Challenging question on impulse (or momentum)

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1. Apr 1, 2016

i_hate_math

1. The problem statement, all variables and given/known data
Pancake collapse of a tall building. In the section of a tall building shown in Figure (a), the infrastructure of any given floor K must support the weight W of all higher floors. Normally the infrastructure is constructed with a safety factor s so that it can withstand an even greater downward force of sW. If, however, the support columns between K and L suddenly collapse and allow the higher floors to free-fall together onto floor K (Figure (b)), the force in the collision can exceed sW and, after a brief pause, cause K to collapse onto floor J , which collapses on floor I , and so on until the ground is reached. Assume that the floors are separated by d = 4.0 m and have the same mass. Also assume that when the floors above K free-fall onto K , the collision lasts 1.7 ms. Under these simplified conditions, what value must the safety factor s exceed to prevent pancake collapse of the building?

2. Relevant equations
I am not sure which laws of physics this question is involved, but I assume it's on conservation of linear momentum. So for complete inelastic collision,
m1*v1i=(m1+m2)vf
Please let me know if theres anything else I should consider

OR the things above could be not relevant at all. The impulse of the falling floors is
J=Favg*∂t --> Favg=J/∂t

3. The attempt at a solution
So if i shall consider the impulse of the falling object, then i have the following
W*s=J/∂t for when the building does not fall
J is simply change in momentum, that is Delta(P)=Pf-Pi=m*vf-m*vi=m*vf
Now 0.5g*t^2=d , g*t=vf
then t=0.9035... and vf=8.8543...m/s
also m=W/9.8
so i got W*s=m*vf/∂t=((W/9.8)*8.8543)/0.00137=659.4948
apparently this is not the correct answer

2. Apr 1, 2016

haruspex

Is it 1.7ms or 1.37ms?

3. Apr 1, 2016

i_hate_math

Soz.. It's 1.7ms.

4. Apr 1, 2016

haruspex

So what answer does that give you?

5. Apr 1, 2016

i_hate_math

OMG im diggin a hole for myself