According to the law of conservation of mechanical energy: Mgh= 12Mv2A, mg? 4h= 12mv2B

Solution, vA=2gh, vB=22gh.

Let the speed of a large object be the minimum after colliding n times, and the minimum speed is zero, then according to the conservation of momentum:

MvA-nmvB=(M+m)v

When v=0 and MvA-nmvB=0, then n=50 times.

(2) When the speed of a large object reaches vA, it can just pass point A. 。

The first collision: MVA-MVB = (m+m) v.

Let it collide k times: (M+m)v+kmvB=(M+m+km)vA,

The simultaneous solution is k=3.

So * * * touched it four times.

(3) collision1:MVA-MVB = (m+m) v1,

Second collision: (M+m)v 1+mvB=(M+2m)v2,

The third collision: (M+m)v2-mvB=(M+3m)v3,

The fourth collision: (M+m)v3+mvB=(M+4m)v4,

…

50th collision: (M+49m)v49+mvB=(M+50m)v50,

Simultaneous solution, v50=23vA.

According to the conservation of mechanical energy, mgh ′ =12m (v50) 2.

The solution is H ′ = 49h.

A:

(1) If every time a large object moves to the right to point O, there are small objects colliding with it, and the speed of the large object is the smallest after 50 collisions.

(2) If a large object collides with a small object when it moves to the right for the first time, it will collide with a small object every time it moves to the left, and the large object can cross the point A after four collisions.

(3) If a small object collides with a large object every time it moves to 0 o'clock, the maximum height of the large object is 49. 5% of h after 50 collisions.