(20 12? As shown in the figure, tie a string with a length of L=0.6m to the fixed point O, and tie a ball with a mass of m= 1.0kg to the lower end of the string (which can be regarded as the mass).

(1) After the collision, the object M makes a flat throwing motion, and the initial velocity of the flat throwing motion is V3.

h= 12gt2 …①

S=V3t …②

Get: v3 = sg2h = 3.0m/s … ③.

When landing, the vertical speed vy = 2gh = 4.0m/s ... ④.

So the speed of the object when it hits the ground is V = V32+VY2 = 5.0m/s ... ⑤.

(2) When the block collides with the ball at point B, let the velocity of the block before collision be V 1 and the velocity of the ball after collision be V2, which is determined by the law of conservation of momentum:

mv 1 = mV2+MV3……⑥

Mechanical energy is conserved when the ball moves from point B to the highest point A after collision. Let the speed of the ball at point A be VA:

12mv 22 = 12mv a2+2 mgl…⑦

When the ball is at the highest point, the conditions are as follows: 2 mg = MVA2L … ⑧

Solution from ⑦ ⑧: V2 = 6.0m/s ...

From 36 pet-name ruby: v1= mv2+mv3m = 6.0? m/s？ …⑩

When the mass m moves from p to b, it is determined by the kinetic energy theorem:

μMgS 1= 12MV 12？ 12MV02

Solution: v 0 = v 12+2 μ gs 1 = 7.0? m/s？

Answer: (1) A 5 m/s M is 5m/s when it lands.

(2) The initial velocity of block M at point P is 7.0m/s. 。