Since you’re in the habit of building little

bits of physics apparatus at home, you decide to build another ramp, but this one is for

a ball to roll down. At the bottom, the track curves back up, launching

the ball into the air. If the ball you use has a mass of 0.15 kilograms,

and the ramp ranges from three meters off the ground at the top to half a meter off

the ground at the lowest point in the path of the ball, what is the velocity of the ball

at the lowest point of the ramp, and when it hits the ground after the jump? To answer this question, we need to know about

kinetic and potential energies, so if that sounds unfamiliar, check out my tutorial on

this subject in the classical physics series. Otherwise, see if you can figure it out. As we said, the name of the game here is kinetic

and potential energy. Kinetic energy is the energy of motion, and

potential energy is the energy of location. At the top of the ramp, before you release

the ball, it’s not moving, so it has zero kinetic energy, but it does have some potential

energy. To find out exactly how much, we use the equation. We can plug in the mass, acceleration due

to gravity, and the height of the ramp, and we get 4.4 kilograms meters squared per second

squared, or 4.4 joules. Now what we also have to understand, is that

as the ball moves down the ramp, the total mechanical energy is always conserved, meaning

the sum of the kinetic and potential energies, in this case 4.4 joules, is constant. So as the ball gets closer to the ground,

potential energy is being converted into kinetic energy. That also means that the instant the ball

hits the ground, where it has zero potential energy, it must have converted all of its

potential energy into kinetic energy, so the kinetic energy of the ball at that moment

must also be 4.4 joules. Now we can use that to find the velocity of

the ball at the moment of impact. 4.4 kilograms meters squared per seconds squared

is equal to one half the mass times velocity squared, so we simplify a bit, take the square

root, and 7.66 meters per second is what we get. Now for the lowest point of the ramp. Let’s get the potential energy first, which

will be the same mass and acceleration as before, but the height is now half a meter. That gives us a potential energy of 0.74 joules. Kinetic and potential energies must add up

to 4.4 joules, so the kinetic energy at this moment must be 3.7 joules. With that, we just plug into the same equation

as before, rearrange and take the square root, and we get 7.02 meters per second at that

point. Not too bad for a little garage-style physics,

if you ask me.

Thanks professor Dave!

it shouldn't be 7.02. it is exactly 7. check it again

What about centripetal velocity? Would it be the same as the calculated one through conservation of energy?

Thanks so much, Professor Dave!

Thank you very much sir.

Great professor

I still dont get it. I need to see all the math equations.

ty daddy

Thank you 😊

That was fun. Thanks. I love energy

Excellent

what about the rotational kinetic energy? do we consider it here?

Most satisfying video. Thank you so much.

This is so hot

God i love this shit

You should've wrote the solutions. Of some of them..

So I can easily get it😢

Thank you Dave

Super sir

Thanks dad