VCE Specialist 3/4 Question Thread!

[Lasercookie]

(1)  Is is true that a complex number is rotated by "θ" (anticlockwise) when it is multiplied by cis(θ)?

Yep. You can see and draw it out it for yourself.

and



(you could also do it in rectangular form if that's more intuitive to you)

(2) If I use the vector formula a.b=a*b*cosθ to find the angle between two vectors,
 
 e.g OA.OB=OA.OB.cosθ (1)   

     OA.BO=OA.BO.cosθ (2) 
 
 The second one is incorrect, isn't it? I think the direction is restricted but some trial exam papers haven't specified that.

Just emphasising that it's the magnitudes of the vectors in that equation. Other than that, I don't see anything wrong with the second one. The is measured to be the angle between u and v. I don't believe there's any domain restriction on that angle.

[martin1106]

(1)  Is is true that a complex number is rotated by "θ" (anticlockwise) when it is multiplied by cis(θ)?

Yep. You can see and draw it out it for yourself.

and



(you could also do it in rectangular form if that's more intuitive to you)

(2) If I use the vector formula a.b=a*b*cosθ to find the angle between two vectors,
 
 e.g OA.OB=OA.OB.cosθ (1)   

     OA.BO=OA.BO.cosθ (2) 
 
 The second one is incorrect, isn't it? I think the direction is restricted but some trial exam papers haven't specified that.

Just emphasising that it's the magnitudes of the vectors in that equation. Other than that, I don't see anything wrong with the second one. The is measured to be the angle between u and v. I don't believe there's any domain restriction on that angle.

for example, OA=i+2j OB=i-j  BO=-i+j ; In this case the dot product of OA and OB vs.OA and BO is different, hence the angle cannot be the same

so the starting point of two vectors must be the same i.e. OA and OB

 I am not sure if i my thinking is correct

[Jenny_2108]

Of course their angles are different. It must be tail-to-tail as well

[Lasercookie]

Oh you were asking if dot products (1) and (2) were equivalent. Sorry, didn't pick up on that.

Yeah what Ennjy said, OB and BO are different vectors, I think forgetting that is where you're getting confused.

[soccerboi]

A hot air balloon is ascending with uniform velocity of 5m/s. When the balloon is 60 m above the ground pone of the occupants drops his watch. Assuming that the watch experiences no air resistance, find the maximum height reached by the watch.

I used v²=u²+2as, but had a=0 as it says uniform velocity but ended up not being able to solve for s. Why is the a=-9.8? I thought that uniform velocity meant velocity stays the same so acceleration is 0.

Thanks

[b^3]

When it is initialy rising before we release it, then it is ascending with uniform velocity. But as soon as we release it (at ), it will start falling under tha influence of gravity alone, and so the acceleration will be the acceleartion due to gravity, .

[soccerboi]

How do i find the time taken for the watch to reach the ground?

I considered the point from when it was at its max height to when it hit the ground. So i had s=61.28,t=?,a=9.8, u=0
And used s=ut+0.5at², but got t=3.54. The answer should be t=4.05, what have i done wrong?

[paulsterio]

(2) If I use the vector formula a.b=a*b*cosθ to find the angle between two vectors,
 
 e.g OA.OB=OA.OB.cosθ (1)   

     OA.BO=OA.BO.cosθ (2) 
 
 The second one is incorrect, isn't it? I think the direction is restricted but some trial exam papers haven't specified that.

Just emphasising that it's the magnitudes of the vectors in that equation. Other than that, I don't see anything wrong with the second one. The is measured to be the angle between u and v. I don't believe there's any domain restriction on that angle.

Hmm, there is a bit of an issue though, the angle between u and v and the angle between v and u will be different (they are supplementary), so yeah, watch what they're asking you I guess.

[b^3]

How do i find the time taken for the watch to reach the ground?

I considered the point from when it was at its max height to when it hit the ground. So i had s=61.28,t=?,a=9.8, u=0
And used s=ut+0.5at², but got t=3.54. The answer should be t=4.05, what have i done wrong?

You will need to add the time it takes to get to max height in the first place aswell. So you found the time for the downwards part of it's motion.
To find the time for the upwards part you will have.
u=5 m/s
v=0 m/s (stationary at max height)
a=-9.8 m/s
t=?
v=u+at
t=(v-u)/a
t=(0-5)/9.8=0.51 seconds

Total time T=0.51+3.54=4.05 seconds
-OR-
If you didn't want to work it out in parts like that, and just do it in one shot you could do this.
We know that it is initially 60 m above the ground. So when it hits the ground it will be 60 m below its starting point.
So we would have
s=-60 m
u=+5 m/s
a=-9.8 m/s^2
t=?
s=ut+0.5at²


Ignore the negative solution, so

EDIT2: Beaten to get the edit in for the easier way.

[paulsterio]

s = 60m
a = 9.8ms-2
u = -5ms-1
t = ?

s = ut + 1/2at^2

60 = -5t + 4.9t^2

CAS gives me: t = -3.03s or 4.05s

Obviously we know it's 4.05s because of the domain of the time variable - might be easier doing it this way as opposed to b^3's, although both methods are right.

[soccerboi]

Ok thanks guys but why is a negative? If its falling downwards, isn't it going in the direction of gravity?

[BubbleWrapMan]

It depends whether you define positive as up or down.

[b^3]

It just depends which way you denote as positive. If you take up as positive, then be consistent with it. I.e. Paul took down as positive, where as I took up as positive (as I was sticking with things from the previous question). It won't matter in the end as long as you are consistent with everything in that part of the question. E.g. if down is positive, then a and s will be positive but the initial velocity u will be negative.

I got used to just always taking up as positive, as it meant I wouldn't make mistakes, but that was just me, what ever works best for you (as long as you have everything relative to each other in the right direction)

EDIT: beaten.

[generalkorn12]

So, for the position vector, r(t) = (4t-20)i + 2t j -+ (80 - 2t) k, the worked solution still confuses me. The question is, 'At what angle does the kite hit the ground?', they seem to be using k still, but Im assuming that k=0....

[b^3]

The angle at which an object hits the ground will be dependent on the velocity vector, not the position vector.
That is

So the angle that it hits the ground at will be given by (draw the triangle out, the hypotenus will be the magnitude of the vector and the Opposite side will be equal to the magnitude of the k component)



EDIT: Added the image, didn't turn out to be what I wanted it to, but it shows it anyway.
EDIT2: Forgot to mention something.
Once you get the velocity vector, you will need to get the velocity vector at time t where it hits the ground, that would be when the k component is 0, then you sub in that t into the velocity vector. But in this case the velocity vector is not dependent on t, as it is moving with constant velocity, so it didn't matter in this situation, but it's just something I thought I would point out