物理答案_2024-2027高三（6-6月题库）_2026年05月高三试卷_260523山东省济宁市全市高考模拟考试（济宁三模）（全科）

2026 年济宁市高考模拟考试
物理试题答案
2026.05
题号 1 2 3 4 5 6 7 8 9 10 11 12
答案 C C D B A B D A AD CD BC BC
13. （1）C （2）1.84 （3）g （每空2分）
b kR
14.（1） （2） 0 （3）无影响 （每空2分）
1k 1k
15. （7分）解析：
H dS
HS
（1）气缸内气体温度升高过程中做等压变化，有  ···········（2分）
T 5
T
4
H
解得d  ··················································································（1分）
5
（2）设气缸内气体压强为p，对气缸由平衡条件得G pS  p S ·············（1分）
0
气缸内气体温度升高过程中，外界对气体做功W pSd ······················· （1分）
由热力学第一定律U W Q··························································（1分）
3
解得U Q p SH ····································································（1分）
25 0
16. （9分）解析：
（1）设粒子做匀速圆周运动的半径为r，A点速度方向与x轴正方向夹角为θ。
v2
由牛顿第二定律qvBm ·····························································（1分）
r
2 3
解得r  L
3
L
由几何关系sin ·······································································（1分）
r
解得60·················································································（1分）
y rrcos···············································································（1分）
A
物理答案 第1页（共3页）3
解得y  L··············································································（1分）
A 3
（2）粒子在O点和A点的速度大小相等，所以OA连线为等势线，
电场强度与y轴正方向所成夹角为30··········································（1分）
设粒子从O点到A点运动时间为t。
L
沿OA连线方向vcost ·······················································（1分）
cos
沿y轴方向 ··························································（1分）
qB2L
解得E  ··············································································（1分）
m
17. （14分）解析：
（1）若小滑块恰好到达圆管轨道最高点E，则v 0
E
1
从D点到E点，有mg2R0 mv2·············································（1分）
2 D
v2
在D点有F mg m D ·································································（1分）
N R
解得F 10N···············································································（1分）
N
从释放到E点过程有mg(h2R)mgL0········································（1分）
解得h=1.8m················································································· （1分）
s
（2）①若只经过C点一次，则有mghmgLmg 0··················（1分）
1 2
s
或mghmgLmg3 0······················································· （1分）
1 2
1
解得 1或  ········································································（1分）
1 1 3
②当滑块第二次到达C点且和传送带速度相等时，有
1
mghmgLmg2s mv2··························································（1分）
1 2 0
解得=0.225················································································（1分）
1
当 0.225时，滑块第二次到达C点后再从传送带离开时，速度大小不变。
1
s
全程由动能定理得mghmgLmgn 0·····································（1分）
1 2
解得
此时<0.225，不符合要求。··························································（1分）
1
物理答案 第2页（共3页）当<0.225时，物体第三次回到C点时速度为v 。
1 0
s 1
此后运动过程根据动能定理有mgn 0 mv2·····························（1分）
1 2 2 0
1
解得  (n1,3,5)································································（1分）
1 10n
18. （16分）解析：
v
（1）对金属棒a、b组成的系统，由动量守恒定律得mv m 0 2mv ······（1分）
0 2 1
v
解得v  0 ··················································································（1分）
1 4
1 1 v 2 1
由能量守恒定律得 mv2  m 0   2mv2Q ·······························（1分）
2 0 2  2  2 1 总
2
金属棒b上产生的热量Q  Q ······················································（1分）
b 3 总
5
解得Q  mv2············································································（1分）
b 24 0
（2）对金属棒b由动量定理得 B BLv a BLv b Lt  2mv···············（1分）
3R
即
B2L2
 x x 2mv ·····································································（1分）
3R a 0 1
3Rmv
解得x  0 x ·······································································（1分）
a 2B2L2 0
对金属棒a、b由动量守恒定律得mv Δt mv Δt2mv Δt （1分）
0 a b ··············
即mv t mx 2mx ·······································································（1分）
0 a 0
3Rm 3x
解得t   0 ········································································（1分）
2B2L2 v
0
（3）设金属棒b进入MN右侧后，整个系统达到稳定时金属棒b、c的速度大小为v。
对金属棒b由动量定理得  BI Lt  2mv
b b
 v 
即BLq 2mv 0 ·····································································（1分）
b  4 
对金属棒c由动量定理得
BI Lt

mv
c c
即BLq mv·················································································（1分）
c
对电容器C有qCBLv···································································（1分）
对MN右侧的电路有qq q ·························································（1分）
b c
CBLmv
解得q 0 ······································································（1分）
2CB2L26m
物理答案 第3页（共3页）