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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
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）
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
【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。 上一页 [1] [2] [3]
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