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Checking a question involving impact/impulse forces 
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#1
Jan2012, 09:44 AM

P: 7

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=P_{final}P_{inital} Fd=ΔKinetic energy PE=mgh KE=1/2mv^{2} 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=1/2mv^{2}=0.25v^{2} and PE=0 FΔt=P_{f}P_{i} However P_{i}=0 as the velocity is 0. equation 1: 20*0.05=P_{f}=mv=0.5v to find v: 1/2mv^{2}=mgh rearrange to give v=√2gh back into equation 1: 1=0.5^{2}*2gh h=1/0.5_{2}*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
Jan2012, 10:20 AM

P: 1,506

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
Jan2012, 10:25 AM

P: 7

That's what I attempted to do with the P_{f}P_{i} with P being momentum (mv) ... is that not the right equation?



#4
Jan2012, 10:34 AM

P: 1,506

Checking a question involving impact/impulse forces
You are correct and if you look at your equation 1 you see that v = 2 m/s ?
So 2 = √2gh. 


#5
Jan2012, 11:03 AM

P: 7

Perfect  thanks :)



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