山西大同高二11月期中考试物理答案(1)

大同市2022——2023 学年度高二测试试题（卷） 物理参考答案 一、选择题：共12 小题，每小题4 分，共48 分。在每小题给出的四个选项中，第1-8 题只 有一个选项符合题目要求，第9-12 题有多项符合题目要求。全部选对的得4 分，选对但不 全的得2 分，有选错的得0 分。 题号 1 2 3 4 5 6 7 8 9 10 11 12 答案 A B A D C B C D AB AD BC CD 二、实验题：共16 分。 13．(1) S (2 分) (3) T (2 分) 0 刻线 (2 分) 14．(1)2.931~2.934 (2 分) 10.02 (2 分) (2) 外接法 (1 分) (3) (3 分) (4) πd2R0 4L (2 分) 三、计算题：共36 分。解答时写出必要的文字说明、方程式和重要演算步骤，只写最后答 案不得分。 15．(1)设电容器上的电压为UC，则UC＝ R2 R1＋R2 U············································· (1 分) 电容器的带电荷量Q＝CUC···········································································(1 分) 解得Q＝1.2×10 －3 C····················································································(1 分) (2)电压表与R2 并联后电阻为R 并＝R2RV R2＋RV ························································(1 分) 则电压表两端的电压为UV＝ R 并 R1＋R 并 U·····························································(1 分) 解得UV＝9V······························································································(1 分) 16．(1)这一电荷必为正电荷，设其电荷量为q，由牛顿第二定律，在A 点时 mg－kQq h2 ＝m·4 5g··························································································(2 分) 在B 点时kQq (0.2h)2－mg＝m·aB·········································································· (2 分) 解得aB＝4g······························································································· (1 分) 即电荷在B 点处的加速度为4g，方向竖直向上。 (2)从A 到B 过程，由动能定理得 mg(h－0.2h)＋qUAB＝0················································································· (2 分) 解得UAB＝－4kQ h ························································································ (1 分) 17．(1)对小物块受力分析如图所示，小物块静止于斜面上， 则mgtan 37°＝qE························································································ (2 分) E＝2.5×104 N/C························································································ (1 分) (2)当场强变为原来的4 5时，小物块沿斜面向下加速运动， 由牛顿第二定律有 mgsin 37°－4 5qEcos 37°＝ma···········································································(2 分) 得a＝1.2 m/s2·····························································································(1 分) (3)由运动学公式，知 x＝1 2at2＝1 2×1.2×22 m＝2.4 m········································································(2 分) 18．(1)因为滑块通过A 点时对轨道的压力恰好为零，所以有 mg＝mv2 A R ··································································································· (2 分) 解得vA＝2 m/s····························································································(1 分) (2)根据动能定理可得qEx－μmgx－mg·2R＝1 2mv2 A··············································· (3 分) 解得x＝4 m·······························································································(2 分) (3) 方法一：离开A 点后滑块在运动过程中只受重力和电场力且均为恒力，设合力为F，重力 mg 与合力F 夹角为θ，如图所示。 将滑块的运动按照图示进行分解，则x 方向做匀速直线运动，y 方向做匀变速直线运动。当 y 方向速度vy 减小为零滑块速度最小，最小为vx。············································ (3 分) vx＝vAcos θ································································································ (1 分) cos θ＝ mg (mg)2+(qE)2·····················································································(1 分) 得：vx＝1.6 m/s·························································································· (1 分) 方法二：离开A 点后在水平方向上做匀变速直线运动，故有 ax＝qE m =7.5m/s2···························································································(1 分) vx＝vA－axt＝2－7.5t···················································································· (1 分) 在竖直方向上做自由落体运动，所以有 vy＝gt＝10t································································································ (1 分) v＝v2 x＋v2 y＝156.25t2－30t＋4＝156.25(t－0.096)2＋2.56································· (2 分) 故vmin＝1.6 m/s·························································································· (1 分)