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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．B 2．C 3．A 4．D 5．D 6．C 7．B 8．C
二、多项选择题：本题共4小题，每小题4分，共16分。
9．AD 10．BC 11．AD 12．ABC
三、非选择题：本题共6小题，共60分。
Ld(t －t )
1 2
13．（6分） （1）12.9（2分）； （2）0.4（2分）； （3） （2分）。
ght t
12
14．（8分） （1）78.9（2分）； （2）37.6（2分）；
1500 1.5
（3）I＝ （2分）（写成I＝ 得1分）；
157.6＋0.38t 157.6＋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＝p V··································（2分）
0
3
PV'
PV 5
（2）对气囊内所有的气体由理想气体状态方程得 ＝ ···························（2分）
T T
3
1
从气囊内排出气体的体积为 ΔV＝V'－V···············································（1分）
3
Δm ΔV 2
排出气体质量与排气前气体总质量的之比为 ＝ ＝ ·························· （2分）
m V' 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＝v t＝0.3m·····················································（2分）
0
（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）
2
P＝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）
m
qEcosθt
垂直分界线方向vsinα－ ＝0··················································· （1分）
m 2
3
解得tanα＝ ············································································（1分）
2
（2）如图，设粒子沿边界线方向的位移为 d
1
qEsinθ
0-(v cosα)2＝-2 d ·······························（1分）
0 1
m
m(v sinα)2
由qvB sinα＝ 0 ································（1分）
1
r
要垂直穿过x轴，那么需要做圆周运动的半径r ＝d
1 1
3mE
解得B ＝ ···············································································（1分）
1
4qr
（3）设粒子圆周运动的半径R ，粒子再次经过分界线时到O点距离为d ，
1 2
根据几何关系(d -R )sinθ＝R ······························································· （1分）
1 1 1
8mv 2
由图可知d ＝2R ＝ 0 ·····································································（2分）
2 1
21qE
（4）第三次穿过分界线的粒子速度依然为v
0
mv
所以在B 中的半径为R ＝ 0······························································（1分）
2 2
qB
2
要想轨迹中心在y轴上，根据几何关系d cosθ＝R cosβ····························· （1分）
2 2
π
cosβ＝cos( -θ-α)
2
(1+sinθ)E
可得 (tanα+tanθ) ····································································（1分）
v cosα
0
5 21E
代入数据B ＝ ，方向垂直纸面向里···············································（2分）
2
8v
0
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）
1
（1）根据机械能守恒，小球摆到最低点时的速度为v
0
满足 mv
0
2＝mgH
2
解得v ＝ 2gH＝10 2m/s···································································· （1分）
0
A碰B时，根据动量守恒mv ＝m v ·······················································（1分）
0 0 1
v m
恢复系数为e＝ 1 解得e＝ ·····························································（1分）
v m
0 0
B碰A时根据动量守恒为m v ＝m v '+mv·······························································（1分）
0 0 0 0
v－v '
恢复系数e'＝ 0 ·············································································（1分）
v
0
联立得v＝v ＝10 2m/s·······································································（1分）
0
（2）设共速时速度为v ，共速的位置距离小车左端距离为，
共
根据动量守恒mv ＝(M+m)v ·································································（1分）
0 共
m 1
解得v ＝ v ＝ v ＝5 2m/s····························································· （1分）
共 M+m 0 2 0
根据能量守恒可得μmgL+F
1
L+F
2
(x-L)+μmg(x-L)＝ΔE
k0
································（1分）
1 1 m MmgH
其中ΔE ＝ mv 2- v 2＝ ＝100J··············································（1分）
k0 2 0 2M+m 共 M+m
5MH 1 253
可得x＝ + L＞L 代入数据可得x＝ ≈4.22m···························（1分）
6(M+m) 2 60
（3）当m再次返回到小车左端的时候，具有的可损失动能为
ΔE
k1
＝ΔE
k0
-μmg(x+x)·········································································（1分）
MmgH
得ΔE
k1
＝ -0.4mgL
3(M+m)
MmgH 1
将 代换回ΔE
k0
得ΔE
k1
＝ ΔE
k0
-0.4mgL
M+m 3
1
其中0.4mgL＝0.8J，所以ΔE
k1
＝ ΔE
k0
-0.8
3
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
1
ΔE
k(n+1)
＝ ΔE
kn
-0.8·················································································（1分）
3
以下列举每次数量关系
1
n＝1时，ΔE
k1
＝ ΔE
k0
-0.8
3
高三物理答案 第3页（共5页）
{#{QQABJYqEogggQBBAAQgCEwFqCgCQkBECCAoORBAMsAAACAFABCA=}#}1 1
n＝2时，ΔE
k2
＝ ΔE
k0
-0.8× -0.8
32 3
1 1 1
n＝3时，ΔE
k3
＝ ΔE
k0
-0.8× -0.8× -0.8
33 32 3
……
1 1 1
n＝n时，ΔE
kn
＝ ΔE
k0
-0.8×( +… +1)
3n 3n-1 3
1 1.2
结合数列求和得ΔE
kn
＝ ΔE
k0
-(1.2- )····················································（1分）
3n 3n
101.2
由ΔE ＝100J，整理得ΔE ＝ -1.2
k0 kn
3n
101.2 4
可知当n＝4时，ΔE ＝ -1.2＝ J
k4
34 81
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x
4
位置处
根据功能关系ΔE ＝(F +f)x ·································································（1分）
k4 1 4
1
带入数据得：x ＝ m≈4mm
4
243
因为x ≈4mm＜0.1m＝L，所以假设正确，
4
综上
1
m物块一共跟左侧挡板碰撞4次，最后停在距离左端x ＝ m≈4mm的位置·（1分）
4
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=}#}
