Linear Momentum l  The linear momentum of a particle, or an object that

can be modeled as a particle, of mass m moving with a velocity is defined to be the product of the mass and velocity: l 

l  The terms momentum and linear momentum will be used interchangeably in the text

vr

!p =m

!v

l  Linear momentum is a vector quantity l  Its direction is the same as the direction of the

velocity l  The dimensions of momentum are ML/T l  The SI units of momentum are kg · m / s l  Momentum can be expressed in component form:

l  px = m vx py = m vy pz = m vz

Newton and Momentum l  Newton’s Second Law can be used to relate the

momentum of a particle to the resultant force acting on it

l  with constant mass Σ!F =m

!a =md

!vdt

=d m!v( )

dt=d!pdt

l  The time rate of change of the linear momentum of a particle is equal to the net force acting on the particle l  This is the form in which Newton presented the

Second Law l  It is a more general form than the one we used

previously l  This form also allows for mass changes

l  Applications to systems of particles are particularly powerful

l  Conservation of momentum can be expressed mathematically in various ways l  l 

l  In component form, the total momenta in each direction are independently conserved l  pix = pfx piy = pfy piz = pfz

l  Conservation of momentum can be applied to systems with any number of particles

l  This law is the mathematical representation of the momentum version of the isolated system model

!ptotal =

!p1 +

!p2 = constant

!p1i +

!p2i=

!p1f +

!p2f

l  Whenever two or more particles in an isolated system interact, the total momentum of the system remains constant l  The momentum of the system is conserved, not necessarily the

momentum of an individual particle l  This also tells us that the total momentum of an isolated

system equals its initial momentum

Conservation of Linear Momentum

Conserva)onofmomentumisrequiredbyNewton’s2ndand3rdlaws:

€

F = ma = m dvdt

=d(mv)dt

=dpdt

(1)

Newton’s2ndlawshowsthenetforceonapar)cleequalstherateofchangeofmomentum:

Considerthegravita)onala?rac)onbetween2objects:

m1 m2

€

F21 =Gm1m2

r2

€

F12 =Gm1m2

r2

€

F12 = −F21 (2)But i.e.

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F12 + F21 = 0 =dp1dt

+dp2dt

=d(p1 + p2)

dt(3)

Thatis,consideringthecaseofagravita)onalinterac)ons,therateofchangeoftotalmomentumiszero,i.e.totalmomentumisconserved.Moregenerally,Newton’s3rdlawrequiresallforcestocomeinpairs–ac)onandreac)onareequalandopposite–i.e.forany2interac)ngbodies,equa)on(2)applies,soequa)on(3)appliesgenerally,notjustforgravity,soconserva)onoflinearmomentumisthusrequiredbyNewton’s3rdlaw.

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r

Archer Example l  The archer is standing on a frictionless surface (ice) l  Approaches:

l  Newton’s Second Law – no, no information about F or a l  Energy approach – no, no information about work or energy l  Momentum – yes

l  The final velocity of the archer (A) is

negative l  Indicates he moves in a direction

opposite the arrow (a) l  Archer has much higher mass than

arrow, so velocity is much lower

mAvAi + mavai = mAvAf + mavaf

0+ 0 = mAvAf + mavaf ⇒ mAvAf = -mavaf So if mA >> ma then vAf << vaf

Impulse and Momentum l  From Newton’s Second Law,

l  Solving for gives l  Integrating to find the change in momentum over some time interval

l  The integral is called the impulse, , of the force acting on an object over Δt

Δ!p =!pf −!pi =

!Fdt

ti

tf∫ =!I

!F = d

!pdt

d!p d

!p =

!Fdt∑

!I

l  This equation expresses the impulse-momentum theorem: The impulse of the force acting on a particle equals the change in the momentum of the particle l  This is equivalent to Newton’s Second Law

Δ!p =!I

(F =ma = mdv/dt = dp/dt)

l  Impulse is a vector quantity l  The magnitude of the

impulse is equal to the area under the force-time curve

l  The force may vary with time

l  Dimensions of impulse are M L / T

l  Impulse is not a property of the particle, but a measure of the change in momentum of the particle

Δ!p =!pf −!pi =

!Fdt

ti

tf∫ =!I

i.e. kg ms-1, same as momentum

l  The impulse can also be found by using the time averaged force l  This would give the same impulse as the time-varying force does

!I =

!FΔt∑

In many cases, one force acting on a particle acts for a short time, but is much greater than any other force present When using the Impulse Approximation, we will assume this is true Especially useful in analyzing collisions The force will be called the impulsive force The particle is assumed to move very little during the collision represent the momenta immediately before and after the collision

!pi and

!pf

dp = Fdt ⇒ Δp = FΔt ⇒ mΔv = FΔt

Crash Test Example l  Assume force exerted by wall is large compared with other forces l  Gravitational and normal forces are perpendicular and so do not

effect the horizontal momentum l  Can apply impulse approximation

Initial momentum, mcvci + mwvwi = mcvci = -15.0mc kg ms-1

Final, mcvcf + mwvwf = mcvcf = 2.60mc kg ms-1

If wall doesn’t move

|Final momentum| < |Initial momentum| ⇒ Impulse imparted to wall

Inelastic collision – kinetic energy

lost Impulse = Δp = pf – pi = 2.60 –(-15.0) = 17.6 kg ms-1

Average force, FΔt = mΔv ⇒ F = 17.6m/Δt If car “weighs” 1500 kg and collision lasts for 0.15 s

⇒ F = 17.6x1500/0.15 = 176,000 N Note, if collision lasted longer, e.g. 1 hour ⇒ F = 17.6x1500/3600 = 7.3 N, which is a lot more gentle. This is why landing on a mattress isn’t as sore as landing on the road and the idea behind crumple zones in cars

(also why it’s nicer to be slapped with a sponge than a brick).

Collisions – Characteristics l  We use the term collision to represent an event

during which two particles come close to each other and interact by means of forces l  May involve physical contact, but must be generalized to

include cases with interaction without physical contact l  The time interval during which the velocity changes

from its initial to final values is assumed to be short l  The interaction forces are assumed to be much

greater than any external forces present l  This means the impulse approximation can be used

Collisions – Example 1 l  Collisions may be the

result of direct contact l  The impulsive forces

may vary in time in complicated ways l  This force is internal

to the system l  Momentum is

conserved

Collisions – Example 2 l  The collision need not

include physical contact between the objects

l  There are still forces between the particles

l  This type of collision can be analyzed in the same way as those that include physical contact

Types of Collisions l  In an elastic collision, kinetic energy is conserved (momentum

is always conserved) l  Perfectly elastic collisions occur on a microscopic level l  In macroscopic collisions, only approximately elastic

collisions actually occur l  Generally some energy is lost to deformation, sound, etc.

•  In an inelastic collision, kinetic energy is not conserved, although momentum is still conserved

•  If the objects stick together after the collision, it is a perfectly inelastic collision

l  In an inelastic collision, some kinetic energy is lost, but the objects do not stick together

l  Elastic and perfectly inelastic collisions are limiting cases, most actual collisions fall in between these two types

l  Momentum is conserved in all collisions

Perfectly Inelastic Collisions l  Since the objects stick

together, they share the same velocity after the collision

l  m1

!v1i +m2

!v2i = m1 +m2( ) !vf

Elastic Collisions l  Both momentum and

kinetic energy are conserved. Typically, there are two unknowns to solve for and so you use both equations:

m1

!v1i +m2

!v2i =m1

!v1f +m2

!v2f

12m1v1i

2 +12m2v2i

2 =12m1v1f

2 +12m2v2f

2

Ballistic Pendulum l  Isolated system of projectile

and block l  Perfectly inelastic collision.

The bullet is embedded in the block of wood

l  Momentum equation will have two unknowns

l  Use conservation of energy from the pendulum to find the velocity just after the collision

l  Then you can find the speed of the bullet

l  Energy is transferred during the perfectly inelastic collision

Conservation of momentum:

m1v1A + 0 = (m1+m2)vB

vB = m1v1A/(m1+m2)

Conservation of mechanical energy: KEi + PEi = KEf + PEf

½ (m1+m2)vB2 = (m1+m2)gh

In ballistic tests want to measure speed of bullet from known masses and measuring height travelled by block -> get rid of vB

⇒½ (m1+m2) [m1v1A/(m1+m2)]2 = (m1+m2)gh

Immediately after the collision KE = ½ (m1+m2)vB2 , all of which

goes into lifting the block to height h, ie.

⇒m12v1A

2 = 2(m1+m2)2gh ⇒ v1A = (2gh)½(m1+m2)/m1

Two-Dimensional Collisions

l  Particle 1 is moving at velocity and particle 2 is at rest

l  In the x-direction, the initial momentum is m1v1i

l  In the y-direction, the initial momentum is 0

!v1i

The momentum is conserved in all directions Use subscripts for identifying the object indicating initial or final values of the velocity components If the collision is elastic, use conservation of kinetic energy as a second equation

For example l  After the collision, the

momentum in the x-direction is

m1v1f cos θ + m2v2f cos φl  After the collision, the

momentum in the y-direction is

m1v1f sin θ + m2v2f sin φl  If the collision is elastic,

apply the kinetic energy equation

l  This is an example of a glancing collision

Two-Dimensional Collisions - tips

l  If the collision is inelastic, kinetic energy of the system is not conserved, and additional information is probably needed

l  If the collision is perfectly inelastic, the final velocities of the two objects are equal. Solve the momentum equations for the unknowns.

l  If the collision is elastic, the kinetic energy of the system is conserved l  Equate the total kinetic energy before the collision to the total

kinetic energy after the collision to obtain more information on the relationship between the velocities

Check to see if your answers are consistent with the mental and pictorial representations Check to be sure your results are realistic

Example

l  Ignore friction l  Model the cars as particles l  The collision is perfectly

inelastic l  The cars stick together

A 1500 kg car travelling east with a speed of 25 m/s collides at an intersection with a 2500 kg van travelling north at a speed of 20 m/s. Find the direction and magnitude of the wreckage after the collision, assuming the vehicles undergo a perfectly inelastic collision (i.e. they stick together).

FIRST! Draw a diagram and write down all the information given

Examine momentum in each component

x: 25mc + 0mv = (mc+mv)vfcosθ Van has no velocity

component in x-direction

Likewise for y: 0mc + 20mv = (mc+mv)vfsinθ

25(1500) = (1500+2500)vfcosθ & 20(2500) = (1500+2500)vfsinθ 37500 = 4000vfcosθ & 50000 = 4000vfsinθ

2 unknowns, but 2 equations – can solve

Divide Eq. 2 by Eq1. ⇒sinθ/cosθ = tan θ = 50000/37500

⇒θ = 53.1o and plugging this into either Eq. 1 or 2 gives vf = 15.6 m/s Could also be 233.1o, but wouldn’t make sense