(1) Let the mass of ball A be m, the charge quantity be q, and the velocities of ball A and ball B before and after collision are V0 and V respectively.

A accelerates uniformly in an electric field. According to the kinetic energy theorem, QEL = 12mv20? ①

Momentum conservation in the collision between A and B: mv0=2mv②

From ① ②: V = QEL2M ③

(2) According to the meaning of the question, two balls do uniform circular motion in the magnetic field after collision.

The coordinate of the center o' is (l, l), as shown in the figure.

So the orbital radius: r = l 4.

According to Newton's second law: qBv=2mv2R⑤ (

From ③ ④ ⑤: k = QM = 2eb2l6.

(3) After passing through the O-point, the initial velocity V of the two balls along the positive direction of the Y-axis makes quasi-flat throwing motion in the electric field.

Let them reach the position Q' on the dotted line MN(x, y) again, and the movement time in the electric field is t 。

The law of motion is: x = 12qe2mt2? ⑦

y=vt？ ⑧

And: yx = tan 45 ⑨

From ⑧ ⑨: x = y = 2l? ⑩

Then, where they reach the dotted line MN is (2l, 2l)?

Answer: (1) The speed at which the two balls A and B collide into the magnetic field is: V = QEL2M.

(2) The specific charge k (i.e. the charge-mass ratio) of the ball A is 2EB2l. ..

(3) After passing through the O point, the two balls stuck together reach the position coordinates (2L, 2l) on the dotted line MN again.