Question

Two identical bowling balls are rolling on a horizontal floor without slipping. The initial speed of both balls is v = 10 m/s. Ball A encounters a frictionless ramp, reaching a maximum vertical height HA above the floor. Ball B on the other hand rolls up a regular ramp (i.e. without slipping), reaching a maximum vertical height HB above the floor. Which ball goes higher and by how much? Show the steps of your calculations.

Answer 1

Answer:

The difference between frictionless ramp and a regular ramp is that on a frictionless ramp the ball cannot roll it can only slide, but on a regular ramp the ball can roll without slipping.

We will use conversation of energy.

$K_A_1 + U_A_1 = K_A_2 + U_A_2\\\frac{1}{2}I\omega^2 + \frac{1}{2}mv^2 + 0 = 0 + mgH_A$

Note that initial potential energy is zero because the ball is on the bottom, and the final kinetic energy is zero because the ball reaches its maximum vertical distance and stops.

For the ball B;

$K_B_1 + U_B_1 = K_B_2 + U_B_2$

$\frac{1}{2}I_B\omega^2 + \frac{1}{2}mv^2 + 0 = 0 + mgH_B$

The initial velocities of the balls are equal. Their maximum climbing point will be proportional to their final potential energy. Since their initial kinetic energies are equal, their final potential energies must be equal as well.

Hence, both balls climb the same point.

Explanation: