Number displayed on an electric scale when an object hits it

[ChessEnthusiast]

Homework Statement

We have an object, whose mass is accumulated in its center. The mass is m.
We let an object positioned at the height of h over the scale free fall.
Let R be the number that will be displayed on the scale right when the item hits it.
Find R(h).

2. The attempt at a solution
v_{terminal} = \sqrt{2gh}
p = m \sqrt{2gh}
R = \frac{F}{g} where F = \frac{dp}{dt}

Since the force the object will exert on the surface of the scale is the change in momentum over the change in time, how am I supposed to determine the time, during which the momentum was "transferred"?

 

[Bystander]

You're certain you've transcribed the problem statement completely and correctly?

 

[lychette]

You're certain you've transcribed the problem statement completely and correctly?

have you seen a fault?

 

[ChessEnthusiast]

You're certain you've transcribed the problem statement completely and correctly?

To be honest if you, I also feel as if some information were lacking here. Anyway, can you understand the point of this problem? Have I explained it precisely enough?

 

[PhanthomJay]

The problem asks for the scale reading when the object first hits it. What is the scales spring deformation at that instant?

 

[ChessEnthusiast]

The problem asks for the scale reading when the object first hits it. What is the scales spring deformation at that instant?

There will be hardly any deformation, because the scale is electric.

 

[Bystander]

Point out to the "poser" of the problem that there's something "missing," and hope for the best.

 

[ChessEnthusiast]

Point out to the "poser" of the problem that there's something "missing," and hope for the best.

What if we assumed that the scale will slightly deformate once the object hits it? Would it, then, be solvable?

 

[PhanthomJay]

There will be hardly any deformation, because the scale is electric.

There won't be any deformation even if the scale is like my old fashioned non trustworthy bathroom spring scale. Either info is missing or if you take the problem as written, the scale reading per F = kx. Is_______?

 

[ChessEnthusiast]

There won't be any deformation even if the scale is like my old fashioned non trustworthy bathroom spring scale. Either info is missing or if you take the problem as written, the scale reading per F = kx. Is_______?

kx / g
?

 

[haruspex]

right when the item hits it.

As others have noted, that does not strictly make sense. But from a practical point of view, what you would see is a large value displayed almost immediately. So I would interpret it as the maximum value displayed.

 

[haruspex]

kx / g
?

That is the mass shown when the spring compression is x, but you do not know what x is. See my preceding post.

 

[ChessEnthusiast]

That is the mass shown when the spring compression is x, but you do not know what x is. See my preceding post.

m\sqrt{2gh} / dt = kx
x(t) = \frac{m}{t} k\sqrt{2gh}

R = \lim_{t \to 0} \frac{x(t)}{g}
Is it the potential solution?

 

[PhanthomJay]

I gather electric scales give max readings at the moment of impact? I think mine reads 888 for some reason. Anyway, imagine it is an old fashioned scale that reads 0 at no load , like a grocery scale for weighing bananas. F= kx and at the very instant of impact, x is _____ so F is?

 

[ChessEnthusiast]

x = \sqrt{\frac{2mgh}{k}}
And hence
F = \sqrt{2mghk}
I think this should work.

 

[haruspex]

x = \sqrt{\frac{2mgh}{k}}
And hence
F = \sqrt{2mghk}
I think this should work.

This cannot be exactly right since it does not tend to 2mg as h tends to zero. In the conservation of energy equation, you need to take into account the total distance which the mass descends (it's a bit more than h). And the final answer should be in terms of the indicated mass, not in terms of the force.