Lmn joke

During the falling process of (1) block c, it is obtained by kinetic energy theorem:

(mg+qE 1)h= 12mv2？ 0,

Solution: h = mv22 (mg+QE1);

(2) The conservation of momentum during the collision of blocks C and D takes the system composed of C and D as the research object, and takes the initial velocity direction of C as the positive direction.

According to the law of conservation of momentum, mv=(m+m)v, which is solved as: v = = v2；;

(3) When C and D just leave the horizontal track, the pressure on the track is zero.

Let their speed be V ′, and in the vertical direction, qv ′ b = 2 mg … ①.

In the process of CD moving to the right together, it is obtained from the kinetic energy theorem: qe2x =12× 2mv ′ 2? 12×2mv2, total ...? ②

X = m2qe2 (4mg2b2b2? v24)；

Answer: (1) The height of box C from the horizontal orbit when it is released from LM is h = mv22 (mg+QE1);

(2) The common speed of C and D at the moment after collision, v = v2.

(3) The distance between blocks C and D and OP is x = m2qe2 (4mg2q2b2? v24)。