Confused about solution to kinetic energy question

[toforfiltum]

Homework Statement

2. Homework Equations [/B]
KE=1/2 mv²

The Attempt at a Solution

To calculate the kinetic energy of the ball just after it leaves the surface, I use the ratio of the centre of gravities of the ball at the two different heights. My working is (0.41/0.76)×0.75 which gives me the answer C. But the answer is B. And according to the solution, I must take the ratio of the bottom of the balls instead. Why is it so? I thought that COG is always used in calculating change in potential energy.

 

[Dr. Courtney]

Use conservation of energy and keep in mind the total energy is kinetic plus potential.

The bounce might be an inelastic collision, so energy is conserved before the bounce and after the bounce, but not necessarily through the bounce.

 

[toforfiltum]

Use conservation of energy and keep in mind the total energy is kinetic plus potential.

The bounce might be an inelastic collision, so energy is conserved before the bounce and after the bounce, but not necessarily through the bounce.

I get what you mean, bu I don't see how is it related to using the bottom of the ball instead of its centre to find the kinetic energy of the ball. Can you explain?

 

[Dr. Courtney]

You can use either the center of mass or the bottom or the top of the ball to define the height to find the potential energy, as long as you are consistent. This is because only the change in potential energy (related to the change in height) is relevant to the answer when properly applying conservation of energy.

I see a lot of students make these kinds of mistakes when they take shortcuts and skip steps. Why not write out the full equation that represents energy conservation after the bounce and go from there? Why not draw a good picture to define the heights right after the bounce and at the top of the trajectory after the bounce?

 

[toforfiltum]

You can use either the center of mass or the bottom or the top of the ball to define the height to find the potential energy, as long as you are consistent. This is because only the change in potential energy (related to the change in height) is relevant to the answer when properly applying conservation of energy.

I see a lot of students make these kinds of mistakes when they take shortcuts and skip steps. Why not write out the full equation that represents energy conservation after the bounce and go from there? Why not draw a good picture to define the heights right after the bounce and at the top of the trajectory after the bounce?

Oh I see where I went wrong now. I didn't take into account the height of the centre of ball from ground during the bounce. Thanks for helping me to point out my mistake.