There are two static wooden blocks A and B on the rough horizontal desktop, and the distance between them is D. Now give A an initial velocity, so that A and B collide elastically, and the collision time is very short. When both blocks stop moving, the distance is still d. ...

Let the moment before the collision, the speed of block A is V;

At the moment after the collision, the velocities of A and B are v 1 and V 2, respectively.

In the process of collision, from the law of conservation of energy and momentum.

mv2 = mv 12+& amp； #8226; 2mv22，

Mv=mv 1+2mv2, where the velocity direction of block A before collision is positive.

Simultaneous solution: v 1=-? 1/2? *v2。

Let the distances between A and B after collision be d 1 and d2, respectively.

From kinetic energy theorem? μmgd 1=？ 1/2? mv 1^2.？

μ(2m)gd2=？ 1/2? 2mv2^2.

D = D2+D 1。

Let the initial velocity of a be v0, μ gd =? 1/2? mv^2-？ 1/2? mv0^2

Simultaneous solution: v0=? Root number 28/5? μgd？

A: What's the initial velocity of A?

Root number 28/5? μgd？ .