Download Momentum and Impulse (updated)

MOMENTUM AND IMPULSE
Chapter 7
Linear Momentum, p
Momentum is a measure of how hard it is to
stop or turn a moving object. It is moving
inertia.


p  mv
(single particle)
ptotal   pn
(system of particles)
Units of momentum: kg(m/s) or (N)(S)
Since velocity, v, is a vector,
momentum, p, is a vector.
p is in the same direction as v
Linear Momentum, p


p  mv
Which car has more momentum? A or B
A
B
The faster car, A.
Linear Momentum, p


p  mv
Which car has more momentum? A or B
A
B
The more massive vehicle, B.
Newton’s 2nd Law
mv m(v  v0 ) mv  mv0
 F  ma  t  t  t
p
 F  t
The rate of change of momentum of a
body is equal to the net force applied to it.
Example - Washing a car: momentum change and
force. Water leaves a hose at a rate of 1.5 kg/s with
a speed of 20 m/s and is aimed at the side of a car,
which stops it (that is we ignore any splashing
back). What is the force exerted by the water on
the car?
In each second, 1.5 kg of water
v  20m / s
leaves hose and has v=20 m/s.
pi  mvi  1.5(20)  30kg m / s
p f  mv f  0
px p f  pi 0  30
 Fx  t  t  1  30 N
By Newton’s 3rd law, force exerted by water on the car
is +30 N
Example - Washing a car: momentum change and
force. What if the water splashes back from the
car? Would the force on the car be more or less?
pi
pf
In each second, 1.5 kg of water
leaves hose and has v=20 m/s.
pi  mvi  30kg m / s
If the water splashes back
p f  mv f  0
Change in momentum would be greater and so the force
should be greater.
The car exerts a force on the water not only to stop it, but an
extra force to give it momentum in the opposite direction
vf
vi
Newtons 2nd Law
v
Fnet
p mv


t
t
Slope ~ Fnet
Fnet
Impulse, J
Contact
begins
Contact
ends
Fnet t  J  p
area = J=p
Impulse-Momentum Theorem
p
Fnet 
t
J  Fnet t  p
Impulse is the product of
a net external force and
time which results in a
change in momentum
Units are N s or kg m/s
Impulsive forces are
generally of high
magnitude and short
duration.
Impulse-Momentum Theorem
J  Fnet t  p
“riding the punch”
Impulse is the area
under an F-t graph.
5
F (N)
4
3
2
1
0
0
2
4
6
t (sec)
8
10
DO NOW – Force to stop a car: momentum
change, force and impulse. A 2200 kg vehicle traveling
at 26 m/s can be stopped in 21 s by gently applying the
brakes. It can be stopped in 3.8 s if the driver slams on the
brakes, or in 0.22s if it hits a concrete wall. What impulse is
exerted on the vehicle in eachof these stops? What net
force is exerted in each case?
v  26m / s
For all three
J x  Fx t  p x  p fx  pix
J x  m(v fx  vix )  2200(0  26)
 57,200 kg m / s  J x
DO NOW – Force to stop a car: momentum
change, force and impulse. What net force is exerted
on the vehicle in each of these stops?
v  26m / s
STOP BY:
Gentle brake
Slam brake
J x  Fnetxt
Jx
 Fx  t
Fnet
J  57,200


 2734 N
t
21
0.13 Gs
Fnet
J  57,200


 15,053N
t
3.8
0.70 Gs
Concrete wall F 
net
J  57,200

 260,000 N
t
0.22
12 Gs
Problem: This force acts on a 1.2kg object moving at
120.0m/s. The direction of force is aligned with
velocity. What is the new velocity of the object?
vf = 328 m/s
3000
Fnet (N)
2000
1000
0
0
0.2
0.4
t (s)
0.6
0.8
1
Problem: This force acts on a 1.2kg object moving at
120.0m/s. The direction of force is aligned with
velocity. What is the new velocity of the object?
0
0.2
0.4
t (s)
0.6
0.8
1
0
F (N)
-1000
-2000
vf = -88.3 m/s
-3000
Impulse
Impulse
EXAMPLE: A 100 g ball is dropped from a height of h = 2.00 m
above the floor. It rebounds vertically to a height of h'= 1.50 m after
colliding with the floor. (a) Find the momentum of the ball
immediately before it collides with the floor and immediately after it
rebounds, (b) Determine the average force exerted by the floor on
the ball. Assume that the time interval of the collision is 0.01 sec.
pA=0
2m
pD=0
1.5m
pC
pB
p=?
Fnet=?
Example
A 100 g ball is dropped from a height of h = 2.00 m above the floor.
It rebounds vertically to a height of h'= 1.50 m after colliding with
the floor. (a) Find the momentum of the ball immediately before it
collides with the floor and immediately after it rebounds, (b)
Determine the average force exerted by the floor on the ball.
Assume that the time interval of the collision is 0.01 seconds.
E A  EB
U GA  K B
2
mghA  1 mvB
2
vB  2 gho  2 * 9.8 * 2  6.26 m / s
EC  ED
K C  U gD
vC  2 gh  2 * 9.8 *1.5  5.4 m / s


p  mv
pbefore  0.100(6.26)  0.626 kg * m / s
pafter  0.100(5.4)  0.54 kg * m / s
Ft  mv  m(v  vo )
F (0.01)  0.100(5.4  (6.26))
F  116.6 N
Problem: A 150-g baseball moving at 40 m/s 15o below the
horizontal is struck by a bat. It leaves the bat at 55 m/s 35o
above the horizontal. What is the impulse exerted by the bat on
the ball?
If the collision took 2.3 ms, what was the average force of the
bat on the ball?
vf = 55
35o
15o
J  Fnet t  p
vi = 40
J x  p x  mv fx  mvix
J y  p y  mv fy  mviy
 0.15(45  (38.6))
 0.15(31.5  10.4)
 12.54 N ( S )
 3.17 N ( s )
I  I x2  I 2y  12.9 N s, 14.20
Fnet  5624 N , 14.2
o
Impulse-Momentum
Theorem
Work-Energy
Theorem
I  Fnet t  p
Wnet  Fnet d  KE
Impulse on a graph:
area under the F-t curve
VECTOR
Work on a graph:
area under the F-x curve
SCALAR
Problem: a tennis player receives a shot with the ball (0.6 kg)
travelling horizontally at 50.0 m/s and returns the shot with
the ball travelling horizontally at 40.0m/s in the opposite
direction. A) what is the impulse delivered to the ball by the
racket? B) what work does the racquet do on the ball?
I x  p x  mv f  mvi
 0.6(40  50)
 54 Ns
vi=50
vf=40
Wnet  KE  12 mv2f  12 mvi2
 0.3(40 2  50 2 )
 270 J
The figure gives the momentum vs time for a
particle moving along an axis. A force directed
along the axis acts on the particle.
a) Rank the four indicated regions according to the
1 > 3 > 2 =4
magnitude of the force, greatest first.
b) In which region is the particle slowing?
3
p
2
1
3
4
t
Conservation of Momentum
1
F12
m1v’1
1
m2v2
m1v1
1
2
F21
2
F
ext
2
0
m2v’2
momentum before = momentum after
m1v1  m2v2  m1v1 'm2v2 '
as long as NO EXTERNAL FORCE ACTS on the SYSTEM
Newtons 2nd Law
p
 F  t
p  Fnet t
BALL 1
BALL 2
1
m1v’1
m1v1
F12
m2v2
1 2
2
F21
1
2
m2v’2
Newtons
2nd Law
p1  m1v1 'm1v1  F12t
p2  m2v2 'm2v2  F21t  F12t
 m1v1 ' m1v1  m2 v2 'm2 v2
m1v1  m2 v2  m1v1 ' m2 v'2
p1  p2  p1 ' p2 '
Conservation of Momentum
Newtons
3rd Law
Conservation of Momentum can be extended
to include any number of interacting bodies
F
F
system

psystem
t
system  0
psystem  0
Total momentum of
system (vector sum of
momenta of all objects)
If there are NO external forces
LAW OF CONSERVATION OF
MOMENTUM – The total momentum
of an isolated system of bodies
remains constant
LAW OF CONSERVATION OF MOMENTUM
The total momentum of an isolated system of
bodies remains constant.
psystem  0
A system is a set of objects that interacts with each
other.
An isolated system in one in which the only forces
present are those between the objects of the
system and those will be zero because of Newtons
3rd law. (  Fisolated system  0 )
Momentum in space
Example. Railroad cars collide: momentum
conserved. A 10,000 kg railroad car traveling at a
speed of 24 m/s strikes an identical car at rest. If the
cars lock together as a result of the collision, what is
their common speed afterward?
v1 = 24
pbefore  m1v1  10000(24)
pafter
 240,000kg(m / s )
 2m1v'  20000v'
pbefore  pbefore
v'  12 m / s
v2= 0
v’
DO NOW Rifle recoil. Calculate the recoil velocity
of a 5.0 kg rifle that shoots a 0.050 kg bullet at a
speed of 120 m/s.
before shooting
after shooting
v’R
v’B
pbefore  0
pafter  mR v' R  mB v' B  5v'R 0.05(120)  5v'R 6
pbefore  pbefore
0  5v' R 6
v' R  1.2 m / s
Types of Collisions
In all collisions where ΣFext = 0, momentum is conserved
Elastic Collisions
No deformation occurs.
Kinetic energy is also
conserved.
Inelastic Collisions:
Deformation occurs.
Kinetic energy is lost.
Explosions
Reverse of perfectly
inelastic collision, kinetic
energy is gained.
Perfectly Inelastic
Collisions
Objects stick together,
kinetic energy is lost.
K
1
2
mv
2
Elastic Collisions
Inelastic Collisions:
http://www.science-animations.com/support-files/collisions.swf
Example. Railroad cars.
A railroad car of mass 3000 kg, moving at 20 m/s
eastward, strikes head-on a railroad car of mass 1000 kg that is moving at 20 m/s
westward. The railroad cars stick together after the impact.
a) What is the magnitude and direction of the velocity of the combined trains after
the collision?
b) What is the impulse exerted on the smaller train by the larger train? On the
larger train by the smaller?
c) Prove that the collision is inelastic by kinetic energy analysis.
v1 = 20
v2 = 20
v’
pbefore  m1v1  m2 v2  3000(20)  1000(20)  40000 N ( s )
pafter  (m1  m2 )v'  (3000  1000)v'  4000v'
pbefore  pafter
40000  4000v'
v'  10.0 m / s
Example. Railroad cars.
A railroad car of mass 3000 kg, moving at 20 m/s
eastward, strikes head-on a railroad car of mass 1000 kg that is moving at 20 m/s
westward. The railroad cars stick together after the impact.
a) What is the magnitude and direction of the velocity of the combined trains after
the collision?
b) What is the impulse exerted on the smaller train by the larger train? On the
larger train by the smaller?
c) Prove that the collision is inelastic by kinetic energy analysis.
v1 = 20
v2 = 20
J1
J2
v’
v'  10.0 m / s
Impulse on smaller train is equal to its change in momentum
J 2  p2  m2 v'm2v2  1000[10  (20)]  3000 Ns
J1   J 2  3000 Ns Newton’s 3rd Law
Example. Railroad cars, inelastic collision.
v1 = 20
v2 = 20
v’
v'  10.0 m / s
Kbefore  K1  K 2  1 2 m1v12  1 2 m2v22  800,000 J
K after  K1 ' K 2 '  1 2 (m1  m2 )v'2  200,000 J
K is reduced so collision is inelastic
Example. Old cannons were built on wheeled carts, both to
facilitate moving the cannon and to allow the cannon to recoil
when fired. When a 150 kg cannon and cart recoils at 1.5 m/s, at
what velocity would a 10 kg cannonball leave the cannon?
v’c = 1.5
v’B = ?
pbefore  0
pafter  mC v'C  mB v' B  (150)(1.5)  10(v' B )
pbefore  pbefore
0  225  10v' B
v' B  22.5 m / s
KEbefore  0
KEafter 
1
2
mC v'  2 mB v'
 2700 J
2
C
1
2
B
Example. Pool or billiards. A billiard ball of mass 0.5 kg
moving with a velocity of 3 m/s collides head on in an elastic
collision with a second ball of equal mass at rest (v2 = 0).
What are the speeds of the 2 balls after the collision?
v1 = 3
v2= 0
pbefore  m1v1  m2 v2  0.5(3)  0  1.5 N ( s )
pafter  m1v'1  m2 v'2  0.5v'1 0.5v'2
pbefore  pbefore
1.5  0.5(v'1 v'2 )
3  v'1 v'2
Example. Pool or billiards.
v1 = 3
from conservation of momentum:
3  v'1 v'2
v2= 0
v'1  0
v'2  3
Since collision is elastic, kinetic energy is also conserved:
K before  K1  K 2  2 m v  2 m v  2.25 J
K after  K1 ' K 2 '  1 2 m v'  1 2 m v'  0.25(v'12 v'22 )
K before  K after
2.25  0.25(v'12 v'22 )
9  v'12 v'22  (3  v'2 ) 2  v'22
1
2
1 1
2
1 1
1
2
2 2
2
2 2
Truck Collision
In a head-on collision:
Which truck will experience the greatest force?
Which truck will experience the greatest impulse?
Which truck will experience the greatest change in
momentum?
Which truck will experience the greatest change in velocity?
Which truck will experience the greatest acceleration?
Which truck would you rather be in during the collision?
Truck Collision
same
same
same
In a head-on collision:
Which truck would you rather be in during the collision?
Example. A nuclear collision. A proton of mass 1.0 u
(unified atomic mass units) traveling with a speed of 20,000m/s
has an elastic, head-on collision with a Helium (He) nucleus
(mHe = 4.00 u) at initially rest. What are the velocities of the
proton and Helium nucleus after the collision?
vP = 20,000
vHe= 0
pbefore  mP vP  mHevHe  1(20000)  0  20,000
pafter  mP v'P  mHev'He  1v'P 4v'He
pbefore  pbefore
20,000  1v'P 4v'He
Example. A nuclear collision.
vP = 20,000
vHe= 0
v'P  12,000
v'He  8000
from conservation of
20,000  1v'P 4v'He
momentum:
Since collision is elastic, kinetic energy is also conserved:
K before  2 m v  2 m v  2 x10
2
2
1
1
K after  2 m v'  2 m v'  0.5v'P 2v'He
K before  K after
8
2
2
2 x10  0.5v'P 2v'He
2
2
 0.5(20000  4v'He )  2v'He
1
2
P P
2
P P
1
2
He He
2
He He
8
Example. Propulsion in space: explosion.
Rocket
F
p fuel
t
J  Ft
Ion Thruster
 5000 N
Rocket thrust:
LARGE force
over short
time
pion
 0.092 N
t
J  Ft Ion Thruster:
F
small force
over LONG
time
Example. Propulsion in space: explosion. An astronaut at
rest in space fires a thruster pistol that expels 35 g of hot gas
at 875 m/s. The combined mass of astronaut and pistol is 84
kg. How fast and in what direction is the astronaut moving
after firing the pistol?
pbefore  0
v’A = ?
v’G = 875
pafter  mAv' A  mG v'G
 84v' A 0.035(875)
pbefore  pbefore
0  84v' A 30.63
v' A  0.36 m / s
K before  0
K after  1 2 mAv'2A  1 2 mG v'G2
 13,404 J
Conservation of Momentum can also be
applied in 2 or 3 dimensions
For 2-dimensional collisions
• Set coordinate system up with x-direction
•
•
•
the same as one of the initial velocities
Label and resolve velocity vectors into x and
y components in a sketch
Resolve momentum vectors into x and y
components when working the problem
Use conservation of momentum
independently for x and y dimensions.
2 Dimensional Problem: A pool player hits the 14- ball in the xdirection at 0.80 m/s. The 14-ball knocks strikes the 8-ball, initially
at rest, which moves at a speed of 0.30 m/s at an angle of 35o angle
below the x-axis. Determine the angle of deflection of the 14-ball.
v’1
v1=0.8
q1
q2=350
q1= 17.20
v’1= 0.58 m/s
v’2=0.3
A 2.5 g bullet is fired into a 215 g wooden block of a ballistic
pendulum. After the collision, the block and embedded
bullet rise to a maximum height of 1.20m. Find the speed of
the bullet.
pbefore  pafter
pb  pP  pb  P
mb vb  0  (mb  mP )v'
(mb  mP )v'
vb 
mb
0.2175v'
vb 
 87v'
0.0025
E B  ET
2
1
m
v
'
 mB gh
B
2
v '  2 gh  4.85
vb  87v'  422m / s
The diagram depicts the beforeand after-collision speeds of a car
that undergoes a head-oncollision with a wall. In Case A,
the car bounces off the wall. In
Case B, the car crumples up and
sticks to the wall.
a. In which case is the change in
velocity the greatest?
b. In which case is the change in
momentum the greatest?
c. In which case is the impulse
the greatest?
d. In which case is the force that
acts upon the car the greatest
(assume same contact times)?