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2024～2025 学年第一学期高三年级期末学业诊断
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 A B B D D D
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 AB BC ABC
三、实验题：共 16分。
11.（7分）
（1） 1.20 （1分） 12.0 （2分） （3）右 （2分） （4）小于（2分）
12.（9分）
（1） （2分）
（3） （2分）
1（4）1.11（1.00-1.30） （1分） 1.31（1.00-1.60）（2分）
（5）小于 （1分） 小于（1分）
四、计算题：共 38分。
13.（8分）
（1）在 t 时间内，cd棒、ab棒均由静止开始运动
cd 棒：Ft -I = mv··················································（1分）
安
ab棒:I = mv ·····················································（1分）
安 ab
v = -v····························································· （2分）
ab
（2）cd 棒由 静止开始做加速度减小的加速运动，ab棒由静止开始做加速
度增大的加速运动。经过足够长的时间，两棒以相同的加速度做匀加
速直线运动
cd 棒：F-BIL=ma····················································（1分）
ab棒：BIL=ma·······················································（1分）
I = ································································ （1分）
△
·······························································（1分）
14.（1 △ 2 分 = ）
（1）质子在直线加速器中做直线运动
6e = mv 2························································· （2分）
0
···························································（2分）
（2）质 子 =在 圆形 区域电场中做类平抛运动
eE = ma·································································（2分）
R +Rcosθ = t······················································ （2分）
Rsinθ = at2··························································· （2分）
2E = ························································· （2分）
( + )
15.（18分）
（1）小球刚要离开地面时
qv B= mg······························································ （2分）
0
v = ······················································· （2分）
0
（2）小球在 OO'由静止开始加速
qEx = mv 2- 0······················································（2分）
0
x= ······························································（2分）
(3)小球重 力 与 电场力的合力为恒力 F ，设 F 与水平方向夹角为θ
合 合
F = ···············································（1分）
合
tanθ=( ··)···+···(· ·· ··)·················································（1分）
小球离地瞬间，将其速度 v 分解为垂直于 F 方向斜向上的速度 v
0 合 1
和另一分速度 v
2
v 与 v 的夹角为 -θ
1 0
2
qv B= F ······························································（1分）
1 合
v = ······················································（1分）
1
( ) +( )
根据小球受 力 情况，另外一分速度 v 方向竖直向下
2
qv B= qE·······························································（1分）
2
v = ····································································（1分）
2
小球的 运动可看作以v 做匀速直线运动与和以 v 做匀速圆周运动
1 2
的合运动
= = cosθ
2 合
1
小球的最大速度为
3v = v + v
max 1 2
v = + ············································（1分）
max
( ) +( )
方向与 v 夹 角 -θ斜向 上·········································（1分）
0
小球的最小速度为
v = v -v
min 1 2
v = - ·············································（1分）
min
( ) +( )
方向与 v 夹 角 -θ斜 向上·········································（1分）
0
4
