From conservation of energy: EP = 12M 1F02? ①
Solution: v0=4m/s②
(2) Let the velocities of A and B after collision be v 1 and v2, respectively, and the momentum of the system is conserved during collision.
Taking the initial velocity direction of A as the positive direction, it is obtained by the law of conservation of momentum: m1v0 = m1v1+m2v2 ③.
According to the law of conservation of mechanical energy:12m1v2 =12m15438+02+12mv22? ④
According to ② ③ ④: v 1=-2m/s, the negative sign indicates that the direction is opposite to the positive direction, and the direction is left? ⑤，v2=2m/s？ ⑥
When the ball B enters the BC section, it moves in a straight line at a uniform speed due to the balance of electric field force and gravity.
After entering the circular orbit, do uniform circular motion, and enter the EFG section to do deceleration linear motion to stop.
When the ball b is at the d point of the circular orbit, the acceleration is set to a,
According to centripetal acceleration formula, acceleration: a=v22R⑦.
Solution: a=50m/s2⑧
(3) Let the distance from the stop point of B to the lowest point of the circular trajectory be S2,
For ball B, it is obtained by kinetic energy theorem: -μm2g(s2-R)=0- 12m2v22? ⑨
After the collision between Party A and Party B, Party A moves to the left at a speed of 2m/s and acts on the spring.
Still entering the circular orbit at a speed of 2m/s, assuming that the speed of reaching the sitting point is V,
From the conservation of mechanical energy:12m1v12 =12m1v 2+m1g? 2R
Solution: v=0.8=gR, just passing through the highest point of the circular orbit.
After entering EFG at a speed of 2m/s, A always slows down and stops. Let the distance between the stop point and the lowest point of the circular trajectory be s 1.
From the kinetic energy theorem:-micron1GS1= 0-12m1v12 attending.
The distance when the two balls finally stop: △s=S2-S 1
Solution: △ s = 0.08 m;
Answer: (1) The velocity of the first two balls before collision is 4m/s;
(2) The acceleration of ball B sliding to point D with the same height as the center of the circle is 50m/S2；; ;
(3) When the two balls finally stand still, the distance is 0.08m. 。