(2005? Huaian second model) shown in Figure A, the ends of a light spring and the mass of mA and mB, respectively, the two blocks A, B connected, and stationary in the smooth water, the spring is in t

(2005? Huaian second model) shown in Figure A, the ends of a light spring and the mass of mA and mB, respectively, the two blocks A, B connected, and stationary in the smooth water, the spring is in the free state. (1) from the B figure can be seen, t1 moment A, B speed is the same and v = 1 m / s; t2 moment, the spring is in a free state,

vA = -1 m / s, vB = 2 m / s.

By the law of conservation of momentum and the law of conservation of energy to get mAv0 = (mA + mB) v?

1

2

mAv02=

1

2

2

mA vA 2+

1

2

mBvB2?

Substitute the data to find v0=3 m/s?

mB=2 kg?

(2) When the two blocks have the same speed, the elastic potential energy of the spring is maximized to Em, which is obtained according to the law of conservation of energy

Em=

1

2<

mAv02-

1

2

(mA+mB)v2=3 J

Answer: (1) The magnitude of the initial velocity v0 of the block A is 3 m/s, and the mass of the block B is 2 kg.

(2) The maximum elastic potential energy of the spring during the interaction of A, B, and the spring is 3 J.

The maximum elastic potential energy of the spring is 3 J.

The maximum elastic potential energy of the spring is 3 J. The maximum elastic potential energy of the spring is 2 J. The maximum elastic potential energy of the spring is 3 J.