2024年高三年级第一次适应性检测物理答案及评分标准一、单项选择题:本题共8小题,每小题3分,共24分。1.B2.C3.A4.D5.D6.C7.B8.C二、多项选择题:本题共4小题,每小题4分,共16分。9.AD10.BC11.AD12.ABC三、非选择题:本题共6小题,共60分。Ld(t1-t2)13.(6分)(1)12.9(2分);(2)0.4(2分);(3)(2分)。ght1t214.(8分)(1)78.9(2分);(2)37.6(2分);15001.5(3)I=(2分)(写成I=得1分);157.6+0.38t157.6+0.38t(4)不变(2分)。15.(7分)(1)充气过程中气囊克服外界大气压强所做的功W=p0V··································(2分)3PV'PV(2)对气囊内所有的气体由理想气体状态方程得=5···························(2分)TT31从气囊内排出气体的体积为ΔV=V'-V···············································(1分)3ΔmΔV2排出气体质量与排气前气体总质量的之比为==··························(2分)mV'5评分标准:第1问,2分;第2问,5分。共7分。16.(9分)(1)设出水口处水的速度为v,由题意可得ΔV=svt········································(1分)解得v=1m/s···················································································(1分)1平抛过程:竖直方向h=gt2································································(1分)2水平方向x=v0t=0.3m·····················································(2分)(2)取水质量m=ρΔV=0.8kg································································(1分)1根据能量守恒定律mv2+mgh=ηPt····················································(2分)2P=1.5W····················································································(1分)评分标准:第1问,5分;第2问,4分。共9分。高三物理答案第1页(共5页){#{QQABJYqEogggQBBAAQgCEwFqCgCQkBECCAoORBAMsAAACAFABCA=}#}17.(14分)(1)由题意可知粒子在电场区域做类斜抛运动轨迹如图qEsinθ沿分界线方向vcosα-t=0···················································(1分)mqEcosθt垂直分界线方向vsinα-=0···················································(1分)m23解得tanα=············································································(1分)2(2)如图,设粒子沿边界线方向的位移为d12qEsinθ0-(v0cosα)=-2d1·······························(1分)m2m(v0sinα)由qvB1sinα=································(1分)r要垂直穿过x轴,那么需要做圆周运动的半径r1=d13mE解得B1=···············································································(1分)4qr(3)设粒子圆周运动的半径R1,粒子再次经过分界线时到O点距离为d2,根据几何关系(d1-R1)sinθ=R1·······························································(1分)28mv0由图可知d2=2R1=·····································································(2分)21qE(4)第三次穿过分界线的粒子速度依然为v0mv0所以在B2中的半径为R2=······························································(1分)qB2要想轨迹中心在y轴上,根据几何关系d2cosθ=R2cosβ·····························(1分)πcosβ=cos(-θ-α)2(1+sinθ)E可得(tanα+tanθ)····································································(1分)v0cosα521E代入数据B2=,方向垂直纸面向里···············································(2分)8v0评分标准:第1问,3分;第2问,3分;第3问,3分;第4问,5分。共14分。高三物理答案第2页(共5页){#{QQABJYqEogggQBBAAQgCEwFqCgCQkBECCAoORBAMsAAACAFABCA=}#}18.(16分)12(1)根据机械能守恒,小球摆到最低点时的速度为v0满足mv0=mgH2解得v0=2gH=102m/s····································································(1分)A碰B时,根据动量守恒mv0=m0v1·······················································(1分)v1m恢复系数为e=解得e=·····························································(1分)v0m0B碰A时根据动量守恒为m0v0=m0v0'+mv·······························································(1分)v-v0'恢复系数e'=·············································································(1分)v0联立得v=v0=102m/s·······································································(1分)(2)设共速时速度为v共,共速的位置距离小车左端距离为,根据动量守恒mv0=(M+m)v共·································································(1分)m1解得v共=v0=v0=52m/s·····························································(1分)M+m2根据能量守恒可得μmgL+F1L+F2(x-L)+μmg(x-L)=ΔEk0································(1分)121m2MmgH其中ΔEk0=mv0-v共==100J··············································(1分)22M+mM+m5MH1253可得x=+L>L代入数据可得x=≈4.22m···························(1分)6(M+m)260(3)当m再次返回到小车左端的时候,具有的可损失动能为ΔEk1=ΔEk0-μmg(x+x)·········································································(1分)MmgH得ΔEk1=-0.4mgL3(M+m)MmgH1将代换回ΔEk0得ΔEk1=ΔEk0-0.4mgLM+m31其中0.4mgL=0.8J,所以ΔEk1=ΔEk0-0.83因为每次从左侧挡板出发都会经过类似的过程,所以递推关系为1ΔEk(n+1)=ΔEkn-0.8·················································································(1分)3以下列举每次数量关系1n=1时,ΔEk1=ΔEk0-0.83高三物理答案第3页(共5页){#{QQABJYqEogggQBBAAQgCEwFqCgCQkBECCAoORBAMsAAACAFABCA=}#}11n=2时,ΔEk2=ΔEk0-0.8×-0.8323111n=3时,ΔEk3=ΔEk0-0.8×-0.8×-0.833323……111n=n时,ΔEkn=ΔEk0-0.8×(+…+1)3n3n-1311.2结合数列求和得ΔEkn=ΔEk0-(1.2-)····················································(1分)3n3n101.2由ΔEk0=100J,整理得ΔEkn=-1.23n101.24可知当n=4时,ΔEk4=-1.2=J3481假设此后m相对小车向右运动过程中会停在小车上的L段,设停在x4位置处根据功能关系ΔEk4=(F1+f)x4·································································(1分)1带入数据得:x4=m≈4mm243因为x4≈4mm<0.1m=L,所以假设正确,综上1m物块一共跟左侧挡板碰撞4次,最后停在距离左端x4=m≈4mm的位置·(1分)243评分标准:第1问,6分;第2问,5分;第3问,5分。共16分。高三物理答案第4页(共5页){#{QQABJYqEogggQBBAAQgCEwFqCgCQkBECCAoORBAMsAAACAFABCA=}#}
2024届山东省青岛市高三一模物理答案
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