A bullet through a pendulum.

For gneill. Ek(pendulum) = Ep(pendulum). Correct, and that will give me its intial velocity that turned into its height?

gneill

Mentor
So, I did Ek(bullet) = Ek(pendulum)<-- found velocity, I used this for the momentum of the pendulum.
That is incorrect. The KE of the bullet is never equal to that of the pendulum. KE is not conserved over the collision. Forget mixing pendulum and bullet KE's.

Use the change in height of the pendulum to find its initial velocity.

Ek(pendulum) = Ep(pendulum), isnt thzat what i said ?

gneill

Mentor
For gneill. Ek(pendulum) = Ep(pendulum). Correct, and that will give me its intial velocity that turned into its height?
Yes.

because with that im not mixing bullet with the pendulum, and there is a form of energy that did give the pendulum height, and thats its initial velocity.

gneill

Mentor
because with that im not mixing bullet with the pendulum, and there is a form of energy that did give the pendulum height, and thats its initial velocity.
Yes.

Because energy is not conserved over the collision. Before and after the collision it is, for both objects separately.?
Can you explain this please ?

I got the answer though :D

And what are some other possibilities of finding the initial velocity of the pendulum. why didnt 300(0.01) = (2.21)V' work? Because for a brief instant that the bullet does hit the block or the slight second its leaving the bullet is in the block, and momentum should be transferred no ?

gneill

Mentor
Because energy is not conserved over the collision. Before and after the collision it is, for both objects separately.
Can you explain this please ?
Before the collision the bullet has a constant KE assuming that there's no air friction.
After the collision the bullet has a constant KE assuming that there is no air friction.

Before collision the pendulum bob has zero velocity and a constant height, so its KE and PE are constant.
After collision the pendulum is moving in a conservative field (gravitation), so total energy is conserved.

gneill

Mentor
And what are some other possibilities of finding the initial velocity of the pendulum. why didnt 300(0.01) = (2.21)V' work? Because for a brief instant that the bullet does hit the block or the slight second its leaving the bullet is in the block, and momentum should be transferred no ?
The bullet and block are never one object moving with the same velocity. Simply occupying the same space does not count

I dont fully understand why EK(bullet) = EK(bullet) + EP(bullet) doesnt work. The kinetic energy before and after the collision is constant, and so it the EP that the pendulum has. Why isnt the energy of the system being conserved.

issacnewton

whats the answer you got finally ?

gneill

Mentor
I dont fully understand why EK(bullet) = EK(bullet) + EP(bullet) doesnt work. The kinetic energy before and after the collision is constant, and so it the EP that the pendulum has. Why isnt the energy of the system being conserved.
Energy is only conserved for perfectly elastic collisions. Otherwise, energy is always lost in the collision to various "loss" pathways such as frictional heating, sound, plastic deformation of materials, breaking of atomic bonds(tearing, breaking), and so on.

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Energy is only conserved for perfectly inelastic collisions. .
Though kinetic energy is only conserved in perfect elastic collision?

gneill

Mentor
Though kinetic energy is only conserved in perfect elastic collision?
D'oh! My bad. Of course it's ELASTIC collisions only. I fixed the text in the original.

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