The two masses are each initially 1.80m above the ground,and
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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.]
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^_^
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^_^
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