Electric field force f = QE = 0.3n
Gravity G=mg=0.3N?
It can be seen that the electric field force and gravity are equal in magnitude and opposite in direction, and the two forces are balanced.
Therefore, it is equivalent to that P only receives Lorentz force in electric and magnetic fields, and makes uniform circular motion.
(2)P is subjected to Lorentz force in magnetic field, which provides centripetal force for P to do circular motion.
So qvB=mv2R?
The radius of circular motion of p R=mvqB? ①
Period t = 2π RV = 2π MQB
T= 12s substitute data.
Judging from the known conditions, there is t =1.0s =112t.
So as shown in the figure below:
P = 30 locus central angle θ?
As can be seen from the figure on the right, the trajectory radius r = dsinθ = 0.2msin30 = 0.4m.
Combined with ① formula, v = 0.2m/s.
(3) When P and Q collide, the momentum of the system is conserved and exists.
mv0=mv+mQvQ
Substituting the data can get VQ = 0.6m/s.
After the collision, the horizontal direction of Q is only affected by friction. Applying Newton's second law, there are
μmg=ma
Get a=0.8m/s2.
Because Q moves in a straight line with uniform speed under the action of friction, the speed is taken as the positive direction and the acceleration a=-0.8m/s2.
So the movement time before q stops t' = 0? vQa=0？ 0.6? 0.8s=0.75s？
Because t' < t means t'