What Are Some Possible Topics for an IB Physics Extended Essay?

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The discussion focuses on potential topics for an IB Physics Extended Essay, highlighting four specific ideas. The first three topics involve kinematics and rotational motion, while the fourth examines the efficiency of Lithium-ion batteries in relation to temperature. It is noted that the kinematics topics share similarities, suggesting a reliance on established formulas. In contrast, the battery efficiency topic may benefit from empirical data and available online resources. Overall, the choice of topic should align with the desired approach and available research material.
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Homework Statement
I am currently doing the IBDP and have chosen physics as my EE subject. I am struggling to decide on a topic.
Relevant Equations
.
These are a few topics I had shortlisted, please advise if anyone of them is EE worthy:
- To what extent does the rotational velocity of a ping-pong ball affect its horizontal travel distance?
-How is the drop time of a parachute affected by variables such as mass, canopy area, size of apex vent, number or length of strings, canopy shape?
- Rolling ball down an inclined plane - At what angle will the horizontal speed of a rolling ball be the fastest?
- To what extent does temperature affect the efficiency of Lithium-ion batteries?
 
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Hi. It just depends what kind of essay you want to write. The first three subjects are very similar kinematics problems, so you will be referring to those formulae and parameters. The last point, the efficiency of Li-ion batteries, is perhaps more readily addressed with reference to empirical data. There will be lots of charts and tables online to help with this.
Good luck
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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