Checking a question involving impact/impulse forces

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To determine the minimum height from which a ceramic pot will break upon impact, the pot's mass is 0.5 kg and it breaks at an impact force of 20 N over a collision duration of 0.05 s. The relevant equations include impulse, kinetic energy, and potential energy. The calculations show that at a height h, the potential energy equals the kinetic energy at impact, leading to the equation h = 0.2 m. However, there is a discussion about the validity of this height, considering real-world implications and the importance of momentum in the calculations. Ultimately, the consensus is that the initial calculations are correct, leading to a velocity of 2 m/s at impact.
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Homework Statement


A ceramic pot of mass 0.5 kg will break if it strikes the floor with an impact force equal
to, or greater than, 20 N. The collision lasts for 0.05 s. Find the minimum vertical height
from which, if dropped from rest, the pot will break upon impact with the floor (you
may ignore air resistance).

Homework Equations


Impulse=FΔt=Pfinal-Pinital
Fd=ΔKinetic energy
PE=mgh
KE=1/2mv2

The Attempt at a Solution


At a height h, the object has KE=0 and PE=mgh=0.5gh
At impact, the object has KE=1/2mv2=0.25v2 and PE=0
FΔt=Pf-Pi
However Pi=0 as the velocity is 0.
equation 1: 20*0.05=Pf=mv=0.5v
to find v:
1/2mv2=mgh
rearrange to give v=√2gh
back into equation 1:
1=0.52*2gh
h=1/0.52*2*9.81
h=0.2m

Is this correct? That doesn't seem big enough in order for the pot to break (thinking in real world terms)
 
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You need to use the fact that force = change in momentum per second.
They give you the force so you need to get momentum information
Do you know how?
 
That's what I attempted to do with the Pf-Pi with P being momentum (mv) ... is that not the right equation?
 
You are correct and if you look at your equation 1 you see that v = 2 m/s ?
So 2 = √2gh.
 
Perfect - thanks :)
 
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