# Checking a question involving impact/impulse forces

1. Jan 20, 2012

### emilypearson

1. The problem statement, all variables and given/known data
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).

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

3. 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)

2. Jan 20, 2012

### technician

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?

3. Jan 20, 2012

### emilypearson

That's what I attempted to do with the Pf-Pi with P being momentum (mv) ... is that not the right equation?

4. Jan 20, 2012

### technician

You are correct and if you look at your equation 1 you see that v = 2 m/s ?
So 2 = √2gh.

5. Jan 20, 2012

### emilypearson

Perfect - thanks :)