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第二次调研考试
物 理
一、单项选择题:本题共7小题，每小题4分，共28分。在每小题给出的四个选项中，只有
一项是最符合题目要求的。
1 2 3 4 5 6 7
B D C A C B C
二、多项选择题:本题共3小题，每小题6分，共18分。每小题有多项符合题目要求，全部
选对的得6分，选对但不全的得3分，有选错的得0分。
8 9 10
BD AD AC
三、非选择题:本题共5小题，共54分。其中第13-15小题解答时请写出必要的文字说明、
方程式和重要的演算步骤，有数值计算时，答案中必须明确写出数值和单位。
11.（8分，每空2分）
𝑏 𝑑2
（1）5.30 （2） （3）偏小
𝑥 2𝑘
12.（8分，除定值电阻和滑动变阻器选择各1分外，其余每空2分）
（1） ③ ⑥
(𝐼2−𝐼1 )𝑅1
𝐼1
（4）176
13.（10分）解：
（1）运功员从A运动到B的过程中，动能定理可得，
𝑚𝑔ℎ= 1 𝑚𝑣 2− 1 𝑚𝑣 2···································（2分）
B A
2 2
𝑣 =30m⁄s·············································（2分）
B
（2）运动员在空中做平抛运动，由运动的分解可得，
水平方向：𝑥 =𝑣 𝑡 ·················································（·2分）
B
竖直方向：𝑦 = 1 𝑔𝑡2·················································（2分）
2
𝑦
tan37°= ···············································（1分）
𝑥
𝑡 =4.5s·················································（1分）
14.（12分）解：
（1）由题意可得，线框恰好匀速进入区域I，有
𝑚𝑔=𝐹 ················································（1分）
安
𝐹 =𝐵𝐼𝐿················································（1分）
安𝐸
𝐼 = ····················································（1分）
𝑅
𝐸 =𝐵𝐿𝑣·················································（1分）
𝑚𝑔𝑅
𝑣 = ·················································（1分）
𝐵2𝐿2
（2）线框下边进入区域II时，上、下边均切割磁感线，线框产生的总电动势为
𝐸′ =2𝐵𝐿𝑣···············································（1分）
此时线框中的感应电流为𝐼′ =
𝐸′
𝑅
此时线框受到的总的安培力为𝐹 ′ =2𝐵𝐼′𝐿·····························（1分）
安
由牛顿第二定律可得𝐹 ′−𝑚𝑔 =𝑚𝑎··································（1分）
安
𝑎 =3𝑔··················································（1分）
（3）令线框离开区域II的速度为𝑣′，由题意可得，线框匀速离开区域II，有
𝑣′ =𝑣···················································（1分）
从线框下边开始进入区域II到线框下边开始离开区域II，由功能关系可得
𝑚𝑔𝐻=𝑄+ 1 𝑚𝑣′2 − 1 𝑚𝑣2································（·1分）
2 2
𝑄 =𝑚𝑔𝐻················································（1分）
15.（16分）解：
带点粒子在组合场中从A点出发，经磁场I和电场再次回到A点的轨迹，如图所示。
（1）带电粒子在电场中做类平抛运动，由运动的分解可得
竖直方向：𝑑 =𝑣 𝑡 ················································（1 分）
0 1
水平方向： 𝑚𝑎 =𝐸𝑞 ················································（1分）
𝑣 =𝑎𝑡 ·················································（1分）
𝑥 1
𝑣 =√𝑣 2+𝑣 2 =√2𝑣 ····································（1分）
0 𝑥 0
方向与直线MN成45°····································（1分）
（2）带电粒子在电场中的水平位移𝑥 = 1 𝑎𝑡 2·······························（1分）
1
2
由几何关系可得，粒子在磁场I中匀速圆周运动的半径 𝑅 =2𝑥sin45°·······（1分）
𝑣2
在磁场I中，粒子运动的向心力由洛伦兹力提供𝑞𝑣𝐵 =𝑚 ················（1分）
𝑅
2𝜋𝑅
圆周运动的周期为𝑇 = ···········································（1分）
𝑣
3
粒子在磁场中的运动时间𝑡 = 𝑇·····································（·1分）
2
4
粒子从A点出发到第一次回到A点所用的时间
2𝑑 3𝜋𝑑 (8+3𝜋)𝑑
𝑡 =2𝑡 +𝑡 = + = ····························（1分）
1 2
𝑣0 4𝑣0 4𝑣0
（3）由题意可得，粒子在磁场I和磁场II中的运动半径之比为1:3，
带电粒子经PQ回到A点，有且只有如图所示的一种情况，由几何关系可得
𝑑 𝑚𝑣′
2 =4 ············································（1分）
tan30° 𝑞𝐵1𝑣′ =4√3𝑣 ···············································（1分）
0
带电粒子经MN回到A点，情况如图所示，令带电粒子在磁场I中运动的次数为n，由几
何规律可得
𝑛𝑑 𝑚𝑣′ 𝑚𝑣′
2 =𝑛 +(𝑛−1) ································（·2分）
tan30° 𝑞𝐵1 𝑞𝐵2
𝑣′ = 16√3𝑛 𝑣 （𝑛=1、2、3···）····························（·1分）
0
4𝑛−3
