四川省广安市高2022级第二次诊断性考试物理答案

广安市高2022级高三第二次诊断性考试物理参考答案一、单项选择题：本题共7小题，每小题4分，共28分。在每小题给出的四个选项中，只有一项是符合题目要求的。1.B2.D3.C4.A5.D6.D7．C二、多项选择题：本题共3小题，每小题6分，共18分。在每小题给出的四个选项中，有多项符合题目要求。全部选对的得满分，选对但不全的得3分，有选错的得0分。8.BD9.AC10.AD第Ⅱ卷非选择题三、实验题：本题共2小题，11题6分，12题10分，共16分。11.（6分，每空2分）①H；푑；③2푔퐻푆②22L푡2√퐻+퐿12.（10分）（每空2分）（1）11.016（11.014—11.018均给满分）2휋푑푅푥（2）（3）①R1③22300④大于4퐿四、计算题：本题共3小题，13题10分，14题12分，15题16分，共38分。解答时应写出必要的文字说明、公式、方程式和重要的演算步骤，只写出结果的不得分，有数值计算的题，答案中必须写出明确的数值和单位。휌0TOvs푃0（푚−휌0푣）13.（10分）答案（1）T1=（2）M=푚휌0Vg【解析】(1)乒乓球刚好能浮起,说明浮力等于重力,即ρ1gV=mg······························1分푚解得ρ1=··········································································1分푉푚푉0푉1气气体发生等压变化时有=,휌气=··················································1分푇0푇1푉气可得：ρ0T0=ρ1T1······································································1分푉휌0푇0解得T1=·······································································1分푚(2)乒乓球刚好也能浮起,说明浮力等于重力,即ρ1gV=mg··························1分푚气气体发生等温变化时p0V0=（p0+Mg）V1；휌气=·····································1分푉气高三物理试卷第1页（共页）{#{QQABLYyUogCAABAAAAgCAwGQCkMQkACACQoOAFAYoAIAwRFABAA=}#}푝푃可得0=0+푀푔······································1分휌0휌1（）解得M=푠푃0푚−푉휌0···························································2分푔푉휌014．（12分）答案：2m/s；1m/s；1.6m/s，1.3m/s解析：（1）设第一次碰后AB两滑块的速度分别为푣퐴和푣퐵，碰前A速度为푣0对B由动能定理可得：퐻1−푚gH−μ푚gcos휃=0−푚푣2·······································2分퐵퐵sin휃2퐵퐵解得푣퐵=2푚/푠······················································1分碰撞过程，由动量守恒：푚퐴푣0=푚퐴푣퐴+푚퐵푣퐵···················································2分解得푣퐴=−1푚/푠·····················································1分′′푣−푣（2）①由恢复系数公式e=21··················································1分푣1−푣2代入第一次碰撞数据解得：e=0.6·············································1分②再次到达水平面时速度为푣퐵1，从H处再次滑下时，由动能定理可得퐻1푚gH−μ푚gcos휃=푚푣2−0·····································1分퐵퐵sin휃2퐵퐵1再次碰撞时：设向左为正方向，碰后AB速度分别为푣퐴2、푣퐵2由动量守恒和恢复系数公式可得：푚퐵푣퐵1+푚퐴푣퐴=푚퐵푣퐵2+푚퐴푣퐴2·····································1分푣−푣퐴2퐵2=0.6······················································1分푣퐵1−푣퐴可得：푣퐴2=1.6푚/푠、푣퐵2=1.3푚/푠············································1分（其它合理解法参照给分)高三物理试卷第2页（共页）{#{QQABLYyUogCAABAAAAgCAwGQCkMQkACACQoOAFAYoAIAwRFABAA=}#}3515.（16分）（1）E=5.0×10푉/푚B=0.1T（2）푣푚푎푥=2×10푚/푠−2（3）푥1=[8+(4푛+2)휋]×10푚（푛=0,1,2,3⋯⋯）−2和푥2=(4+4푛휋)×10푚（푛=1,2,3⋯⋯）解：（1）(6分)如图，粒子在I区域从A到B做类平抛运动，设粒子运动的时间为t1O1푣푦푣r푐푣푥L=푣0푡1112퐿=푎푡2211qE=m푎1解得E=5.0×103푉/푚······································3分22粒子运动到B点的速度为푣퐵=√푣0+(푎푡1)5解得푣퐵=√2×10푚/푠如图方向与x轴正方向成450如图，粒子在Ⅱ区域从B到C做匀速圆周运动푚푣2q푣퐵=퐵퐵푟2퐿r=√2解得B=0.1T······································3分（2）(4分)如图，粒子在Ⅱ区域逆时针转过900运动到C点5푣퐶=√2×10푚/푠方向与x轴成450速度分解如图5푣푥=1×10푚/푠5푣푦=1×10푚/푠퐵因q푣=푞퐸푥22퐵푚푣푦q푣=푦2푅2휋푚T=퐵푞2高三物理试卷第3页（共页）{#{QQABLYyUogCAABAAAAgCAwGQCkMQkACACQoOAFAYoAIAwRFABAA=}#}解得R=L=0.02m，T=4π×10−7푠则粒子的运动可分解成在x轴正方向以푣푥做匀速直线运动和在xoy平面做半径为R、顺时针方向的匀速圆周运动，则两个分运动的速度大小不变，均为v=1×105푚/푠푇粒子从C点开始经t=+푛푇，两个分运动速度方向都为x轴正方向，4即最大速度为푣푚푎푥=2푣5푣푚푎푥=2×10푚/푠······································4分(3)（6分）如图，由粒子在Ⅲ区域的运动由在x轴正方向以푣푥做匀速直线运动和xoy平面做半径为R的匀速圆周运动合成，运动轨迹如图：1当匀速圆周运动t=nT+푇时,粒子恰好运动到x轴21푥=2L+2R+푣(푛푇+푇)1푥2−2解得푥1=[8+(4푛+2)휋]×10푚（푛=0,1,2,3⋯⋯）···························3分当匀速圆周运动t=nT时，粒子也恰好运动到x轴푥2=2L+푣푥푛푇−2解得푥2=(4+4푛휋)×10푚（푛=1,2,3⋯⋯）··································3分(单位用cm,表达式正确也得分)其它解法正确合理同样给分高三物理试卷第4页（共页）{#{QQABLYyUogCAABAAAAgCAwGQCkMQkACACQoOAFAYoAIAwRFABAA=}#}