{#{QQABDQaEogCAQJAAABgCQQVACgCQkACCCKoGBAAMMAABSBFABAA=}#}{#{QQABDQaEogCAQJAAABgCQQVACgCQkACCCKoGBAAMMAABSBFABAA=}#}{#{QQABDQaEogCAQJAAABgCQQVACgCQkACCCKoGBAAMMAABSBFABAA=}#}{#{QQABDQaEogCAQJAAABgCQQVACgCQkACCCKoGBAAMMAABSBFABAA=}#}大庆市2024届高三年级第三次教学质量检测物理试题答案及评分标准1.B2.A3.D4.C5.B6.C7.D8.AD9.BC10.ABD11.（6分）（每空2分）（）（或电阻箱的阻值）理想电流表（或电流表，或）的读数仍为（）2R0AI3R0+R（分）（前个空每空分，中间个空每空分，最后个空每空分12.8212221）k（1）34.51.21（3）2.0（1.9-2.1均给分）1.0（0.9-1.1均给分）（4）2�13.（10分）解：（1）木块mgsinθ-μmgcosθ=ma····························2分代入数据解得a=2m/s21前2s内的位移x=at2······························2分2代入数据解得x=4m·································2分π（2）前2s内重力做的功为W=mgxcos(-θ)=48J···················2分2－W重力在前2s内的平均功率为P==24W····················2分t14.（13分）解：（1）根据动量守恒定律可知(M+m)v0=mv····································2分M+mv=v0·································2分m（2）根据能量守恒定律可知1212Q=mv-(M+m)v0·································2分22(Mm)M2Q=v································2分2m0（3）设发射炮弹过程中，炮弹、轮式加榴炮的位移大小分别为s1、s2显然有s1-s2=d··································1分1212F·s1=mv-mv0································1分22{#{QQABDQaEogCAQJAAABgCQQVACgCQkACCCKoGBAAMMAABSBFABAA=}#}12-F·s2=0-Mv0·································1分2m代入数据解得sd··························2分2(Mm)或：Q=F·d······································2分12-F·s2=0-Mv0···································1分2m代入数据解得sd··························2分2(Mm)15.(17分）解：（1）带电粒子进入电场中，在电场加速过程中，由动能定理有11qUmv2mv2································2分2206qU解得v···································2分m（2）粒子在磁场中做圆周运动，洛伦兹力充当向心力，有v2mvqvBm可得r······························1分rqB轨道圆半径r3R·······························1分粒子垂直上边界射出磁场，则粒子轨迹圆的圆心必然在上边界线上，且根据几何关系，轨迹圆圆心到坐标原点的距离d满足d2r2R2·····································1分如图所示当r3R时，d2R，所以轨迹圆的圆心在O＇点············1分3所以，该粒子入射速度方向与y轴正方向夹角为θ，则，sin·······1分2{#{QQABDQaEogCAQJAAABgCQQVACgCQkACCCKoGBAAMMAABSBFABAA=}#}则θ=60°·······································1分（3）当磁感应强度为3B时，粒子运动圆半径3rrR··································2分13如图所示①粒子轨迹圆与上边界相切时，粒子运动到最左边，有xR1·······································2分②当粒子轨迹圆与上边界的交点、运动轨迹圆圆心以及坐标原点O点三点共线时，粒子打在最右边，此时有x2(2R)2(2RR)22·····························1分解得分x2221R·······························1可知粒子能从磁场上边界射出粒子的边界宽度为Lxx2211R分12··························1{#{QQABDQaEogCAQJAAABgCQQVACgCQkACCCKoGBAAMMAABSBFABAA=}#}