Question

如图所示，匀强电场方向竖直向上，匀强磁场方向水平且垂直纸面向里，有两个带电小球a和b，a恰能在垂直于磁场方向的竖直平面内做半径r=0.8 m的匀速圆周运动，b恰能以v=2 m/s的水平速度在垂直于磁场方向的竖直平面内向右做匀速直线运动．小球a、b质量m_a=10 g，m_b=40 g，电荷量q_a=1×10^－2C，q_b= 2×10^－2C，g=10m/s²。求：

（1）小球a和b分别带什么电？电场强度E与磁感应强度B的大小？

（2）小球a做匀速周周运动绕行方向是顺时针还是逆时针？速度大小v_a是多大？

（3）设小球b的运动轨迹与小球a的运动轨迹的最低点相切，当小球a运动到最低点即切点时小球b也同时运动到切点，a、b相碰后合为一体，设为c，在相碰结束的瞬间，c的加速度a_c=？

Answer

（1）小球a在互相垂直的匀强电场和匀强磁场中做匀速圆周运动，电场力和重力的合力为零，电场力方向向上，所以小球a带正电，且有

mag=qaE··································································①

E=10 N / C································································②

小球b做匀速直线运动，合力为零，带正电，且有

mbg=qbvB+qbE·····························································③

B=5T·····································································④

（2）小球a做匀速圆周运动绕行方向是逆时针方向。

由

得······························································⑤

va=4 m / s·································································⑥

（3）a、b相碰前速度方向相同，设碰后的共同速度为vc，则

ma va+mb v=(ma+mb) vc······················································⑦

vc=2.4 m / s································································⑧

由牛顿第二定律得

mcac=qcE+qc vcB – meg······················································⑨

mc=ma+mb

qc=(qa+qb)

解得ac=3.2 m / s2··························································⑩

评分标准：本题满分18分。其中判断a、b带电性质各1分，a绕行方向1分，①式2分，②式1分，③式2分，④式1分，⑤式2分，⑥式1分，⑦式2分，⑧式1分，⑨式2分，⑩式1分。

【解析】略