物理参考答案_全国高考模拟卷_2026年2月_260205山东省青岛市2026届高三第一学期期末学业水平检测（全科）_山东省青岛市2026届高三第一学期期末学业水平检测物理

2025-2026 学年度第一学期期末学业水平检测 高三物理答案及评分标准 2026.02 一、单项选择题：本大题共8个小题，每小题3分，共24分。 1．D 2．B 3．C 4．A 5．B 6．C 7．D 8．B 二、多项选择题：本大题共4个小题，每小题4分，共16分，选不全得2分，错选不得分。 9．AB 10．AD 11．BD 12．AC 三、非选择题 13．（6分） 2d 2g （1）1.035（2分）；（2） （2分）；（3） （2分）。 Δt 11d2 14．（8分） （1）AD（2分）；（2）3000（１分）；（3）左（１分），3030（2分），等于（2分）。 15．（7分） （1）光路图如图甲所示，根据几何关系，可得入射角θ ＝60°，折射角θ ＝30° 1 2 sin 根据n 1 ····················································（1分） D sin 2 得折射率n 3 单色光在材料内传播的速度v c ····························（1分） θ1 O C n M 3 θ2 解得v c······················································（1分） 3 A B （2）光路图如图乙所示，由几何关系可知，MO与PO垂直 甲 OM 2 3R 解得∠MPO＝30°，MP  ··················（1分） sin30 3 D Q 根据几何关系分析可知PQ∥BD N PQ2Rcos30 3R··········································（1分） 2QNcos30R···················································（1分） O C 3 解得QN  R M 3 光传播的总光程sMPPQQN 2 3R A B P s 6R 乙 光传播所用的时间t   ·································（1分） v c 评分标准：第1问，3分；第2问，4分。共7分。 16．（9分） （1）设猫向上的加速度为a ，木杆向下的加速度为a 1 2 对猫 f－mg＝ma ···············································································（1分） 1 对杆 f＋2mg＝2ma ············································································（1分） 2 3 9 解得a  g a  g 1 2 2 4 物理答案 第1 页 共3页 {#{QQABAYSUggCgAoBAABhCQQUICEOYkBGACIgGwAAcsAIAgRNABAA=}#}1 1 猫加速到木杆上端过程中两者的位移关系 L at2 a t2·······················（1分） 11 21 2 2 8L 解得 t  ·················································································（1分） 1 15g 6gL （2）猫与杆分离时的速度 v at  ················································（1分） 1 11 5 v 猫与杆分离后做竖直上抛运动的时间 t  1 ··········································（1分） 2 g 猫与杆分离后杆的速度 v ＝a t ····························································（1分） 2 21 1 12 从猫与杆分离到猫抓住细绳末端过程中，杆的位移hv t  gt2  L·······（1分） 2 2 2 2 5 1 从杆与细绳末端脱离到猫抓住细绳末端，杆下降的距离H h a t2 3L····（1分） 21 2 评分标准：第1问，4分；第2问，5分。共9分。 17．（14分） （1）设物块1与物块2碰撞前的速度为v，由动能定理 L 1 m gssinθμm gcosθsμm g  mv2··············································（2分） 1 1 1 1 2 2 碰撞过程中动量守恒、能量守恒，设碰后物块1的速度为v ，物块2的速度为v 1 2 mvmv m v ··············································································· （1分） 1 1 1 2 2 1 1 1 mv2  mv2 m v2······································································（1分） 1 1 1 2 2 2 2 2 解得 v ＝5m/s················································································（1分） 2 （2）设物块2运动到C点时的速度为v ，由动能定理 2C L 1 1 μm g  m v2  m v2·····································································（1分） 2 2 2C 2 2 2 2 2 解得v ＝4m/s 2C 设物块2运动到圆弧轨道D点时，水平速度为v ，竖直速度为v x y 物块2与圆弧槽水平方向动量守恒，有 m v  m m  v ············································································（1分） 2 2C 3 2 x 1 m v2  1 m m  v2 1 m v2m gR·················································· （2分） 2 2C 2 3 x 2 y 2 2 2 2 解得v ＝1m/s，v ＝2m/s x y v  v2v2 ···················································································（1分） 2D x y 解得v 2D ＝ 5m/s···············································································（1分） v （3）物块2从离开D点到落回D点的时间t 2 y ······································（1分） g 圆弧槽运动的距离x＝v t·····································································（1分） x 解得x＝0.4m····················································································（1分） 评分标准：第1问，5分；第2问，6分；第3问，3分。共14分。 物理答案 第2 页 共3页 {#{QQABAYSUggCgAoBAABhCQQUICEOYkBGACIgGwAAcsAIAgRNABAA=}#}18．(16分) （1）带电粒子在第二象限做类平抛运动 2Lv t····························································································（1分） 0 3L 1qE t2······················································································（1分） 2 m E 3mv 0 2 ························································································（1分） 2qL （2）带电粒子到达O点速度大小为v 1 1 1 qE 3L mv2 mv2··········································································（1分） 2 1 2 0 带电粒子到达O点竖直速度v  v2v2  3v ········································（1分） y0 1 0 0 粒子在0xx 区域内垂直于xOy平面内做匀速圆周运动，设半径为r，周期为T 0 v2 qv B m y0 ····················································································（1分） y0 1 r L d 2r ·························································································（1分） π 2πr L T   ···················································································（1分） v y0 3v 0 x v nT 1qE nT 2··········································································（1分） 0 0 2 m 3nL  4n  解得x  （n＝1，2，3…）····················································（1分） 0 12 （3）带电粒子到达Q点 qE 3 v v  T  v ··········································································· （1分） Qx 0 0 m 2 21 v  v2 v2  v ··········································································（1分） Q y0 Qx 2 0 在匀强磁场B 中，设粒子距离y轴最远时的速度大小为v，方向竖直向下 2 沿x轴方向由动量定理 qv B t0mv ·············································································· （1分） y 2 Qx v t y·····························································································（1分） y 3 解得y＝ L 2 1 1 由动能定理qEy＝ mv2 mv2··························································· （1分） Q 2 2 3 3 解得v＝ v ·················································································（1分） 0 2 评分标准：第1问，3分；第2问，7分；第3问，6分。共16分。 物理答案 第3 页 共3页 {#{QQABAYSUggCgAoBAABhCQQUICEOYkBGACIgGwAAcsAIAgRNABAA=}#}