Q (2 Viewers)

[CrashOveride]

a particle of mass m moves along a straight line under the action of a constant propelling force P, and a resistive force mkv, where k is a constant, v is the speed at any time t. Show that if the speed increases from 2m/s to 4m/s over a time interval of 5 seconds,

P = 2km [ (2e^(5k) - 1) / (e^(5k) - 1)] . Done.

Find the corresponding distance moved. I thought i went about this the right way but at the back they have this semi-long log thing.

 

[CM_Tutor]

the equation of the normal to the ellipse of standard form (a^2 > b^2) at the point P(x1,y1) is a^2y1.x - b^2x1.y = (a^2 - b^2)x1.y1

This normal meets the major axis of the ellipse at G. S is the focus of the ellipse. Show that GS = e.PS (where e is the eccentricity of the ellipse)

Ok i found the co-ords of G and doing found GS but for the PS part i went manually with the distance formula and it looks like it gets messy any easier ways to do this one/ cheers

1. Show that G is at (e²x₁, 0) by putting y = 0 into the equation of the normal, and using b² = a²(1 - e²).

2. Noting that G must lie between S and O, the origin, it follows that GS = ae - e²x₁

3. Now, PS = ePM = e[(a / e) - x₁] = a - ex₁

4. It follows that GS = ePS

 

[CM_Tutor]

The acceleration due to gravity at a distance r from the earth's centre is directed towards the centre and equal to k/r^2 when r>R, equal to kr when r < R and equal to g when r = R. Imagine a narrow tunnel along a diameter XY of the earth and the particle is projected from X with initial velocity U towards Y. (R = radius of earth)

If U^2 < 2gR, prove that motion is oscillatory and amplitude is given by R / (1 - U^2/2gR)

Just started resisted motion, little confused.

Firstly, you have the right to be confused - this is a tricky question.

OK, take the origin of motion (x = 0) at the centre of the Earth, which is at the midpoint of XY. We have acceleration governed by:

x'' = -k₁ / x², for x > R
x'' = -g, for x = R
x'' = -k₂x, for -R < x < R
x'' = g, for x = -R
x'' = k₃ / x², for x < -R

For the sake of continutiy, we need lim_x-->R+x'' = lim_x-->R-x'' = -g, and lim_x-->-R+x'' = lim_x-->-R-x'' = g.
From this, it follows that k₁ = k₃ = gR² and k₂ = g / R. Thus, we have:

x'' = -gR² / x², for x > R
x'' = -g, for x = R
x'' = -gx / R, for -R < x < R
x'' = g, for x = -R
x'' = gR² / x², for x < -R

The motion starts at X (x = +R) with v = -U. Using x'' = -gx / R for the motion between X and Y, we get that
v² = U² + gR - gx² / R.

From this, it follows that v = -U at Y (x = -R), and so the particle travels through the tunnel to Y, emerging with speed U. (Note that the particle cannot stop inside the tunnel, and so the sign of v cannot change inside the tunnel.)

We now must use x'' = gR² / x² with v = -U at x = -R, and we find that
v² = U² - 2gR - 2gR² / x
Provided U² < 2gR (which we are given), the particle is stationary (v = 0) at
x = 2gR² / (U² - 2gR) = -R / (1 - U² / 2gR).
At this point, x'' > 0, so the particle rises back to Y, where v = +U. It then travels through the tunnel, governed by
v² = U² + gR - gx² / R (derived above), returning to X but with v = +U.

We now must use x'' = -gR² / x² with v = +U at x = +R, and we find that
v² = U² - 2gR + 2gR² / x
Provided U² < 2gR (which we are given), the particle is stationary (v = 0) at
x = 2gR² / (2gR - U²) = R / (1 - U² / 2gR).
At this point, x'' < 0, so the particle falls back to X, where v = -U, which is where it started.

The cycle repeats, and so the motion is oscillatory, with amplitude R / (1 - U² / 2gR), as required.

 

[CM_Tutor]

a particle of mass m moves along a straight line under the action of a constant propelling force P, and a resistive force mkv, where k is a constant, v is the speed at any time t. Show that if the speed increases from 2m/s to 4m/s over a time interval of 5 seconds,

P = 2km [ (2e^(5k) - 1) / (e^(5k) - 1)] . Done.

Find the corresponding distance moved. I thought i went about this the right way but at the back they have this semi-long log thing.

I didn't check the 'P' bit, but for the second part, I got the distance was:

x = (5P - 2m) / mk = [10(2e^5k - 1) / (e^5k - 1)] - (2 / k) = 2[e^5k(10k - 1) - 5k + 1] / k(e^5k - 1)

Can you confirm that one of these forms is the required answer? If so, I can post an explanation. If not, what is the answer they got? (My answer was a 'semi-long log thing'.)

 

[CrashOveride]

answqer

Answer:

1/K [-2 + (P/(mk))log[(P-2mk)/(P-4mk)]]

 

[CM_Tutor]

Answer:

1/K [-2 + (P/(mk))log[(P-2mk)/(P-4mk)]]

Yep, that's the same answer I got. The (P - 2mk) / (P - 4mk) bit simplifies to e^5k, which gives:

(1 / k) * [-2 + (P / mk) log_e[(P - 2mk) / (P - 4mk)]]
= (1 / k) * [-2 + (P / mk) log_ee^5k]
= (1 / k) * [-2 + (P / mk) * 5k]
= (1 / k) * (-2 + 5P / m)
= (5P - 2m) / mk

which was the first of my answers. The other answers come from substituting in the value of P and trying to simplify.

My method: We know from the question that &Sigma;F = P - mkv = mx''
So, x'' = (P - mkv) / mv.

Using x'' = v * dv/dx and rearranging, you get:

dx/dv = mv / (P - mkv) = (1 / k) * [P / (P - mkv) - 1]

We want the distance travlled when the speed increases from v = 2 to v = 4, which is the definite integral
&int; (2-->4) dx/dv dv
= (1 / k) &int; (2-->4) P / (P - mkv) - 1 dv
= (-2 / k) + (P / mk²) * ln |(P - 2mk) / (P - 4mk)|, upon evaluating.

 

[CrashOveride]

Ok makes sense just how did you see this mv / (P - mkv) = (1 / k) * [P / (P - mkv) - 1]

I can see how it works by expanding...how did u get to there?

 

[CM_Tutor]

Ok makes sense just how did you see this mv / (P - mkv) = (1 / k) * [P / (P - mkv) - 1]

I can see how it works by expanding...how did u get to there?

You need to rewrite the numerator until in matches the denominator. So:

mv / (P - mkv)
= (1 / -k) * -mkv / (P - mkv)
= (-1 / k) * (P - mkv - P) / (P - mkv)
= (-1 / k) * [(P - mkv) / (P - mkv) - P / (P - mkv)]
= (-1 / k) * [1 - P / (P - mkv)]
=(1 / k) * [P / (P - mkv) - 1]

or, do the long division.

 

[KeypadSDM]

Man, I'm so out of it. Looking at that stuff makes my head swim now ...

 

[ToO LaZy ^*]

easy conics q

i know the formula for a tangent to the ellipse is x*x1/a² + y*y1/b² = 1

but when i work it out manually, i can't seem to get it into that form.

 

[ngai]

was that some past catholic trial?
looks familiar
anyway, forget about the 2cp, cp2 for the moment, and prove ur xx1/a^2 + yy1/b^2 = 1
use implicit diff, sub in (x1,y1), etc etc
and u should get tangent is:
xx1/a^2 + yy1/b^2 = x1^2/a^2 + y1^2/b^2
and RHS of that = 1, coz x1 y1 is on ellipse
then use the formula for ur 2cp cp2
if u try 2 do it using (2cp,cp2) straight away, then u'll get a mess for the equivalent of x1^2/a^2 + y1^2/b^2, which looks even worse if u put everything on common denominator

 

[ToO LaZy ^*]

yep. its the 2002 CSSA paper
ohhhkk..yeah, i used your way after i couldnt' figure out my first way and it worked out ok.
so basically you should just let x1 = 2cp and y = cp2 and then sub them back in once you've done the differentiation.
cheers

 

[ToO LaZy ^*]

what was that rule again for the red dots?...was it something like...angle between the chord and tangent is equal to the opp angle in the alternate segment?..

 

[unmentionable]

yup thats right

 

[ToO LaZy ^*]

any ideas for part iii) (the one circled in red)

 

[unmentionable]

< aep = < adf (ext. < = to opp. int. < in cyclic qaud AEFD)
< pba = < adf (" " in cyclic quad ABCD)

.'. < aep = < pba
.'. PBEA is cyclic quad (= <'s standing on same arc AP)

 

[ToO LaZy ^*]

but how does that tell me it's a cyclic quad?

 

[unmentionable]

the angles are subtended by the same arc and they are equal
that means those points all stand on the circumference of the same circle
because the angles on the circumference of a circle subtended by the same arc are equal

o.o i hope that made sense

 

[ToO LaZy ^*]

ohhk..kinda get it.
thanks man.

 

[CrashOveride]

a projectile is fired vertically upwards from the earths surface with velocity U m/s. The retardation due to gravity is given by the law k/x^2 where x is the distance of the projectile from the centre of teh earth, and k is a constant. The acceleration due to gravity on the earths surface is g. The earths radius is R. Neglecting the air resistance show that if U^2 = gR, then the projectile reaches the height R above the earths surface. What is the time for this journey?

Ok i do all the fundamental stuff in deriving this but i get R(gR-v^2) / (v^2 + gr). Wont cancel :/