Reply To: 2021 Section B Q9a

[Man Lam]

#8498

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.

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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
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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.  

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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