Chapter 9Momentum

• Impulse

• Momentum

• The impulse-momentum theorem

• Conservation of momentum

• Inelastic collisions

Topics:

Sample question:

Male rams butt heads at high speeds in a ritual to assert their dominance. How can the force of this collision be minimized so as to avoid damage to their brains?

Slide 9-1

Interactions

• Collision – a short duration interaction between two objects colliding

• Explosion – an interaction that forces two objects apart

Time

• Collision occur over a time or duration

• The duration of a collision depends on the material from which the objects are made

• The harder the objects the shorter the contact time

• Example – two steel balls – 1ms

• Example – tennis ball and racket -10ms

Impulse

The force of the foot on the ball is an impulsive force.

Slide 9-8

Impulse

• The amount by which the soccer ball is compressed is a measure of the magnitude of the force the foot exerts on the ball

• More compression greater force

Impulse

Impulse

• Impulsive force – A large force exerted during a small interval of time.

• Example – hammer on a nail

• Example – bat on a ball

Impulse

• The area under the curve of the force vs time graph is called the impulse of the force

• A harder kick or a kick of longer duration yields a larger impulse and a higher speed

Impulse

Impulse

• Impulse J = area under the force curve

• J = FaveΔt

• Can think of the collision in terms of the average force defined as the constant force that has the same duration and the same area under the curve as the real force

Units of Impulse

• Units are N s

• Equivalent to kg m/s

• Impulse has a sign + or -

• Impulse is a vector quantity pointing in the direction of the average force vector

• J = Fave Δt

How are impulse, mass, and velocity related

Derivation of momentum equation

• a = Fave/m

• a = Δv/ Δt =vf –vi / Δt

• a = Fave/m = vf –vi / Δt

• Fave Δt = m vf - m vi

• The product of an object’s mass and velocity is called

momentum

Momentum

Momentum is the product of an object’s mass and its velocity:

Slide 9-10

p = mv

Units

• Units of momentum are kg m/s

• These are also the units of impulse

The Impulse-Momentum Theorem

Impulse causes a change in momentum:

Slide 9-11

J =pf - pi = ∆p

Impulse momentum Theorum

• An impulse delivered to an object causes the object’s momentum to change

• How can we slow down an object in the gentlest way possible?

• If the duration of the collision can be increased, the force of the impact will be decreased.

Conservation of Momentum

Conservation of Momentum

• For the collision of the two balls

• Δp1x = J1x and Δp2x = J2x

• But since J1x = -J2x

• Δp1x = -Δp2x or

• Δp1x + Δp2x = 0

• p1x + p2x = constant

• Then (p1x)f + (p2x)f = (p1x)i + (p2x)i

Conservation of Momentum

• The x component of total momentum is conserved

• (px)f = (px)i

Momentum of a System is Conserved

• Total change in the

total momentum P

of the system is zero

• Therefore the total

momentum of the

system remains

constant

Isolated System

• Internal Forces – act only between particles within the system

• Isolated system is subject only to internal forces

• The total momentum of an isolated system is conserved

Non isolated system

• External Forces – act on the system from agents outside the system

Conservation of Momentum

• External forces can change the momentum of the system

• ΔP = Δp1+ Δp2 + Δp3

• Then by the impulse momentum theorem

• (Fexton1Δt) + (Fexton2Δt) + (Fexton3Δt)

• (Fexton1 + Fexton2+ Fexton3) Δt

• Fnet Δt

Conservation of Momentum

• If Fnet =0 the total momentum of the system does not change

• An isolated system is one in which

Fnet = 0 The total momentum of an isolated system

is a constant. Interactions within the system do not change the system’s total momentum

Pf =Pi

Conservation of Momentum

• The total momentum after an interaction is equal to the total momentum before the interaction

Example p 277

• Two skaters Sandra and David stand facing each other on frictionless ice. Sandra has a mass of 45 Kg. David has a mass of 80 Kg. They then push off from each other. After the push Sandra moves at a speed of 2.2 m/s. What is David’s speed?

Example p 277

• mS(vSx)f +mD(vDx)f = mS(vSx)i +mD(vDx)i = 0

• Solving for (vDx)f

• David moves backward with a speed of

1.2 m/s

sm

sm

kg

kgv

m

mv fSx

D

sFDx 2.12.2

80

45

Inelastic Collisions

For now, we’ll consider perfectly inelastic collisions:

A perfectly elastic collision results whenever the two objects move off at a common final velocity.

Slide 9-21

Elastic Collisions

• Two objects bounce apart after the collision• Example p281• A 200 g air track glider and a 400g air track

glider are pushed toward each other. The gliders have velcro tabs on the front so they will stick together when they collide. The 200 g glider is pushed with an initial speed of 3.0 m/s. The collision causes it to reverse direction at .50 m/s. What was the initial speed of the 400 g. glider?

Example p282

• (m1 +m2)(vx)f = m1(v1x)i +m2(v2x)i

m

ivmvmmv

xfx

ix

1121

2

sm

kgs

mkgsmkg

v f

ix 3.240.

0.320.50.60.2

Momentum and Collisions in 2 dimensions

• Motion in a plane

• The total momentum P is a vector sum of the momenta p = mv of the individual particles.

• Momenta is conserved only if each component of P is conserved

Example p 283

Peregrine falcons grab their prey from above while both are in flight. A falcon flying at 18m/s, swoops down at a 45° angle from behind a pigeon flying horizontally at 9.o m/s. The falcon has a mass of .80 kg. The pigeon has a mass of .36 kg. What are the speed and direction of the falcon (now holding the pigeon) immediately after impact?

Example p 183

• This is a perfectly inelastic collision.• The total momentum of the falcon + pigeon

system is conserved• The x component of the total momentum before

the collision must equal the x component of the total momentum after the collision

• The y component of the total momentum before the collision must equal the y component of the total momentum after the collision

Example p283

Example p 283

• First find the x and y components of momentum before the collision

• For x• (Px)i = mF(vFx)i + mP(vpx)i =mF(-vF cosθ) +mP(-vP)• = (.80kg)(-18m/s)(cos 45°) + (.36kg)(-.90m/s)• =-13.4 kg m/s• For y• (Py)i = mF(vFy)i + mP(vPy)i =mF(-vF sinθ) +0• = (.80kg)(-18m/s)(sin 45°) = -10.2 kg m/s

Example p 283

• After the collision the birds move with common velocity at angle β from horizontal.

• The x component of the final momentum is

• (Px)f = (mF +mP)vx)f

• Pxf = Pxi because momentum is conserved

s

mkgkgs

mkg

mm

Pv

pF

ixfx 6.11

36.80.

4.13

Example p283

• Similarly Pyf = Pyi

• Find the angle β

s

mkgkgs

mkg

mm

Pv

pF

iy

fy 79.836.80.

2.10

37

6.11

79.8tantan 11

sm

sm

v

v

fx

fy

Example p283• Use the Pythagorean theorem to find the final velocity

• The falcon with its meal is moving 37° below the horizontal at a speed of 15 m/s

22

fyfx vvv

sm

sm

smv 1579.86.11

22

Momentum and Collisions in 2 dimensions