# 2013绵阳一诊物理试题及答案

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 ?

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

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) ··········· ·········· ·· ·········· ··········· ·

mg ? m

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

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

··································(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

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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