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太原市 2023 年高三年级模拟考试（一）
物理参考答案及评分建议
二、选择题:本题共8小题，每小题6分，共48分。在每小题给出的四个选项中，第14～
18题只有一项符合题目要求，第19～21题有多项符合题目要求。全部选对的得6分，选对
但不全的得3分，有选错的得0分。
题号 14 15 16 17 18 19 20 21
选项 C C D B D BD BD BD
三、非选择题：共62分。
22.（6分）
(1) AB
(2)
1
评分 标准：4每 空各2分。
23.（12分）
（1）正 1.1
（2）○
1
○
2 0.96
○
3 2.9 4.8
评分标准：每空（或每图）各2分。24.（10分）
（1）设“冰壶”被推出时瞬间速度的大小为v，继续滑行时加速度
的大小为 ，位移大小 =40m，时间t=10s，由运动学规律有
= t ·· ·· 1 ······················ ·· 1 ··························································（1分）
1 2
0=v- t························································································（1分）
解得 ：1 =0.8m/ ，v=8m/s
2
设“冰壶 ” 1与冰面 的动摩擦因数为μ，由牛顿运动定律有
μmg=m ······························································（1分）
解得：μ = 1 0.08 ························································（1分）
运动员推“冰壶”的过程中位移大小 =5m，加速度的大小为 ，
水平恒力的大小为F， 2 2
=2 ······························································（1分）
F 2 -μm g 2 = m 2 ···························································（1分）
解得：F= 1 2 44N ·······················································（1分）
（2）当“冰壶”与冰面的动摩擦因数为μ´=0.1时，设“冰壶”被水平
恒力推出时瞬间速度的大小为v´
，
+ = t
解 1 得 ： 2 v´=29m/s ·······················································（1分）
“冰壶”被推出后滑行的加速度为 ´，“冰壶”被水平恒力推出后滑
行的距离为 ´ 1
μ´mg=m ´ 1
=2 ´ 1 ´
‘2
解得： 1 ´1=40.5m ····················································（1分）
水平恒力 1的大小变为F´，其作用距离为 ´，加速度的大小为 ´
+ = ´+ ´ 2 2
解1得 ：2 1 ´=4 . 2 5m
=2 ´ 2
‘2
F´-μ´m 2g= m2´´ ´
解得：F´=2 00 2 N ······················································（1分）
另：用动能定理或动量定理求解，方法更优。25．（14分）
v sin30°
（1）粒子在电场中运动时间t =2 0 ····················（1分）
1
a
qE
a= ································································（2分）
m
mv 2
粒子在磁场中运动半径为R， qv B 0 ···················（1分）
0
R
αR
粒子在磁场中运动时间t = ································（1分）
2
v
0
π
由几何关系可知α= ··············································（1分）
3
粒子从a点运动到b点所用的时间t=t ＋t （1分）
1 2 ··························
mv πm
联立 可得t= 0＋ ··········································（1分）
qE 3qB
（2）Ob两点间距离l =2Rsin30° ·······························（1分）
1
粒子第二次在a点速度方向与x轴正方向夹角为θ
v sinθ
v cos30°t =2v cosθ 0 ···········································（2分）
0 1 0
a
Oc两点间距离l =2Rsinθ ························································（1分）
2
bc两点间距离l=l -l （1分）
1 2 ········································································
mv
联立 得l=( 3-1) 0 ················································（1分）
qB
26.（20分）
（1）Q下落：设在最低点速度为 ，
·················· ·· 0 ·····························（1分）
1 2
1 ∙ 0 =2 1 0
在最低点 ， ····································（1分）
2
0
解得 − 1 = 1 0 ···································（1分）
（2） Q 0 =和4Pm相/s碰： 设 =碰1后8NQ、P速度分别为 、 ，以左为正
···························· ·· 1 ···· ·· 2 ········ （1分）
1 0 = 1 1+ 2 2 ······················································（1分）
1 2 1 2 1 2
2解 得 1 0 =2 1 1+2 2 2 ································（2分）
碰后： 1 P=在−滑1m板/s上滑动 2，=P3与m滑/s板共速时，相对AB的高度最大，
设此高度为H，比较H与R即可判断能否从C点飞出。
设此时P和滑板的速度大小均为
3······························································（1分）
2 2 = 2+ 3 3 ·····················（1分）
1 2 1 2
2解 得 2 2 =2× 2 ； + 3 3+ 2 + 2
R 3故=不1能m飞/s出···· ···=···0·.·0··5·m······································（2分）
H设<小球碰撞后能摆到的最高点与 的连线和竖直方向的夹角为θ，
′
由机械能守恒定律得
1
mv2 mgL (1cos) ··········································（1分）
2 1 1 1 0
解得 =0.9375， <
0
可知cθo>s ，故碰后小co球s Qc的os运5动不能视为简谐运动········（2分）
0
（3）计5算P两次经过B时的速度，以左为正
···························································（1分）
′ ′
2 2 = 2 2+ 3 3
································（1分）
1 2 1 ′ 2 1 ′ 2
2 2 2 =2 2 2+2× 3 3+ 2
解得 ·················································（1分）
′ 3± 6
2 = 3 m/s
第一次经过B点速度
′ 3+ 6
2+ = 3 m/s
第二次经过B点速度
′ 3− 6
故物体P相对滑板滑动 2 时 − ，= 3 m/s>0
没有相对地面向右运动的速度····································（1分）
在AB段相对滑板从 时间
′
2− 2+ 6− 6
→ 1 = = 15 s
在AB段相对滑板从 相对静止的时间
′
3− 2− 6
物体P相对滑板的水 平→部分 2 = = 15s
AB运动的总时间 ····························（2分）
= 1+ 2 =0.4s
