Rod's Physics 3/4 Questions Thread

[lzxnl]

Technically, if the Earth revolved around an axis through its centre of mass, changing its rotation speed would not change the centre of mass's momentum, so changing the rotation speed would not compensate for the lost momentum
I agree with potentially changing its trajectory through space though.

Thank you so much PB!

I'm going to follow everything you have said apart from the questions part. I thought doing questions would consolidate my knowledge and prove that I've studied that chapter well and can move on to the next one. So I think I will be doing questions, but not as much. Initially, I've been doing the chapter review questions, chapter questions, A+ note questions and heinemann text book questions but might just do one, not four. I'm up to chapter 3 with 6.5 weeks remaining.

So this is what I'm doing:
- Reading the whole chapter
- Summarising it and taking notes
- Doing four lots of questions
- Then moving on

Is that okay? The only thing I'll change after reading your advice is probably do only one lot of questions, maybe the chapter review. Thanks for all your help, I joined Atarnotes yesterday and I can't believe how awesome it is. How VCE specialists like you are sparing their time and helping others in need.

Your plan seems quite good, certainly more work than what I did for any of my maths or sciences. Remember though, you don't need to know everything in the holidays. You just need to know enough to be able to get 100% on SACs after the teacher teaches everything in class. There's no need to stress right now; if you have a solid understanding 9f everything now, that'll be enough. But by all means, do as much work as you want (:

[SocialRhubarb]

Angular momentum is conserved as well though.

[Rod]

Hey guys just need some help on another question that I've been stuck on for a while

So this guy has been squeezing on his rubber balls. The rubber ball weighs 0.1 g. He throws it a brick wall at a speed of 4 m/s. What is the maximum compression produced in the ball after this impact.

This is what I did

Change in Ek = work

so 1/2mv2=fx

0.8=fx

what do I do next?

[PB]

Hmm, I think something is missing from the question. In particular k (the spring constant), which you may not have come across yet in your studies
If thats the case, then the question is unsolvable.

[Rod]

Hmm, I think something is missing from the question. In particular k (the spring constant), which you may not have come across yet in your studies
If thats the case, then the question is unsolvable.

Here I'll copy the whole question. I could have read it wrong and missed out something important;

A gymnast decides to improve the strength of her grip
by repeatedly squeezing a small rubber ball of mass
100 g whose force–compression graph is shown.

7 a How much work must the gymnast do to compress
the ball by 10 mm?

b Discuss the energy changes that occur in the ball
during one cycle of compression and release.

c How much power is expended by the gymnast if
she performs 120 identical compressions of 10 mm
in a 60 s interval? (Assume that no work is done
on release.)

8 After the gymnast finishes her exercises, she throws
the ball at a solid brick wall. The ball strikes the wall
horizontally with a speed of 4.0 m s–1. Calculate the
maximum compression produced in the ball during
this impact.

Question 8 is what I need

PS - There was also a f verses compression graph which I couldn't copy

[PB]

Ok that clears things up. First things first. You ultimately want to use this formula Us(spring potential energy) = 1/2kx^2 to find the answer  x (the compression of the ball). How, you might ask?
1.Well, k can be found from the gradient of the F-x graph. This is just a thing you have to remember. I am sure you know how to find the gradient of a linear graph (rise over run etc.)
2. Us=Ek
        =1/2mv^2
        =1/2 *.1 *4^2
        =...
 Spring potential energy (Us) is actually equal to the kinetic energy of the ball. This is because all the kinetic energy of the ball is converted into the ball's spring potential energy when it comes to a temporary stop against the wall.

Now, having Us and k, simply sub into the formula Us =1/2kx^2 to find x  
 I think thats right and I hope that helps.

[Rod]

Ok that clears things up. First things first. You ultimately want to use this formula Us(spring potential energy) = 1/2kx^2 to find the answer  x (the compression of the ball). How, you might ask?
1.Well, k can be found from the gradient of the F-x graph. This is just a thing you have to remember. I am sure you know how to find the gradient of a linear graph (rise over run etc.)
2. Us=Ek
        =1/2mv^2
        =1/2 *.1 *4^2
        =...
 Spring potential energy (Us) is actually equal to the kinetic energy of the ball. This is because all the kinetic energy of the ball is converted into the ball's spring potential energy when it comes to a temporary stop against the wall.

Now, having Us and k, simply sub into the formula Us =1/2kx^2 to find x  
 I think thats right and I hope that helps.

Yep easy! Thank you!

I wasn't expecting any question from that lot to ask me anything about strain potential energy or hooke' law, because those two topics come after the lot I did just then.

I'm really enjoying physics at the moment. I'm going to start projectile motion and circular motion tomorrow, looks really interesting. Chemistry has also been great, but a bit harder. I read the ATARnotes physics guide and I don't understand why the guide neglected physics so much. It's an awesome subject, and I'm going to expect the subject to be really fun and interesting next year.

PS - How hard is electricity and introductory photonics? If I can fly through motion how easily do you think I can fly through electricity?

Mod Edit: Fixed quote

[archenemy]

PS - How hard is electricity and introductory photonics? If I can fly through motion how easily do you think I can fly through electricity?

It's different for everyone but I think electronics/photonics U3 was relatively easy. You should have no problems with it

[Homer]

PS - How hard is electricity and introductory photonics? If I can fly through motion how easily do you think I can fly through electricity?

 electricity and introductory photonics is quite different from motion. Although relatively easy, it doesn't necessarily mean that you would be able to fly through it. However if you have a strong foundation from unit 1&2, it shouldnt be too hard  

[Rod]

Hey guys, so with nearly 2 and a half weeks gone in the holidays I've been able to go through a lot of motion theory and work. So at the moment I've been doing lots of chapter review questions and I'm getting a lot of them wrong, and am finding them hard. I was able to do the chapter questions easily. Should I be worried? I know a lot of past physics students tended to not do any questions at all, I'm thinking of doing that know considering the time I am wasting doing trying to do them and the confidence I'm losing.

Thanks

[Rod]

I just stuffed up four questions in a row, getting really frustrated. Ahhhhhhhhhhhhhhhh I hate springssssssssssssssssss. Might just move on to electricity.

[Aurelian]

Hey guys, so with nearly 2 and a half weeks gone in the holidays I've been able to go through a lot of motion theory and work. So at the moment I've been doing lots of chapter review questions and I'm getting a lot of them wrong, and am finding them hard. I was able to do the chapter questions easily. Should I be worried? I know a lot of past physics students tended to not do any questions at all, I'm thinking of doing that know considering the time I am wasting doing trying to do them and the confidence I'm losing.

Thanks

Definitely don't be worried - it's not even 2014 yet! My advice to you would be to stop doing questions and go over the theory the difficult questions are concerned with. Make sure you fully understand all the concepts involved and then retry the questions in a day or two =)

[Rod]

Merry Christmas everyone, just need help on a projectile motion question here

A ball was thrown horizontally at a speed of 40ms from the top of a cliff into the ocean below and takes 4 seconds to land in the water. Air resistance can be ignored.

d) What angle to the horizontal does the ball strike the water?

I have no Idea where to start, here are the vertical and horizontal component values I calculated prior to this question, if it helps;

Vertical - V=40ms, u=0, a=10, t=4, x=78

Horizontal - V=40, u=40, a=0, t=4, x=160

Thank you! 

[lzxnl]

Merry Christmas everyone, just need help on a projectile motion question here

A ball was thrown horizontally at a speed of 40ms from the top of a cliff into the ocean below and takes 4 seconds to land in the water. Air resistance can be ignored.

d) What angle to the horizontal does the ball strike the water?

I have no Idea where to start, here are the vertical and horizontal component values I calculated prior to this question, if it helps;

Vertical - V=40ms, u=0, a=10, t=4, x=78

Horizontal - V=40, u=40, a=0, t=4, x=160

Thank you! 

Just saying, for vertical, we normally use y instead of x

OK, so my table looks like this:
Vertical           y=? u=0 v=? a=-10 t=4
Horizontal      x=?  u=40  v=40  a=0  t=4

Firstly you can work out x; x=160
Then, let's work out stuff in the vertical column. What we want is v; y doesn't actually matter, because the angle the water strikes the water is the direction of its velocity vector.
So, v=u+at=0-40=-40
If it is moving down at 40 m/s and moving across at 40 m/s, draw up the triangle and you'll find that the angle it makes with horizontal is 45 degrees.

[Rod]

Just saying, for vertical, we normally use y instead of x

OK, so my table looks like this:
Vertical           y=? u=0 v=? a=-10 t=4
Horizontal      x=?  u=40  v=40  a=0  t=4

Firstly you can work out x; x=160
Then, let's work out stuff in the vertical column. What we want is v; y doesn't actually matter, because the angle the water strikes the water is the direction of its velocity vector.
So, v=u+at=0-40=-40
If it is moving down at 40 m/s and moving across at 40 m/s, draw up the triangle and you'll find that the angle it makes with horizontal is 45 degrees.

Ohhh I see, thank you so much lzxnl.

And I'll be using y instead of x for the vertical from now on!!

PS - How hard is electricity/phonetics compared to motion? I'll be starting them on Sunday.