{#{QQABaQChxwIwghQACY7aBQkOCEmQsIGTJSomBRCQOAQCSRNABCA=}#}
{#{QQABaQChxwIwghQACY7aBQkOCEmQsIGTJSomBRCQOAQCSRNABCA=}#}
{#{QQABaQChxwIwghQACY7aBQkOCEmQsIGTJSomBRCQOAQCSRNABCA=}#}
{#{QQABaQChxwIwghQACY7aBQkOCEmQsIGTJSomBRCQOAQCSRNABCA=}#}2024～2025学年第二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 7小题，每小题 4分，共 28分。
题号 1 2 3 4 5 6 7
选项 C D A D B C A
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 AD BD AD
三、实验题：共 16分。
11.（8分）
1 （ ） （2分）
2 mgh 2 1m
2 2
（ ） （ 分） （2分） 2gh= （2分）
2
12.（8分）
1 + （ ） （4分） （2）6mg （4分）
四、计算题：共 38分。
13.（10分）
（1）篮球做自由落体运动
·······························································（1分）
=0.5s
重力的冲量
=mgt1································································（2分）
=3N s······················································································ （1分）
1
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方向竖直向下···············································································（1分）
（2）篮球落地时的速度为
=gt1········································································································································（1分）
篮球反弹上升时的速度为
2gh = 0················································································（1分）
地面对篮球的平均作用力为 ，以竖直向上为正方向，篮球与地面接触过程中，
受重力 mg和地面的平均作用力
= ·························································· (2分)
F = 60N······················································································ （1分）
14.（13分）
（1）机械能的变化量大小等于滑动摩擦力做功大小，物体运动 0~5m过程中
°x =E2 E1········································································································（4分）
=0.5························································································· （2分）
（2）根据图像可知传送带的长度为 16m，物体下降高度 h=9.6m由能量守恒
mgh Wf= Ek··························································································································（5分）
Ek= 3600J···················································································（2分）
15.（15分）
（1）小球做圆周运动，在 E点
······················································· （3分）
······························································（2分）
（2）小球由 A到 E
mg2r = ··················································································（3分）
= 3mgr············································································································（2分）
2
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（3）小球在 D点脱离轨道，轨道对小球的弹力为 0，重力沿半径方向的分力提供向心力
·····················································（2分）
小球由 A到 D
+ + ···············································（2分）
W= mgr （ cosθ 2）··································································（1分）
3
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