(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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?

2. Relevant equations

PE=mgh

KE=mv^2/2

mv=mv

E(total energy in spring)=kA^2/2

3. 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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# Conservation of momentum/mass when dropping an object on a spring.

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