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

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=？

解：（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

m_ag=q_aE··································································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

m_bg=q_bvB+q_bE·····························································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

v_a=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v_c，则

m_a v_a+m_b v=(m_a+m_b) v_c······················································⑦

v_c=2.4 m / s································································⑧

由牛顿第二定律得

m_ca_c=q_cE+q_c v_cB – m_eg······················································⑨

m_c=m_a+m_b

q_c=(q_a+q_b)

解得a_c=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。