高一学年第一模块考试物理答案

1.D   2.A   3.D  4.C  5.C  6.A  7.C  8.BCD  9.BD  10.BD  11.CD  12.AD

13.D(3分)      14.C（3分）       15.（1）0.50（3分）（2）加速（2分）（3）AB（3分）

16．解析：(1)根据运动学公式v2－v＝2ax.     ·············································2’

代入数据v0＝20 m/s.     ···········································2’

(2)刹车开始后1s的位移                                                          ············································2’

刹车开始后2s的位移

                                  ············································2’

最后1s内的位移                   ············································1’

       ·············································1’

17．(12分)解：小球在液面上方运动的运动是自由落体运动，设运动时间是t1，进入时的速度是v，在作用力区域内运动是匀减速运动，设加速度大小是a，运动时间是t2，则

自由下落：v＝g t1        ·············································2’

        ·············································2’

进入液体后：F-mg=ma    ·····································2’

a＝    

0＝v－a t2         ··················································2’

        ············································2’

得：      h＝H/4        ············································2’

18．（共16分，第1问5分；第2问4分；第3问7分求出木板位移得3分，求出木块水平位移得3分，求出木块总位移得4分）

①、木块下滑过程：mgsinα    -µ0mgcosα=ma0        ···················································2’

a0=g(sina-µ0cosα)=2.5m/s2

vA2-0=2a0l0                                     ···················································1’

得：vA =5.0m/s                                     ···················································2’

②、滑块在木板上滑行时

m：μ1mg=ma1        得：a1=2m/s2                 ···················································1’

M：μ1mg-μ2(m+M)g=Ma2     得：a2=3m/s2             ···················································1’

M、m达到共速的时间为t1有：v2=v1- a1 t1= a2 t2    ···················································1’

 得：v2=3.0m/s，    t1=1.0s                        ···················································1’

(3)、此过程中各自位移：s1=( v1+ v2) t1/2=4.0m         ···················································1’

s2=(0+ v2) t1/2=1.5 m                             ···················································1’

共速后滑块与木板以相同的加速度减速运动：

μ2(m+M)g=(m+M)a3    得：a1=1m/s2                ···················································1’

减速位移：0-v22=2 (-a3)s3     得：s3=4.5m            ···················································1’

故：木板总位移：s木板= s2+ s3=6.0m                ···················································1’

木块在水平面上的位移：s木块1= s1+ s3=8.5m        ···················································1’

木块的总位移：s木块=[(l0sinα)2+(l0cosα+ s1+ s3) 2]1/2=13m    ···································1’下载本文