Physics Predictions/Thoughts (5 Viewers)

[Qiaochu Chen]

@Qiaochu Chen this is how I did it.









Correct me if I'm wrong though.

I think you can't cancel out the masses because in the centripetal force the mass is the mass of the car (since it is the car that spinning) and in gravitational force the mass is the mass of the bob?
Correct me too if I'm wrong.

 

[Arrowshaft]

I think you can't cancel out the masses because in the centripetal force the mass is the mass of the car (since it is the car that spinning) and in gravitational force the mass is the mass of the bob?
Correct me too if I'm wrong.

Oh that’s right! Damn, I didn't consider that.

 

[Balajanovski]

@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.









Correct me if I'm wrong though.

That's exactly my calculations too.
However, I argued from the perspective of an opposite fictitious force from inside the car's non-inertial frame of reference.
Same result though.

For @Qiaochu Chen's argument, that's right.
However, if you argue from the perspective of a ficticious force from inside the car, the ficticious acceleration on the bob must be the car's centripetal acceleration yet in the opposite direction.
So, .

Doing a free body diagram on the bob, then solving using simultaneous equations you still get @Arrowshaft's equation.
Hence, from there you can cancel the bob's mass.

Also, @Arrowshaft is still right if you argue from the perspective of the bob itself.
Ignore the motion of the car. The bob is still moving in uniform circular motion. To do so, it must have a centripetal acceleration and force, which can be calculated based on its velocity (car's velocity) and its own mass.
The net force on the bob is the sum of the tension of the string and its mass. And the net force on the bob is its centripetal force.

 

[4u Boy]

Can someone send the paper plz

 

[Arrowshaft]

That's exactly my calculations too.
However, I argued from the perspective of an opposite fictitious force from inside the car's non-inertial frame of reference.
Same result though.

Fictitious force as in inertial force or centrifugal? Because I saw the Dot Point book use inertia for a similar question.

 

[Arrowshaft]

That's exactly my calculations too.
However, I argued from the perspective of an opposite fictitious force from inside the car's non-inertial frame of reference.
Same result though.

For @Qiaochu Chen's argument, that's right.
However, if you argue from the perspective of a ficticious force from inside the car, the ficticious acceleration on the bob must be the car's centripetal acceleration yet in the opposite direction.
So, .

Doing a free body diagram on the bob, then solving using simultaneous equations you still get @Arrowshaft's equation.

Hmm. Seems like Jimmy got the right answer though, we didn't consider the individual masses.

 

[Balajanovski]

Hmm. Seems like Jimmy got the right answer though, we didn't consider the individual masses.

Wrong. Look at my edit.

 

[Arrowshaft]

Hang on... that can't be right actually @Qiaochu Chen, I saw the dot point book had a pretty much identical question, and they obtained the same result as using . Did they do it wrong as well? I'm confused now.

 

[Arrowshaft]

Also @Qiaochu Chen you have to look at from the point of the mass

 

[Arrowshaft]

Wrong. Look at my edit.

Yep. realised that, I didn't realise you commented before I posted

 

[Balajanovski]

Hang on... that can't be right actually @Qiaochu Chen, I saw the dot point book had a pretty much identical question, and they obtained the same result as using . Did they do it wrong as well? I'm confused now.

Jimmy is wrong.

As said in my previous post, ignore the motion of the car and look at the bob as its own independent object in uniform circular motion.
It must have a centripetal force.
This force is the net force acting on it (sum of string tension & its weight).

It would be unreasonable to say that the centripetal force of the bob alone is the net force / centripetal force of the car.

 

[Arrowshaft]

Jimmy is wrong.

As said in my previous post, ignore the motion of the car and look at the bob as its own independent object in uniform circular motion.
It must have a centripetal force.
This force is the net force acting on it (sum of string tension & its weight).

It would be unreasonable to say that the centripetal force of the bob alone is the net force / centripetal force of the car.

yes lol, I corrected myself in the above post. I was too quick and doubted myself.

 

[Arrowshaft]

I was going to mention tension, however I didn't because I felt it was unecessary to the net forces, because a component of the centripetal force really is a manifestation of the tension in the string. Although, the oblique linear force (net force) is due to tension.

 

[Qiaochu Chen]

Well fuck. I was having a good day and now I just want to kms.

 

[Arrowshaft]

Well fuck. I was having a good day and now I just want to kms.

Don't worry man, happens to the best of us. Also, you probably still are getting a band 6 anyway. And I doubt you’ll lose more than 1 mark.

 

[Arrowshaft]

Also, did anyone feel the sample questions were a lot harder than the actual questions provided?

 

[Balajanovski]

Don't worry man, happens to the best of us. Also, you probably still are getting a band 6 anyway. And I doubt you’ll lose more than 1 mark.

@Arrowshaft is right @Qiaochu Chen. It is only 1 mark. It is a very minor error.
I bet a majority of the state were not able to complete it fully.

 

[Balajanovski]

Also, did anyone feel the sample questions were a lot harder than the actual questions provided?

No. Around the same difficulty honestly.
But that's just my personal opinion.

 

[Arrowshaft]

I was expecting two 9 markers, maybe that’s why. The actual nine marker provided also had a calculation element to it so it withdrew on the theory a bit. I just feel it was easier, because I hate waffling about theory, I prefer actual mathematical analysis.

 

[Husky]

I keep forgetting that everyone on this forum is a state ranker lol. I felt like that this exam was really difficult, a lot harder than anything in the excel books and previous years... like wtf even was multiple choice I think I got like 10/20 for it lmao. But I think I did okay on everything else except the second last question