2024届四川省攀枝花市高三上学期第一次统一考试理科综合试卷高三物理答案

攀枝花市高2024届高三第一次统一考试物理理综(物理)参考答案及评分标准二、选择题(每小题全对6分，对而不全3分，有错或不答的0分，满分48分)14．B15．B16．C17．C18．D19．AB20．AC21．BD22．（5分）（1）入射小球的质量大于被碰小球的质量(2分)（2）碰前入射小球(1分)mmm112OBOAOC=+（2分）23．（10分）（1）BE（2分）（2）1.0（2分）xnn7654321nhahx22+（3）−（2分）012345678xgx（4）如图（2分）0.27（0.26-0.28）（2分）24．（12分）解：（1）设圆筒在下端刚落地时的速度为v0，对圆筒和小球，由自由落体运动规律有：2v0=2gh··············································································（2分）得：v0=3m/s······································································（2分）（2）对圆筒落地后，设地面对其的支持力为N，由平衡关系有：N=Mg+f········································································································（2分）由题意可知：f=3mg得：N=50N······································································（2分）（3）当小球速度刚减为零时刚好与地面接触，此时圆筒的长度最小。对小球，由动能定理，有：1−−=−()0fmgLmv2····························································（2分）20得：L=0.225m····································································（2分）25．（15分）解：（1）Q球在C点时，由牛顿第二定律及圆周运动的规律有：mv2N+=mgQC···································································（2分）QR高三物理答案共3页，第1页{#{QQABKYSUogAoAAAAAAgCUwHACgAQkBECCIoOAEAMIAABQBFABAA=}#}由牛顿第三定律及题上条件，可知Nm=gQ·············································································（1分）Q球从C点飞出后做平抛运动，由平抛运动的规律有12Rg=t2，xt=v·······························································（2分）2C联立以上格式，得：xR=22·························································（1分）（2）P球从出发到B点过程，由动能定理有：11−=−mgRmm4vv22·················································（1分）PPP022P球与Q球发生碰撞，设碰后P、Q的速度分别为vvPQ、，由动量守恒定律有：mmmPPPQQvvv=+································································（1分）P球从B点返回出发位置的过程，由动能定理有：1−=−mgRm40v2························································（1分）PPP2得：vP=2gRQ球从B点运动C点的过程，设Q球到C点时的速度大小为vC，由机械能守恒定律有：11mgRmm=−2vv22···················································（1分）QQQPC22m3联立以上各式并代入数据得：P=或3······································（1分）mQ3P、Q碰撞前后的总动能之差应不增加，有111=−=−+EEEmmmvvv222()································（1分）KkkpPPQQ222mP3若=得EKQ=(3−3)mgR0···············································（1分）mQ3mP若=3得=−EmgRKQ(333)0··············································（1分）mQm故P、Q两球的质量之比P=3····················································（1分）mQ26．（20分）解：（1）设木板A和小滑块C一起运动时的加速度为a，对木板A与滑块C，由牛顿第二定律，有F−2()()mA+mCg=mA+mCa················································①对滑块C，有1CCmgma=········································································②联立两式，得：F=9N·······························································③（2）从木板A和小滑块C一起运动到刚与木板B相撞的过程中A对C施加了支持力和摩22擦力，设该段时间为t1，则A对C的冲量I=()()mCg+1mCgt1······④高三物理答案共3页，第2页{#{QQABKYSUogAoAAAAAAgCUwHACgAQkBECCIoOAEAMIAABQBFABAA=}#}1da=t2·············································································⑤21得：I=26Ns·········································································⑥（3）设A刚与B相撞时的速度为v0，A、B撞后的速度分别为vA1、vB1，由运动学规律、动量守恒定律和能量守恒定律有v01=at···············································································⑦mmmA0AA1BB1vvv=+····························································⑧111mv2=+mv2mv2··················································⑨2A02AA12BB1A、B碰后，A、C将出现相对滑动，此时对A进行受力分析，地面对A的摩擦力、拉2力、C对A的摩擦力恰有Fmgmmg+=+=()3N·················⑩91C2AC故A将做匀速直线运动，C做匀减速直线运动。假设C未滑离A，A、C过段时间后将会共速，设从A、B碰撞到A、C共速这段时间为t2，这段时间内A、C间的相对位移为x，根据牛顿运动定律，有vvA102=−at········································································○111xtat=−v2···································································○12C0222xtAA12=v············································································○13=−xxxCA·········································································○14得：==xL0.5625m，即A、C共速时C恰好滑到A的最右端。·······○15对B，设B碰后到停止的位移为xB，1mgxm=v2·······························································○163BBBB12得：xxBA=0.375m=····································································○17故A、B会第二次碰撞，由动量守恒定律，有mmmAA1AA2BB2vvv=+··························································○18111mmmvvv222=+················································○19222AA1AA2BB2得：vA2=0.125m/s，vB2=0.625m/svC=vA2=0.5m/s····················○20故木块C不能滑到木板B上，会从木板A上掉落到水平地面上。·········○21高三物理答案共3页，第3页{#{QQABKYSUogAoAAAAAAgCUwHACgAQkBECCIoOAEAMIAABQBFABAA=}#}