[Man Lam]

Hi, Dean

You may contact Halid Takimoglu, De La Salle College. He has a session in Physcon 2024 this year. Below is the abstract from this year’s (2024) program. I did attended his session but couldn’t locate my own document that he gave me on the day.

Man Lam

A4 Hybrid
New Study Design – VCE Physics
Halid Takimoglu, De La Salle College
As the new study designs will be commencing
for Unit 3 and 4 VCE Physics in 2024, most of
us are focused on adjusting our assessments to
the new requirements. However, adjusting our
content delivery to reflect the new study design
is more important. No one wants to keep
teaching an extinct dot point. More importantly,
no one wants to miss teaching a new one.
In this workshop, we will be comparing the dot
points of the new study design with the old one
and classify each dot point as ‘exactly the same’,
‘completely new’, ‘modified’ or ‘deleted’.

[Man Lam]

I asked a friend who is an retired engineer. His Ph.D. project was in string theory. I think it is worth to share his comments in here.

Message 1.
==============
Hi Man Lam,

I hope you are well.

I think I see what the examiner is getting at.
As you say v^2 = u^2 + 2as comes from conservation of energy.
After all it just derives from the integral of the force in the direction of motion over distance.

However, the examiner wishes to draw the distinction that the ‘a’ experienced by the cart is not constant, as you noticed.
I.e. What is ‘a’ of this equation in this problem?

Now, energy conservation means that however you slice and dice it, you will get the right values.

Nevertheless, I agree with the examiner preferring demonstration that the understanding of the problem is in terms of CoE against
using the other equation in which the meaning of the terms in this context is unclear.

Kind regards,

Keith.

——————-

Message 2
=========
Hi Man Lam,

To be more succinct, in the equation

v^2 = u^2 + 2as

‘a’ represents a constant acceleration over a distance ‘s’.
At no point is the truck subject to an acceleration ‘g’.

If a body were to be falling freely, then certainly a = g and s = h.

One could therefore:
– apply the equation to a freely falling body.
– Note that in the rolling truck case with negligible friction the speed would have to be the same as for the falling body to conserve energy.

[And strictly, we also need to assume that the wheels of the trolley are also of negligible mass.]

I can’t really speak for examiners or teachers, but if a student uses an equation in a context to which it does not obviously apply,
then they ought to include the explanation as to why it can in fact be used if they are to be credited with understanding the actual physics.

Regards,

Keith

[Man Lam]

Thank you Barbara and Colin prompt responses.

I think it really depends on how you view the equation or how you derive it.  There are two very distinctive approaches.

———————-
Method 1. 
Just using definition of a= dv/dt  and v=ds/dt
a=dv/dt
= dv/ds * ds/dt                    
a= v  * dv/ds
a ds= v dv
then integrate both sides by putting when s=0, v=u  and when s=s  v=vwe will have  as = 1/2 v^2 -1/2 u^2
————————–

Method 2.
Work = force * distance
dW = F* ds                    (in here, the * means the vector dot product)
input work = output work.   (conservation of energy)
Gravitational energy = work done to the particle with mass m.

(mg) * ds = F *ds             
= ma  *ds             
= m (dv/dt) * ds             
= m dv* (ds/dt)               
= m v * dv

The left hand side is just mgh.
Canceled the m from both sides, this will also result in exactly the same equation.   
—————–
Please note in the method 2, mg and s are in vector form.  Although the vector acceleration a in method 1 is not clear for the track, I still can use g as the acceleration and h as displacement.  But as Barbara mentioned, it really doesn’t matter as g will always parallel to h.  
The first approach used the definition of a and v.  The result from method 1 is totally just a mathematical results of manipulation of symbols and chain rule. 

Method 2 showed the conservation of energy that KE = Force* distance in different form.  

==========

That is why I puzzled about the examiner report which regarded students using v^2=u^2 +2as is wrong. In my opinion, this kinematic equation or KE=PE are just the same.

Man

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