Chapter 6: Momentum and Collisions

Momentum and Impulse Linear momentum

• Linear momentum p of an object of mass m moving with velocity v is the product of its mass and velocity:

vmp SI unit: kilogram-meter per second (kg m/s)

),(),( yxyx mvmvppp

Chang of momentum and force

constant. are and m where)(netnet F

tp

tvm

tvmamF

const. with lim)(limlim 000 mtp

tvm

tvmamF tttnet

Momentum conservation

0 if 0 netFtp

Momentum and Impulse

The (linear) momentum of an object isconserved when Fnet = 0.

Impulse• If a constant force F acts on an object, the impulse I delivered to the object over a time interval t is given by :

tFI

SI unit: kilogram-meter per second (kg m/s)

if vmvmptF

• When a single constant force acts on an object,

• When the force is not constant, then

)(limlimlim 000 ifttt vmvmptF

Impulse-momentumtheorem

Impulse-momentum theorem

Momentum and Impulse

)(limlimlim 000 ifttt vmvmptF

The impulse of the force acting on an object equals the change inmomentum of that object as long as the time interval t is takento be arbitrarily small.

An example of impulse

ptFav

avF

force Average

The magnitude of the impulse delivered by a force during the time intervalt is equal to the area under the force vs. time graph or, equivalently, toFavt.

Examples

Momentum and Impulse

• Example 6.1 : Teeing offA golf ball is struck with a club. Theforce on the ball varies from zero atcontact and up to the max. value.(a) Find the impulse.

m = 5.0x10-2 kgvi = 0, vf = 44 m/s

m/s kg 2.2 fif ppppI

(b) Estimate the duration of the collision and the average force.

N 104.2

s 101.9

3

2

tpF

vxt

av

av

Examples

Momentum and impulse

• Example 6.2 : How good are the bumpers(a) Find the impulse delivered to the car.

m/s kg 1064.2

m/s kg 10390.0

m/s kg 1025.2

4

4

4

if

f

i

pppI

p

p

(b) Find the average force.

N 1076.1 5

tpFav

t=0.150 s

Injury in automobile collisions

Momentum and Impulse

• A force of about 90 kN compressing the tibia can cause fracture.

• Head accelerations of 150g experienced for about 4 ms or 50g for 60 ms are fatal 50% of the time.• When the collision lasts for less than about 70 ms, a person will survive if the whole-body impact pressure (force per unit area) is less than 1.9x105 N/m2. Death results in 50% of cases in which the whole-body impact pressure reaches 3.4x105 N/m2.

Consider a collision involving 75-kg passenger not wearing s seat belt, traveling at 27 m/s who comes to rest in 0.010 s after striking an unpadded dashboard.

25

2

22

5

N/m 104/

280m/s 8.9m/s 2700m/s 2700

N 100.2

AF

ggtva

tmvmv

F

av

ifav

AF

a

F

av

av

/N/m 104.3

g150

kN 90

25

Fatal

Conservation of momentum

Conservation of Momentum

if

if

vmvmtF

vmvmtF

222212

111121

average force on 1 by 2

average force on 2 by 1

1221 FF

ffii vmvmvmvm 22112211

ifif vmvmvmvm 22221111

Conservation of momentum

When no net external force acts on a system, the total momentumof the system remains constant in time

An example

Conservation of Momentum

• Example 6.3 : The archerm1=60 kg (man + bow)m2=0.500 kg (arrow)

speed ofarrow v2=50.0 m/s

m/s 417.0

0

21

21

2211

ff

ff

fi

vmmv

vmvm

pp

The archer is movingopposite the directionof the arrow

Three types of collisions

Collisions

• Inelastic collisionA collision in which momentum is conserved, but kineticenergy is not.

• Perfectly inelastic collisionA collision between two objects in which both stick togetherafter the collision.

• Elastic collisionA collision in which both momentum and kinetic energyare conserved.

Perfectly inelastic collisions

Collisions

• Consider two objects with mass m1 and m2 moving with known initial velocities v1i and v2i along a straight line.• They collide head-on and after the collision, they stick together and move with a common velocity vf.

fii vmmvmvm )( 212211

21

2211

mmvmvmv ii

f

Examples of perfect inelastic collision

Collisions

• Example 6.4 : An SUV vs. a compact

m1=1.80x103 kgv1i=15.0 m/s

(a) Find the final speed after collision.

fii vmmvmvm )( 212211

m2=9.00x102 kgv2i=-15.0 m/sm/s 00.5

21

2211

mmvmvmv ii

f

(b) Find the changes in velocity.m/s 0.1011 if vvv

m/s 0.2022 if vvv(c) Find the change in kinetic energy of the system.

J 1070.2)(21,

21

21 52

21222

211 KEvmmKEvmvmKE ffiii

Examples of perfect inelastic collision

Collisions

• Example 6.5 : Ballistic pendulum

m1=5.00 gm2=1.00 kgh = 5.00 cm

Find the initial speed of bullet.

ffii PEKEPEKE

ghmmvmm f )(00)(21

212

21

m/s 990.0222 ghvghv ff

fii vmmvmvm )( 212211

fi vmmvm )(0 2111

m/s 199)(

1

211

mvmm

v fi

At the height h

Right after collisionBefore collision

Right after collision

Elastic collisions

Collisions

• Consider two objects with mass m1 and m2 moving with known initial velocities v1i and v2i along a straight line.• They collide head-on and after the collision, they leave each other with velocities v1f and v2f .

(2) )()( 22211122112211 fififfii vvmvvmvmvmvmvm

21

21

21

21 2

22211

222

211 ffii vmvmvmvm

)()( 22

222

21

211 iffi vvmvvm

(1) ))((

))((

22222

11111

ifif

fifi

vvvvm

vvvvm

ffiiiffi vvvvvvvv 21212211:)2/()1(

An example of elastic collision

Collisions

• Example 6.7 : Two blocks and a spring(a) Find v2f when v1f=+3.00 m/s.

m2=2.10 kgv2i=-2.50 m/sm/s 74.1

2

1122112

mvmvmvm

v fiif

(b) Find the compression of the spring.

m1=1.60 kgv1i=+4.00 m/s

k=6.00x102 N/m

ffii vmvmvmvm 22112211

2222

211

222

211 2

1 21

21

21

21 kxvmvmvmvm ffii

m 173.0x

Collisions in 2-dimension

Glancing Collisions

• Momentum conservation in 2-D

ffii vmvmvmvm 22112211

fxfxixix vmvmvmvm 22112211

fyfyiyiy vmvmvmvm 22112211

coscos0 221111 ffix vmvmvm

sinsin00 2211 ff vmvm

An example of a collision in 2-D

Glancing Collisions

• Example 6.8 : A perfect inelastic collision at an intersection

mcar=1.50x103 kg

mvan=2.50x103 kg

Find the magnitude and direction ofthe velocity of the wreckage.

m/s kg 1075.3 4carcarix vmp

cos)( fvancarfx vmmp

fxix pp

coskg) 1000.4(m/s kg 1075.3 34fv

m/s kg 1000.5 4vanvaniy vmp

sin)( fvancarfy vmmp

fyiy pp

sinkg) 1000.4(m/s kg 1000.5 34fv

An example of a collision in 2-D (cont’d)

Glancing Collisions

• Example 6.8 : A perfect inelastic collision at an intersection (cont’d)

mcar=1.50x103 kg

mvan=2.50x103 kg

Find the magnitude and direction ofthe velocity of the wreckage.

coskg) 1000.4(m/s kg 1075.3 34fv

sinkg) 1000.4(m/s kg 1000.5 34fv

33.1m/s kg 1075.3m/s kg 1000.5tan 4

4

1.53

m/s 6.151.53sin)kg 1000.4(

m/s kg 1000.53

4

fv

Principle (hand-waving argument)

Rocket Propulsion

• The driving force of motion of ordinary vehicles such as cars and locomotives is friction. A car moves because a reaction to the force exerted by the tire produces a force by the road on the wheel.

• What is then driving force of a rocket? When an explosion occurs in a spherical chamber with fuel gas in a rocket engine the hot gas expands and presses against all sides of the chamber uniformly. So all forces are in balance-no net force.

If there is a hole as in (b), part of the hot gas escapes from the hole (nozzle), which breaks the balance of the forces. This unbalance create a net upward force.

Principle (detailed argument)

Rocket Propulsion

• At time t, the momentum of the rocket plus the fuel is (M+m)v.

M : mass of rocketm : mass of fuel to be ejected in t

• During time period t, the rocket ejects fuel of mass m whose speed ve relative to the rocket and gains the speed to v+v. From momentum conservation:

)()()( evvmvvMvmM

mvvM e• The increase m in the mass of the exhaust corresponds to an equal decrease in the mass of the rocket so that m=-M.

MvvM e

Principle (detailed argument)

Rocket Propulsion

• Using calculus:

M : mass of rocketm : mass of fuel to be ejected in t

MvvM e

f

ieif M

Mvvv ln

Thrust

• is defined as the force exerted on the rocket by the ejected exhaust gases.

Instantaneous thrust

tMv

tvMMa e

An example

Rocket Propulsion

• Example 6.0 : Single stage to orbit M : mass of rocket 1.00x105 kgm : burnout mass 1.00x104 kgve : exhaust velocity 4.50x103 m/st : blast off time period 4 min

(b) Find the thrust at liftoff.

kg 1000.9kg 1000.1kg 1000.1 454

if MMM

kg/s 1075.3 2tM

N 1069.1 6

tMvT ehThrust

(c) Find the initial acceleration.

2m/s 10.7gMTaMgTFMa h

h

fv