Home Forums Exams and revision 2021 Section B Q9a

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• The official answer is using KE=PE; mgh=1/2m^v2 to calculate v

In the examiner report,
“The most common error was to use v^2=u^2+2as. While this gives the same result as the conservation of energy approach it shows that students are not aware that the kinematics equations can only be used in situations where the acceleration is constant and that the shape of the track shows that the gradient, and hence the acceleration, will vary as the car travels down the slope.”

However, v^2=u^2 +2as is a manifest of conservation of energy.

v^2=u^2 +2as multiply both sides by 1/2m You will get
1/2mv^2= 1/2mu^2 =mgs
which is equivalent to 1/2mv^2-1/2mu^2 = mgs The change of KE equals to the change in PE.

I agree that the vector acceleration is not constant but v^2 =u^2 +2as is just another form of conservation of energy of constant acceleration manifested in different form. If g is not constant, then mgh=1/2mv^2 will not hold either. In fact, this equation is different from all the other 3(4) kinematic equations that v^2 is a scalar; the other equations are in vectors. It just indicates some sort of conservation law and in this case, it is energy. And that is why applying both kinematic equation and conservation of energy will give the same answer.

Am I missing something?

Man Lam
(Mt Alexander College)

Hi Nam,
v^2 = u^2 + 2as, is an equation suitable for constant acceleration. In this scenario the acceleration cannot be assumed to be constant, so this formula is not appropriate. The student is not able to find the value for either ‘a’ or ‘s’ to use in this formula.

In my opinion v^2 = u^2 + 2as, is a manifestation of conservation of energy, only when the acceleration is constant.

I hope that this goes some of the way towards explaining what the exam report is saying.
Colin Hopkins

Dear Colin and Nam,
This question illustrates the conservative nature of the gravitational field. Because the track is frictionless, the only external force on the cart is gravity, so all the track is doing is controlling the path taken by the cart in the field, and because the field is conservative, the change in kinetic energy of the cart has to equal the product of the magnitude of the field and the vertical displacement of the cart, no matter what shape the track takes to get there. Conservation of energy is the correct principle to apply because a conservative force is applied to the cart.

I agree with Nam that the scalar kinematic equation has a different status from the other vector kinematic equations. I think it is perhaps overlooked that the product of acceleration and displacement in the equation is actually a vector dot product, corresponding to the twice the work done on the particle by the average net external force experienced by the particle as it moves from its initial to final positions divided by the mass of the particle. I would argue that using this kinematic equation implies conservation of energy, work done by a net force on a free particle changes its kinetic energy.

Feedback welcome.
Barbara

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

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

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