Jim's maths and physics problems

[JimC]

Just for fun, I will pose a series of maths, physics and possibly chemistry problems aimed at the upper secondary level. Feel free to attempt solutions, or add problems of a comparable level yourself...

Problem 1

An object is moving in a straight line, with a velocity/time function of:

v = 3t2 + 5t - 28

and we know that its displacement from the origin at t = 0 is 5 metres

(in the following answers, negative answers for time are fine...)

a) At what times will the object have a velocity of 0?

b) At what time will the object have an acceleration of 23 m/s2?

c) Calculate the total displacement of the object after 20 seconds

d) At what times will the object be at the origin?

[JimC]

Problem 2

A 2 kg mass of copper at 250 degrees C is added to 20 litres of water at 20 degrees C.

Assuming no loss of heat energy to the surroundings, calculate the temperature at thermal equilibrium, given that the specific heat capacity of water and copper are 4200 j/kg/degree and 390 j/kg/degree respectively.

[Twiglet]

a) Factorising:

v=(3t+7)(t-2)

hence v=0 when t=-7/3 or t=2

b) a=dv/dt= 6t+5 hence 6t+5=23 simplifying, 6t=18 hence t=3

c) Not sure of this as just been to the pub!

v=dx/dt so dx/dt=3t^2+5t-28. Hence x(5) = Integral (3t^2 +5t-28).dt between limits of 0 and 20.

Performing the integral gives t^3 +5/2t^2 -28t +constant.

We know that when t=0 distance from origin is 5, so the constant is too.

anyway calculating the integral between limits just substitute t=20 into the above equation, the constant cancels anyway gives

x= [(20^3) + 5/2 (20^2) - 28.20 +5] -5 = 8440

d) at origin, x=0 so 0= t^3 +5/2 t^2 -28t +5
buggered if I can do that easily, but factorising the above gives the times.

I expect c & d are wrong but it's been a long time since I did A level physics

[JimC]

a) Factorising:

v=(3t+7)(t-2)

hence v=0 when t=-7/3 or t=2

It actually factorises to v=(3t+7)(t-4), so the answers are t = -7/3 s or t = 4 s

b) a=dv/dt= 6t+5 hence 6t+5=23 simplifying, 6t=18 hence t=3

Correct

c) Not sure of this as just been to the pub!

v=dx/dt so dx/dt=3t^2+5t-28. Hence x(5) = Integral (3t^2 +5t-28).dt between limits of 0 and 20.

Performing the integral gives t^3 +5/2t^2 -28t +constant.

We know that when t=0 distance from origin is 5, so the constant is too.

anyway calculating the integral between limits just substitute t=20 into the above equation, the constant cancels anyway gives

x= [(20^3) + 5/2 (20^2) - 28.20 +5] -5 = 8440

I get 8445 seconds... (the 5 is still added, I think...)

d) at origin, x=0 so 0= t^3 +5/2 t^2 -28t +5
buggered if I can do that easily, but factorising the above gives the times.

I cheated, and used my CAS calculator to solve the cubic, getting t -= -6.75, + 0.18 and + 4.07
I expect c & d are wrong but it's been a long time since I did A level physics

[Twiglet]

Sweet, just goes to show you should never differentiate (or factorise) in public, at least not with a beer in you!

You fooled me on c & d by having such nice numerical solutions to a & b I thought it would all nip nicely into place!

At least my method is still sound, even if my arithmetic sucks balls.

Cheers for the challenge Jim, I'll leave the second to someone else

I'm pretty sure the constants cancel each other out, as you have to calculate x(t)-x(0) which both contain the constant.

[JimC]

Now for a little chemistry...

a) Write a balanced equation for the complete combustion of octane

b) 1 kg of octane is burned. Calculate the mass of oxygen consumed, and the masses of carbon dioxide and water vapour produced, given the following relative atomic masses (accurate to 2 decimal places)

Oxygen 16.00
Hydrogen 1.01
Carbon 12.01

I have given such a problem to my advanced Year 10 science class, and damn well expect them to get it right!

[JimC]

Sweet, just goes to show you should never differentiate (or factorise) in public, at least not with a beer in you!

One G & T brings me to my optimum, more and the curve (at least for maths) goes down...

You fooled me on c & d by having such nice numerical solutions to a & b I thought it would all nip nicely into place!

Very hard to produce from scratch a cubic with whole number solutions which, when differentiated, makes a quadratic with also whole number solutions...

At least my method is still sound, even if my arithmetic sucks balls.



Cheers for the challenge Jim, I'll leave the second to someone else

I'm pretty sure the constants cancel each other out, as you have to calculate x(t)-x(0) which both contain the constant.

On reflection, I think you are right...

[Twiglet]

Here's a simple relativity one:

An object has been speeded up to 2.7x10^8 m/s relativity to its starting point.
The object weighs 1kg
Assume c=3x10^8m/s

Calculate the energy required to to propel the object to its final velocity using relativity.
Calculate the energy which would be required in purely classical terms.
Obtain the difference.

Perform the same calculation for a final velocity of 2.99 x 10^8 m/s

[JimC]

Here's a simple relativity one:

An object has been speeded up to 2.7x10^8 m/s relativity to its starting point.
The object weighs 1kg
Assume c=3x10^8m/s

Calculate the energy required to to propel the object to its final velocity using relativity.

2.86x10^17 J

Calculate the energy which would be required in purely classical terms.

4.5x10^16 J

Obtain the difference.

1.61x10^17 J (which is the energy equivalent of the increase in mass, from rest mass to relativistic mass)

Perform the same calculation for a final velocity of 2.99 x 10^8 m/s

I'll let some other bugger do that!

[Twiglet]

Here's a simple relativity one:

An object has been speeded up to 2.7x10^8 m/s relativity to its starting point.
The object weighs 1kg
Assume c=3x10^8m/s

Calculate the energy required to to propel the object to its final velocity using relativity.

2.86x10^17 J

Calculate the energy which would be required in purely classical terms.

4.5x10^16 J

Obtain the difference.

1.61x10^17 J (which is the energy equivalent of the increase in mass, from rest mass to relativistic mass)

Perform the same calculation for a final velocity of 2.99 x 10^8 m/s

I'll let some other bugger do that!

It's just fascinating to get a feel for the figures, when each extra bit of energy is just going "into the mass" rather than speeding the bugger up, and just how close to c one needs to get for there to be much difference from the classical picture.

If anyone wants a crack at seeing what I mean, try with 300m/s and 30,000 m/s.

I didn't check your calculations Jim, but from how you described them, you've used exactly the right method.

[Don't Panic]

Problem 2

A 2 kg mass of copper at 250 degrees C is added to 20 litres of water at 20 degrees C.

Assuming no loss of heat energy to the surroundings, calculate the temperature at thermal equilibrium, given that the specific heat capacity of water and copper are 4200 j/kg/degree and 390 j/kg/degree respectively.

Hmmm,

2*390(T-250)=20*4200(T-20)
780T-195,000=84,000T-1,680,000
1,485,000=83,220T
T=17.84

That makes no sense, think I fucked up somewhere.

[colubridae]

a) Factorising:

v=(3t+7)(t-2)

hence v=0 when t=-7/3 or t=2

It actually factorises to v=(3t+7)(t-4), so the answers are t = -7/3 s or t = 4 s

b) a=dv/dt= 6t+5 hence 6t+5=23 simplifying, 6t=18 hence t=3

Correct

c) Not sure of this as just been to the pub!

v=dx/dt so dx/dt=3t^2+5t-28. Hence x(5) = Integral (3t^2 +5t-28).dt between limits of 0 and 20.

Performing the integral gives t^3 +5/2t^2 -28t +constant.

We know that when t=0 distance from origin is 5, so the constant is too.

anyway calculating the integral between limits just substitute t=20 into the above equation, the constant cancels anyway gives

x= [(20^3) + 5/2 (20^2) - 28.20 +5] -5 = 8440

I get 8445 seconds... (the 5 is still added, I think...)

d) at origin, x=0 so 0= t^3 +5/2 t^2 -28t +5
buggered if I can do that easily, but factorising the above gives the times.

I cheated, and used my CAS calculator to solve the cubic, getting t -= -6.75, + 0.18 and + 4.07
I expect c & d are wrong but it's been a long time since I did A level physics

Jimc and twiglet

twiglet gives 8440m as his answer. this integration is the distance covered from t=0s to t=20s

the displacement is already 5m at t=0 s so the displacement ( not the distance covered) is 8445 m (Units guys tsk tsk)

[Twiglet]

Quite so, thanks col.

Farsight and mistermack - please take note - this is what an admission of a mistake looks like.

I stand corrected. I got the maths a bit wrong.

Oddly, I don't feel stupider or diminished for admitting it

[colubridae]

Quite so, thanks col.

Farsight and mistermack - please take note - this is what an admission of a mistake looks like.

I stand corrected. I got the maths a bit wrong.

Oddly, I don't feel stupider or diminished for admitting it

no worries dude. You are far ahead of me on these subjects. and I'm sober.

I've just come from o level physics invigilation.
one of the questions. "Why should a downhill skier wear a crash helmet?" dumbing down anyone?

How did you know you would expose farsight's twaddle so easily. We've been trying for nigh on 50 pages?

Edit and I failed at the cubic solution.

[JimC]

Problem 2

A 2 kg mass of copper at 250 degrees C is added to 20 litres of water at 20 degrees C.

Assuming no loss of heat energy to the surroundings, calculate the temperature at thermal equilibrium, given that the specific heat capacity of water and copper are 4200 j/kg/degree and 390 j/kg/degree respectively.

Hmmm,

2*390(T-250)=20*4200(T-20)
780T-195,000=84,000T-1,680,000
1,485,000=83,220T
T=17.84

That makes no sense, think I fucked up somewhere.

The error was that the copper is losing heat energy, and so it is a negative term, leading to:
2*390(250 - T) = 20*4200(T - 20)
When solved by the same method you used, the temperature at thermal equilibrium comes to 22.12 degrees C...

(is it sad that I have made an Excel sheet which calculates problems like this?)