青岛期末-物理答案

高三物理试题答案第1页（共4页）学科网（北京）股份有限公司γ2022—2023学年度第一学期期末学业水平检测高三物理答案及评分标准一、单项选择题：本大题共8小题，每小题3分，共24分。1．A2．C3．C4．B5．D6．A7．B8．C二、多项选择题：本大题共4小题，每小题4分，共16分，选不全得2分，有选错得0分。9．AD10．AD11．CD12．BCD三、非选择题（60分）13．（6分）（1）2.10（2分）；（2）0.48（2分）；（3）①木板的倾角要适中；②A点与传感器距离适当大些。（2分）（给出其中一种说法即可）14.（8分）（1）AC（2分）；（2）不变（1分）；变长（1分）（3）图像与x轴所围图形的面积与电容器的电荷量Q数值相等，由C＝QU求出电容（2分）（4）B（2分）15．（8分）（1）璃砖转过30°角时，折射光路如图，由几何关系可知入射角i＝30°又因为tanθ＝10103＝33则θ＝30˚折射角γ＝60°·························（2分）由折射定律可知sinisinγ＝1n解得n＝3··························（2分）（2）发生全反射时有sinC＝1n······（2分）所以sinα＝sinC＝33··········（2分）评分标准：第1问，4分；第2问，4分。共8分。16.（9分）（1）滑船从A点滑到C点时，由机械能守恒定律可知212CmgHmv············································（1分）在C点时由牛顿第二定律可得2CNCvFmgmR······（1分）解得H＝0.4R＝5m·························································（1分）高三物理试题答案第2页（共4页）学科网（北京）股份有限公司（2）划船到达D点时速度mg(H－h)＝12mvD2解得vD＝5g················································（1分）滑船在斜面上只受重力和斜面的支持力，则运动的加速度大小a＝mgsin37°m＝0.6g····················（1分）运动最高点J到水平底边ad的距离s＝(vDsin53°)22a＝83m·······································（1分）（3）滑船从D点开始到进入接收平台的时间为t＝2vDsin53°a·········································（1分）则x＝vDcos53°t···············································（1分）解得：x＝8m··················································（1分）评分标准：第1问，3分；第2问，3分；第3问，3分。共9分。17．（13分）（1）粒子在0x空间中做匀速圆周运动，由qv0B＝mv02R················································（1分）得R＝mv0qB······················································（2分）（2）由已知可得，粒子在x0范围中偏转，磁场为一圆柱体,如图可得磁场垂直y方向的截面半径：r＝Rsin30˚·········································（1分）根据V＝πr2h可得：V＝πhm2v024q2B2············································（2分）（3）由分析知最低点的粒子x0，y0区域内向x轴正方向做螺旋前进，即yOz平面的圆周运动与沿x轴正向的匀速直线运动的合运动，其半径为r1＝mv1qB1由于两粒子在x轴相遇，可得：r1＝h4················（1分）其中速度v1＝v0cosθ·········································（1分）联立可得：B1＝23mv0qh·····································（1分）xORrθαv0v0高三物理试题答案第3页（共4页）学科网（北京）股份有限公司（4）由分析知最高点释放粒子在y方向为匀加速运动可得：h2＝12at2······································································（1分）由两粒子恰好在x轴第一次相遇，可知：t＝πmqB1····（1分）又因为a＝qEm且B1＝23mv0qh·····························（1分）联立可得：E＝12mv02π2qh·······································（1分）评分标准：第1问，3分；第2问，3分；第3问，3分；第4问，4分。共13分。18．（16分）（1）设B到达水平位置时的速度为v，根据机械能守恒定律：mBgL＝12mBv2·················································（1分）C击中B的过程中二者动量守恒，击中后BC的速度为v1：mcv0+mBv＝(mB+mC)v1·······································（1分）由牛顿第二定律得：Tm－(mB+mC)g＝(mB+mC)v12L·（1分）根据牛顿第三定律得：Tm＝90N·························（1分）（2）对M受力分析得：μ(mB+mC)g＝Ma解得：a＝1m/s2········································································（1分）由v2A＝2ax解得A碰P前的速度：vA＝2m/s························（1分）由于碰撞挡板P之前A和BC总动量守恒，由(mB+mC)v1＝(mB+mC)vB+MvA解得：vB＝4m/s··············································（1分）可求出碰撞P之后：(mB+mC)vB＝－MvA因此：Ａ与Ｐ仅碰撞一次··································（1分）（3）由(mB+mC)vB＝－MvA可知，碰撞一次后木板和木块最终会停下来有能量守恒可得：μ(mB+mC)gd＝12(mB+mC)v12··············（1分）解得d＝16m··················································（1分）高三物理试题答案第4页（共4页）学科网（北京）股份有限公司（4）若木板A与挡板恰好发生了8次碰撞，最后，物块B和木板A都停下来。而每次木板发生x大小的位移所用时间t相同，则木块在第8次碰撞后：15μ(mB+mC)gt＝(mB+mC)v1－(mB+mC)vB··························（1分）木板A每次与挡板碰撞的速度均满足：μ(mB+mC)gt＝MvA····································································（1分）由于恰好发生了8次碰撞：(mB+mC)vB＝MvA联立解得：vA＝14m/s········································（1分）根据v2A＝2ax解得：x8＝132m·····························（1分）同理：恰好发生了7次碰撞联立解得：vA＝27m/s根据v2A＝2ax解得解得：x7＝249m······················（1分）因此能碰8次的条件是132m≤x＜249m··················（1分）评分标准：第1问，4分；第2问，4分；第3问，2分；第4问，6分。共16分。