山东省济宁市2025年高考模拟考试物理答案_2025年5月_250525山东省济宁市2025年高考模拟考试（济宁三模）（全科）

2025 年高考模拟考试 物理试题参考答案及评分标准 题号 1 2 3 4 5 6 7 8 9 10 11 12 答案 B C B B A C A D AD BC BD ACD 13.(6分)(1)C (2)9.150 (3)750 (每空2分) 14.(8分)(1)b (3)R R (每空2分) G 1 15.(8分)解析：(1) 气泡在池底时压强 p  p gh1.5105Pa …………………(1分) 1 0 pV pV 由理想气体状态方程得 1 1  0 2 ···········································（2分） t 273 t 273 1 0 解得 t 17oC················································································（1分） 1 p  p (2) 气泡上升过程中，外界对气体做功为W  0 1(V V)···············（2分） 2 2 1 由热力学第一定律U W Q·························································（1分） 解得Q=0.39J················································································ （1分） 16.(8分) 解析： (1) 金属棒恰好静止，有mgsinILB 1分 E I  ························································································（1分） 2R E nk ························································································（1分） 2mgRsin 解得磁通量的变化率k  ···················································（1分） nLB B2L2v (2) 速度最大时，有mgsin m ··············································· （1分） 2R 金属棒加速运动过程中，有mgsintILBt mv 0························（1分） m E BLv I   ···············································································（1分） 2R 2R vtx 2mR B2L2x 解得t  ································································（1分） B2L2 2mgRsin 高三物理答案 第1页（共3页）17.(14分) 解析： v2 (1) 根据洛伦兹力提供向心力qvBm 1 ……………………………………（2分） 1 r 1 由题意可知最小半径为r L··························································· （1分） 1 qBL 解得v  ··············································································· （1分） 1 m 5qBL (2)由题意可知最大速度v  2 3m 5 最大半径r  L···········································································（1分） 2 3 r 设v 与y轴夹角为，则sin 1 ···················································（1分） 2 r 2 在O点，粒子的最大速度在y轴方向上的分速度v =v cos·················（1分） y 2 v2 粒子穿过小孔后在平行于yOz平面上做匀速圆周运动，由qv Bm y 得 y R 粒子做匀速圆周运动的最大半径 ············································（1分） 8 穿过小孔的粒子到达接收屏前离开x轴的最大距离d 2R L············· （1分） 1 3 (3)设在P点发射的速度为v介于v 与v 间的粒子与y轴的夹角为θ，则 1 2 L mv 粒子在磁场中的运动半径r   ·············································（1分） sin qB qBL 粒子穿过小孔后沿x轴方向的分速度v =vsin ··························（1分） x m 2m 粒子做圆周运动的周期T  ·······················································（1分） qB 当粒子运动时间为整数个周期时，这些粒子打在接收屏上同一点。 1 qE d v nT  (nT)2··································································（1分） 2 x 2 m 解得 (n=1,2,3,…)··········································（1分） ， 高三物理答案 第2页（共3页）18.(16分)解析： (1)由图乙可知， 时，A与C恰好一起运动。 m 1kg 1 对C由牛顿第二定律可得m g m a ···············································（2分） 2 2 1 解得a 2m/s2············································································（1分） 1 (2)B的质量 m 1kg时， 1 对A、B、C由牛顿第二定律可得m g m m Ma ························（1分） 1 1 2 1 B的质量 m  2kg 时，由图乙可知A的加速度 a 4m/s2·····················（1分） 1 2 对A、B由牛顿第二定律可得mgm g mMa ························· （1分） 1 2 1 2 联立解得M 2kg， m 2kg···························································（2分） 2 (3)当m 2kg时，设B运动时间t后落地。 1 则此时C的速度v at2t····························································（1分） C 1 A的速度v a t4t·····································································（1分） A 2 从B落地到C恰好到达Q点的过程中，A与C组成的系统在水平方向上动量守恒，可 得m v Mv m Mv·····························································（1分） 2 C A 2 解得 v3t 1 1 t时间内，A与C的相对位移Δx a t2  at2···································（1分） 2 2 2 1 解得 Δxt2 从B落地后到C恰好到达Q点的过程中，由能量守恒得 1 1 1 m v2  Mv2 m gLΔxm gr m Mv2···························（2分） 2 2 C 2 A 2 2 2 2 1 B下落的高度 h a t2···································································（1分） 2 2 解得h=4m····················································································（1分） 高三物理答案 第3页（共3页）