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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）
14. 1 2 b kR（ ） （ ） 0 （3）无影响 （每空 2分）
1 k 1 k
15. （7分）解析：
1 H d S HS（ ）气缸内气体温度升高过程中做等压变化，有 ···········（2分）
T 5T
4
解得 d H ·················································································· （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 p0S ·············（1分）
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 p0SH ····································································（1分）25
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
解得 r 2 3 L
3
L
由几何关系 sin ·······································································（1分）
r
解得 60 ·················································································（1分）
yA r rcos ···············································································（1分）
物理答案 第 1页（共 3 页）
y 3解得 A L ··············································································（1分）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。
沿 OA连线方向 vcos L t ······················································· （1分）
cos
沿 y轴方向 ··························································（1分）
qB2E L解得 ··············································································（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 vE 0
从 D点到 E点，有 mg 2R 0
1
mv2D ·············································（1分）2
2
在 D点有 FN mg
v
m D ·································································（1分）
R
解得 FN 10N ···············································································（1分）
从释放到 E点过程有mg(h 2R) mgL 0 ········································（1分）
解得 h=1.8m················································································· （1分）
2 s（ ）①若只经过 C点一次，则有mgh mgL 1mg 0 ··················（1分）2
或mgh mgL 1mg
s
3 0 ······················································· （1分）
2
1
解得 1 1或 1 ········································································（1分）3
②当滑块第二次到达 C点且和传送带速度相等时，有
mgh mgL mg2s 1 1 mv
2
0 ··························································（1分）2
解得 1=0.225················································································（1分）
当 1 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。
s
全程由动能定理得mgh mgL 1mg n 0 ·····································（1分）2
解得
此时 1<0.225，不符合要求。··························································（1分）
物理答案 第 2页（共 3 页）
当 1<0.225时，物体第三次回到 C点时速度为 v0。
s 1
此后运动过程根据动能定理有 1mg n 0 mv
2
0 ·····························（1分）2 2
解得 11 (n 1,3,5 ) ································································（1分）10n
18. （16分）解析：
1 a b v（ ）对金属棒 、 组成的系统，由动量守恒定律得mv 00 m 2mv1 ······（1分）2
v
解得 v 01 ·················································································· （1分）4
1 1 v
2
1
由能量守恒定律得 mv 20 m 0 2mv
2
1 Q ·······························（1分）2 2 2 2 总
2
金属棒 b上产生的热量Qb Q ······················································（1分）3 总
Q 5解得 2b mv0 ············································································（1分）24
2 b BLva BLv（ ）对金属棒 由动量定理得 B b L t 2m v ···············（1分）3R
B 2L2
即 xa x0 2mv1 ·····································································（1分）3R
x 3Rmv解得 0a x2 2 0 ·······································································（1分）2B L
对金属棒 a、b由动量守恒定律得 mv0Δt mvaΔt 2mvbΔt ··············（1分）
即mv0t mxa 2mx0 ······································································· （1分）
t 3Rm 3x解得 0 ········································································（1分）
2B 2L2 v0
（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 bL t 2m vb
BLq 2m
v
v 0 即 b ·····································································（1分）
4
对金属棒 c由动量定理得 BI cL t m vc
即 BLqc mv ·················································································（1分）
对电容器 C有 q CBLv ···································································（1分）
对 MN右侧的电路有 q qb qc ·························································（1分）
CBLmv
解得 q 0 ······································································（1分）
2CB 2L2 6m
物理答案 第 3页（共 3 页）

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