高一物理參考答案第物理一、單項選擇題:本題共7小題每小題4分，共28分。二、多項選擇題:本題共4小題，每小題5分，共20分。全部選對得5分，選對但不全的得3分，有選錯的得0分。題號1234567891011答案CBCBDDCABDBCCDAC實驗：本大題共2小題，共16分。12.（6分，每空2分）（1）10；（2）1000；（3）F=1000(L-0.1)(10分)重物會落在桌面上(或“紙帶打點過短”等類似的答案)；應先接通電火花計時器電源，再釋放紙帶。(4分)（2）b（2左分）；9.75分）四、計算題(本題共3小題，共36分)14（8分）.(1)冰車和小孩受力如圖所示豎直方向的合力為零，則有FN＋Fsinθ＝mg··············································（2分）解得支持力FN＝188N·······················································································（1分）(2)在水平方向，根據牛頓第二定律得Fcosθ-Ff＝ma·····························（2分）滑動摩擦力Ff＝μF壓由牛頓第三定律得FN＝F壓·········································（2解得加速度a＝0.33m/s2·············································································（1分）15.（12分）解：據題意小轎車初速度v0=30m/s，加速度a=—0.5m/s2；駕駛員的視距d=60m，反應時間(1)從剎車到停止時間為t2，則：··········································（3分）(2)反應時間內做勻速運動，則：··················································（2分）從剎車到停止汽車做勻減速運動直到速度為0，則：········（3分）小轎車從發現物體到停止的全部距離為：····················（2分）三角警示牌距車的最小距離為：······································（2分）16.(16分)解：（1）（6分）對小物塊和小車分別使用牛頓第二定律得：μmg=mam··················································（2分）F-μmg=MaM····················································（2分）代入數據解得：am=2m/s2·················································（1分）aM=0.5m/s2··················································（1分）（2）（4分）對小物塊和小車分別使用運動學公式得：v共=amt1，v共=v0+aMt1·····解得：t1=1s··········································································（2分）（3）（6分）當兩者達到相同的速度后，假設兩者保持相對靜止，以共同的加速度a做勻加速運動對小物塊和小車整體，由牛頓第二定律有：F=(M+m)a解得：a=0.8m/s2此時小物塊和小車之間的靜摩擦力f=ma=1.6N而小物塊和小車之間的最大靜摩擦力fm=μmg=4Nf<fm，所以兩者達到相同的速度后保持相對靜止。························（2分）在開始的t1=1s內，小物塊的位移sm=eq\f(1,2)amteq\o\al(2,1)=1m末速度v=amt1=2m/s·····（2分）在接下來的t-t1=0.5s內，小物塊與小車相對靜止，以共同的加速度a=0.8m/s2做勻加速運動，這0.5s內小物塊通過的位移s=v(t-t1)+eq\f(1,2)a(t-t1)eq\o\al(2,)代入數據解得：s=1.1m從小物塊放上小