物理答案及评分标准一、二：选择题题号123456789101112答案CADCDBABCCDACBCAB三、解答题13．84.5080.0无影响......................................................................每空2分214．C0.388斜面倾角过大.......................................................每空2分M15．（9分）【答案】（1）25N，15N；（2）0.25【详解】（1）对O点进行受力分析可知o..............................................................................................................2分TDcos37mgo.................................................................................................................2分TTDsin37B代入数据可得DO，BO线的拉力大小分别为...........................................................................................................................1分TD25NTB15N............................................................................................................................1分（2）由于整个系统恰好处于静止状态，则TAMg..........................................................................................................................2分解得0.25.....................................................................................................................1分116（9分）(1)hgt2..............................2分2得：t2s........................................1分vgt.........................................2分得：v20m/s......................................1分（2）由题意知，窗口的高度为：h2m12hvtgt......................................2分12v19ms......................................1分17（9分）（1）设最大速度为vm，匀速阶段所用时间为t2，根据题意有：tt1t212s................................................................................................................1分vxmt16m..............................................................................................................1分121试卷第1页，共3页x2vmt264m......................................................................................1分联立解得：t14s..............................................................................................................................1分vm8m/s......................................................................................................................1分（2）小明匀加速运动阶段的加速度大小为vm2a2m/s.............................................................................................................1分t1根据牛顿第二定律可得，对m有分Ncos37°fsin37°mg............................................................................................1分Nsin37°fcos37°ma.............................................................................................1联立解得：N9.2N.....................................................................................................................1分18（11分）解：（1）设物块的加速度大小为a1，由牛顿第二定律有1mgma1.......................................................................................................................1分解得：2a11g2m/s.............................................................................................................1分因为2s时达到共速，此时速度大小：v1v0a1t12m/s........................................................................................................1分长木板由静止做匀加速直线运动，加速度大小：v12a21m/s................................................................................................................1分t1（2）对长木板由牛顿第二定律有：1mg2mMgMa2.............................................................................................1分解得：m4kg..............................................................................................................1分2s内物块对地位移大小为：vvx01t8m1212s内长木板对地位移大小为：vx1t2m221则t2s时物块到长木板左端的距离为：试卷第2页，共3页dx1x26m...............................................................................................................1分（3）共速后，对物块和长木板整体，由牛顿第二定律有：2mMg()mMa3..........................................................................................1分解得：2分a31m/s................................................................1与挡板碰撞前瞬间，整体具有速度：v2v1a3t2....................................................................................................................1分可得:v21.5m/s碰撞后，物块m运动到停下有：20v22a1s解得：s0.56mLd0.5m因此，物块会脱离长木板22vv22a1(Ld)解得：v0.5m/s…...............................................................................................1分试卷第3页，共3页