141.Combining Concepts in Calculations

Physics Factsheet www.curriculum-press.co.uk

Number 141

Combining concepts in calculations In Physics, it is often necessary to combine equations involving different concepts e.g. kinetic energy gained = gravitational energy lost, gravitational force = centripetal force, electrical force = weight, etc. The intention of this factsheet is to provide practise in combining different topics mathematically.

v2 = 2gh v = √(2gh) putting the numbers in: v = √(2×10×20) v = √400 v = 20m/s

The most important thing when combining equations is to keep what you write down logical. The examiner must be able to follow your reasoning. Also, you might be confident you can skip a line or two but you will be amazed how often this doesn’t work or leads to problems. This is particularly annoying when you actually know the physics but because you’re in a hurry you get the calculation wrong.

Now you try this Exam question: (Take g = 10m/s2)

Worked example 1

Worked example 2

An object of mass 5kg is dropped vertically. It hits the ground at a speed of 40m/s. How high was it dropped from? (Answer = 80m. Question: Did you really need to know the mass of the object?)

A spy satellite orbits the Earth in a close in pole-to-pole orbit. How fast is it moving?

An object of mass 6kg falls from a height of 20m. Ignoring air resistance, with what velocity does it hit the ground? (Take acceleration due to gravity to be 10m/s2)

We are dealing with a satellite so we can assume it is moving in a circle around the Earth. This means we can use circular motion equations.

The principle of conservation of energy states that the gravitational potential energy of the object will be turned into kinetic energy. You don’t need to say this, you can just go straight in with:

If it moves in a circle it must have a centripetal force acting on it. The size of this centripetal force is given by F = mv2/r

gpe = mgh k.e. = 1/2mv2

m = mass of satellite (kg) v = velocity of satellite (speed in orbit) in m/s r = radius of satellite orbit (m)

Now its up to you. You can work out both then put them equal to each other. This is the first method.

but we know the satellite is close in, so we can say that the radius of its orbit is the same as the radius of the Earth. (This isn’t quite true but it is near enough. If you want the exact value find out how high above the surface of the Earth these satellites actually are and make the appropriate correction.)

gpe = 6×10×20 = 1200J And k.e. = ½ × 6 × v2 = 3v2 So: 3v2 = 1200 So: v2 = 1200/3 = 400 v = √400 = 20m/s So:

So we will take r = radius of Earth = 6,400km.

Notice how the answer is written logically. Each step follows on from the previous step and it’s easy for you to spot if there is any mistake. You will get full marks.

Exam Hint:- A common mistake is to use this value directly but we can’t because it is in km. We must change this into metres. The easiest way to do this is simply to multiply by the conversion factor.

Exam Hint:- Don’t worry about the space provided on the question paper for answering a calculation question. Take as much as you need. Don’t squash it all up into one line and lose marks because you make a calculation error.

There are 1000m (103m) in a km so: 6,400km = 6400 × 103 m And this can be entered directly into your calculator. Doing the conversion this way avoids the common error of getting the number of zeros wrong.

There is a quicker way. If you are confident with equations then you can do this:

Now, the centripetal force must be caused by something. In this case it is provided by the gravitational attraction between the satellite and the Earth. The equation for this is:

mgh = 1/2mv2 (because of conservation of energy) but we notice that m appears on both sides of the equation so it can be cancelled out, so:

F = GMm/r2 G = gravitational constant (6.67 x 10-11 Nm2kg-2) M = mass of Earth = 6 x 1024 kg m = mass of satellite, also in kg, but lets not worry about this just yet. Can you see why? r = radius of orbit (see above)

gh = 1/2v2 We want to find v on its own so we rearrange the equation. Be careful with the ½, its easy to get this rearranged incorrectly.

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Physics Factsheet

141. Combining concepts in calculations Worked example 3

The centripetal force is provided by the gravitation attraction between the satellite and the Earth so they must be equal to each other.

In a hydrogen atom an electron orbits a proton. How many times bigger is the electrostatic force between these two particles than the gravitational force between them?

So: mv2/r = GMm/r2

This question is asking you to work out the ratio of the two forces.

This is the crucial step. Now notice why we weren’t bothered by not knowing the mass of the satellite. It appears on both sides of the equation so we can cancel it out. Now we have:

Exam Hint:- You need to use two equations, one for the electrostatic force and one for the gravitational force and then divide them.

v2/r = GM/r2 r also appears on both sides so we can cancel again. Now we have

Electrostatic force between two charged particles is given by F = kQq/r2 k = electrostatic force constant = 9 × 109 Nm2C-2

v2 = GM/r So now we can find the speed of the satellite

Q and q are the charges on the particles. Overall, the hydrogen atom is neutral so the charges on the proton and electron must be equal and opposite. Your data sheet gives the charge on the electron as q = -1.6 × 10-19C. so the charge on the proton must be +1.6 × 10-19C.

v =√(GM/r) = √(6.67 × 10 × 6 × 10 /6400 × 10 ) -11

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v = 7.9 × 103 m/s Exam Hint:- Sometimes the question asks for the time period of such a satellite - that is how long it takes to make one orbit around the Earth. As we know its velocity, this is easy:

Exam Hint:- In a calculation question you can ignore the sign of the charge. So don’t get mixed up trying to take account of the minus sign of the electron’s charge. Ignore it and take both charges as positive.

velocity = distance travelled/time taken (v = d/t) for one orbit the distance travelled is the circumference of the orbit (a circle) so d = 2πr so t = d/v = 2 x 3.14 x 6400x103 / 7.9x103 t = 5,087s

r = distance between the proton and the electron. Again, you can assume the electron orbits in a circle so this distance is constant. Take r = 5.3 x 10-11m.

Or, t = 5087/60 = 85 minutes. So a close in spy satellite can orbit the Earth in about one and a half hours.

As seen above, the Gravitational force between two masses is given by: F = GMm/r2 M = mass of proton = 1.67 x 10-27 kg m = mass of electron = 9.1 x 10-31 kg

Exam Question A geostationary satellite is one which orbits the Earth in exactly the same time as it takes the Earth to spin once on its axis. This has the effect of the satellite always appearing to be in the same place in the sky as viewed from the Earth. How high above the surface of the Earth must such a satellite be? Does the mass of the satellite make any difference at all to your calculation?

The question asks you for the ratio between these two forces. You could just go ahead and work out the two forces from the two equations then divide you answers and this would give the correct answer. Exam Hint:- A quicker way is to divide the equations before you put numbers into them. Like this. Ratio = (kQq/r2) / (GMm/r2)

(Answer = 35,912km. If you got this, well done. If not, here are some hints: The Earth spins once on its axis in 24 hours. You must convert this to seconds.

Using the ‘invert and multiply’ rule for dividing fractions this becomes kQq/r2 × r2/GMm

You can assume the orbit is a circle so you can use the equations given above.

Notice that having done this the r2 cancels out. This leaves the ratio as kQq/GMm and this is quicker to calculate. Putting the numbers in the ratio is:

You will end up with an answer for r. But this is the radius of the orbit. To find the height above the Earth you must subtract the radius of the Earth from your answer.

(9 × 109 × 1.6 × 10-19 × 1.6 × 10-19) / (6.67 × 10-11 × 1.67 × 10-27 × 9.1 × 10-31) Ratio = 2.3 × 1039

Finally, don’t worry about getting the exact answer above. I used π = 3.14, your calculator may give a more exact value. So long as you are close to the answer given you will gain full marks.)

This is a ratio so it has no units. Exam Hint:- Notice how much bigger the electrostatic force is compared to the gravitational force. Gravity is a weak force. It only becomes important when very big masses are involved, the masses of planets for instance. On a small scale where the masses are tiny the gravitational force is negligible.

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Physics Factsheet

141. Combining concepts in calculations Worked example 4

Exam Question Deuterium (‘heavy hydrogen’) consists of a single electron orbiting a nucleus of a proton bound to a neutron. The neutron can be taken to have the same mass as the proton. Calculate the ratio of the electrostatic force to the gravitational force between the electron and the nucleus of deuterium.

The Earth has density d and radius r. The gravitational field strength at the surface is g. What is the gravitational field strength at the surface of a planet of density 2d and radius 2r? Gravitational field strength for a planet (assumed to be a sphere) is given by the equation g = GM/r2 where M = mass of planet. But in the question we are not given the mass we are given the density of the planet. So we need to relate the mass of the planet to its density. This requires a new equation. You should know that:

Answer: You can redo the calculation or you could realise that the charges haven’t changed because the neutron has no charge. The mass of the nucleus is however twice what it was. Since the ratio is kQq/GMm Everything is the same except M, which is twice its original value. So the answer is half of the previous answer. Answer = 1.1 × 1039

Density = mass/volume and for a sphere, volume = (4/3)π r3. Now because G is a constant, the equation says that g is proportional to M/r2. (g ∝ M/r2) Rearranging the density equation we get mass = density × volume

Possible full exam question

So M = d × 4/3 × π × r3. So g ∝ (d × 4/3 × π × r3/ r2).

Electron energy orbits in a Hydrogen atom have the following values:

Leaving out the constants this gives: g ∝ (d × r).

Electron energy levels in the hydrogen atom

Now we can answer the question. The new planet has twice the old density and twice the old radius, so the new g is proportional to: 2d x 2r = 4dr

0 ev (ionisation) - 0.54 ev - 0.85 ev

So the answer is that the new planet has 4 times the value of g.

- 1.51 ev

Exam Question A moon of this new planet has a density of twice the Earth’s but a radius only a quarter of the Earth’s. What is the value of the gravitational field strength at the surface of this moon? (Take the value of the Earths gravitational field strength to be g)

- 3.39 ev

Answer: Using the argument from above g is proportional to d × r. So moon’s g is proportional to 2d × r/4 = dr/2 So the new planets moon has half the value of g.

- 13.6 ev

Why do the energy levels have negative values? Because we define the potential energy of the electron when it is far away from the proton to be equal to zero. Far away means that the electron has been ionized. This means that the electron and the proton are far enough away from each other that there is zero force between them.

Worked example 5

B into page

V F

But the electron is attracted to the proton. So it’s just like when a mass is attracted to the Earth - when the mass gets closer to the Earth it loses gravitational potential energy. So when the electron gets closer to the proton it loses electric potential energy. Since it started off with zero potential energy in the first place, as it gets closer to the proton it loses potential energy, so its potential energy becomes less than what it was. Less than zero means its potential energy must be written as a negative number. This does not mean there is such a thing as negative energy. It just means that we have chosen a place to define as zero. Try to understand this – it is important.

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The diagram shows a positively charged particle moving with velocity v at right angles to a magnetic field directed away from you. The magnetic field has a field strength of B. The particle is a proton. If its velocity is 6 × 106 m/s and it moves in a circle of radius 50cm what is the strength of the magnetic field in Tesla? Answer: The particle is a proton so its mass is 1.67 × 10-27 kg. Its charge is +1.6 x 10-19C

The radius of its orbit must be changed into metres so this is r = 0.5m It is moving in a circle so there must be a centripetal force acting on it which is given by F=mv2/r. This centripetal force is provided by the magnetic force on the proton which is given by F=BQv.

Exam Hint:- In questions you will never be interested in negative values for the electron’s energy, only in the difference between the levels.

So: mv2/r = BQv Rearrange the equation to make B the subject: B = mv/rQ Put the numbers in: B = (1.67 × 10-27 × 6 ×106) / (0.5 × 1.6 × 10-19) B = 0.125T.

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141. Combining concepts in calculations Exam Question What would the magnetic field strength be if the charged particle was an electron moving in the same radius orbit at the same speed? Answer: The electron has the same charge as the proton but its mass is 9.1 × 10-31 kg. so: B = (9.1 × 10-31 × 6 × 106) / (0.5 × 1.6 × 10-19) B = 6.825×10-5T. (or 68.25 microTesla) Exam Question The orbit of the Moon, which has a mass m, is a circle of approximate radius 60r, where r is the radius of the Earth. Show that the gravitational attraction between the Earth, mass M, and the Moon is given by F = GMm/3600r2. The mass of the Earth is 6.0 × 10

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kg and its radius is 6400 km.

What is the speed of the moon in orbit around the Earth? How many days does it take the moon to orbit once around the Earth? Is this consistent with the idea of a ‘month’? Exam Question An oil drop of mass 3.26 × 10-11 g and carrying a charge equal to that of one electron is stationary between the plates of a charged parallel-plate capacitor (the weight of the oil drop is balanced by the electrostatic force on it). The distance between the plates is 1mm. To what voltage is the capacitor charged? (Hint: careful with the units) (Answer = 2,000V) (Further hint: remember what was said about significant figures in calculations!)

Acknowledgements: This Physics Factsheet was researched and written by Paul Dutton The Curriculum Press,Bank House, 105 King Street,Wellington, Shropshire, TF1 1NU Physics Factsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber. No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission of the publisher. ISSN 1351-5136

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