Could you help me solve a physics problem? An axis of length L and mass M is balanced on the side of the table at its center of gravity. A ball with a relative height h is dropped onto its most protruding end. What is the speed of the ball immediately after impact?
user
The speed of the ball immediately after impact can be found using the conservation of energy principle. The initial potential energy of the ball at height h is converted into the kinetic energy of the axis and ball after impact.

PE = mgh (Potential energy of ball)
KE = 1/2 mv^2 (Kinetic energy of ball and axis)

where m is the mass of the ball, g is the acceleration due to gravity, and v is the velocity of the ball and axis after impact.

Setting the initial PE equal to the final KE, solving for v, and using the fact that the center of mass of the axis is at L/2:

mgh = (m + M)(1/2)(v^2 + (L/2)^2 * (v/t)^2)

where t is the time it takes for the axis to come to a stop after impact.

Approximating t as 0, we can simplify the equation to:

mgh = 1/2 (m + M) v^2

Solving for v:

v = sqrt(2gh/(1 + (m/M)))

So the speed of the ball immediately after impact is equal to the square root of (2 * g * h / (1 + (m/M))), where m is the mass of the ball and M is the mass of the axis.
assistant
What about the angular velocity of the axis?
user