### Video Transcript

A ball with an initial velocity of 20 meters per second rolls along a curved surface, as shown in the diagram. The mass of the ball is 100 grams. Assume that the only energy conversions that take place are between the kinetic energy and the gravitational potential energy of the ball and calculate the height of the ball at different positions to the nearest meter. Find ℎ one. Find ℎ two. Find ℎ three. Find ℎ four.

Okay, so in this question, we’ve been given a ball which is rolling up and down a curved surface. We’ve also been told that as the ball moves up and down this curve, it gains or loses both gravitational potential energy and kinetic energy as one form gets converted to the other or vice versa. We need to use the velocities of the ball given at different points along the curve to work out the heights ℎ one, ℎ two, ℎ three, and ℎ four.

Okay, so the first thing that we can do is to work out the ball’s total energy. The easiest way to do this is to work out the ball’s kinetic energy over here. Because we can say that when the ball is at this height, its gravitational potential energy is zero. Therefore, the only form of energy that the ball has is kinetic energy. So in general, the total energy of the ball, which we will call 𝐸 sub tot, is equal to the kinetic energy 𝐸 sub kin plus the gravitational potential energy 𝐸 sub grav.

However, at this point, there is no gravitational potential energy because as we said earlier this is the point at which we say the gravitational potential energy is zero. Any height above this point and the ball will have gravitational potential energy, but here it doesn’t. So for this location which we’ll call location zero, the total energy is equal to the kinetic energy at location zero, which we’ll call 𝐸 sub kin, zero. And of course, we have a plus zero here because 𝐸 sub grav, zero — the gravitational potential energy of this location — as we’ve already seen is zero.

So let’s work out the kinetic energy at position zero. We can recall that the kinetic energy of an object is given by multiplying half by the mass of that object 𝑚 by the velocity of that object 𝑣 squared. Happily for us, we already know the mass of the ball and the velocity of the ball at location zero. We know that the mass of the ball is 100 grams and we know the velocity at location zero which we’ll call 𝑣 sub zero is 20 meters per second. This is because we’ve been told this in the diagram — 20 meters per second.

However, let’s not work with our mass in grams. Let’s convert our mass into the standard units of mass, which is kilograms. To do this, we can recall that one kilogram is equal to 1000 grams. So if we divide this equation by 1000 on both sides, we see that the 1000s on the right-hand side cancel out, which leaves us with one thousandth of a kilogram is equal to one gram.

However, the mass of our ball is 100 grams. So we can multiply both sides of the equation by 100. On the right side of the equation, we’re left with 100 grams as we needed it and on the left, we’re left with 0.1 kilograms. So we know that the mass of the ball is 0.1 kilograms. And hence, we can write it down as 0.1 kilograms here.

Okay, let’s now work out the kinetic energy of the ball at location zero. So we say that 𝐸 sub kin, which is the kinetic energy at location zero, is equal to half multiplied by the mass of the ball multiplied by the speed at location zero squared. We can substitute in all the values now, which for the mass is 0.1 kilograms and for the velocity is 20 meters per second. So when we evaluate this, we find that the kinetic energy of the ball is 20 joules.

And since we use standard units for the mass of kilograms and for the velocity of meters per second, we know that what we’ve worked out in terms of the kinetic energy is going to be in its standard units, which is joules. So the kinetic energy of the ball at the bottom of the slope is 20 joules.

So why is this relevant? Well, as we said earlier, this kinetic energy is also equal to the total energy of the ball because remember at this point it has no gravitational potential energy. So its total energy is only going to be its kinetic energy. And hence, the total energy of the ball is going to be 20 joules.

Now, we can use the law of conservation of energy. What that tells us is that the total energy of the ball 𝐸 sub tot stays constant. This means that even though there are conversions between gravitational potential energy and kinetic energy for the ball as it travels along the curve the total energy is still going to be 20 joules. It’s going to stay the same.

So the total energy here in location one is 20 joules as well. The total energy here is 20 joules. The total energy in location three here is 20 joules and same for location four. In other words, the total energy of the ball does not change. It stays constant. It’s conserved. And for locations one, two, three, and four, we can no longer say that the gravitational potential energy is zero because at these locations it’s not zero. So we see that at these locations, they have some gravitational potential energy.

In other words, some of the kinetic energy that the ball had at location zero gets converted to gravitational potential energy as the ball moves along the curve. However, the total energy as we’ve already seen is still going to be 20 joules. So let’s calculate the gravitational potential energy at each of these locations. Because using this, we can work out the height of the ball at these locations.

So let’s recall that the gravitational potential energy 𝐸 sub grav is given by the mass of an object multiplied by the gravitational field strength of the Earth multiplied by the height of that object above the location that we define zero gravitational potential energy to be. And as we’ve said earlier, it said this height here, where the gravitational potential energy of the ball is zero. So at any height above this, the gravitational potential energy is not zero. And height is measured relative to the height of the ball at this location — location zero.

Okay, so let’s first look at location one now. Well, we know that the total energy of the ball is given by the kinetic energy of the ball plus the gravitational potential energy. So at location one, we can say that the total energy of the ball is equal to the kinetic energy at location one, which is given by half multiplied by the mass of the ball multiplied by the velocity of the ball at location one squared plus the gravitational potential energy, which is given by the mass of the ball multiplied by the gravitational field strength of the Earth multiplied by the height at location one, which is ℎ one.

So this is the expression for the total energy of the ball. But remember we already know that the total energy must stay the same throughout the ball’s motion. So the total energy must be 20 joules. As well as this, we already know the mass of the ball and we know the velocity of the ball at location one. We’ve been given this. It’s 12 meters per second.

So the only thing we don’t know in this equation is the height of the ball ℎ one. This means we can rearrange this equation to solve for ℎ one. To do this, we can first subtract half 𝑚𝑣 squared from both sides of the equation so that it cancels on the right-hand side. And what we’re left with is 𝐸 sub tot the total energy minus one-half 𝑚𝑣 one squared is equal to 𝑚𝑔ℎ one. Then, we divide both sides of the equation by 𝑚𝑔 so that the 𝑚𝑔s on the right-hand side cancel. This way we’re left with just ℎ one on the right-hand side.

And at this point, we can substitute in all the values. Like we’ve already said, we know what 𝐸 sub tot is, we know what 𝑚 is, we know what 𝑣 one is, we know what 𝑚 is again, and we know what 𝑔 is. Remember 𝑔 is the gravitational field strength of the Earth which is 9.8 meters per second squared. So let’s find ℎ one.

So ℎ one is equal to 𝐸 sub tot which is 20 joules minus one-half multiplied by the mass which is 0.1 kilograms multiplied by 𝑣 one squared. In this case, 𝑣 one is 12 meters per second. So we square this value and then we divide this whole thing by the mass which is 0.1 kilograms multiplied by the gravitational field strength which is 9.8 meters per second squared. So this expression is going to give us the value of ℎ one. And when we evaluate this, we find that ℎ one comes out to be something like 13.06... meters. But remember we’ve been asked to give our answers to the nearest meter. So we round this value to 13 meters. And hence, we have our value for ℎ one, now. We can say that ℎ one is equal to 13 meters.

Now, luckily for us, we can use the same analysis as what we’ve just done to work out the values of ℎ two, ℎ three, and ℎ four because the same logic applies. The only thing that we have to change from this equation is the velocity at location one, which will now become velocity at location two or three or four depending on which location we’re looking at. And this way, we’ll find an expression for ℎ two or three or four depending on which location we’re looking at. And the reason for this is that the total energy still doesn’t change — remember conservation of energy — and so neither does the rest of the expression apart from the velocity at that location and the height of that location.

So using this equation, let’s now work out what ℎ two is. This time we say that ℎ two is equal to 𝐸 sub tot which is 20 joules minus half multiplied by the mass of the ball which stays the same multiplied by 𝑣 squared. Now, here velocity is 5.6 meters per second. So we take this value and square it and we know that this is 5.6 because again we’ve been told this in the diagram.

At location two, the velocity is 5.6 meters per second. And we once again divide this by 0.1 multiplied by 9.8 because that’s the mass multiplied by the gravitational field strength of the Earth. And using this equation, we find that ℎ two is 18.808 meters. However, once again, we have to round to the nearest meter. And so we see that this number — the one after the decimal point — is an eight. This value is therefore larger than five. So the value before the decimal place is going to round up to a nine. So to the nearest meter, ℎ two becomes 19 meters. And hence, that’s our answer for the second part of the question.

Applying the same logic then for ℎ three, so here we’ve substituted 𝑣 two with 𝑣 three now and ℎ two with ℎ three. We see that ℎ three is equal to 20 joules minus one-half multiplied by 0.1 multiplied by 1.5 squared because here the velocity is 1.5 meters per second. We find that ℎ three is equal to 20.29 meters. But once again, we have to round to the nearest meter and this value is less than a five because it’s a two. And so the value before the decimal point stays the same. So we can say that ℎ three is 20 meters to the nearest meter.

Okay, so moving on to ℎ four, we see that ℎ four is equal to 20 minus one-half multiplied by 0.1 multiplied by 3.1 squared because here we’ve been given a velocity of 3.1 meters per second. And once again, we divide this by 0.1 multiplied by 9.8. When we evaluate this, we find that ℎ four ends up being 19.91 meters. And yet again, we round. Now, this value is a nine. So it’s larger than a five. So this value is going to round up to a zero. But remember it doesn’t actually round up to a zero. It rounds up to 10. So we carry over the one into the tens column, which means that 19.91 actually rounds to 20 meters. And so we find that ℎ four is actually also equal to 20 meters to the nearest meter.

Now, this doesn’t mean that ℎ three and ℎ four are exactly the same because if they were, then the ball couldn’t be travelling at different speeds when it was at ℎ three and ℎ four. However, to the nearest meter, they are the same. In other words, they both round to 20 meters.

And so at this point, we have our final answers. ℎ one is 13 meters, ℎ two is 19 meters, ℎ three is 20 meters, and ℎ four is 20 meters to the nearest meter.