山东省青岛市四区统考2023-2024学年高三上学期期中高三物理答案

2023－2024学年度第一学期期中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．B2．A3．D4．C5．C6．B7．C8．D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．ACD10．BC11．BD12．BC三、非选择题13．（6分）F0（1）C（2分）；（2）（2分）、C（2分）。g14．（8分）（1）天平（1分）；（2）C（1分）；（3）ks1＝ks2+s3（2分）；ks1＝ks2+s3（2分）；（4）小于（2分）。15．(8分)（1）由已知条件得OA3T只有A点振动时，t1＝+·····················································（1分）v4OBT只有B点振动时，t2＝+·······················································（1分）v4解得周期T=8s··········································································（2分）（2）由（1）得v＝10m/s,···························································（1分）可得波长λ＝vT＝80m·································································（1分）1OA-OB＝40m＝λ······································································（1分）2由已知条件得O点为振动加强点··················································（1分）评分标准：第1问，4分；第2问，4分。共8分。16．(10分)（1）在BC段，餐盘收到的向心力是静摩擦力提供v2由牛顿第二定律：μmg＝m·····················································（2分）R解得：μ＝0.1·········································································（2分）L1+L2（2）AC段速率不变，t1＝＝4s··············································（1分）v物理答案第1页共3页{#{QQABBQKAggiAAhBAABgCEwXCCkKQkAGACKoGBEAMsAAAARFABCA=}#}由μmg＝ma···········································································（1分）解得机器人减速的最大加速度：a＝1m/s2由v2＝2ax解得机器人减速的位移为：x＝2m·············································（1分）CD段匀速运动的位移为：L3＝12－x＝10mL3时间为：t2＝＝5s·································································（1分）v匀减速阶段由v＝at3得：t3＝2s············································································（1分）配送最短时间为：t＝t1+t2+t3＝11s··············································（1分）评分标准：第1问，4分；第2问，6分。共10分。17．（12分）（1）研究Δt时间内出射的水柱12由动能定理得水枪对出射水柱做的功W＝mv0－0························（1分）2Δt时间内出射的水柱质量为m＝ρv0SΔt········································（1分）W由P＝···············································································（1分）Δt解得：P＝0.22W·····································································（1分）（2）水柱在空中做斜上抛运动，水平方向vx＝v0cos37°；L＝vxt····················································（1分）12竖直方向vy＝v0sin37°；y＝vyt－gt·············································（1分）2解得y＝－0.35m······································································（1分）由已知得H＝h+y＝0.15m··························································（1分）（3）vx＝vcos37°；L＝vxt··························································（1分）2vy＝vsin37°；2gy＝vy·······························································（2分）解得：v＝5m/s········································································（1分）评分标准：第1问，4分；第2问，4分；第3问，4分。共12分。18．(16分)(1)滑块A在传送带上运动时，重力和摩擦力做功由动能定理可得12mgsin30°d－μ1mgdcos30°d＝mv0················································（2分）2物理答案第2页共3页{#{QQABBQKAggiAAhBAABgCEwXCCkKQkAGACKoGBEAMsAAAARFABCA=}#}16解得d＝m＝2.29m································································（1分）7(2)A在B板上滑动的过程中，A、B系统动量守恒由mv0＝(m+m)v1······································································（1分）得v1＝2m/s············································································（1分）由能量守恒定律1212μ2mgs＝mv0-2mv1·································································（1分）22解得s＝4m·············································································（1分）A给B的滑动摩擦力等于B受到的合外力由动能定理12μ2mgL＝mv1－0·····································································（1分）2L＝2m···················································································（1分）(3)滑块C与挡板碰撞前，A、C系统动量守恒，有mv1＝mvA+MvB···································································（1分）滑块C与挡板碰撞后到A、C达到共速的过程中，A、C系统动量守恒，有mvA+M(-vB)＝(m+M)v共···························································（1分）2解得A、C共速时v＝(vA－1)共3全过程A、C与弹簧组成的系统机械能守恒，A、C共速时弹簧弹性势能最大，则有121212Ep＝mv1－(m+M)v＝1－(vA－1)··········································（1分）22共3在C与挡板碰撞前，若从A开始压缩弹簧至弹簧再次恢复原长的过程中C与挡板未碰撞，121212根据机械能守恒有mv1＝mvA+MvB······································（1分）222又根据mv1＝mvA+MvB2解得vA＝－m/s32故C与挡板碰撞前A的速度vA介于－m/s~2m/s之间312根据Ep＝1－(vA－1)3vA＝1m/s时，Ep＝1J································································（1分）22vA＝－m/s时，Ep＝J····························································（1分）3272可知弹簧弹性势能最大值取值范围为J≤Ep≤1J··························（1分）27评分标准：第1问，3分；第2问，6分；第3问，7分。共16分。物理答案第3页共3页{#{QQABBQKAggiAAhBAABgCEwXCCkKQkAGACKoGBEAMsAAAARFABCA=}#}