【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
转载请保留链接:https://www.doc5u.com/view-7c134e3affe2b9c1044141568b4cafd4.html
以下为本文档部分文字说明:
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)1m;(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)