MOMENTUM AND IMPULSE
Chapter 8
Linear Momentum, p
ntotal pp
vmp
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
Momentum is a measure of how hard it is to stop or turn a moving object. It is moving inertia.
(single particle)
(system of particles)
Linear Momentum, p
vmp
Which car has more momentum? A or B
A
B
The faster car, A.
Linear Momentum, p
vmp
Which car has more momentum? A or B
A
B
The more massive vehicle, B.
Newton’s 2nd Law
t
pF
t
mvmv
t
vvm
t
vmmaF 00 )(
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?
Nt
pp
t
pF ifxx 30
1
300
smv /20In each second, 1.5 kg of water leaves hose and has v=20 m/s.
0
/30)20(5.1
ff
ii
mvp
smkgmvp
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?
ipsmkgmvp ii /30
0 ff mvp
If the water splashes back
In each second, 1.5 kg of water leaves hose and has v=20 m/s.
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
fp
vi
t
vm
t
pFnet
pItFnet Fnet
vNewtons 2nd Law
Slope ~ Fnet
Impulse, I
area = I=Dp
Contact begins
Contact ends
vf
ptFIt
pF
net
net
Impulse-Momentum Theorem
Impulse is the product of a net external force and time which results in a change in momentum
Impulsive forces are generally of high magnitude and short duration.
Units are N s or kg m/s
ptFI net
Impulse-Momentum Theorem
0 2 4 6 8 100
1
2
3
4
5
t (sec)
F (
N)
“riding the punch”
Impulse on a graph: area under the curve
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? smv /26
ixfxxxx ppptFI
For all three
x
ixfxx
Ismkg
vvmI
/200,57
)260(2200)(
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?
t
IF xx
smv /26
Nt
IFnet 2734
21
200,57
Nt
IFnet 053,15
8.3
200,57
Nt
IFnet 000,260
22.0
200,57
Gentle brake
Slam brake
Concrete wall
tFI netxx
STOP BY: 0.12
Gs
0.68 Gs
11.8 Gs
http://www.walter-fendt.de/ph14e/collision.htm
http://www.nbclearn.com/portal/site/learn/nfl/cuecard/50974/
“Newtons 2nd law of motion”: Football kick, Newtons laws and Impulse
Collision simulation
http://www.youtube.com/watch?v=yUpiV2I_IRI&feature=related car crash 22 min
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
1000
2000
3000
0 0.2 0.4 0.6 0.8 1
Fnet
(N)
t (s)
vf = 328 m/s
0 0.2 0.4 0.6 0.8 1
-3000
-2000
-1000
0
t (s)
F (
N)
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 = -88.3 m/s
Impulse
Impulse
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?
vi = 40
vf = 55
)(54.12
))6.38(45(15.0
SN
mvmvpI ixfxxx
35o
15o
)(17.3
)4.105.31(15.0
sN
mvmvpI iyfyyy
022 2.14,9.12 sNIIIyx
ptFI net
onet NF 2.14,5624
ptFI net
Impulse-Momentum
Theorem
Impulse on a graph: area under the F-t curve
KEdFW netnet
Work on a graph: area under the F-x curve
Work-EnergyTheorem
VECTOR 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?
vi=50
vf=40Ns
mvmvpI ifxx
54)5040(6.0
J
mvmvKEWifnet
270)5040(3.0 22
2212
21
DEMO 1: A ball is dropped to the ground
Forces acting on the ball as it is falling
FG
Is there a net force?
Is the momentum of the ball conserved (constant) as it falls?
mv1
mv2
Dp ≠ 0
0F
Describe a system in which the total momentum is conserved.
DEMO 1: A ball is dropped to the ground
FG,ball,earth
Dp = 0
0F
A system in which the total momentum is conserved – BALL + EARTH
FG,earth,ball
In this system of the Ball + Earth, there are NO EXTERNAL FORCES. Only forces are those between the objects in the system
Conservation of Momentum
1m1v1 2
m2v2
2m2v’2
1m1v’1
1 2F21F12 0extF
'' 22112211 vmvmvmvm momentum before = momentum after
as long as NO EXTERNAL FORCE ACTS
Conservation of Momentum
1m1v1 2m2v2
2m2v’21
m1v’1
1 2F21F12
tFvmvmp 1211111 '
tFtFvmvmp 122122222 '
''''
''
2121
22112211
22221111
ppppvmvmvmvmvmvmvmvm
tFpt
pF
net
BALL 1
BALL 2
Newtons 2nd Law
Newtons 3rd Law
Newtons 2nd Law
0
0
system
system
systemsystem
p
F
t
pF
If there are NO external forces
Conservation of Momentum can be extended to include any number of interacting bodies
Total momentum of system (vector sum of momenta of all objects)
LAW OF CONSERVATION OF MOMENTUM – The total momentum of an isolated system of bodies remains constant
0 systemp
LAW OF CONSERVATION OF MOMENTUM
The total momentum of an isolated system of bodies remains constant.
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. ( )
0systemisolatedF
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 v2= 0
v’
smv
pp
vvmp
smkg
vmp
beforebefore
after
before
/12'
'20000'2
)/(000,240
)24(10000
1
11
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.
v’B
smv
v
pp
vvvmvmp
p
R
R
beforebefore
RRBBRRafter
before
/2.1'
6'50
6'5)120(05.0'5''
0
v’R
before shooting
after shooting
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.
Perfectly Inelastic CollisionsObjects stick together, kinetic energy is lost.
ExplosionsReverse of perfectly inelastic collision, kinetic energy is gained.2
21 mvKE
Elastic Collisions
Inelastic Collisions:
http://www.science-animations.com/support-files/collisions.swf
Example. Railroad cars, inelastic collision. 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. What is the magnitude and direction of the velocity of the combined mass after the collision? Prove that the collision is inelastic by KE analysis.v1 = 20
v’
v2 = 20
smv
v
pp
vvvmmp
sNvmvmp
afterbefore
after
before
/0.10'
'400040000
'4000')10003000(')(
)(40000)20(1000)20(3000
21
2211
Example. Railroad cars, inelastic collision.
v1 = 20
v’
v2 = 20
smv
v
pp
vvvmmp
sNvmvmp
afterbefore
after
before
/0.10'
'400040000
'4000')10003000(')(
)(40000)20(1000)30(3000
21
2211
JvmmKE
JvmvmKE
after
before
000,200')(
000,8002
2121
2222
12112
1
KE is reduced so collision is inelastic
Bouncing ball
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’B = ?
smv
v
pp
vvmvmp
p
B
B
beforebefore
BBBCCafter
before
/5.22'
'102250
)'(10)5.1)(150(''
0
v’c = 1.5
J
vmvmKE
KE
BBCCafter
before
2700
''
02
212
21
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?v2= 0
v1 = 3
21
21
212211
2211
''3
)''(5.05.1
'5.0'5.0''
)(5.10)3(5.0
vv
vv
pp
vvvmvmp
sNvmvmp
beforebefore
after
before
Example. Pool or billiards. v2= 0
v1 = 3
21 ''3 vv
22
22
22
21
22
21
22
21
2222
12112
1
2222
12112
1
')'3(''9
)''(25.025.2
)''(25.0''
25.2
vvvv
vv
KEKE
vvvmvmKE
JvmvmKE
afterbefore
after
before
from conservation of momentum:
Since collision is elastic, kinetic energy is also conserved:
3'2 v0'1v
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
In a head-on collision:
Which truck would you rather be in during the collision?
Truck Collision
same
same
same
Example. A nuclear collision. A proton of mass 1.01 u (unified atomic mass units) traveling with a speed of 3.60 x 104 m/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? vHe= 0vP = 3.6 x 104
HeP
beforebefore
HePHeHePPafter
HeHePPbefore
vvx
pp
vvvmvmp
xxvmvmp
'4'1106.3
'4'1''
106.30)106.3(1
4
44
Example. A nuclear collision.
vHe= 0vP = 3.6 x 104
HeP vvx '4'1106.3 4
224
228
222212
21
82212
21
'2)'4106.3(5.0
'2'5.01048.6
'2'5.0''
1048.6
HeHe
HeP
afterbefore
HePHeHePPafter
HeHePPbefore
vvx
vvx
KEKE
vvvmvmKE
xvmvmKE
from conservation of momentum:
Since collision is elastic, kinetic energy is also conserved:
41045.1' xv He 41015.2' xv P
Example. A nuclear collision. A proton of mass 1.01 u (unified atomic mass units) traveling with a speed of 3.60 x 104 m/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? vHe= 0vP = 3.6 x 104
HeP
beforebefore
HePHeHePPafter
HeHePPbefore
vvx
pp
vvvmvmp
xxvmvmp
'4'1106.3
'4'1''
106.30)106.3(1
4
44
Example. Propulsion in space: explosion.
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?
v’A = ?
smv
v
pp
v
vmvmp
p
A
A
beforebefore
A
GGAAafter
before
/36.0'
63.30'840
)875(035.0'84
''
0
J
vmvmKE
KE
GGAAafter
before
404,13
''
02
212
21
v’G = 875
Conservation of Momentum can also be applied in 2 or 3 dimensions
For 2-dimensional collisions• Use conservation of momentum
independently for x and y dimensions.
• You must resolve your momentum vectors into x and y components when working the problem
v1=0.8
v’2=0.3
v’1
q1
q2=350
2 Dimensional Problem: A pool player hits the 14- ball in the x-direction 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.
q1= 17.20
v’1= 0.58 m/s
The diagram depicts the before- and after-collision speeds of a car that undergoes a head-on-collision 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)?
http://www.walter-fendt.de/ph14e/collision.htm
http://www.nbclearn.com/portal/site/learn/nfl/cuecard/50974/
“Newtons 2nd law of motion”: Football kick, Newtons laws and Impulse
Collision simulation
“Newtons 3rd law of motion”: Football kick, conservation of momentum and collisions
http://www.youtube.com/watch?v=yUpiV2I_IRI&feature=related car crash 22 min
http://www.youtube.com/watch?feature=endscreen&v=OuA-znVMY3I&NR=1
Mythbusters giant Newtons cradle
interactive quiz