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Impulse-momentum theorem Momentum and Impulse The impulse of the force acting on an object equals the change in momentum of that object as long as the time interval t is taken to be arbitrarily small. An example of impulse The magnitude of the impulse delivered by a force during the time interval t is equal to the area under the force vs. time graph or, equivalently, to F av t.
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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