山东省济宁市全市高考模拟考试（济宁三模）物理+答案

2026年济宁市高考模拟考试

物理试题答案2026.05

题号123456789101112

答案CCDBABDAADCDBCBC

13.（1）C（2）1.84（3）g（每空2分）

bkR

14.（1）（2）0（3）无影响（每空2分）

1_k1_k

15.（7分）解析：

（1）气缸内气体温度升高过程中做等压变化，有（2分）

H

解得d=··················································································（1分）

5

（）设气缸内气体压强为，对气缸由平衡条件得（分）

2pG+pS=p0S·············1

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ΔU=W+Q··························································（1分）

解得ΔU=Qp0SH····································································（1分）

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。

2

v

由牛顿第二定律qvB=m·····························································（1分）

r

解得rL

L

由几何关系sinθ=·······································································（1分）

r

解得θ=60o·················································································（1分）

yA=r_rcosθ···············································································（1分）

物理答案第1页（共3页）

解得yAL··············································································（1分）

（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“=30o··········································（1分）

设粒子从O点到A点运动时间为t。

沿OA连线方向vcos“.t·······················································（1分）

沿y轴方向··························································（1分）

解得E··············································································（1分）

17.（14分）解析：

（）若小滑块恰好到达圆管轨道最高点，则

1EvE=0

从D点到E点，有一mgx2Rmv·············································（1分）

在D点有FN一mg=m·································································（1分）

解得FN=10N···············································································（1分）

从释放到E点过程有mg(h一2R)一μmgL=0········································（1分）

解得h=1.8m·················································································（1分）

（2）①若只经过C点一次，则有mgh一μmgL一μ1mgx··················（1分）

或mgh一μmgL一μ1mgx3x·······················································（1分）

解得或1（分）

μ1=1μ1=········································································1

3

②当滑块第二次到达C点且和传送带速度相等时，有

mgh一μmgL一μ1mg2smv··························································（1分）

解得μ1=0.225················································································（1分）

当μ1≥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。

全程由动能定理得mgh一μmgL一μ1mg.n·····································（1分）

1

解得μtn5,7,9)

此时μ1<0.225，不符合要求。··························································（1分）

物理答案第2页（共3页）

当μ1<0.225时，物体第三次回到C点时速度为v0。

此后运动过程根据动能定理有_μ1mg.nmv·····························（1分）

解得n=1,3,5…)································································（1分）

18.（16分）解析：

（1）对金属棒a、b组成的系统，由动量守恒定律得mv0=mmv1······（1分）

解得v··················································································（1分）

由能量守恒定律得mvmvQ总·······························（1分）

金属棒b上产生的热量QbQ总······················································（1分）

解得Qbmv············································································（1分）

（2）对金属棒b由动量定理得LΔt=Σ2mΔv···············（1分）

即（分）

=2mv1·····································································1

解得xax0·······································································（1分）

对金属棒a、b由动量守恒定律得mvΔtmvΔt2mvΔt（1分）

Σ0=Σa+Σb··············

即mv0t=mxa+2mx0·······································································（1分）

解得t········································································（1分）

（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。

对金属棒由动量定理得BILΔt=2mΔv

b_ΣbΣb

即_BLqb=2m·····································································（1分）

对金属棒c由动量定理得BILΔt=mΔv