Transcript Lecture6

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:


p  mv
SI unit: kilogram-meter per second (kg m/s)

p  ( p x , p y )  (mvx , mv y )
 Chang
of momentum and force




v (mv ) p
Fnet  ma  m


where m and Fnet are constant.
t
t
t




v
(mv )
p
Fnet  ma  m lim t 0
 lim t 0
 lim t 0
with m const.
t
t
t
Momentum and Impulse

Momentum conservation


p
 0 if Fnet  0 
t

The (linear) momentum of an object is
conserved 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 :
 
I  Ft
SI unit: kilogram-meter per second (kg m/s)
• When a single constant force acts on an object,


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Ft  p  mv f  mvi
• When the force is not constant, then
Impulse-momentum
theorem

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

lim t 0 Ft  lim t 0 p  lim t 0 (mv f  mvi )
Momentum and Impulse

Impulse-momentum theorem

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

lim t 0 Ft  lim t 0 p  lim t 0 (mv f  mvi )
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
Average force Fav


Favt  p
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
Favt.
Momentum and Impulse

Examples
• Example 6.1 : Teeing off
A golf ball is struck with a club. The
force on the ball varies from zero at
contact and up to the max. value.
(a) Find the impulse.
I  p  p f  pi  p f  2.2 kg m/s
(b) Estimate the duration of the collision
and the average force.
x
t 
 9.110  2 s
vav
Fav 
p
 2.4 103 N
t
m = 5.0x10-2 kg
vi = 0, vf = 44 m/s
Momentum and impulse

Examples
• Example 6.2 : How good are the bumpers
(a) Find the impulse delivered to the car.
pi  2.25 10 4 kg m/s
p f  0.390 10 4 kg m/s
I  p  p f  pi  2.64 10 4 kg m/s
(b) Find the average force.
Fav 
p
 1.76 105 N
t
t=0.150 s
Momentum and Impulse

Injury in automobile collisions
• 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.
Fatal F  mv f  mvi  2.0 105 N
90 kN
150g
 Fav
a
3.4 105 N/m 2  Fav / A
av
t
v
2700 m/s 2
2
a
 2700 m/s 
g  280 g
2
t
9.8 m/s
Fav / A  4 105 N/m 2
Conservation of Momentum

Conservation of momentum
average force on 1 by 2
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F21t  m1v1 f  m1v1i
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F12t  m2 v2 f  m2 v2i
average force on 2 by 1

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F21   F12
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m1v1 f  m1v1i  m2 v2 f  m2 v2i

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

m1v1i  m2 v2i  m1v1 f  m2 v2 f
Conservation of momentum
When no net external force acts on a system, the total momentum
of the system remains constant in time
Conservation of Momentum

An example
• Example 6.3 : The archer
m1=60 kg (man + bow)
m2=0.500 kg (arrow)
pi  p f
0  m1v1 f  m2 v2 f
m2
v1 f  
v2 f  0.417 m/s
m1
The archer is moving
opposite the direction
of the arrow
speed of
arrow
v2=50.0 m/s
Collisions

Three types of collisions
• Inelastic collision
A collision in which momentum is conserved, but kinetic
energy is not.
• Perfectly inelastic collision
A collision between two objects in which both stick together
after the collision.
• Elastic collision
A collision in which both momentum and kinetic energy
are conserved.
Collisions

Perfectly inelastic 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.
m1v1i  m2 v2i  (m1  m2 )v f
m1v1i  m2 v2i
vf 
m1  m2
Collisions

Examples of perfect inelastic collision
• Example 6.4 : An SUV vs. a compact
(a) Find the final speed after collision.
m1v1i  m2 v2i  (m1  m2 )v f
m1v1i  m2 v2i
vf 
 5.00 m/s
m1  m2
m1=1.80x103 kg m2=9.00x102 kg
v1i=15.0 m/s
v2i=-15.0 m/s
(b) Find the changes in velocity.
v1  v f  v1i  10.0 m/s
v2  v f  v2i  20.0 m/s
(c) Find the change in kinetic energy
of the system.
KEi 
1
1
1
m1v12i  m2 v22i , KE f  (m1  m2 )v 2f  KE  2.70 105 J
2
2
2
Collisions

Examples of perfect inelastic collision
• Example 6.5 : Ballistic pendulum
Find the initial speed of bullet.
Right after collision At the height h
KEi  PEi  KE f  PE f
1
(m1  m2 )v 2f  0  0  (m1  m2 ) gh
2
v 2f  2 gh  v f  2 gh  0.990 m/s
Before collision
Right after collision
m1v1i  m2 v2i  (m1  m2 )v f
m1v1i  0  (m1  m2 )v f
v1i 
(m1  m2 )v f
m1
 199 m/s
m1=5.00 g
m2=1.00 kg
h = 5.00 cm
Collisions

Elastic 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 .
1
1
1
1
m1v12i  m2 v22i  m1v12f  m2 v22 f
2
2
2
2
m1 (v12i  v12f )  m2 (v22 f  v22i )
m1 (v1i  v1 f )(v1i  v1 f )
 m2 (v2 f  v2i )(v2 f  v2i ) (1)
m1v1i  m2 v2i  m1v1 f  m2 v2 f  m1 (v1i  v1 f )  m2 (v2i  v2 f ) (2)
(1) /( 2) : v1i  v1 f  v2 f  v2i  v1i  v2i  v1 f  v2 f 
Collisions

An example of elastic collision
• Example 6.7 : Two blocks and a spring
(a) Find v2f when v1f=+3.00 m/s.
m1v1i  m2 v2i  m1v1 f  m2 v2 f
v2 f 
m1v1i  m2v2i  m1v1 f
m2
 1.74 m/s
(b) Find the compression of the spring.
1
1
1
1
1
m1v12i  m2 v22i  m1v12f  m2 v22 f  kx2
2
2
2
2
2
x  0.173 m
m1=1.60 kg
m2=2.10 kg
v1i=+4.00 m/s
v2i=-2.50 m/s
k=6.00x102 N/m
Glancing Collisions

Collisions in 2-dimension
• Momentum conservation in 2-D
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m1v1i  m2 v2i  m1v1 f  m2 v2 f
m1v1ix  m2 v2ix  m1v1 fx  m2 v2 fx
m1v1iy  m2 v2iy  m1v1 fy  m2 v2 fy
m1v1ix  0  m1v1 f cos   m2 v2 f cos 
0  0  m1v1 f sin   m2 v2 f sin 
Glancing Collisions

An example of a collision in 2-D
• Example 6.8 : A perfect inelastic collision at an intersection
Find the magnitude and direction of
the velocity of the wreckage.
4
p

m
v

3
.
75

10
kg m/s
 ix car car
 p  (m
p p
fx
car
ix
 mvan )v f cos
mcar=1.50x103 kg
fx
3.75 104 kg m/s  (4.00 103 kg) v f cos
4
p

m
v

5
.
00

10
kg m/s
 iy van van
 p  (m
p p
fy
iy
car
 mvan )v f sin 
fy
5.00 104 kg m/s  (4.00 103 kg) v f sin  mvan=2.50x103 kg
Glancing Collisions

An example of a collision in 2-D (cont’d)
• Example 6.8 : A perfect inelastic collision at an intersection (cont’d)
Find the magnitude and direction of
the velocity of the wreckage.
3.75 104 kg m/s  (4.00 103 kg) v f cos
5.00 104 kg m/s  (4.00 103 kg) v f sin 
mcar=1.50x103 kg
5.00 104 kg m/s
tan  
 1.33
4
3.75 10 kg m/s
  53.1
5.00 104 kg m/s
vf 
 15.6 m/s
3
(4.00 10 kg) sin 53.1
mvan=2.50x103 kg
Rocket Propulsion

Principle (hand-waving argument)
• 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.
Rocket Propulsion

Principle (detailed argument)
• At time t, the momentum of the
rocket plus the fuel is (M+m)v.
• 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:
( M  m)v  M (v  v)  m(v  ve )
Mv  ve m
• The increase m in the mass of the
exhaust corresponds to an equal
decrease in the mass of the rocket
so that m=-M.
Mv  ve M
M : mass of rocket
m : mass of fuel to be ejected
in t
Rocket Propulsion

Principle (detailed argument)
• Using calculus:
Mv  ve M
 Mi
v f  vi  ve ln 
M
 f





Thrust
• is defined as the force exerted on
the rocket by the ejected exhaust
gases.
Instantaneous thrust
v
M
Ma  M
 ve
t
t
M : mass of rocket
m : mass of fuel to be ejected
in t
Rocket Propulsion

An example
• Example 6.0 : Single stage to orbit
(b) Find the thrust at liftoff.
M  M f  M i
M
m
ve
t
: mass of rocket 1.00x105 kg
: burnout mass 1.00x104 kg
: exhaust velocity 4.50x103 m/s
: blast off time period 4 min
 1.00 104 kg  1.00 105 kg  9.00 104 kg
M
 3.75 10 2 kg/s
t
vf 
Thrust Th  ve
M
 1.69 10 6 N
t
(c) Find the initial acceleration.
Ma   F  Th  Mg  a 
Th
 g  7.10 m/s 2
M