Question

8.a bowling ball rolls up a ramp 0.5 m high without slipping to storage. it has an initial velocity of its center of mass of 3.0 m/s. (a) what is its velocity at the top of the ramp? (b) if the ramp is 1 m high does it make it to the top?

Answer 1

The velocity at the top of the ramp is approximately 3.13 m/s. The ball does not have enough velocity to reach the top of a 1 m ramp.

Step-by-step explanation:

(a) Velocity at the top of the ramp:

When the bowling ball rolls up the ramp without slipping, its kinetic energy is converted into its potential energy at the top of the ramp. By applying the principle of conservation of mechanical energy, we can equate the initial kinetic energy to the final potential energy:

1/2mv2 = mgh

Simplifying the equation, we have:

v2 = 2gh

Given that the ramp height is 0.5 m and acceleration due to gravity (g) is approximately 9.8 m/s2, the velocity at the top of the ramp is:

v2 = 2 * 9.8 * 0.5

v2 = 9.8

Therefore, the velocity at the top of the ramp is approximately 3.13 m/s.

(b) Reaching the top of a 1 m ramp:

To determine if the ball can reach the top of a 1 m ramp, we need to compare the calculated velocity at the top with the velocity required to reach that height. Using the same equation for kinetic energy:

v2 = 2gh

Given that the ramp height is 1 m and acceleration due to gravity (g) is approximately 9.8 m/s2:

v2 = 2 * 9.8 * 1

v2 = 19.6

Since 3.13 m/s is less than 19.6 m/s, the ball does not have enough velocity to reach the top of a 1 m ramp.