.5s将幡稳稳接在手中。已知一根中幡质量约为20kg,重力加速
o.
度g取10ms2,忽略幡运动过程中所受的空气阻力，演员接幡过程
D.2向右
中，幡对手的平均作用力为
是
6.如图，直线A为某一电源的路端电压与电流的关系图线，直线B为某一电阻R的伏特性
A.300N
曲线，将该电源直接与电阻R相连组成闭合电路，下列说法正确的是
B.400N
AU/v
C.600N
A.电源的电动势为4V,内阻为2
4
D.800N
B.电源的输出功率为4W
3.受哭斯特发现电流磁效应实验的启发，某兴趣小组用铁钉与漆包线绕制成电磁铁，电磁铁正
C.电阻R的阻值为1Q
2 3 IA
上方的小磁针稳定后如图所示，下列说法正确的是
D.电源的效率为75%
A.导线B端接电源正极
.用如图所示的电路研究微型电动机的性能时。闭合开关，调节滑动变阻器R并控制电动机
B.铁钉内磁场方向向右
停止转动，电流表和电压表的示数分别为1.0A和3.0V。重新调节R使电动机恢复正常运转
C.电磁铁右端为电磁铁的N极
wowg
D.小磁针所在位置的磁场方向水平向左
B
电流表和电压表的示数分别为4.0A和30.0V,此时电动机的输出功率为
A.64W
B.72W
C.80W
D.88W
M
高二物理试题，第1页（共7页）
高二物理试题第2页（共7页）
