Jim's maths and physics problems

[colubridae]

I never went beyond the french and taylor. So you can guage mye level from that.

Nice explanation. Thanks.

[Pappa]

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.

B3TA has a lot to answer for.

[lpetrich]

I'll suggest one:

A central force has the form F = (r/r)*f(r)

What possible forms of the force function f(r) give closed noncircular orbits? That is, orbits where r(a=2*pi) = r(a=0), where a is the position angle.

[JimC]

I'll suggest one:

A central force has the form F = (r/r)*f(r)

What possible forms of the force function f(r) give closed noncircular orbits? That is, orbits where r(a=2*pi) = r(a=0), where a is the position angle.

I'll look forward to an exposition on this one, it is somewhat beyond this Year 11 physics teacher, though I hope that when I see the process explained, I will understand (and enjoy) the steps...

[JimC]

No one has had a go at this one yet:

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! )

If no one has in the next day or so, I will give my solutions...

Hint - the inclined plane has to be very, very long...

[Twiglet]

I've just done the solution to this, but I am going to hang back on posting it, because I'd like to see if farsight can solve it. I'm happy to PM the solution to you Jim, within 24 hours and suggest anyone else does the same.

This problem is a great test, because unlike most of the quantum ones like solving hydrogen atom orbits, the solution can't be wiki'd easily lets see if farsight can put his money where his mouth is.

NB my solution is now PM'd to Jim.

[JimC]

I've just done the solution to this, but I am going to hang back on posting it, because I'd like to see if farsight can solve it. I'm happy to PM the solution to you Jim, within 24 hours and suggest anyone else does the same.

This problem is a great test, because unlike most of the quantum ones like solving hydrogen atom orbits, the solution can't be wiki'd easily lets see if farsight can put his money where his mouth is.

NB my solution is now PM'd to Jim.

Got it mate, a little different to mine...

Will check it tomorrow, to eliminate the faint chance that I made a mistake...

[lpetrich]

I've put all the problems and their solutions into a Mathematica notebook, so I can easily get results. If using Excel is cheating, then using Mathematica is even worse cheating. But it's easy to see the equations that one's using in it.

In the first problem, the velocity factorizes as (4 + t) (-7 + 3 t)

In the copper-and-water one, the reason that the copper seems so wimpy is its smaller volume and its smaller heat capacity.

In the octane-combustion one, its' 1 kg of octane and 3.50 kg of oxygen making 3.08 kg of carbon dioxide and 1.42 kg of water, the opposite of what JimC had posted earlier.

I have a solution to the inclined-plane-and-blocks one.

[JimC]

I've put all the problems and their solutions into a Mathematica notebook, so I can easily get results. If using Excel is cheating, then using Mathematica is even worse cheating. But it's easy to see the equations that one's using in it.

In the first problem, the velocity factorizes as (4 + t) (-7 + 3 t)

In the copper-and-water one, the reason that the copper seems so wimpy is its smaller volume and its smaller heat capacity.

In the octane-combustion one, its' 1 kg of octane and 3.50 kg of oxygen making 3.08 kg of carbon dioxide and 1.42 kg of water, the opposite of what JimC had posted earlier.

I have a solution to the inclined-plane-and-blocks one.

Just checked my previous answer for the octane one:

for every 1 kg of octane, 3.50 kg of O2 is consumed, and 1.42 kg of CO2 and 3.08 kg of H2O are produced

I wrote the damn things in the wrong order...

And please give you solutions to the plane problem, Twiglet can give his, and I'll give mine (which I must re-check). There's even a chance they may be the same!

[lpetrich]

I first calculate the accelerations of the two objects:

a = a(gravity) - a(friction)
a(gravity) = a(vertical)*sin(elevation angle) = 8.48705 m/s2
a(friction) = force(friction)/mass = {100 newtons / 20 kg, 50 newtons, 20 kg} = {5, 2.5} m/s2

a = {3.48705, 5.98705} m/s2

Each velocity is v = a*(t-t0) and each position is x = (1/2)*a*(t-t0)2, where x = v = 0 at t = t0.

Setting the two x's equal gives an equation for t with two solutions: 2.8358 s and 21.1124 s. The first one is for before the second object is released, so the appropriate time is the second one. At that time, both distances are 777.147 m.

When they meet, the two objects' velocities are 73.62 and 96.4657 m/s, and their difference is 22.8458 m/s.

[Twiglet]

My solution agrees with Ipetrichs:

Block 1.

F=ma

In the plane, F(body a)=mgSin60 - 100=69.74
F(body b)=119.74

a(body a)=3.49m/s^2
a(body b)=5.99m/s^2

s=ut+1/2at^2

we want to find the distance travelled & time when the particles catch each other,

B is released 5 seconds after A, so effectively,
Initial velocity for both particles is 0, as they start at rest

s(a)=s(b)

1/2xa(body a) (t+5)^2=1/2a(body b)t^2
1.745(t^2+10t+25)=2.995t*2
1.25t^2-17.45t-43.63=0

provides solutions for t from the standard quadratic formula:

x= [-b +/-(b^2-4ac)^1/2]/2a

a=1.25
b=-17.45
c=-43.63

so

t= [17.45 +/-22.86 ]/2.5

clearly negative time is unphysical so t must be 16.124s seconds (this is the time particle B has been travelling, a will have been going 5 seconds longer)

The distance travelled by both particles is obviously identical and should yield the same result for s=1/at^2
s(a) =0.5x3.49*(21.124)^2=778.66m
s(b)=0.5x5.99x(16.124)^2=778.65m which is good enough

v=u+at
v(a)=3.49x21.124=73.72m/s
v(b)=5.99x16.124=96.58m/s

[PsychoSerenity]

I keep meaning to get in on these. I'll see if I can find the time tomorrow, to find out if I can actually do any of them. I did physics and maths A-level, but it all seems rather a long time ago.

[Twiglet]

I'll suggest one:

A central force has the form F = (r/r)*f(r)

What possible forms of the force function f(r) give closed noncircular orbits? That is, orbits where r(a=2*pi) = r(a=0), where a is the position angle.

I could use a hint on this. Are you asking for all possible forms f(r) can take, or just the Keplers law/perihelion-apehelion stuff and conditions for escape from orbit?

[JimC]

After re-checking my plane solutions, I found a silly error; when I recalculated, got the same as lpetrich and Twiglet...

I first did a quick sketch of a velocity time graph with 2 lines; at a time t, you will have 2 triangles, whose area represents the displacement of each object. I found an expression for the 2 areas, set them to be equal, and ended with the following quadratic in t:

2.5t2 - 59.9t +149.75 = 0

Of which 1 solution is under 5 seconds, and so is ignored, the other is 21.12 s

The rest is easy... (it did turn out to be a bit of a ski slope, didn't it... )

This level of problem would be at the hard end of secondary physics, but certainly doable...

I like problems that involve simultaneous equations; they appeal to me, for some reason...

[lpetrich]

I'll suggest one:

A central force has the form F = (r/r)*f(r)

What possible forms of the force function f(r) give closed noncircular orbits? That is, orbits where r(a=2*pi) = r(a=0), where a is the position angle.

I could use a hint on this. Are you asking for all possible forms f(r) can take, or just the Keplers law/perihelion-apehelion stuff and conditions for escape from orbit?

All possible forms. I don't need conditions for escape from orbit - just which central-force functions give closed noncircular orbits.