Momentum, Impulse and ImpactIf there is no force on a mass, acceleration is zero

mv = 0 = mv is constantIf two particles with masses m1 and m2 interact with equal and opposite forces F and F (and noexternal force) then

F = m1v1,F = m2v2Adding:

m1v1+m2v2 = 0so

m1v1+m2v2 = is constant

Definition The momentum p of a mass m travelling with velocity v is

p = mv.

Newtons Second Law for the case of variable mass is that force is rate of change of momentumF = p. Of course for a particle of constant mass p = mv.If we consider a complete mechanical system the forces sum to zero (that is there are no forces externalto the system). Often we dont do this, for example we regard the earth, or the ceiling from whicha pendulum is hung, as static and not part of the system. This means that the sum of the forces inthe system we are considering is balanced by the external forces. If we considered the whole system,earth and pendulum for example, the earth swings a little bit with the pendulum!Conservation of Momentum If there are no external forces on any number of interacting particlesthen the sum of the momentums of the masses is constant. Interacting here means that there may beforces between the particles due for example to strings, gravity, springs or collisions. For n particlesp1+p2+ +pn is constant.Unlike energy momentum is a vector, so the x and y (and z) components of momentum are conserved.Think for example of billiard balls before and after a collision. Conservation of momentum alone doesnot give enough equations to solve for two colliding particles. For example we have velocities v1 andv2 of particles before a collision, and v1 and v2 after the collision. In three dimensions conservationof momentum gives us three equations for the velocities after the collision but there are six variables.Clearly we need to know more about the collision.Impulse and impactSome forces are very large and very brief (eg in the collision of billiard balls). Such forces are difficultto measure, but theri overall effect of a change in momentum is not.Definition:The impulse of a force F on a particle of mass m over a time interval from t1 to t2 is

I =t2Z

t1

Fdt =Z t2

t1m

dvdt dt = mv(t2)mv(t1)

It follows that the impulse is exactly the same as the change in momentum.In a collision F(t) (the force exerted by one object on the other) is non-zero for a very short time.A very rapid change in momentum of each object results, although the sum of their momenta staysconstant.

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Coefficient of RestitutionIn general energy is lost on an impact so that colliding particles move apart with a smaller reltivespeed than the speed in coming together. We define the coefficient of restitution as

R =relative speed of moving apart

relative speed of moving together

We will see below that 0 R 1 always.There are two extreme cases: R = 1 perfectly elastic collisions where no energy os lost, and R = 0colliding masses which stick together.Example Taking all movements to be along the x-axis (say) a mass m1 moving at velocity s hits amass m2 at rest. If the coefficient of restitution is R find the velocities v1 and v2 of the masses aftercollision as a fraction of the initial kinetic energy 12m1s

2?Momentum is conserved so m1s = m1v1+m2v2. Coefficient of restitution is R = (v2 v1)/s.Solving (check this for yourself) gives

v1 =m1R m2m1+m2

s

v2 =(1+R )m1

m1+m2s

With the ratio of energy after to energy before given by (check this for yourself)12m1v

21+

12m2v

22

12m1s

2 =m21+(1+R 2)m1m2+R 2m22

m21+2m1m2+m22

Note that the ratio is one (energy stays the same), if R = 1 and is less than one if 0 R < 1.Bouncing ball under gravity(a) Without air resistance a ball falls back to same level with the same speed that it left that level; thisis because

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mv2+mgy = E (a constant)

(b) Consider a ball hitting a flat rigid surface with speed s0. After impact, the speed at which it risesfrom the surface is s1 = R s0. After rising and falling back it returns with speed s1 and bounces withspeed s2 = R s1 = R 2s0.If sn is the speed after the nth impact we have

sn = R sn1.

(c) Height of rise after nth impact: We have (exercise) v = sn gt (so v = 0 at t = sn/g) and y =snt 12gt2 (so at v = 0, y = 12s2n/g).We see that the maximum of y after n impacts is

yn =12

s2n/g =12R 2s2n1/g

andyn = R 2yn1

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(d) Time of flight after nth impact. This is when v =sn, from above time of flight after n impacts is

tn = 2sn/g = R tn1.

We can solve each of these to givesn = R ns0yn = R 2ny0tn = R nt0

Do all the collisions happen in a finite interval of time? For that we want

n=0

tn

to be finite. This is a sum of a geometric series which for R < 1.Recurrence relations and return mapsWe have seen that the time evolution of a system can be modelled by either differential equations(continuous time) or by difference equations, also called recurrence relations where time is a discretevariable. In population models we saw both. The bouncing ball gives an example of a recurrencerelation in a mechanical system.Both differential equations x = f (x) and difference equations xn = f (xn1) are studied in the mathe-matical area called dynamical systems. This is a active area of mathematical research and our depart-ment at UMIST has a particularly strong dynamical systems group. You can learn mode about thistopic in 361 Dynamical Systems and Chaos.You may have met difference equations, or recurrence relations in numerical methods for solvingequations such as Newton-Raphson method. To solve F(x) = 0 we might use the recurrence relation

xn = f (xn1) = xn1 f (xn1)/ f (xn1).

If it converges to a fixed point where x = f (x) this is a solution of F(x) = 0.In our bouncing ball example, Newtons equations of motion give us a differential equation governingthe system. We made a discrete time dynamical , or recurrence relation, by looking at the state of thesystem each time the solution to the differential equation returned to y = 0. More generally returnmaps are discrete time systems which simplify a continuous time systems.Rockets systems with changing massSystems where the mass changes can be understood in terms of momentum. Rockets are a goodexample as they eject the products of combustion backwards to propel them forwards.Example A rocket moving in a straight line ejects a mass m in a small time t at a speed s relativeto the rocket. If the mas of the rocket is m(t) what is the acceleration of the rocket?At time t: rocket has mass m and velocity v.At time t + t the mass of the rocket has changed (decreased) to m+ m (note m is negative). Thevelocity of the rocket is v+v, and the ejected matter with massm is travelling with velocity v s.Now we apply conservation of momentum at time t and t+t.Before mass is ejected, time t, momentum is mv.

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After mass is ejected, time t+t, momentum is m(v s)+(m+m)(v+v).Momentum is conserved so these are equal. Dividing by t we have

mvt + s

mt +

mvt = 0.

Now we have to assume that v and m are differentiable functions of t. A differentiable function is alsocontinuous (as will be proved in a later course). So we can take the limit as t 0 and we know thatv 0 and m 0 (this is what continuous means), but the ratios of these small quantities becomederivatives. We have

mdvdt + s

dmdt = 0

We now have acceleration:dvdt =

s

m

dmdt

(Question: why is this always positive for a realistic model of a rocket?)We now need some extra assumptions about m and s to find the acceleration, and hence the motion ofthe rocket.(a) Suppose that at t = 0, m = m0 and v = v0, with s a constant, find v as a function of m: Integratinggives Z

dv =sZ dm

mso

v =s lnm+CUsing v = v0 when m = m0 gives C = v0+ s lnm0 so that (check this)

v = v0+ s lnm0m.

Note as m decreases,v increases. But we still dont know m as a function of time. More assumptions:(b) If dm/dt is a constant (dm/dt =B, B > 0) find (i) m and v as functions of time (ii) the distancetravelled from time t = 0.(i) dm/dt =B = m =Bt+D with D = m0+B0 so m = m0Bt and hence

v = v0+ s lnm0

m0Bt(ii)

v = dx/dt = v0+ s lnm0

m0Btso

x = v0t+ sZ

ln m0m0Bt dt+ const

leading to (exercise)x = (v0+ s)t+

s

B(m0Bt) ln m0Bt

m0

Reality checks: dimensions ok? Do x and v go the right way as t increases? What happens for largetime when we take the log of a negative number? (x lnx 0 as x 0)

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