Applied Maths: Mechanics

Scroll to see replies

Reply 20

abdult93

Now to wait for Meteor to hurry up with section A... anyday now meteor :P

Reply 21

TheRabbit

ukdragon37

Solutions to section B

Thanks, I can see I made a few small mistakes - for instance I did integration by parts on one side but said the integral of y dy was y :facepalm:

Reply 22

abdult93

Ouch "/ I'm sure you've done well :smile:

Meteor please put up your solutions! =D

Reply 23

Meteorshower

I am sorry, I had a lot on today that I didn't foresee happenning. Am getting going now :p:

Thank you =D

How long will you take before you finish and upload the solutions? Cos if it's gonna be more than half an hour I may go to sleep lol :L

Reply 26

Meteorshower

abdult93

How long will you take before you finish and upload the solutions? Cos if it's gonna be more than half an hour I may go to sleep lol :L

Go to sleep. I think i'll finish them in the morning.

Reply 27

scotty dog

Yes i agree solutions please and if you're busy then just do Q8??!!!! Thanks for section b that was good

Reply 28

Meteorshower

Ok I don't know how to do a fancy template like Ukd does, so you'll have to have them in crappy form. I'd like to sress that this is the way I did them. I recognise there are various other ways possible, but they should all get the same answer. If you spot a mistake feel free to point it out. Posting incomplete now to save work, will edit as I complete. Will do some more this afternoon, it's taking long to type up and i'm busy at the moment. Sorry about that.

1. At the origin (traffic light stop), the car is at rest but has acceleration a and the lorry has speed u. So doing two separate integrals

For car acc = a Int(acc)dt = adt = 0 + at (0 is initial speed)
v = at Int(v)dt = 0 + 0.5at^2 (choose traffic lights as origin)
For lorry velocity = u integrating gives you 0 + ut

Equating distances will be at the time they meet - ut = 0.5at^2 Gives you u =at/2 and consequently t = 2u/a

Put this into either expression for distance you derived and you get S = 2U^2/a

2. Draw a triangle, and you will see that the vertical component of force is mg acting down, and horizontal component is mw^2r. You can therefore represent the tension in the string like this Tsin(a) = mw^2r and Tcos(a) = mg (gonna use a as angles) Dividing Tsin(a) by Tcos(a) leaves you with tan(a) = w^2r/g which if you evaluate with the given values you get a = 78.9 degrees

3. Equating forces perpendicular to the plane (u shall be mu)

R = mgcos(a)
ma + uR = mgsin(a)

ma = mgsin(a) - umgcos(a)
a = gsin(a) - ugcos(a)

Then using the equations V^2 = U^2 - 2as we rearrange to s = V^2/2a given that U = 0
Substitute in and you get the required answer. I have a more lengthy calculus derivation if you aren't satisfied with the rigour there, but for AH mechanics you are allowed to take for granted the equations of motion as long as a is constant.

4. Change in momentum required for her to stop is 400kgms^-1 This is the equivalent to force integrated with respect to time. So int(8t)dt = 4t^2

4t^2 = 400 so t = 10 seconds. You can get this easily using newtons second law and some calculus aswell.

using a = -8t/m = -2t/25

v = 4 - t^2/25 this is a integrated with respect to t. 4 is the initial velocity

s = 4t - t^3/75 this is v integrated with respect to t. Initial displacement is 0 when t = 10 s = 40 - 1000/75 = 26 and 2/3 m

5. F(t) = 100cos((pi/2)t) Change in momentum is force integrated with respect to time.
This gives change in momentum = 200/pi(sin((pi/2)t) and when you substitute in t = 0 and t = 1 you get 200/pi kgms^-1

6. F = -kx F= -yx/l where y is the modulus of elasticity given as y = kl and l is the length
Subsitute l = 1 and y = 4 to get F=-4x newtons
Since F= ma a = -4x/0.25 = -16x. A is dx^2/dt^2 and this gives you what you need with w = 4.

Maximum speed is v = wa which gives 0.2*4 = 0.8ms^-1

7.

Fnet = -mkv^2 - mg as both gravity and resistance are acting downwards

F = ma so a =-kv^2 - g

dv/dt = -kv^2 - g dv/dt = (dv/dx)*(dt/dt) = (dv/dx)*(dx/dt) = vdv/dx

-vdv/(kv^2+ g) = dx

-(1/2k)[ln(kv^2+ g)] = h over v = u to v = 0

so -(1/2k)(lng - ln(kv^2 + g) = h

1/2k * ln((kv^2+g)/g) = h Substitute k = 25 and you will get the right answer

8. Ok this is a fun one :biggrin:

Energy conservation is a very good way to do it, as Rabbit says.

We have energy before = to the energy stored in the spring and energy after is the energy left in the spring + work done against friction and gain in potential

Energy stored in spring is (yx^2)/2l since l is one and x is 1 before hand energy = y/2

Energy left in spring is obtained using the same equation, but x = 3/4 This gives (y*(3/4)^2/)/2 = 9y/32

Work done against friction. Force is constant over the distance 0.25 m. Equating forces perpendicular to plane gives R = mgcos(a) so since force due to friction is uR, we get u*cos(a)*M*G/4 and since a = 45degrees and m = 2 this all works out to ug/2root2

Potential energy change = 0.25sin(a)*g*m = g/2root2

This gives y/2 = 9y/32 +(g/2root2)(1+u)

y/2 = 16y/32 and 16y/32 -9y/32 = 7y/32.

So 7y/32 = (g/2root2)(1+u)

and y = (8root2*g/7)(1+u)

9. Pretty standard derivation, just integrating and putting initial conditions in each time.

a = -g
v = Vsin(a)-gt
y = Vsin(a)t -0.5gt^2

x = Vcos(a)t so t = x/Vcos(a)

Substituting in gives y = (Vsin(a))(Vcosa))*x - gx^2/V^2*cos^2a
y = xtan(a) -(gx^2/V^2)(1+tan^2a) using the identity given

When V = 3(gh/2)^0.5 x = 3h and y = h

h = 3htan(a) - (18gh^2/2*9gh)(1+tan^2a)

h = 3htan(a) - h -htan^2a

tan^2a - 3tana + 2 = 0

(tana - 2)(tana-1) = 0 So tana = 1 or 2 giving 45 degrees and 63.4 degrees as possible angles of projection (don't need to state the angles)

So when tana = 2 we have the same v, but we want x = R and y = 0

0 = 2R - 2gR^2/2*9gh(1+4) = 2R - 10R^2/18h

So 2R(1-5R/18h) = 0 so either x = 0 or 1 = 5R/18h 5R= 18h so R = (18/5)h metres

10. Hate these questions! But anyway...

If it's going 45 degrees and the velocities for x and z components are respectively Vsina and Vcosa then we have

VA = (V/root2)i +(V/root2)k so rA = (V/root2)t i + (V/root2)t k as it starts at the origin

RB is just the planes position when t = 0 + it's velocity times time. So rB = Li +((V/root2)t - L)j +4Lk

To get the distance between the planes, we need to consider the positions relative to each other, this is found by the formula
position of B relative to A = rB-rA = (L-Vt/root2)i + (Vt/root2 - L)j +(4L-Vt/root2)k

If we want to find the magnitude of this, we have to use pythagoras

D^2 = (L-Vt/root2)^2 +(Vt/root2 - L)^2 +(4L-Vt/root2)^2 = V^2t^2(1/2+1/2+1/2) +L^2(1+1+16) +VLt(root2 +root2 +4root2) which gives you the correct answer

Differentitaing with respect to time we get 3V^2t -6root2VL = 0 when there is a turning point in D
solving this you get VT = 2root2L

Substituting this into the equation you obtain for the last part, you get D^2 = 12L^2 - 24L^2 + 18L^2 = 6L^2 So D= root6L at minimum

11. Ok, we know that when the rod is at the bottom the total energy is 0.5MU^2 and this will be equivalent to the change in potential energy when it has no kinetic energy.

(MU^2)/2 = mgh
h=R(1-cosa) a = pi/4 so h=R(1-1/root2)

so U^2 = Rg(2-root2)

For the next bit i'm going to put the disclaimer that i'm not sure and would like you guys to correct me if i'm wrong :p:

My exams have just finished and i'm tired so cut me some slack

Total energy = 0.5mU^2

Kinetic = 0.5mV^2

Potential = mgR(1-cosa)

So V^2 = U^2 -2gR(1-cosa)

Not confident enough on the last bit to post it up, well at least not at this time of night. If anyone could let me know how they started/did it that would be helpful.

Reply 29

TheRabbit

scotty dog

Yes i agree solutions please and if you're busy then just do Q8??!!!! Thanks for section b that was good

Can't write out in full as revising for bio:

PE(spring, before) = PE(spring, after) +mgh +WD(friction)

lambda/2 = 9/32 lambda + mgh + WD(friction)

PE(spring) = lambda x^2 / 2l

If you work through you come out with the correct line.

Reply 30

scotty dog

THanks for the solutions and thanks rabbit for Q8- i get it now....not sure where i went wrong in exam b/c i did a different method- resolving forces which i think was completely wrong but came out with the same answer just 8/6 instead of 8/7 :frown:

oh well the rest seems to have gone pretty well- off to do more biology now

Reply 31

Meteorshower

Updated solutions - will get the finished before tea time, busy again now :p:

Reply 32

TheRabbit

Can't really remember my answers but I did 2 a different way- getting the same answer- and 4 i did differently - but I can't remember my answer (I did something like a=8t^2 /100 and went down that route). Did 5 the same but didn't know to put 0 and 1 in - hopefully I will get marks for recognising you had to integrate it.

Hoping for a high B or A, even though it doesn't matter in terms of university I will probably appeal if I get a B as I got over 90% in the prelim and 100% in the second prelim.

Good luck for tomorrow Scottydog, i'm pretty worried about it (I hate unit 1 and essay questions)

Reply 33

SEljamel

Damn hard exam for me.

Half the stuff I revised wasn't in there. The stuff I got told to forget about was in there.

I did all the past papers there were and found them considerably easy. Then the exam comes and it's like... what the... did they suddenly turn the knob up to extreme difficulty?!

Reply 34

scotty dog

Thanks for solutions- think i've done alright except for q8.....thanks rabbit- you 2. I'm worried about essays too but there was a bitch of a DNA one last year so at least that won't come up....best case scenario- essay on nitrogen cycle/succession (which i used to hate)/ something like grazing + parasitism etc.

Also I want the Data Handling to be nice - 13 marks would be good not 13 and also what did u do for unit 3??

And re Q5 i did the same thing...put limits as t and 0 instead of 1 and 0 because i thought the force could be for anything between 0 and 1??

Reply 35

Meteorshower

scotty dog

And re Q5 i did the same thing...put limits as t and 0 instead of 1 and 0 because i thought the force could be for anything between 0 and 1??

You've gone from time equal to 0 and "summed up" all the force elements times time. If the force is constant then you just multiply it by the change in time to get the change in momentum, but the integration takes care of it potentially being anything, as the principle is that you've summed it continuously.