UK A level Physics question & Marking scheme question

[ChrisXenon]

This is question 24 from the OCR AS Level Physics A H156/01 Breadth in physics paper from May 24th, 2016.

Question: State Newton's second law

Marking Scheme answer: "(Resultant) force is (directly) proportional / equal to the rate of change of momentum"
Marking scheme guidance notes: NOT force = mass x acceleration, NOT force x change in momentum over time

Does anyone have any comments?

 

[Mark44]

This is question 24 from the OCR AS Level Physics A H156/01 Breadth in physics paper from May 24th, 2016.

Question: State Newton's second law

Marking Scheme answer: "(Resultant) force is (directly) proportional / equal to the rate of change of momentum"
Marking scheme guidance notes: NOT force = mass x acceleration, NOT force x change in momentum over time

Does anyone have any comments?

The marking scheme answer is closer to what Newton wrote, although I believe he wrote it in Latin.
The second marking scheme guidance note is incorrect. Instead of "force x change in momentum over time" it should have said "force = change in momentum over time" or $F = \frac d {dt}\left(mv\right)$

 

[ChrisXenon]

The marking scheme answer is closer to what Newton wrote, although I believe he wrote it in Latin.
The second marking scheme guidance note is incorrect. Instead of "force x change in momentum over time" it should have said "force = change in momentum over time" or $F = \frac d {dt}\left(mv\right)$

Thanks for your time, Mark. I am wondering - what is wrong with f=ma?

 

[mjc123]

F = d(mv)/dt. This is not equal to mdv/dt when dm/dt is not zero. Common example is a rocket whose mass is continually decreasing due to burning of fuel.

 

[Mark44]

I am wondering - what is wrong with f=ma?

I believe the marking scheme wants you to state what the Second Law actually says, which is about force being proportional to the time rate of change of momentum. The formula F = ma is a special case, when mass m is constant.

 

[PeroK]

F = d(mv)/dt. This is not equal to mdv/dt when dm/dt is not zero. Common example is a rocket whose mass is continually decreasing due to burning of fuel.

So, if something is losing mass but traveling at constant speed. E.g. if bits are falling off, then it must be subject to a negative force?

For that reason Newton's second cannot be applied to something that is losing or gaining mass. Since the momentum change in the mass that is lost or gained must also be taken into account.

 

[PeroK]

This is question 24 from the OCR AS Level Physics A H156/01 Breadth in physics paper from May 24th, 2016.

Question: State Newton's second law

Marking Scheme answer: "(Resultant) force is (directly) proportional / equal to the rate of change of momentum"
Marking scheme guidance notes: NOT force = mass x acceleration, NOT force x change in momentum over time

Does anyone have any comments?

Mass is conserved in Newtonian physics. I'd vote for $F = ma$. If the mass changes, then it's not the same object before and after.

Ps Khan Academy agrees with me. And Wikipedia has a discussion of the difficulties when the mass is changing.

 

[gmax137]

Mass is conserved in Newtonian physics. I'd vote for $F = ma$. If the mass changes, then it's not the same object before and after.

Isnt it a relationship that holds instant by instant? Then $F = ma$ is really $F(t) = m(t) a(t)$. The (t)s are implied in the first form of the relationship.

 

[PeroK]

Isnt it a relationship that holds instant by instant? Then $F = ma$ is really $F(t) = m(t) a(t)$. The (t)s are implied in the first form of the relationship.

The mass can't change if you want to use the second law.

 

[tech99]

The mass can't change if you want to use the second law.

Surely if a bit falls off a rocket, the mass decreases, the force of the engine is unaltered and the acceleration will increase to still satisfy F=ma. I teach A-level and the guidance seems very unfair on pupils in my opinion.

 

[PeroK]

Surely if a bit falls off a rocket, the mass decreases, the force of the engine is unaltered and the acceleration will increase to still satisfy F=ma. I teach A-level and the guidance seems very unfair on pupils in my opinion.

The point is that if there is no force on the rocket and a bit falls off (and you redefine your rocket with a new mass) then that does not constitute a change in momentum as defined by the second law.

In this case the "object" has lost momentum not because of any impulse but because you have redefined your object.

 

[PeroK]

I teach A-level and the guidance seems very unfair on pupils in my opinion.

I agree it's absurd. I checked Kleppner & Kolenkow and they give the second law as $F = ma$.

 

[Andy Resnick]

For that reason Newton's second cannot be applied to something that is losing or gaining mass. Since the momentum change in the mass that is lost or gained must also be taken into account.

I had to read this a few times to make sure I understood what you meant, and I think you are correct- another way to state your point is that ΣF = dp/dt applies to a *closed system*. For example, in the 'rocket equation', the momentum of the exhaust is included rather than simply allowing the mass of the rocket to decrease.

 

[PeroK]

I had to read this a few times to make sure I understood what you meant, and I think you are correct- another way to state your point is that ΣF = dp/dt applies to a *closed system*. For example, in the 'rocket equation', the momentum of the exhaust is included rather than simply allowing the mass of the rocket to decrease.

Yes. There are some problems, like sand being poured into a moving truck, where you get an increasing mass of the "system" and you get an equation that involves $\frac{dm}{dt}$. The equation reflects the fact that the sand is gaining momentum as it is transported by the truck. And a force is needed to maintain a steady speed of truck plus increasing load.

But, you have to be very careful. If, for example, sand is leaking out of a moving truck, then the truck plus load is losing mass, hence losing momentum, but there is no retarding force needed to stop the truck from accelerating.

In general, therefore, $F = \frac{d}{dt}(mv) = \frac{dm}{dt}v + m\frac{dv}{dt}$ is an equation that you must be very careful with. And, moreover, calling that Newton's Second Law is absurd, IMO.

IMO, Newton's Second Law assumes a particle, or system of particles, of fixed mass. In which case:

$F = m\frac{dv}{dt} = \frac{d}{dt}(mv)$

Allows force to be interpreted as the rate of change of momentum (of a fixed mass system).

 

[ChrisXenon]

Thanks for your observations people. I am a private tutor and I teach A level Physics, but I do not work in a school and have no insights into how markers follow marking schemes. I have seen many examples of idiotic marking guidance from OCR. I had a long but ultimately pointless dialogue with them on this when they speciefically disallowed "rate of acceleration" claiming "rate" implied growth (which it doesn't) and they didn't accept that "deceleration" is a special case of "acceleration" and the latter is acceptable where the former is.
We went aorund in circles but no change of min on either side.

What bothers me is the sheer number of these anomolies. Sometimes the question is far too general and the required answer required one of many valid interpretations of the quesiton.

This stuff matters - people's A levels matter.

Do any A level markers want to comment?

Thanks
Chris

 

[anorlunda]

I had a long but ultimately pointless dialogue with them on this when they speciefically disallowed "rate of acceleration" claiming "rate" implied growth (which it doesn't) and they didn't accept that "deceleration" is a special case of "acceleration" and the latter is acceptable where the former is.

That's unbelievable. Where to they find those people?

 

[Dale]

In this case the "object" has lost momentum not because of any impulse but because you have redefined your object.

To give an extreme example suppose that you have a stress-free rod traveling inertially not interacting with anything at all nor changing internally, and suppose that you define your “system” as including none of the rod at the beginning, the back 1% of the rod at t=1 s, the back 2% of the rod at t=2 s, and so on. According to the definition of force this system is experiencing a “force” towards the front.

 

[jbriggs444]

One might argue that F=ma is not an apt synopsis because it assumes a coherent system of units. F=kma would be correct as it would correctly indicate proportionality rather than equality.