2024～2025学年第二学期高二年级期中学业诊断
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 A B B D C D B
二、多项选择题：本题包含 3小题，每小题 6分，共 18分。
题号 8 9 10
选项 AC BD ABD
三、实验题：共 16分。
11.（7分）
（1）不会 （2分） 右 （3分） （2）向左 （2分）
12.（9分）
（1）1.26 （3分） （2） （3分） （3）AB（3分）
四、计算题：共 38分。
13.（8分）
（1）M、N在 1s内各完成一次全振动
周期 T = 1s ················································································· （1分）
从图中可知
波长 = 0.4m··············································································（1分）
可得波速 v = = 0.4m/s·································································（1分）
机械波传播一个半波长，二者分开
t = = 1.5s················································································ （1分）
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（2）O为振动加强点····································································（1分）
振幅 A=0.2m················································································（1分）
2.5s内 O只完成一次全振动····························································（1分）
S = 4A =0.8m················································································（1分）
14.（14分）
（1）当 v = 0时，由牛顿第二定律得
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当 v = v0时，加速度 a = 0，此时导体棒受力平衡
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（2）依题意，由能量守恒
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R与 r串联，电流相同
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15.（16分）
（1）导体棒减速上滑，电容器放电，
电流的方向为由 N到M···············································································(1分)
导体棒沿斜面向下加速，电容器充电，根据右手定则，
电流的方向为由 N到M···············································································(1分)
（2）撤去外力后，对MN应用动量定理，以沿斜面向上为正方向
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当导体棒到达最高位置，电容器放电完毕
初始时 ········································································· (1分)
电荷量全部放出 ····························································· (1分)
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（3）依题意，所有电阻不计，电容器两端电压与导体棒两端电势差时刻相等
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导体棒在斜面轨道上做匀变速直线运动
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