# General Thoughts: Physics (1 Viewer)

#### spatula232

##### Active Member
Atar cannot be created nor destroyed but can be transformed. It usually transforms into something called a mystery mark
I'm sig-ing this

#### Mr_Kap

##### Well-Known Member
RIP for that copper question.

I drew a straight line with negative gradient and then from C to D it was FAIRLY straight (but not exactly).

lol. at least my explanation was alright, maybe 2/5 lol.

#### Drsoccerball

##### Well-Known Member
RIP for that copper question.

I drew a straight line with negative gradient and then from C to D it was FAIRLY straight (but not exactly).

lol. at least my explanation was alright, maybe 2/5 lol.
Exactly this is why I used Extension 2 since I saw Integrand do it once and it actually worked

#### PhysicsMaths

##### Active Member
Exactly this is why I used Extension 2 since I saw Integrand do it once and it actually worked
They can't award you marks for doing this.

#### Drsoccerball

##### Well-Known Member
They can't award you marks for doing this.
I didn't think they would... It was for me not them.

#### Mr_Kap

##### Well-Known Member
Biggest regret: Dropping 4u. Should've kept it just for physics.

was this the mechanics topic where you learnt this..because that is the ONLY 4u topic i didn;t learn.

or was it the 4u integration topic ---> because i already know 4u integration so if i could have used it i'm such a dickhead for not using it.

#### Drsoccerball

##### Well-Known Member
Biggest regret: Dropping 4u. Should've kept it just for physics.

was this the mechanics topic where you learnt this..because that is the ONLY 4u topic i didn;t learn.

or was it the 4u integration topic ---> because i already know 4u integration so if i could have used it i'm such a dickhead for not using it.
Mechanics helped me alot for physics... Mainly resistance questions...

#### SlimiJay

##### Member
no one has the paper??

#### IamKirby

##### Member
I didn't think they would... It was for me not them.
Hey man, I did 4u but I don't know how it could be used here. Can you explain? Thanks

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#### apurba

##### Member
Okay haha thought so, thanks

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Okay this was my reasoning for my graph:

I said as eddy currents are induced on the copper plate, causing it to slow down.
but then, the trolley passes through the magnet, and the direction of the eddy currents are then switched, causing it to experience a force in the opposite direction (i.e. accelerating it to the right slightly).

Where did I go wrong?

#### Drsoccerball

##### Well-Known Member
Hey man, I did 4u but I don't know how it could be used here. Can you explain? Thanks

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acceleration = -k

$\bg_white v \frac{dv}{dx} = -k$

$\bg_white \frac{dx}{dv}= -\frac{v}{k}$

$\bg_white [x]_0^x =[ \frac{-v^2}{2k}]_v^W$

Where W is the final speed even though k is changing it gave me a rough estimate of what the graph looked like.

#### Thjv

##### Member
acceleration = -k

$\bg_white v \frac{dv}{dx} = -k$

$\bg_white \frac{dx}{dv}= -\frac{v}{k}$

$\bg_white [x]_0^x =[ \frac{-v^2}{2k}]_v^W$

Where W is the final speed even though k is changing it gave me a rough estimate of what the graph looked like.
4u motion..

#### Drsoccerball

##### Well-Known Member
Okay this was my reasoning for my graph:

I said as eddy currents are induced on the copper plate, causing it to slow down.
but then, the trolley passes through the magnet, and the direction of the eddy currents are then switched, causing it to experience a force in the opposite direction (i.e. accelerating it to the right slightly).

Where did I go wrong?

As it leaves it is also slowed down remember how a similar experiment where a magnet dropped through a metal induces an m field to always slow down the magnet whether it enters or leaves.

#### Mr_Kap

##### Well-Known Member
As it leaves it is also slowed down remember how a similar experiment where a magnet dropped through a metal induces an m field to always slow down the magnet whether it enters or leaves.
if i explained all the shit correctly with CORRECT REASONING (i did) but drew the graph that i did (negative straight line from A to C, and then a fairly straight line from C to D), can i get 2/3 out of 5???

#### InteGrand

##### Well-Known Member
acceleration = -k

$\bg_white v \frac{dv}{dx} = -k$

$\bg_white \frac{dx}{dv}= -\frac{v}{k}$

$\bg_white [x]_0^x =[ \frac{-v^2}{2k}]_v^W$

Where W is the final speed even though k is changing it gave me a rough estimate of what the graph looked like.
$\bg_white This is basically a derivation of the well-known formula v^2 = u^2 + 2a\Delta s for \textsl{constant} a.$

#### Drsoccerball

##### Well-Known Member
$\bg_white This is basically a derivation of the well-known formula v^2 = u^2 + 2a\Delta s for \textsl{constant} a.$
Agahahahaahah how many things did I derive in the exam...

#### Drsoccerball

##### Well-Known Member
if i explained all the shit correctly with CORRECT REASONING (i did) but drew the graph that i did (negative straight line from A to C, and then a fairly straight line from C to D), can i get 2/3 out of 5???
I think so. Also the question would've been more amazing if there was friction from the floor aswell... But since V didnt change there was no friction...

#### Thjv

##### Member
I think so. Also the question would've been more amazing if there was friction from the floor aswell... But since V didnt change there was no friction...
Pls no... this is hsc physics