233d=v0t 1
d= 12at2 1
According to Newton's second law, = horse.
Solve with the above three formulas: u = 3mv202q, t 1 = 23d3v0.
(2) Let the velocity of a particle passing through A be V, and the angle between the direction and the X axis be α. According to the kinetic energy theorem:
qU= 12mv2？ 12mv20
cosα=vv0
Solution: v=2v0, α = π 3.
Let the particle pass through the y axis for the first time at point d, and the trajectory according to the meaning of the question is shown in the figure. According to the geometric relationship, the heights of point D and point A are equal, △C 1DO is an equilateral triangle, R=d,
According to Newton's second law, qvB=mv2R.
Finishing: b = 2mv2qd
(3) The trajectory of the particles on the right side of the Y axis is as shown in the figure, which is obtained from the geometric relationship: de = 2cosθ = d.
That is, the ordinate of point E is: yE=2d.
(4) As can be seen from the figure, the time from a to d is T2 = 13t.
Time from D to E: T3 = 56t.
And: t = 2π mqb
So: t = t 1+t2+t3 = (43+7π) d6v0.
Answer: (1) The voltage between plates M and N is 3mv202q; (2) The magnetic induction intensity of uniform magnetic field is 2mv2qd；; (3) The ordinate value of the particle passing through the Y axis for the second time is 2D; (4) The time from the particle entering the plate to the second passing through the Y axis is (43+7 π) d6v0.