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2013绵阳一诊物理试题及答案


绵阳市高 2013 级第一次诊断性考试 理科综合能力测试 物理部分参考答案及评分标准
第Ⅰ卷(选择题,共 42 分)
在每题给出的四个选项中,只有一个选项是最符合题目要求的。共 7 题,每题 6 分。 1.C 2.D 3.A 4.B 5.B 6.AC 7.BD

第Ⅱ卷(选择题,共
8.(16 分) (1)①AC(3 分) ;②A(3 分) 。 (2)①CD(2 分) ;②1.56(3 分) ;能(2 分); ③9.70~9.90 都给分(3 分) 。 9. (15 分) (1)设“嫦娥一号”的质量是 m1,则 Mm1 4? 2 ??????(4 分) G ? m1 ( R ? H ) 2 2 T ?R ? H ?

分)

4? 2 ( R ? H )3 ??????(2 分) GT 2 (2)设月球表面的重力加速度为 g,则 M?
G Mm ? mg R2

??????(3 分) ···································· (4 分) ··········· ·········· ··········· ···· ·········· ··········· ··········· ····

F ? mg ? m

?2
r

解得 F ?

4? 2 ( R ? H )3 m ?2 ··························· 分) ·························· (2 ·········· ··········· ····· ?m R 2T 2 r

10. (17 分) (1)设小滑块上滑的加速度大小是 a1,则 ma1= mgsinθ+μmgcosθ ·······························(2 分) ··········· ·········· ·········· ·········· ··········· ········· v0=a1t ·········································(2 分) ··········· ·········· ··········· ········· ·········· ··········· ··········· ········ 解得 a1=7.2 m/s2,t =2.5 s ······························ (2 分) ··········· ·········· ········· ·········· ··········· ········· (2)小滑块从斜面底端上滑到最高处的过程中,设沿斜面上滑的距离是 x,根据动能定 理有 0-

1 2 ··························· (2 ·········· ··········· ······ m?0 =-mgh-μmgxcosθ ···························· 分) 2

h= xsinθ ········································ 分) ······································· (2 ·········· ··········· ··········· ······· 解得 h=13.5 m ···································· (2 分) ··········· ·········· ··········· ···· ·········· ··········· ··········· ···· 另解:2a1x= ? 0
2

···································(2 分) ··········· ·········· ··········· ··· ·········· ··········· ··········· ··

h= xsinθ ········································ 分) ······································· (2 ·········· ··········· ··········· ······· 解得 h=13.5m ····································(2 分) ··········· ·········· ··········· ···· ·········· ··········· ··········· ··· (3)根据功能关系有△E=2μmgxcosθ ·······················(3 分) ··········· ·········· ·· ·········· ··········· ·

解得△E = 10.8 J ···································(2 分) ··········· ·········· ··········· ··· ·········· ··········· ··········· ·· 11.(20 分) 解: (1)设小滑块能够运动到 C 点,在 C 点的速度至少为 vc,则
mg ? m

? c2 ······································ (2 分) ··········· ·········· ··········· ······ ·········· ··········· ··········· ······ R
·························· (2 分) ··········· ·········· ····· ·········· ··········· ·····

1 1 2 m?c2 ? m?0 ? ?2mgR ? ?mgL 2 2

解得 v0= 2 15 m/s

··································(1 分) ··········· ·········· ··········· ·· ·········· ··········· ··········· ·

(2)设传送带运动的速度为 v1,小滑块在传送带上滑动时加速度是 a,滑动时间是 t1,滑 动过程中通过的距离是 x,则 v1=rω ········································ 分) ······································· (1 ·········· ··········· ··········· ······· ma=μmg ·······································(1 分) ··········· ·········· ··········· ······· ·········· ··········· ··········· ······ v1=at1 ········································ (1 分) ··········· ·········· ··········· ········ ·········· ··········· ··········· ········
x? 1 2 ······································· (1 分) ·········· ··········· ··········· ······· at 1 ······································· 2

解得 v1=2m/s,a=4m/s2,t1=0.5s,x=0.5m 由于 x<L,所以小滑块还将在传送带上与传送带相对静止地向 B 点运动,设运动时间为 t2,则 L-x= v1t2 ······································(2 分) ··········· ·········· ··········· ······ ·········· ··········· ··········· ····· 解得 t2=2.25s 则 t= t1+t2=2.75s ·································· (2 分) ··········· ·········· ··········· ·· ·········· ··········· ··········· ·· (3)轮子转动的角速度越大,即传送带运动的速度越大,小滑块在传送带上加速的时间 越长,达到 B 点的速度越大,到 C 点时对圆轨道的压力就越大。 小滑块在传送带上一直加速,达到 B 点的速度最大,设为 vBm,对应到达 C 点时的速度 为 vcm,圆轨道对小滑块的作用力为 F,则
2 ? Bm ? 2aL

······································(2 分) ··········· ·········· ··········· ······ ·········· ··········· ··········· ····· ······························ 分) ····························· (2 ·········· ··········· ········

1 1 2 2 m?cm ? m? Bm ? ?2mgR 2 2

mg ? F ? m

Fm=F 解得 Fm=5N

··································· (1 分) ··········· ·········· ··········· ··· ·········· ··········· ··········· ··· R ········································(1 分) ··········· ·········· ··········· ········ ·········· ··········· ··········· ······· ·····································(1 分) ··········· ·········· ··········· ····· ·········· ··········· ··········· ····

2 ?cm


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