How Does Energy Conservation Apply in a Pendulum-Object Collision?

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Cathr
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1. A ball hanging on a pendulum hits an object standing on the table. The interaction is elastic and linear. After that, the object falls on the floor.


Homework Equations


From state 1 to 2, we have the conservation of the potential energy of the pendulum to its kinetic energy, right before it hits the object.
Then it is the kinetic energy conservation, so mv^2=mv1^2+mv2^2, right?[/B]3. How to write the conservation laws of the next states?
 
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Hello Cath, :welcome:

So: state 1 is the ball held at a certain position away from the object. Let go and the moment just before collision is state 2.
You correctly write (the collision is fully elastic): kinetic energy before collision = kinetic energy after collision or $$ {1\over 2} m_1v_2^2 = {1\over 2} m_1v_3^2 + {1\over 2} m_2 v_4^2$$ where I added some subscripts:
##m_1 \quad## mass of ball
##m_2 \quad## mass of object
##v_2 \quad## speed of ball before collision
##v_3 \quad## speed of ball after collision
##v_4 \quad## speed of object after collision

And you are interested in "the next states". Could you describe them ? And perhaps already write down some of the energies that are associated with these staets ?
 
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Thank you!
I am interested to know how to write the conservation law when the object just got hit by the ball, and it started falling down.
I guess there's another equation to write before saying that the kinetic energy of the object right before it hits the floor is equal to the potential energy+its initial kinetic energy (I am not even sure that this is correct).
 
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Right! Excuse me, it's my mistake.
 
There is conservation of energy until the object hits the floor. From that point, because we cannot account for the energy transferred two the floor (and earth) there is no longer conservation of energy. But you can use both potential energy and kinetic energy of the object the instant before it hits the floor when its height is that of the floor.