Wenzhou Yu Xiang photography

(1) Let 1 The speed at which the ball and the wooden box move together is v 1. According to the law of conservation of momentum, mv0-Mv=(m+M)v 1.

Substituting the data, the result is: v1= 3m/s.

(2) Let the distance between the point where the 1 th ball meets the wooden box and the left end of the conveyor belt be s, and the 1 th ball meets the wooden box through t0.

Then: t0 = sv0

Let the acceleration of the 1 th ball moving together after entering the wooden box be a, and according to Newton's second law: μ(m+M)g=(m+M)a, a=μg=3m/s2.

Let the deceleration time of the wooden box be t 1 and the acceleration time be t2, then: t1= t2 = △ va =1s.

So the displacement of the wooden box in 2s is zero.

According to the meaning of the question: s = v0 △ t1+v (△ t+△ t1-t1-t2-t0).

Substituting the data, the solution is: s = 7.5mt0 = 0.5s

(3) When the wooden box meets the 1 th ball to the second ball, the displacement of the conveyor belt is S, and the displacement of the wooden box is s 1, so: S = V (△ t +△ t1-t0) = 8.5 ms1= V (△

Therefore, the displacement of the wooden box relative to the conveyor belt is △s=S-s 1=6m.

Then the heat generated by the friction between the wooden box and the conveyor belt is: Q=f△s=54J.

Answer: (1) At the moment when 1 the ball meets the wooden box, the speed at which they move together is 3m/s;

(2) The first 1 ball meets the wooden box after 0.5s;

(3) From the time the wooden box meets the 1 th ball to the second ball, the heat generated by the friction between the wooden box and the conveyor belt is 54J. ..