Newton’s Laws are often misquoted, largely because F is meant to be a resultant force. Quote instead ∑ F = ma.
There is further confusion over acceleration being a, because that implies it is a constant. Far better to recognise that acceleration is dv/dt = dv/ds . ds/dt = v dv/ds = d²s / dt².
Assuming you recognise s as displacement, v as velocity and t for time, then you ought to use the continuous and variable versions of the various definitions of Newtonian mechanics, Work = ∫ F ds (and F is a resultant, therefore probably a combination of forces possibly including a P/v from a driving force). Typical cars working uphill and against a rolling resistance would have m dv/dt = P/v – R – Mgsinø and the work done would be the integral ∫(P/v – R – Mgsinø) ds.
Newton said that momentum was the fundament, not force as we generally teach. Applying the chain rule to a rate of change of momentum d/dt (mv) = m dv/dt + v dm/dt , which gives an accelerating force and a thrust from chucking mass away, such as a rocket does.
Issue here: is d/dt (mv) the accelerating force, or is that m dv/dt? If so, what is the proper label for v dm/dt?
Answer: Newton’s Laws apply to particles or to systems of particles and there is an assumption that these systems are closed. Rocket systems are not closed. So the differentiation is (of course correct) but what is wrong is the rubric for Newton’s Second Law being quoted (if you ever do so): the net force on a particle is the rate of change of momentum (d/dt (mv) in an inertial reference frame (e.g. on this planet’s surface) and which is only applicable to constant-mass systems – because outside that restriction it wouldn’t be a closed system.
The sneaky way around having the 2nd Law continue to work is to call the v dm/dt term the thrust caused by the rocket and to consider it as a force. If something that accretes [gains mass], like a raindrop, the gain in mass becomes some sort of drag.
There is a similar issue for some people with the concept of Kinetic Energy. I have come to prefer (at school level) the idea that the sum of all the F.s makes for the work that causes a change in KE, so ∫ ∑ F ds = KE because ∫ m dv/dt ds = ∫ mv dv = [mv²/2] = KE
Issue: what about variable force delivering power in a system that loses mass?
∫ P dt = Work = ∫ d/dt (mv) ds = ∫ m dv/dt ds + ∫ v dm/dt ds
= ∫ mv dv + ∫ v² dm
This result sort-of checks dimensionally as K.E. changes, definitely so for constant mass.
Impact and restitution
The law of conservation of momentum, using sigma for summation, ∑(mu) = ∑(mv) is easily applied. Put this with restitution (generally energy is lost in impacts, largely from deforming the two particles), which says
separation speed = e x (approach speed)
and you get a pair of simultaneous equations. This gives rise to a collection of predictable questions at A-level. Those of you who don’t check Sim Eqns automatically deserve the bad results that follow ! Diagrams are essential and one recommends (this one does) making as many vectors go in the same direction as possible, to avoid nonsense results. Which leads us to:
Oblique impact
Where the only thing you need to remember, while you have particles hitting walls, is that the horizontal component is unaffected by the impact. Typically initial velocity u has a component ucosø parallel to the impacted surface. The perpendicular component usinø becomes vsinß=eusinø.
Since ucosø = v sin ß, it follows that tanß = e tanß. Diagrams help a good deal, but I can’t make them happen here.
Issues to explore: Don’t lose track of the fact that this is a model. The process of impact is treated as instantaneous or not understood. The bodies compress and distort and energy goes into that. There are other forces applied, such as friction during impact and that consumes energy too. The result is what we cover by using the idea of restitution.
Issue: We generally ignore cases with bodies decelerated by a force. We assume that the body under consideration is a particle.
Rotating bodies – rolling spheres – are not the same as sliding particles. A snooker ball hitting a cushion is rolling in the direction of travel (it isn’t, often, but assume that it is behaving ‘nicely’ at first) and, once the bounce is completed, is rolling in a different direction. Between these two states the axis of rotation must change and that requires some sliding on the table surface (and you might separate out the idea of spinning as the axis rotates). When we look at oblique impacts, the assumption that the horizontal component is unaffected ignores any friction reducing that component during the impact. We also ignore inertial effects, or that the ball surface hits the cushion, not the centre.
All this points to the truth (sorry !) that the mechanics models we construct are very simple. What we should be doing is exploring how well these models describe the situations we want to reflect. What we often find is that we run up against two problems; we cannot record good data so as to discover what is (more nearly) going on and we can’t do the more difficult maths. It is quite amazing how little more difficult the model is before it disappears beyond even the FM student.
Projectiles
A projectile is assumed to be a particle, a point mass, m. It is assumed to be projected in a uniform gravity field, free of Coriolis force and air resistance. Given that, then using the assumption that m dv/dt = ma = -mg, and measuring positively upwards:
m dv/dt = -mg => ∫ dv = -g ∫ dt => v-u = -gt => v = u- gt
but v = ds/dt, so separating variables and integrating again gives
∫ ds = ∫(u – gt) dt => s = ut – ½ gt² (integral signs don’t transfer from Word easily; must use Unicode characters).
Going back to v dv/ds = dv/dt means we could start again and have
∫ v dv = -g ∫ ds => v² = u² -2gs as expected.
Working from a projectile initially at u and at ø to the horizontal, the ucosø component is unaffected and the vertical component is dealt with as above. Vocabulary is often from the military, firing projectiles from a firing point which make an impact down-range.
Thus displacement is described by x and y in an obvious manner, then
x = u cosø; y = ut sinø – ½ gt².
Substituting to eliminate t gives the ‘big’ equation,
y = x tanø – g x² (1 + tan²ø) / 2u². This is quadratic in both tanø and in x. There are some fairly hard questions that require you to be very clear which quadratic you should be working on, such as those for maximum range under constraints.
Range, from the firing point back to the same vertical level, is R = u².sin2ø /g and the maximum height is given by H = u.sin²ø /2g. You need the big equation to solve cases where the impact is not at the same height as the firing point. There are some ‘nice’ equations formed by expressing R and H together, often found in extension papers.
Problems are made complicated by having two projectiles in flight, or by firing up a slope; both of these cases are solved by setting up simultaneous equations to reflect the motion.
Whenever you are having difficulty, separate the horizontal and vertical components and take care to look at where things are at a point in time.
For further maths, air resistance is added. The manageable version has this as F = mkv and the more difficult one mkv². The handling of the differential equations is as at the top of this page, and good practice. Modelling something like a tennis serve is valid for a rubbish player with only the gravity considered: for class players, only air resistance is sufficient. A good investigation to do.
Exercises included above:
Develop the big equation for yourself.
Find at least one equation connecting H and R.
If you can throw a shot (in athletics) at 12 m/s and your release point is 2.4m above the landing point, what is the longest throw you can make? At what angle is this throw? This is an example of solving one or both quadratics within the big equation. You are encouraged to use 2.4 = 12/5 and g = 10.
Develop the new equation systems for F=mkv. There is a neat form mixing gt and ks and v, almost worth remembering.
Develop the new equation systems for F = mkv². The integration is only a little stretch, but the algebra gets messy; use the experience of the previous problem to keep it manageable.
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