【文档说明】广东省茂名市电白区2022-2023学年高二上学期期末考试物理试卷(图片版) 含答案.doc,共(11)页,2.233 MB,由管理员店铺上传
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2022~2023学年度第一学期期末考试高二物理参考答案1.B2.B3.D4.B5.C6.D7.D8.AC9.BD10.BD11.(10分)C44.79(44.76~44.82范围内均可)112mONmOMmOP=+水平位移小球下落高度相同,做平
抛运动的时间相同,碰撞前后的速度正比于水平位移12.(6分)2010PURUka()01Rk-13.(9分)(1)5m;(2)6.5N·s;(3)9J(1)小物块由A到B过程做匀减速运动,由动能定理可得22101122mgxmvmv−=−........
...................................................................................(2分)解得x=5m..................................................
.........................................(1分)(2)选初速度方向为正方向,由动量定理得I=-mv2-mv1..................................
.........................................................(2分)解得I=-6.5N·s,即冲量大小为6.5N·s.................
.............................................(1分)(3)物块反向运动过程中,由动能定理2212Wmv=−........................................................................
...................(2分)得W=-9J,即克服摩擦力所做的功为9J....................................................(1分)14.(15分)(1)10m/s;(2)90N;(3)15kgm/s(1)设滑块从C
点飞出时的速度为Cv,从C点运动到D点时间为t,滑块从C点飞出后,做平抛运动,竖直方向2122Rgt=.........................................................................
..................(1分)水平方向1csvt=.........................................................................................
..(1分)解得10m/scv=...........................................................................................(1分
)(2)设滑块通过B点时的速度为Bv,两滑块从B到C的过程中,根据机械能守恒定律()()()2211222BCMmvMmvMmgR+=+++..........................................................(2分)解得
102m/sBv=...........................................................................................(1分)设在B点滑块受轨道的运动力为
N,根据牛顿第二定律()()2BvNMmgMmR−+=+................................................................................
...........(1分)解得90NN=..........................................................................................
.(1分)(3)设滑块从A点开始运动时的速度为Av,刚运动到B点时的速度为Av,两滑块碰撞过程中根据动量守恒()ABmvmMv=+.............................................................
..............................(1分)解得202m/sAv=...........................................................
................................(1分)滑块从A到B的过程中,根据动能定理2221122AAmgsmvmv−=−..........................
...................................................(2分)解得30m/sAv=...................................................................................
........(1分)设滑块在A点受到的冲量大小为I,根据动量定理AImv=..........................................................................
.................(1分)解得15kgm/sI=...........................................................................................(1分)15.(14分)(1)21m/s;(2)1
m;(3)2.32m(1)A和B一起匀加速运动,由牛顿第二定律2qEmgma−=.................................................................
..........................(2分)代入数据解得21m/sa=..............................................................
.............................(1分)(2)当B木板与挡板P相碰时速度22vad=.......................................................................................
....(1分)木板B与挡板P撞后停止,物块A继续滑动加速,由牛顿第二定律11mgqEma−=..........................................................................
.................(2分)代入数据解得212m/sa=...........................................................................................(1分)当A恰好滑到挡板P处停
止时有2102vaL=...........................................................................................(1分)距离关系为00dLS+=..........................
.................................................................(1分)代入得01mL=....................................................................
.......................(1分)即木板最小长度为1m(3)因为L小于L0,故物块与挡板碰撞,然后原速返回,与木板B共同反向匀减速运动,直到速度为零,再共同加速向右滑动,不断往复,最终A、B都最终停在挡板P处。物块A和木板B间产生的热
量11QmgL=...........................................................................................(1分)木板与
水平面间产生热量22QmgS=...........................................................................................(1分)整个
过程由能量守恒定律012qESQQ=+...........................................................................................(
1分)代入数据得2.32mS=...........................................................................................(1分)