How High Will the Unattached Ball Rise After the Collision?

[boredaxel]

Homework Statement

A 5.00-kg ball is dropped from a height of 12.0 m above
one end of a uniform bar that pivots at its center. The bar has mass
8.00 kg and is 4.00 m in length. At the other end of the bar sits
another 5.00-kg ball, unattached to the bar. The dropped ball sticks
to the bar after the collision. How high will the other ball go after
the collision?

Homework Equations

L= r X p
L = I $$\omega$$
mgh =1/2 mv^2

The Attempt at a Solution

The solution seems to require the conservation of angular momentum. But i am not sure how angular momentum could be conserved? Doesnt the weight of the falling ball provide a net external torque to the system?

 

[tiny-tim]

Welcome to PF!

A 5.00-kg ball is dropped from a height of 12.0 m above one end of a uniform bar that pivots at its center.
The bar has mass 8.00 kg and is 4.00 m in length. At the other end of the bar sits another 5.00-kg ball, unattached to the bar.
The dropped ball sticks to the bar after the collision.
How high will the other ball go after the collision?
…
The solution seems to require the conservation of angular momentum. But i am not sure how angular momentum could be conserved? Doesnt the weight of the falling ball provide a net external torque to the system?

Hi boredaxel! Welcome to PF!

Yes, it does provide a torque, but that doesn't matter, because it isn't an external torque …

at least, if you consider the whole system together, it isn't external!

Angular momentum, like momentum, is always conserved.

Hint: treat this exactly the same way as you would if the ball was hitting a block with another ball directly the other side …

you'd use conservation of (linear) momentum, and v1f = v2f, wouldn't you?

Well, do the same, except … "angularly"!