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JustinLiang
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Homework Statement
A spring of negligible mass and force constant k=400 is hung vertically and a 0.2kg pan is suspended from its lower end. A butcher drops a 2.2kg steak onto the pan from a height of 0.4m. The steak makes a totally inelastic collision with the pan and sets the system into vertical SHM. What is the speed of the pan and steak immediately after the collision?
Homework Equations
PE=mgh
KE=mv^2/2
mv=mv
E(total energy in spring)=kA^2/2
The Attempt at a Solution
First I found out the energy produced from dropping the steak 0.4m:
PE=mgh
=2.2kg(9.8)(0.4m)
=8.624J
Then I used this to find the speed of the steak before the collision:
PE=KE
8.624J=(2.2kg)(v^2)/2
v=2.8m/s
When the steak goes into the pan by conservation of momentum we have:
mv=mv
(2.2kg)(2.8m/s)=(2.4kg)(v)
v=2.56666666m/s
My problem: However if I plug 2.566666m/s back into the KE equation I do not get 8.624J, instead I get 7.095J which means that the energy is not conserved? Do you guys know why? If I have 7.095J as the energy and I plug it into the E(total energy in spring) equation I do not get the right amplitude, instead plugging in 8.624J will get me the right answer.
HERE IS A SIMPLER PROBLEM
I found the same problem when solving for the final velocity of a block m=0.992kg when a bullet with v=280m/s and m=0.008kg strikes it and embeds itself into it.
By conservation of momentum:
(0.008)(280)=(1kg)(v)
v=2.24m/s
Where initially the KE in the bullet initially is
KE=0.008(280)^2/2=313.6J
But the KE after the collission is
KE=1kg(2.25m/s)^2/2=2.5088J
So clearly the energy is not conserved? I am confused.
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