A solid and a hollow ball, each with a mass of 1 kg and radius of 0.1 m, start from rest and roll down a ramp of length 3 m at an incline of 35 degrees. An ice cube of the same mass slides without friction down the same ramp. Which ball will reach the bottom first? Does the ice cube travel faster or slower than the solid ball at the base of the incline? Which is the speed of the slid ball at the bottom of the incline?

To determine which object will reach the bottom of the ramp first, we need to consider their moments of inertia. The moment of inertia for a solid ball and a hollow ball are different.

The moment of inertia for a solid sphere is given by the formula: I_solid = (2/5) * m * r^2, where m is the mass of the object and r is the radius of the object.

The moment of inertia for a hollow sphere is given by the formula: I_hollow = (2/3) * m * r^2.

Now, let's calculate the moments of inertia for both the solid and hollow balls:
- For the solid ball: I_solid = (2/5) * 1 kg * (0.1 m)^2 = 0.004 kg m^2.
- For the hollow ball: I_hollow = (2/3) * 1 kg * (0.1 m)^2 = 0.00667 kg m^2.

Since the hollow ball has a larger moment of inertia, it will take more time for it to accelerate and reach the bottom of the ramp compared to the solid ball. Therefore, the solid ball will reach the bottom of the ramp first.

Now let's compare the ice cube and the solid ball at the base of the incline. Since both objects have the same mass and the incline is frictionless, the only force acting on them is due to gravity.

The force due to gravity can be decomposed into two components: one parallel to the incline (mg*sin(theta)) and one perpendicular to the incline (mg*cos(theta)), where m is the mass of the object and theta is the angle of the incline.

Since the mass and angle of the incline are the same for both the ice cube and the solid ball, the forces acting on them are also the same. Therefore, both objects will accelerate down the incline at the same rate.

The speed of the solid ball at the bottom of the incline can be calculated using the principles of conservation of energy. The potential energy at the top of the incline is converted into kinetic energy at the bottom, neglecting any energy losses due to friction.

The potential energy at the top of the incline can be calculated as m * g * h, where m is the mass, g is the acceleration due to gravity, and h is the height of the incline.

The height of the incline can be calculated as h = l * sin(theta), where l is the length of the incline and theta is the angle of the incline.

Plugging in the given values:
- l = 3 m
- theta = 35 degrees
- m = 1 kg
- g = 9.8 m/s^2

The height can be calculated as h = 3 m * sin(35 degrees) = 1.711 m.

The potential energy at the top of the incline is then m * g * h = 1 kg * 9.8 m/s^2 * 1.711 m = 16.7846 J.

Since the potential energy is converted entirely into kinetic energy at the bottom of the incline, the kinetic energy at the bottom is also 16.7846 J.

The kinetic energy can be calculated as (1/2) * m * v^2, where v is the velocity of the solid ball at the bottom. Rearranging the equation, we get v = sqrt((2 * KE) / m).

Plugging in the given values:
- KE = 16.7846 J
- m = 1 kg

The velocity of the solid ball at the bottom is then:
v = sqrt((2 * 16.7846 J) / 1 kg) = sqrt(33.5692) ≈ 5.8 m/s.

Therefore, the speed of the solid ball at the bottom of the incline is approximately 5.8 m/s.