# When does the mass leave the spring-connected pan? [closed]

Upper mass: $M$, lower mass: $m$, normal force between them: $N$.

When the masses are in contact, $$\frac{N-Mg}M =\frac{kx-N-mg}m$$ $$N=\frac{Mkx}{m+M}$$

At the moment they lose contact, $N=0$. $$\frac{Mkx}{m+M}=0$$

Solving it gives $x=0$. However, the answer should be $3.9$ cm.

Actually, I predicted that I have to take a limit of $m\to 0$ at last. It turns out that my amswer is independent of $m$.

What mistakes did I make?

I'm not even sure how you have managed to divide by $m$ in the first equation and if $m$ zero (notice - light pan). I'm unsure what the first equation represents at all. I can just give some hints, which are easily solving this task (I've checked).

So, what's happening here? Once the pan passes the point of equilibrium while going up, the Hooke's force is equal to zero and nothing is slowing the pan; as the pan approaches the topmost position, the pan is slowing with an increasing rate since the Hooke's force is growing. The ball is also slowing, though it's slowing at the constant rate, since $g=9.8~m/s$ is constant.

At some point of time, the Hooke's force becomes smaller than the gravity force, and at this time the pan is already slowing with a greater rate than the ball, so the ball leaves the pan.

• but $m$ is the one for the light pan...I attached my work because I just hoped others can point out my mistake clearly...hope that wouldn’t make my question getting closed. Commented Apr 9, 2018 at 10:39
• Well, I thought considering non zero mass then taking the limit of $m\to 0$ wouldn’t affect the answer. Commented Apr 9, 2018 at 10:42
• Solving the problem with $m>0$ is legitimate and also $m \rightarrow 0$ should work fine. Your approach can be applied directly to an horizontal spring for example. For the vertical spring, you should carefully examine your definition of $x$. You are explicitly accounting for $mg$ and $Mg$, so $x$ cannot be defined relative to the equilibrium point. Rather, it should be defined relative to the un-streched point. With both $m$ and $M$, the equations may become messy and @nicael approach is easier to solve. Commented Apr 9, 2018 at 18:30