Question

a solid uniform marble and a block of ice, each with the same mass, start from rest at the same height h above the bottom of a hill and move down it. the marble rolls without slipping, but the ice slides without friction. Find the speed of each of these objects when it reaches the bottom of the hill. Ice and marble

Answer 1

Applying the conservation of energy, the marble's final speed is determined by considering both translational and rotational kinetic energy, while the ice's final speed is solely due to translational kinetic energy.

To determine the final speeds of the solid marble and the block of ice as they move down a hill, we can apply the principles of conservation of energy. Both objects start from rest at the same height h above the bottom of the hill.

The conservation of energy equation can be expressed as:

`\[ mgh = (1)/(2)mv^2 \]`

where:

• m is the mass of the object,

• g is the acceleration due to gravity,

• h is the initial height above the bottom of the hill,

• v is the final velocity.

For the marble, which rolls without slipping, it possesses both translational and rotational kinetic energy. The total kinetic energy (translational + rotational) is given by:

`\[ \text{KE}_{\text{total, marble}} = (1)/(2)mv^2 + (1)/(2)I\omega^2 \]`

where:

• I is the moment of inertia,

• 
  `\(\omega\)` is the angular velocity.

Since the marble rolls without slipping,
`\(\omega\)` can be expressed as
`\(\omega = (v)/(R)\)`, where R is the radius of the marble. Substituting this into the kinetic energy equation, we get:

`\[ \text{KE}_{\text{total, marble}} = (1)/(2)mv^2 + (1)/(2)\left((2)/(5)mR^2\right)\left((v)/(R)\right)^2 \]`

For the block of ice, which slides without friction, all the initial potential energy is converted into translational kinetic energy:

`\[ mgh = (1)/(2)mv^2 \]`

Solve both equations for v to find the final speeds of the marble and the block of ice when they reach the bottom of the hill.