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物 理
一、选择题：
题号 1 2 3 4 5 6 7 8 9 10
答案 C C B D B A B BD BC AD
二、非选择题：本题共5小题，共60分。
11.（1）BC （2分）； （2） 1.73 （2分）；（3） 不能 （2分）
12.（1）最大（1分） ， 50.0 （1分） ；（2）丙（2分）
R R b R R
（3）E A 0 （2分） ， r A 0 （2分）
kR k R R
0 A 0
13.（1）图乙到图丙的过程为等温变化，以左侧气体为研究对象，设其图丙中的压强为p ，
1
L L
根据玻意耳定律有 p S =p S ①·································2分
0 1
2 4
L 3L
同理对右侧气体 p S =p S ②·································2分
0 2
2 4
联立①②解得 p :p =3:1 ··································1分
1 2
（2）以活塞为研究对象，图乙中
L
Mg=k ③·································2分
2
L
图丙中 p S=p S+k ④································2分
1 2
4
8p S
联立①②③④解得 m= 0 ··································1分
3g
14．（17分）
（1）由x-t图像可知，以竖直向上为正方向；碰前，甲球的最大高度为h =5m，设起抛速度为v ，根
0 0
据匀变速直线运动规律有 2gh =v 2 ①·································1分
0 0
两球碰撞的时刻t =1.5s,设高度为h,碰前甲球速度为v ，乙球为v
1 1 1 2
v =v-gt ②·································1分
1 0 1
1
h=v t - gt 2 ③···························1分
1 01 1
2
2gh =v 2-v 2 ④···························1分
1 0 2
解得 v =-5m/s 方向竖直向下 ·································2分
1
v =5m/s 方向竖直向上 ·································2分
2
（2）碰后，甲球的最大高度为h =8.75m，设碰后的速度为v '，位移为Δh，则
2 1
Δh=h -h ⑤································1分
2 1
v '2=2gΔh ⑥································1分
1
联立⑤⑥解得 v '=10m/s ·································1分
1
两球弹性正碰，由动量守恒定律和能量守恒定律有
m v +m v =m v '+m v ' ⑦································2分
1 1 2 2 1 1 2 2
1 1 1 1
m v 2+ m v 2= m v '2+ m v '2 ⑧······························2分
1 1 2 2 1 1 2 2
2 2 2 2
m 1
联立⑦⑧解得 1 = ···································2分
m 3
215．（19分）
（1）设线框刚进入磁场时速度大小为v
1
E=Blv ①·······································1分
1
E
I= ②·······································1分
R
F =IlB ③·······································1分
安
线框匀速通过磁场区域Ⅰ的过程，以重物为研究对象
T=Mg ④······································1分
以线框为研究对象 F +mg=T ⑤······································1分
安
联立①②③④⑤解得 v =2.5m/s ·······································1分
1
（2）由线框匀速通过区域Ⅰ，可知 d =l ···········································1分
1
方法一：根据能量守恒定律 Q =(M-m)g(d +l) ···································2分
1 1
解得 Q =0.2J ·········································1分
1
2l
方法二：匀速通过区域Ⅰ的时间 t = ·············································1分
1
v
1
Q =I2Rt ··········································2分
1 1
解得 Q =0.2J ·········································1分
1
（3）从上边框进入区域Ⅱ到下边框离开，设时间为t，克服安培力做功W ，安培力的冲量为I
克 安。
根据动能定理 (M-m)g(d +l)-W =0 ⑥·······························1分
2 克
W =P t ⑦··········································1分
克
根据动量定理 (M-m)gt-I =0 ⑧····································2分
安
BΔS B2lΔS
其中 I =BlI ∆t=Bl( )∆t= ⑨····················1分
安
RΔt R
线框在区域Ⅰ、Ⅱ间的无场区做加速运动，则进入区域Ⅱ时即开始减速运动，若要上边框进入时和下
边框离开时线框的速度相同，则下边框出区域Ⅱ前一定有一段加速运动过程，所以d ≠l。
2
ⅰ若d ＜l ∆S=2ld ⑩·······································1分
2 2
联立⑥⑦⑧⑨⑩解得 d =0.1m ········································1分
2
ⅱ若d ＞l ∆S=2l2 ⑪·········································1分
2
联立⑥⑦⑧⑨⑪解得 d =0.4m ········································1分
2
综上，区域Ⅱ的宽度d 为0.1m或0.4m
2
