2024届黑龙江省大庆市高三下学期第三次教学质量检测物理试题及答案

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.ABD6分）（每空2分）2）（或电阻箱的阻值）理想电流表（或电流表，或A）的读数仍为I3）0+R12.（8分）（前2个空每空1分，中间2个空每空2分，最后2个空每空1分）k2134.51.213）2.01.9-2.1均给分）1.0（0.9-1.1均给分）4）ꢀ13.（10解：（1）mgsinθ-μmgcos····························2分代入数据解得2m/s212前2s内的位移x=at2······························2分代入数据解得=4m·································2分分分π22）前2s内重力做的功为W=mgx-θ)=48J···················2－W重力在前2s内的平均功率为P==24W····················2t14.（13解：（1）根据动量守恒定律可知M+m)v=mv····································2分mv=v0·································2分2）根据能量守恒定律可知1212Q=mv2-(M+m)v2·······························2分0(Mm)M2mv2································2分0Q=3）设发射炮弹过程中，炮弹、轮式加榴炮的位移大小分别为s、s2显然有s-s=d··································1分121212F·s=mv-mv2································1分10{#{QQABDQaEogCAQJAAABgCQQVACgCQkACCCKoGBAAMMAABSBFABAA=}#}12-F·s=0-Mv2·································1分20ms2d··························2代入数据解得分(Mm)或：Q=·d······································2分12-F·s=0-Mv2···································1分20ms2d··························2代入数据解得分(Mm)15.(17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112qUmv202································2分26qUv···································2分m2）粒子在磁场中做圆周运动，洛伦兹力充当向心力，有v2rqvBm······························1分1分r轨道圆半径r3R·······························系，轨迹圆圆心到坐标原点的距离dd2r2R2·····································1分如图所示当r3Rd2R，所以轨迹圆的圆心在O············11分分3所以，该粒子入射速度方向与y轴正方向夹角为θsin·······2{#{QQABDQaEogCAQJAAABgCQQVACgCQkACCCKoGBAAMMAABSBFABAA=}#}则θ=°·······································1分3）当磁感应强度为3B时，粒子运动圆半径31rR··································2分3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1R·······································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x22(2R)2(2RR)2·····························1分222R·······························1分可知粒子