Bowling Ball Rise: Find Speed at Top w/Conservation of Energy

In summary, the problem involves a bowling ball encountering a 0.662 m vertical rise with a translational speed of 3.70 m/s at the bottom. We need to find the translational speed at the top, assuming no friction and uniform distribution of mass. Using conservation of energy and momentum, the initial energy is equal to the change in potential energy plus the final kinetic energy. However, simply setting kinetic energy equal to potential energy does not yield the correct answer.
  • #1
scheng12
3
0

Homework Statement



A bowling ball encounters a h = 0.662 m vertical rise on the way back to the ball rack, as the figure below illustrates.

Ignore frictional losses and assume that the mass of the ball is distributed uniformly. The translational speed of the ball is 3.70 m/s at the bottom of the rise. Find the translational speed at the top.

Homework Equations



KE= 1/2mv^2

U=mgh

Conservation of energy and momentum?


The Attempt at a Solution


I tried basic conservation of energy but that didn't really work out. I'm not sure where to start on this one since it doesn't give the mass of the ball. Though that probably cancels out anyway. When I set KE = U the answer is incorrect. I'm not sure on what else I can do.
 
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  • #2
The initial kinetic energy is not equal to the change in potential energy. The initial energy is equal to the change in potential energy plus the final kinetic energy.
(KE + U)_1 = (KE + U)_2
 
  • #3
So:

1/2mvi2=mgh + 1/2mvf2?


When I plug in the values for this the answer is incorrect as well.
 
  • #4
The answer is not 0.837 m/s?
 

1. How does conservation of energy play a role in the bowling ball rise?

Conservation of energy states that energy cannot be created or destroyed, only transferred or converted from one form to another. In the case of the bowling ball rise, the potential energy of the ball at the top of its trajectory is equal to the kinetic energy it had at the bottom of the trajectory, which follows the law of conservation of energy.

2. What factors affect the speed of the bowling ball at the top of its trajectory?

The speed of the bowling ball at the top of its trajectory is affected by several factors, including the initial speed of the ball, the mass of the ball, the height of the release point, and the angle of release. These factors determine the amount of potential energy the ball has at the top of its trajectory, which then converts to kinetic energy as it falls back down.

3. Can we accurately calculate the speed of the bowling ball at the top of its trajectory?

Yes, using the principles of conservation of energy and the equations for potential and kinetic energy, we can accurately calculate the speed of the bowling ball at the top of its trajectory. This calculation requires knowing the initial speed, mass, height, and angle of release of the ball.

4. Is the speed of the bowling ball at the top of its trajectory affected by air resistance?

Yes, air resistance does affect the speed of the bowling ball at the top of its trajectory. Air resistance acts as a force that opposes the motion of the ball, causing it to lose some of its kinetic energy. This means that the actual speed of the ball at the top of its trajectory may be slightly less than the calculated speed.

5. How can we use the speed of the bowling ball at the top of its trajectory in practical applications?

The speed of the bowling ball at the top of its trajectory can be used in practical applications such as designing amusement park rides, calculating the trajectory of projectiles, and studying the motion of objects in freefall. It can also be used in sports such as bowling, where understanding the speed and trajectory of the ball can improve performance.

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