Mechanics Help! (1 Viewer)

[Beaconite]

a) (ii)

 

[jet]

We know we'll need an expression for the terminal velocity, so we have to calculate it:


This makes sense - it should depend on the strength of gravity and the resistive force of the medium (i.e. )

The force on the second particle is when our axis goes vertically upwards from the origin. Hence,


The second particle is momentarily at rest when . This means


So when we substitute into the equation for velocity of the first particle (and writing for exponential), we get

 

[math man]

 

[Beaconite]

Thanks guys

for terminal velocity, we could also say acceleration =0 right?

 

[math man]

Yes that's what I was implying

 

[jet]

Thanks guys

for terminal velocity, we could also say acceleration =0 right?

That's the more physical way of interpreting it (and technically more correct). I just know this means it will have eventual constant speed so I take the limit, there's no difference.

 

[Beaconite]

how did it go from second last step tp last step?

 

[jet]

4 = 2² so you move one of the twos into the 2^k, and 2*a > a when a is not zero or negative (a is 2^k+1 in this case so it can never be 0 or negative).

 

[jet]

They're not... Can't you see the > sign?

 

[barbernator]

I know that, but how could you write 2.2^k+1 as 2^k+1

shouldn't it be 2^k+2 ?

they arent writing it as that. They are saying that one is less than the other for the purpose of the proof.

 

[Beaconite]

they arent writing it as that. They are saying that one is less than the other for the purpose of the proof.

lol yeah I get it

 

[Beaconite]

iii) and iv)

 

[nightweaver066]

View attachment 26205

iii) and iv)

Suppose z = 2(cos@ + isin@)

Conjugate z = 2(cos@ - isin@), so just reflected across the x-axis.

-z/2 would be reflected across the x and the y axis where the modulus is halved (so now 2)

1/z - it is reflected across the x-axis and the modulus is now 1/4 (shrinked to 25% original)

sqrt(z) - modulus is now 2, and argument is halved (De Moivre's theorem).