# ATAR Notes: Forum

## VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: Bri MT on February 22, 2019, 03:19:43 pm

Post by: Bri MT on February 22, 2019, 03:19:43 pm
VCE PHYSICS

If you want feedback on your answers so you can get more marks for your knowledge, this is the place!

Step 1. Share a question you didn't get full marks on, including your answer and how many marks the question was worth
Step 2. Wait for another user (maybe even multiple!) to provide feedback on how they might've approached the question and how you could improve your answer
Step 3. Apply the feedback you've learnt to aim for higher marks more confidently :)

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Everyone is welcome to contribute; even if you're unsure of yourself, providing different perspectives is incredibly valuable.

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There'll be a whole bunch of other high-scoring students with their own wealths of wisdom to share with you, including TuteSmart tutors! So you may even get multiple answers from different people offering their insights - very cool.

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OTHER PHYSICS RESOURCES
Post by: Brittank88 on May 15, 2019, 09:34:13 am
Alright, here's my question:
(https://media.discordapp.net/attachments/219480949312847872/578000896156368919/unknown.png?width=567&height=126)

(https://media.discordapp.net/attachments/219480949312847872/578001917972578324/JPEG_20190515_093038.jpg?width=1383&height=702)

Honestly really wondering how I could actually fully answer this let alone set it out properly.
Post by: fiona_atarnotes on May 19, 2019, 10:09:48 am
Alright, here's my question:
(https://media.discordapp.net/attachments/219480949312847872/578000896156368919/unknown.png?width=567&height=126)

I believe for part a), you could use the formula $qV=\frac{1}{2} mv^2$. Where $q$ is the charge of an electron $1.6\times 10^{-19}C$ and $V=12N/C$. This will enable you to find the speed $v=2.05\times10^6 m/s$.
For part b), we can again use the same formula $qV=\frac{1}{2} mv^2$ to understand the situation. The idea before the formula is that kinetic energy is being converted to electric potential energy or vice versa. Therefore, in the context of this question since the proton is moving from a negative plate to a positive plate, the energy transformation we are interested in is conversion from kinetic energy to electric potential energy. Hence, qV tells us the amount of electric potential energy that is required for the proton to move and cross the positive plate. If the amount of kinetic energy energy that the proton possesses is less than qV, it won't be able to cross the plate but if it exceeds qV, it will be able to cross the plate.
But we know that the magnitude of the electrical charge of protons and electrons are the same so qV from part a) is still the same, and we know that the speed found in part a) is the speed of the proton initially. What we can see is that since the mass of the proton is greater than that of an electron, $\frac{1}{2} mv^2$ for the proton will be greater than $qV$ so the proton will make it across the positive plate.