# Physics Predictions/Thoughts (1 Viewer)

#### StudyOnly

##### Active Member
I found MC difficult (last few questions). Some of the short answers were also hard for me. Overall hoping for a low to mid B6.
EDIT: On second thoughts: Might be a B5

#### akkjen

##### Active Member
what did people get for the fan question? and also the protractor one?
i found it quite hard

#### psmao

##### Member
How do you think you went?
I was switching between “wtf is this” and “okay I got it” during the exam lol

#### psmao

##### Member
The fan question was stolen from atar notes
No u, I did every problem from atar notes and it’s not on it iirc

#### Arrowshaft

##### Well-Known Member
I was switching between “wtf is this” and “okay I got it” during the exam lol
Same! With string question, duuuude I didn’t realise that the Picture showed the front of the car until 20 mins before the exam Hahahah, I thought it was linear motion and was so confused for a while. I hate how subtle they are with the diagrams. Got it in the end though

#### psmao

##### Member
Same! With string question, duuuude I didn’t realise that the Picture showed the front of the car until 20 mins before the exam Hahahah, I thought it was linear motion and was so confused for a while. I hate how subtle they are with the diagrams. Got it in the end though
Did u deduce a relationship between acceleration and angle?

#### psmao

##### Member
I did the same lol What did u get

#### Idkwhattoput

##### Member
I did the same lol What did u get
I let theta be the angle between the string and the normal, and then saud tan(theta)=(gravity force)/(centripetal force). This was assuming the car was turning. I then reaarranged to find v, but im not sure if its right. I have to try it out

#### Arrowshaft

##### Well-Known Member
I let theta be the angle between the string and the normal, and then saud tan(theta)=(gravity force)/(centripetal force). This was assuming the car was turning. I then reaarranged to find v, but im not sure if its right. I have to try it out
Yep. Got tan theta = v^2/rg

#### Arrowshaft

##### Well-Known Member
Its all over now my dudes! Anyone up to play minecraft?

#### AHafza

##### Member
Its all over now my dudes! Anyone up to play minecraft?
90 raw my guy?

#### Arrowshaft

##### Well-Known Member
Maybe if I’m lucky but idek at this point

#### Qiaochu Chen

##### New Member
Yep. Got tan theta = v^2/rg
Did you have mass in you formula at all?

#### Drdusk

##### π
Moderator
Were there lots more calculations compared to the Old syllabus?

Looks like everyone did well. Well done!

#### Arrowshaft

##### Well-Known Member
Were there lots more calculations compared to the Old syllabus?

Looks like everyone did well. Well done!
Yep, a lot more! there was a good balance between theory, skill and calculations

#### Arrowshaft

##### Well-Known Member
@Qiaochu Chen this is how I did it.
By resolving forces, we see that the centripetal force points inwards, and the bead experiences a downward acceleration.

$\bg_white \tan\theta=\dfrac{F_c}{F_g}$

$\bg_white \tan\theta=\dfrac{\left(\dfrac{mv^2}{r}\right)}{mg}$

$\bg_white \tan\theta=\dfrac{v^2}{rg}$

$\bg_white \theta=\tan^{-1}\left(\dfrac{v^2}{rg}\right)$

Correct me if I'm wrong though.