Jim's maths and physics problems

[Twoflower]

Problem 4

An illuminated object is 23 cm in front of a convex lens with a focal length of 17 cm. Calculate the position and magnification of the image produced, and comment intelligently on the nature of the image.

1/u +1/v = 1/f

u=0.23, f=0.17

1/v= (1/0.17) - (1/0.23)

v=0.651 metres

The image is inverted.

Both you and colubridae were correct, the image would be 65.17 cm on the other side of the lens. The image is real and inverted, and has a magnification of x2.83

I always approach this with my students via lots of prac work (curved mirrors likewise), making a large range of measurements of object and image distances, and teasing out the relationships using the data. Plus lots of ray diagrams, of course...

You know Jim if you had taught me maths I might actually understand it.

[Azathoth]

I used to be able to do this stuff. Sadly it has been so long since I needed to I have forgotten the lot and replaced it with other random junk.

[JimC]

I used to be able to do this stuff. Sadly it has been so long since I needed to I have forgotten the lot and replaced it with other random junk.

The old "use it or lose it" thing is very true...

I use this stuff on a daily basis; but some of the stuff I could do at uni has vanished into the netherworld... Probably still there in my memory, but the access process is all rusted up...

[colubridae]

This one might be quite fun!

Calculate escape velocity from the Earths surface

You may use the following assumptions:

radius of earth = 6400km
mass of earth=5.97 × 10^24 kilograms
gravitational constant = 6.673 × 10-11 m3 kg-1 s-2

In practical terms, explain why a higher velocity would be needed to achieve orbit.

Not done yet, but have you played with this orbiter simulator?

http://orbit.medphys.ucl.ac.uk/


edit: it's free and there is a whole subculture running add-ons Scifi 2001, silent running, starwars, star trek, gemini apollo mercury.
the saturn five apollo add-on is awesome hubble

[JimC]

Problem 5 (a hard problem, not conceptually, but fiddly...)

Two 20 kg blocks are held, side by side, at the top of a long inclined plane (60 degrees to the horizontal), ready to slide down when released.

Block 1 will have a frictional force opposing motion when it slides of 100 N
Block 2 will have a frictional force opposing motion when it slides of 50 N

Block 1 is released, then Block 2 is released 5 seconds later. Calculate:

a) the time after the release of block 1 when block 2 catches up with it
b) the distance down the slope where this happens
c) the difference in velocity between the 2 blocks at that point

Take g = 9.8 m/s/s (none of this wussy approximation to 10! )

Problem 6

A parabola exists that passes through the following 3 points: (3, 8) , (-1, 16) & (5, 40)

Write the equation tor that parabola in the form y = ax2 + bx + c

Good luck!

[Xamonas Chegwé]

Problem 6:

Substituting the given values of y and x into the equation gives: -

i { 8 = 9a + 3b + c
ii { 16 = a - b + c
iii { 40 = 25a + 5b + c

i + 3ii ==> 56 = 12a + 4c ==> 14 = 3a + c ==> c = 14 - 3a

Substituting this into the equations above gives: -

iv { -6 = 6a + 3b
v { 2 = -2a - b
vi { 26 = 22a + 5b

5v + vi ==> 36 = 12a ==>

a = 3
and a bit of substituting gives...
b = -8
c = 5

so the equation of the parabola is y = 3x2 -8x + 5

[JimC]

And XC wins a bag of kudos for the correct solution to problem 6!

Matrices are a useful alternative method for solving multiple simultaneous equations - I'd certainly use them for the next level of this sort of problem - 4 points lying on a cubic...

[lpetrich]

This one might be quite fun!

Calculate escape velocity from the Earths surface

You may use the following assumptions:

radius of earth = 6400km
mass of earth=5.97 × 10^24 kilograms
gravitational constant = 6.673 × 10-11 m3 kg-1 s-2

In practical terms, explain why a higher velocity would be needed to achieve orbit.

In general, the satellite's velocity is v = sqrt(G*M*(2/r - 1/a)) for gravitational constant G, Earth's mass M, distance from the center r, and semimajor axis a. For an elliptical orbit between distances r1 and r2, a = (r1 + r2)/2. For a hyperbolic orbit, a is negative, and for barely escaping (parabolic), it is infinite.

That last result is easily shown from v = 0 and r = infinity. The velocity at radius r is sqrt(2*G*M/r)

For a circular orbit, r1 = r2 = r, and the velocity is sqrt(G*M/r)

Note: (escape velocity) = sqrt(2) * (circular-orbit velocity)

For celestial objects, G*M is sometimes known to much greater accuracy than G, so one often uses a standard gravitational parameter. That Wikipedi article has a table of values, and I'll use Earth's here, 398,600.442 km3/s2.

Circular-orbit velocity: 7.9 km/s
Escape velocity: 11.2 km/s

The velocity formula I'd presented earlier is derived from conservation of energy, where the energy per unit mass is - G*M/(2*a) Not surprisingly, it is negative for a bound orbit and zero for a barely-escaping one. For a flyby, it is positive.

[Twiglet]

This one might be quite fun!

Calculate escape velocity from the Earths surface

You may use the following assumptions:

radius of earth = 6400km
mass of earth=5.97 × 10^24 kilograms
gravitational constant = 6.673 × 10-11 m3 kg-1 s-2

In practical terms, explain why a higher velocity would be needed to achieve orbit.

In general, the satellite's velocity is v = sqrt(G*M*(2/r - 1/a)) for gravitational constant G, Earth's mass M, distance from the center r, and semimajor axis a. For an elliptical orbit between distances r1 and r2, a = (r1 + r2)/2. For a hyperbolic orbit, a is negative, and for barely escaping (parabolic), it is infinite.

That last result is easily shown from v = 0 and r = infinity. The velocity at radius r is sqrt(2*G*M/r)

For a circular orbit, r1 = r2 = r, and the velocity is sqrt(G*M/r)

Note: (escape velocity) = sqrt(2) * (circular-orbit velocity)

For celestial objects, G*M is sometimes known to much greater accuracy than G, so one often uses a standard gravitational parameter. That Wikipedi article has a table of values, and I'll use Earth's here, 398,600.442 km3/s2.

Circular-orbit velocity: 7.9 km/s
Escape velocity: 11.2 km/s

The velocity formula I'd presented earlier is derived from conservation of energy, where the energy per unit mass is - G*M/(2*a) Not surprisingly, it is negative for a bound orbit and zero for a barely-escaping one. For a flyby, it is positive.

I was really looking for the derivation from the starting point of F=GMm/r^2 worked through with appropriate integration and limits Perhaps I should have said so!

[lpetrich]

I was really looking for the derivation from the starting point of F=GMm/r^2 worked through with appropriate integration and limits Perhaps I should have said so!

I can do that also.

First, the case of a circular orbit. The distance from the center point:

r = r*{cos(p), sin(p), 0}
for distance r and angle p.

The velocity:
v = r*w*(-sin(p), cos(p), 0}
where w = dp/dt
It has magnitude v = r*w.

The acceleration:
a = r*w2*{-cos(p), -sin(p), 0} = - r*w2

This must equal the acceleration of gravity: - G*M*r/r3

Thus giving the Kepler-Newton 1-2-3 law:
w = sqrt(GM/r3)

or the circular-orbit velocity:
v = sqrt(GM/r)


Now for the escape velocity. We must start with the general expression for the acceleration:

dr/dt = v
dv/dt = - G*M*r/r3

Multiply the second expression by each term of the first one:
dv/dt . v = - G*M*r/r3 . dr/dt

Integrate over time:
(1/2)v2 = G*M/r + E

where E is the integration constant, which is the total energy per unit mass.

For a barely-escaping object, v = 0 when r = infinity, and thus, E = 0. Thus, the escape velocity is

v = sqrt(2*G*M/r)

[Twiglet]

Perfect solution as far as I can tell Ipetrich, wtg!

[JimC]

Excellent! This thread has become somewhere for people to share learning about physics and maths...

I must find a problem involving factorising polynomials with complex coefficients...

And then something complicated involving the addition of vectors in 3 dimensions...

[redunderthebed]

Excellent! This thread has become somewhere for people to share learning about physics and maths...

I must find a problem involving factorising polynomials with complex coefficients...

And then something complicated involving the addition of vectors in 3 dimensions...

You are truly a sadistic person foisting such things on poor children.

[JimC]

Excellent! This thread has become somewhere for people to share learning about physics and maths...

I must find a problem involving factorising polynomials with complex coefficients...

And then something complicated involving the addition of vectors in 3 dimensions...

You are truly a sadistic person foisting such things on poor children.

Just think of me as the cognitive equivalent of a personal fitness trainer...

Unless I make their brains hurt, I'm not doing my job!

[colubridae]

If anyone is interested.
On the way maths and science interacts.


This is the general form of the schrodinger equation for a particle in a potential well.

schrod.jpg (5.14 KiB) Viewed 1880 times

The term on the l.h.s. contains the symbol i.

This stand for the square root of -1.

This number does not exist.

Can you explain why it is there?

Five or six senteces.[/quote]


I've posted my answer to myself.