(2) Let the spring force do work as w, mgh+W= 12mv2? 12mv02
Data brought in: W=-2.05J, △ EP =-w = 2.05j;
(3) Let the velocities of P and N after collision be v 1 and V respectively, and the horizontal direction to the right is the positive direction.
By: qv 1B=mv 12r
v 1 = bqrm = 5m/s；
If the velocity directions of P and N are the same after impact: mv=mv 1+MV.
Calculate V = 1.25m/s < V 1, and this collision cannot be realized.
Let p and n instantaneously move in opposite directions, as follows: mv=-mv 1+MV.
It is obtained that v = 2.5m/s;
Assuming that the speeds of P and N are equal, the passing time of N is tN, and the passing time of P is tP. At this time, the angle between the direction of speed N (V 1) and the horizontal direction is θ, including:
cosθ=VV 1=Vv 1
Solution: θ = 60.
gtN=V 1sinθ=v 1sinθ
Substitution data: tN = 34s；;
For the ball P, the period of its circular motion is t, that is, t = 2π rv 1 = 2π mqb = π 5s.
That is to say TN < t
When P passes through tP, the corresponding central angle is α, including: TP = α 2 π t.
When the direction of magnetic induction intensity B is perpendicular to the paper surface, if the velocities of P and N are the same, as shown in the figure, α 1=π+θ.
Simultaneous correlation equation: tP 1=2π 15s.
Comparison: tN≠tP 1, in this case, the speeds of P and N cannot be the same at the same time;
When the direction of magnetic induction intensity B is perpendicular to the paper surface, if the velocities of P and N are the same, as shown in the figure, α2=π-θ.
Same as above: tP2=π 15s.
Comparison: tN≠tP2, in this case, the speeds of P and N cannot be the same at the same time;
Answer: (1) Ball P is positively charged;
(2) When the spring swings from horizontal position to vertical position, its elastic potential energy changes by 2.05 J;
(3) The balls P and N can't have the same speed at some time after the collision.