山东省青岛市2025年高三年级第三次适应性检测物理答案_2025年5月_250529山东省青岛市2025年高三年级第三次适应性检测（青岛三模）（全科）

2025 年高三年级第三次适应性检测 物理答案 一、单项选择题：本大题共8小题，每小题3分，共24分。 1．A 2．D 3．B 4．C 5．C 6．A 7．D 8．C 二、多项选择题：本题共4小题，每小题4分，共16分。 9．AC 10．CD 11．AC 12．ABC 三、非选择题：本题共6小题，共60分。 13．（7分）（1）1.60（1分）； 7.5×10-7（2分）； （2）S （2分）； 3.0×10-6（2分） 2 14．（7分）（1）D（1分）； （2）1.0×10-7（2分）；（3）0.87（2分） 调小（2分） 15．（7分） p p （1）加热过程气体经历等容变化： 1  2 ·········································（1分） T T 1 2 p S  p S f ················································································· （1分） 2 0 解得： p 0.96105Pa·····································································（1分） 1 充气过程中： pV  p V  p V ···························································（1分） 1 0 2 2 V V V 1 液 ···········································································（1分） 解得：V 375ml 液 （2）瓶内原有气体， pV  p V ························································（1分） 1 0 3 注入气体与原有气体质量的比值 m 注  V 2  1 ········································（1分） m V 4 原 3 评分标准：第1问，5分；第2问，2分。共7分。 16．（9分） （1）设弹簧弹开两棒过程任一时刻流过两棒的电流为i，有F 2BiLF Bi2L a b 故两棒系统动量守恒。mx 2mx ·······················································（1分） a b x x  x ·····················································································（1分） a b 0 2x x 解得：x  0 ，x  0 ··································································（1分） a 3 b 3 （2）弹簧恢复原长时，有mv 2mv ··················································（1分） a 0 1 1 1 kx2  mv2 2mv2Q Q ···························································（1分） 2 0 2 a 2 0 a b 物理答案 第 1 页 共 4 页Q r a  ·························································································（1分） Q 2r b kx2 解得：Q  0 mv2········································································（1分） a 6 0 （3）闭合开关S后，设b棒继续运动的距离为x， b B2Lv 4B2L2 有B 2Ltp，即： xp R2r R2r 4B2L2 求和得： x 2mv ···································································（1分） R2r b 0 x mv (R2r) 解得：d  x x  0  0 ·····················································（1分） b b 3 2B2L2 评分标准：第1问，3分；第2问，4分；第3问，2分。共9分。 17.（14分） 1 （1）粒子在加速电场中有：qU= mv2················································（1分） 2 0 mv2 在磁场中有：qv B = 0 ··································································（1分） 0 0 r 1 2πr T= 1 ························································································· （1分） v 0 当粒子在圆形磁场中轨迹对应弦长最长，为圆形磁场直径2a时， 粒子在磁场中运动的时间最长 轨迹如图： 对应轨迹圆运动圆心角60°， T πm 在磁场中运动的最长时间t   ··················································（2分） 6 3qB 0 （2） （i）①d=0，粒子直线运动经过O点···················································（1分） ②粒子在右侧磁场中匀速圆周运动，由牛顿第二定律得： v2 2 3qv B m 0 ··············································································（1分） 0 0 r 2 O 3a a 解得：r  2 3 r 2 由几何关系得：当粒子指向圆心进入无磁场区域时，粒子轨迹与对称轴相切（1分） 2 3a 所以，粒子到对称轴的距离：d 2r  ·········································（2分） 2 3 物理答案 第 2 页 共 4 页（ii）取临界状态，粒子在右侧磁场中运动轨迹圆恰好与无磁场区域圆相切，如图所示。 O1 r1 0.75a a O O2 3 题目要求有一半的粒子圆形区域，所以上下两轨迹圆圆心距离为 a········· （2分） 2 3a 由几何关系得：a2 ( )2 (ar )2···················································（1分） 3 4 解得：r =0.25a 3 v2 由牛顿第二定律得：qv B m 0 0 2 r 3 解得：B 8B ···············································································（1分） 2 0 评分标准：第1问，5分；第2问，9分。共14分。 18．（16分） （1）C斜上抛运动分解成水平匀速和竖直方向竖直上抛运动。 竖直方向：2ghv2，······································································ （1分） y 解得：v 4m/s y v 另有：tan53 y ···········································································（1分） v x v  v2v2 5m/s································································· （1分） 0 x y （2）C弹出时，AC系统水平方向动量守恒：m v m v ·····················（1分） A A C x 解得：v 0.6m/s A C斜上抛运动，竖直方向：v =at ························································（1分） y 1 解得：t =0.4s 1 水平方向A匀速运动：Δd=v t···························································（1分） A 解得：Δd=0.24m·············································································（1分） （3）规定向左为正方向，水平方向BC第一次共速过程动量守恒， m v  m m  v ，······································································· （1分） C x C B 1 解得：v 1.5m/s 1 1 1 由能量转化和守恒得：m gx  m v2  (m m )v2························（1分） C 1 C x C B 1 2 2 物理答案 第 3 页 共 4 页9 解得：x  m 1 20 B与A第一次弹性碰撞， m v m v m v m v ······························································· （1分） A A B 1 A A1 B B 1 1 1 1 m v2  m v2  m v2  m v2 ·····················································（1分） A A B 1 A A1 B B 2 2 2 2 解得，v 0.1m/s，v 2m/s A B BC第二次共速过程，动量守恒， m v m v  m m  v ， C 1 B B C B 2 1 解得：v  m/s 2 4 1 1 1 由能量转化和守恒得：m gx  m v2 m v2  (m m )v2 C 2 C 1 B B C B 2 2 2 2 17 解得：sx x  m·································································（1分） 1 2 16 （4）整个过程中，C相对于B向左运动的距离为Δx，由能量转化和守恒： 1 1 m gx m v2  m v2···································································（1分） C A A C x 2 2 解得Δx=1.08m 对AC整体和B运用人船模型得： （m m ）x m x A C A1 B B x x 30.241.082.16m A1 B 1 2.16 所以A向右运动的位移为：x  2.16m m································（1分） A1 7 7 对AB整体和C运用人船模型得： （m m ）x m x A B A2 C C x x 30.241.24.44m A2 C 1 4.44 所以A向右运动的位移为：x  4.44m m·······························（1分） A2 7 7 33 x x x  m0.94m······························································（1分） A1 A2 35 评分标准：第1问，3分；第2问，4分；第3问，5分；第4问，4分。共16分。 物理答案 第 4 页 共 4 页