a marble is dropped into a U-tube. If the tube is semicircular with mean radius 5cm, mass of ball is 2g, find its velocity at bottom of tube?

Simple mechanical energy conservation does the trick when friction is irrelevant, but you have to assume something about the rolling - or not - of the marble when it reaches the bottom. If it is purely rolling, the potential energy of gravity will be transformed into translational and rotational kinetic energy and the sum will equal

1/2 mv^2 + 1/5 mv^2 = 7/10 mv^2

Then energy conservation gives

7/10 mv^2 = mgR

v = sqrt(10gR/7).

If the marble is only sliding then you get the “standard” answer where only translational kinetic energy is involved

1/2 mv^2 = mgR

So

v = sqrt(2gR)

As the marble moves from the top to the bottom of the semicircular U-tube, it has moved a vertical distance that is equal to the radius of the circle. Let’s use the following equation to determine its velocity at bottom of tube.

vf^2 = vi^2 + 2 * a * d, vi = 0 m/s, a = 9.8 m/s^2, d = 0.05 m

vi^2 = 2 * 9.8 * 0.05

vf = √0.98

This is approximately 1 m/s.

I have no idea what this problem is supposed to be but the equation is conservation of energy:

KE + PE = constant, if there is no friction involved.

So it follows that the decrease in potential energy of something falling in a gravitational field is equal to the _increase_ in kinetic energy, and vice versa.

If a ball of mass m is dropped from height h, then its potential energy decreases by mgh.

This is equal to the increase in kinetic energy mv^2 / 2, if the ball starts from rest and friction is negligible.

v = root(2gh)

Take whatever height the ball dropped from start to the "bottom" and run it in the equation.

A U tube has straight sides joined by a semicircle at the bottom (the shape of a "U"). You do not list the length of the straight sides.