Post by: LinusX on June 12, 2011, 09:57:13 am
Thanks in advance,
LinusX
Oh yeah, the solutions dont do this
Post by: golden on June 12, 2011, 11:15:37 am
Question 3
If the hanging mass, mH
Worked solution , is increased to 1000 kg and the friction force stays the same, what is the magnitude of the container’s acceleration now?
ΣF = ma
10 000 – (1000 + 7000) = 1400a
Therefore, a = 1.43 m s–2.
I don't get why they took off the 1000 at all.
Post by: LinusX on June 12, 2011, 11:45:11 am
Post by: golden on June 12, 2011, 12:12:52 pm
Shouldn't it be C not D?
Post by: Lasercookie on June 12, 2011, 12:15:41 pm
Question 7 from Relativity on Insight 2010:
Shouldn't it be C not D?
I got C as my answer as well.
The worked solutions for Insight 2010 also show the answer for C...yet they wrote D in the box.
Post by: r_damico on June 12, 2011, 12:16:18 pm
ΣF = Fdriving - Fretarding
where Fretarding = weight component + friction
Post by: golden on June 12, 2011, 12:27:56 pm
Question 7 from Relativity on Insight 2010:
Shouldn't it be C not D?
I got C as my answer as well.
The worked solutions for Insight 2010 also show the answer for C...yet they wrote D in the box.
And I suppose question 8 is wrong too? I got 2.98 x 10^-14 J.
Could someone please explain Q12.
Two ships go at 0.65 c away from each each other.
What is the speed of one ship as measured by the other?
Post by: LinusX on June 12, 2011, 10:09:09 pm
Post by: thatricksta on June 12, 2011, 10:20:49 pm
Wait a minute how did the solutions even get this?
Question 3
If the hanging mass, mH
Worked solution , is increased to 1000 kg and the friction force stays the same, what is the magnitude of the container’s acceleration now?
ΣF = ma
10 000 – (1000 + 7000) = 1400a
Therefore, a = 1.43 m s–2.
I don't get why they took off the 1000 at all.
this is the answer I got
the previous question indicated that when fnet = 0 and mh is the only force acting upon the object, the friction force is 8000N. when increasing the applied force on the object to 10,000 N, the frictional force is still the same (8000N)
Question 7 from Relativity on Insight 2010:
Shouldn't it be C not D?
I got C as my answer as well.
The worked solutions for Insight 2010 also show the answer for C...yet they wrote D in the box.
And I suppose question 8 is wrong too? I got 2.98 x 10^-14 J.
Could someone please explain Q12.
Two ships go at 0.65 c away from each each other.
What is the speed of one ship as measured by the other?
This one is not as complicated as it seems.
the formula for this is
So, lets say Freddie is going at w and Khokho is going at velocity v
because velocity is a vector, in respect to Freddie, Khokho is going at a velocity of -v
therefore you will get
= 0.91c :)
I also got C for question 8.
Post by: Vincezor on June 12, 2011, 10:21:54 pm
Can anyone explain this question? I still dont understand why you dont use both the masses in the calculation as the net force has to act on both the objects to give a shared acceleration.
Yeah I'm wondering as well...
I get
As F = ma
Post by: thatricksta on June 12, 2011, 10:37:36 pm
Can anyone explain this question? I still dont understand why you dont use both the masses in the calculation as the net force has to act on both the objects to give a shared acceleration.
Yeah I'm wondering as well...
I get
As F = ma?
In relation to tension, components are isolated.
I don't believe this question was designed to act as a pulley, this is just regular motion with a net force similar to an engine if you want to think of it like that.
Post by: Vincezor on June 12, 2011, 10:43:10 pm
In relation to tension, components are isolated
Sorry I still don't get it. Why aren't both masses accelerating at the same rate? The question doesn't ask for tension? :S
Post by: thatricksta on June 12, 2011, 10:44:49 pm
In relation to tension, components are isolated
Sorry I still don't get it. Why aren't both masses accelerating at the same rate? The question doesn't ask for tension? :S
Just take the falling mass as the driving force acting on the object... I do this every time, the net force comes from the tension of the string caused by the falling object.
Post by: Vincezor on June 12, 2011, 10:47:06 pm
Just take the falling mass as the driving force acting on the object... I do this every time, the net force comes from the tension of the string caused by the falling object.
Oh I think I get it now. So does this mean the hanging block is moving at a constant speed down? (a=0)?
Post by: thatricksta on June 12, 2011, 10:53:28 pm
Just take the falling mass as the driving force acting on the object... I do this every time, the net force comes from the tension of the string caused by the falling object.
Oh I think I get it now. So does this mean the hanging block is moving at a constant speed down? (a=0)?
a = gravity, so F=ma
I tend not to over think it too much, rather just look at it and be like OK, gravity, mass, F=ma.
Post by: LinusX on June 13, 2011, 12:27:59 am
Oh well I think I'll sleep on it and see what happens.
Post by: thatricksta on June 13, 2011, 12:29:37 am
Oh yes I kinda get it now - just kinda :P
Oh well I think I'll sleep on it and see what happens.
id love to say just do more questions on it, but the questions don't come up frequently... don't stress :) put it on your cheat sheet if your worried
Post by: golden on June 13, 2011, 10:56:24 am
We have to add the masses in that case. Why is it different?
Post by: thatricksta on June 13, 2011, 01:18:46 pm
Lol what about 2010 VCAA Physics question 3?
We have to add the masses in that case. Why is it different?
there is no slope?
otherwise idk what the answer is
Post by: cranberry on June 13, 2011, 06:32:59 pm
anyone know why they bias the Vout graph to have the middle Vin as the starting value??(<----lol...)
see attached