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=}#}