江苏省淮阴中学2007-2008学年度第二学期高一物理期中试卷参考答案

江苏省淮阴中学2007－2008学年度第二学期期中考试

高一物理试卷参考答案

一、单项选择题 二、多项选择题

三、实验题 (每空2分，共20分)

10．； (2) (3) ； (4) 2、图略 11．(1) (2) (3) 、

。 四、计算或论述题 12．(1) UAB

QAB 6 10 4

························· (3分) V 200V · 6

q 3 10

电场线方向如右图所示 ··········································· (3分)

(2) UAB A B；所以 B A UAB 100V ········································(2分) EpB q B 2 10

4

·············································································(2分) J·

13．（1）从A到B的过程中损失的机械能为：

1212

mvA mgh mvB 13000J ········································ （3分） 22

1

（2）B到C的距离为：lBC (vB vC) tBC 36m ····························· （3分）

2

12

（3）B到C过程中，根据动能定理： flBC 0 mvB ······················· （2分）

2

E损

代入数据可解得：f=2000N ································································ （2分） 14．（1）设两颗星的转动半径分别为r1、r2则有：

m1m22 2

m()r1 ········································································ （2分） 12

LTmm2

G122 m2()2r2 ······································································· （2分）

LTG

r1＋r2＝L ······················································································

······· （2分） 可解得：r1

m2

L ····················································

················ （2分）

m1 m2

（2）同样可解得：T 2················································· （4分）

15．（1）物块下落至最低点过程中，有：mgx 可解得：x

12

kx ····································· （2分） 2

2mg

·············································································· （2分） k

(kx mg)

g，竖直向上 · 此时物块的加速度 a ····················· （2分） m

（2）当物体的质量为2m，仍下落x时，

121

kx 2mv2 ························································· （3分） 22

可解得：v ····························································· （3分）

有：2mgx

16．（1）小物块的加速度为：a1=μg＝2m/s2， 水平向右 ······························· （1分） 长木板的加速度为：a2

F mg

0.5m/s2，水平向右 ·············· （1分） M

令刚相对静止时他们的共同速度为v，

对小物块有：v= a1t ············································································ （1分）

对木板有：v= v0＋a2t ········································································· （1分） 代入数据可解得：t=2s；共同速度为：v=4m/s ··································· （2分） （2）此过程中小物块的位移为：x1=(v+0)t/2=4m ········································ （1分） 长木板的位移为：x2=(v0+v)t/2=4m=7m ··············································· （1分） 所以长板的长度至少为：L= x2-x1=3m ··············································· （1分） （3）此过程中由于摩擦而产生的内能为：Q=μmgL=12J ·························· （3分） 17．（1）令此时杆的角速度为ω，则：vA=ωl；vB=ω3l；所以vB=3 vA ········ （1分） 在杆转过90º的过程中系统机械能守恒： mg 3l 2mgl

解得：vA

1122

2mvA mvB ··············································· （2分） 22

·········································································· （2分） ·

2vA

（2）令此时杆对A球有向上的支持力，2mg N 2m ··················· （2分）

l

18

mg，竖直向上 解得：N ·························································· （2分） 11

12

（3）对B球，mg 3l W杆 mvB－0 ················································ （2分）

2

24

mgl · 解得：W杆 ········································································ （2分） 11

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