The two masses are each initially 1.80m above the ground,and the massless frictionless pulley is 4.8m above the ground.What maximum height does the lighter object reach after the system is released?[Hint:First determine the acceleration of the lighte

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Thetwomassesareeachinitially1.80mabovetheground,andthemasslessfrictionlesspulleyis4.8mabovetheground

The two masses are each initially 1.80m above the ground,and the massless frictionless pulley is 4.8m above the ground.What maximum height does the lighter object reach after the system is released?[Hint:First determine the acceleration of the lighte
The two masses are each initially 1.80m above the ground,and the massless frictionless pulley is 4.8m above the ground.What maximum height does the lighter object reach after the system is released?[Hint:First determine the acceleration of the lighter mass and then its velocity at the moment the heavier one hits the ground.This is its "launch" speed.Assume it doesn't hit the pulley.]

The two masses are each initially 1.80m above the ground,and the massless frictionless pulley is 4.8m above the ground.What maximum height does the lighter object reach after the system is released?[Hint:First determine the acceleration of the lighte
设两个物体一大一小,质量分别为M m
小物体的加速度a=（M-m）g/m
依据机械能守恒：
大物体落地时势能减少：Mg*1.8
小物体此时势能增加：mg*1.8
势能改变的差值=（M-m）g*1.8
势能改变的差值转化为动能
系统动量守恒：设一大一小两个物体,速度分别为V1,V2
mV1=MV2
0.5mV1^2+0.5MV2^2=（M-m）g*1.8
V1=（这个方程是个二元一次方程组,有些复杂,我就不算了）
大物体落地后小物体继续上抛,此时上升的高度h‘=V1^2/2g
综上：小物体可以上升的最大高度：H=1.8+1.8+h’=?
大致就是这样,没有给出最终答案,我也不指望奖励喽^_^

我来翻译一下。两物体均离地1.8米，无阻力滑轮离地4.8米。滑轮释放后，轻的物体能到达的高点是多少？（提示：首先确定轻的物体的加速度和重物到达地面时的速度。这就是轻物的初速度。假设轻物不碰到滑轮）

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