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2009届高考理科综合物理部分押题卷(二)

查询理科综合的详细结果

 

在Q点平行于OM方向的分速度:

vy=at···························································································· (2分)

SQ方向做匀速运动:=v0t ···················································· (2分)

且v0=vytanθ·················································································· (3分)

解得:=cosθ ······································································ (2分)

显然P点为OS的中点,故P离O点的距离S==cosθ·········· (1分)

(2)如图所示·············································································· (5分)

25.解:(1)因为m<M<2m,所以开始时弹簧处于伸长状态,其伸长量x1,则

(2m-M)g=kx1····················································································································· (2分)

得x1=g······················································································································ (1分)

若将B与C间的轻线剪断,A将下降B将上升,当它们的加速度为零时A的速度最大,此时弹簧处于压缩状态,其压缩量x2,则

(M-m)g=kx2······················································································································· (2分)

得x2=g······················································································································· (1分)

所以,A下降的距离为x=x1+x2=时速度最大··································································· (2分)

(2)A、B、C三物块组成的系统机械能守恒,设B与C、C与地面的距离均为L,A上升L时,A的速度达到最大,设为v,则

2mgL-MgL=(M+2m)v2 ··································································································· (4分)

当C着地后,A、B两物块系统机械能守恒。

若B恰能与C相碰,即B物块再下降L时速度为零,此时A物块速度也为零,则

MgL-mgL=(M+m)v2········································································································ (4分)

解得:M=m···················································································································· (3分)

由题意可知,当M>m时,B物块将不会与C相碰。··························································· (1分)

 


 

备用题

 

O

x

y

v0

d

如图所示,在第四象限,存在竖直向下的匀强电场E,一带负电小球(电量为q,质量为m),从y=d的P点以沿x轴正方向的速度v0开始运动。已知Eq=2mg ,则小球运动轨迹与x轴的各个交点中,任意两相邻交点之间的距离L为(  )

A.2v0

B.v0

C.2v0

D.2d

 

 

 

1932年,著名的英国物理学家狄拉克,从理论上预言磁单极子是可以独立存在的。他认为:“既然电有基本电荷——电子存在,磁也应该有基本磁荷——磁单极子存在。”假设在真空玻璃盒内有一固定于地面上空附近的S极磁单极子,其磁场分布与负电荷电场分布相似,周围磁感线呈均匀辐射式分布,如图所示。一质量为m,电荷量为q的点电荷P在磁单极子的正上方h高处的水平面内做半径为r的匀速圆周运动。对该电荷的分析正确的是(已知地球表面的重力加速度g)

S

 

P

A.该电荷一定带负电荷

B.该电荷一定带正电荷

C.可确定电荷运动的速率v

D.可确定电荷运动的圆周上各处的磁感应强度B

 

 

 

S

 

P

θ

Bqv

mg

【答案】CD

【解析】如图,电荷做匀速圆周运动,受到重力和洛伦兹力的合力始终指向圆心,若电荷带正电,根据左手定则可知电荷在P处的速度方向垂直纸面向里,同理若电荷带负电,电荷在P处的速度方向垂直纸面向外,故A、B均错;由于圆半径r及圆心到磁单极子的高度h已知,故θ可确定,由水平方向的牛顿第二定律、向心力公式可确定粒子的向心加速度a=gtanθ=m ,因此可确定电荷运动的速率,C对;由竖直方向的平衡条件Bqvsinθ=mg,因此可确定电荷运动的圆周上各处的磁感应强度B ,D对。

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