柳州市2023届高三第三次模拟考试物理参考答案二、选择题：本大题共8小题，每小题6分。在每小题给出的四个选项中，第14~18题只有一项是符合题目要求，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1415161718192021答案BDCBCADBCDBD三、非选择题：xL22.（6分）（1）12N/m（2分）（2）0.2kg（2分）（3）（2分）h23.（9分）（1）B（1分）（2）rg=1200Ω（2分）R2=360Ω（2分）（3）9.75mA（1分）Rx=rg+R2或Rx=38R1（3分）。24.（12分）解：（1）A自由下落过程，机械能守恒，1有，mgh=mv2.......................................................................................................2分20A与B相碰，动量守恒，有,mv0=2mv.........................................................................................................2分1212系统损失的机械能E损mv2mv..............................................................2分2021解得：E=mgh....................................................................................................1分损2（2）B静止时，mg=kx...................................................................................................1分AB反弹至最高点的过程，与弹簧组成的系统机械能守恒，1×2mv2=2mgh...........................................................................................................2分2又C刚好离地，有，h=2x.....................................................................................1分8mg解得：k.........................................................................................................1分h25.（20分）解：（1）O到P的过程，设抛出初速度为v0，到P点的速度为v1，水平方向v1=at1........................................................................................................1分竖直方向v0=gt1.......................................................................................................1分又F=ma....................................................................................................................1分1Emv216J....................................................................................................1分k0201Emv29J......................................................................................................1分kP211F3解得：=...........................................................................................................1分G4(2)P到Q的过程，根据竖直方向的对称性，t2=t1，....................................................................................1分水平方向vx=v1+at2=2v1........................................................................................1分竖直方向vy=gt2=v0.................................................................................................1分22vQvxvy......................................................................................2分1解得：Emv252J...................................................................................1分kQ2Q(3)当速度与合力垂直时，速度最小，.............................................................................1分F设速度与水平方向成θ，有tan==0.75.............................................................1分G此时水平方向vx1=v1－at.................................................................................1分竖直方向vy1=gt..................................................................................1分v又tany10.75.........................................................................1分vx122vminvx1vy1...........................................................................2分1解得：Emv25.76J............................................................................1分kmin2min33.(1)（5分）CDE(2)（10分）解：(I)甲、乙两部分气体发生等压变化，设活塞M上升的距离为x，对理想气体甲有V甲0h2shs3hs............................................................................1分V甲1(hx)s.............................................................................................1分VV又甲0甲1.......................................................................................1分T0T1对理想气体乙有V乙0hs.....................................................................................................1分V乙1(hx)s.............................................................................................1分VV又乙0乙1T0T111解得：TTxh1202(II)气体甲的压强22mgP甲p.................................................................................................1分02S气体乙的压强mg2mgP乙p甲-p-.................................................................................1分S0S气体乙被压缩过程中，对气体做的功为Wp乙Sx.........................................................................................................1分气体乙内能的增加量为UWQ.................................................................1分解得：U0.5p0hsmghQ则气体乙内能减少量为Qmgh0.5p0hs.......................................................1分34.(1)（5分）ABE(2)（10分）解：(I)当光线在BC边上G点恰好发生全反射时，设光线发生全反射的临界角为C则有1sinC=........................................................................,.........................1分nBG又=tanC..............................................................................................1分OB23R解得：BG=即为所求...........................................................................1分3(II)当光线在BC边恰好发生全反射时，设光线发生全反射的临界角为C，在柱形光学元件中的路程最大为S1，速度为v，有cv........................................................................................................1分n6RSR...............................................................................................1分1sinCSt=1.............................................