MOMENTUM AND IMPULSE
Chapter 7
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 ifx
x 301
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
pJtFnet Fnet
vNewtons 2nd Law
Slope ~ Fnet
Impulse, J
area = J=Dp
Contact begins
Contact ends
vf
ptFJt
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
ptFJ net
Impulse-Momentum Theorem
0 2 4 6 8 100
1
2
3
4
5
t (sec)
F (
N)
“riding the punch”
Impulse is the area under an F-t graph.
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 ppptFJ
For all three
x
ixfxx
Jsmkg
vvmJ
/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
JF x
x
smv /26
Nt
JFnet 2734
21
200,57
Nt
JFnet 053,15
8.3
200,57
Nt
JFnet 000,260
22.0
200,57
Gentle brake
Slam brake
Concrete wall
tFJ netxx
STOP BY: 0.13
Gs
0.70 Gs
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?
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
pB
pC
2m
1.5m
Dp=?Fnet=?
pA=0
pD=0
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.
ExampleA 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.
smghv
mvmgh
KU
EE
oB
BA
BGA
BA
/26.62*8.9*22
21 2
smghv
UK
EE
C
gDC
DC
/4.55.1*8.9*22
smkgp
smkgp
vmp
after
before
/*54.0)4.5(100.0
/*626.0)26.6(100.0
NF
F
vvmvmFt o
6.116
))26.6(4.5(100.0)01.0(
)(
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
mvmvpJ ixfxxx
35o
15o
)(17.3
)4.105.31(15.0
sN
mvmvpJ iyfyyy
022 2.14,9.12 sNIIIyx
ptFJ 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
t
p
1
2
3
4
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 magnitude of the force, greatest first.
b) In which region is the particle slowing?
1 > 3 > 2 =4
3
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 on the SYSTEM
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
Ft
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
Types of Collisions In all collisions where ΣFext = 0, momentum is conserved
Elastic Collisions No deformation occurs.Kinetic energy is also conserved.
ExplosionsReverse of perfectly inelastic collision, kinetic energy is gained.
Inelastic Collisions: Deformation occurs.Kinetic energy is lost.
Perfectly Inelastic CollisionsObjects stick together, kinetic energy is lost.
22
1 mvK
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
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. 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
v’
v2 = 20
NsJJ
NsvmvmpJ
smv
3000
3000)]20(10[1000'
/0.10'
21
22222
Newton’s 3rd Law
Impulse on smaller train is equal to its change in momentum
J2J1
Example. Railroad cars, inelastic collision.
v1 = 20
v’
v2 = 20
smv /0.10'
JvmmKKK
JvmvmKKK
after
before
000,200')(''
000,8002
2121
21
2222
12112
121
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’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
121
2222
12112
121
')'3(''9)''(25.025.2
)''(25.0''''
25.2
vvvvvv
KKvvvmvmKKK
JvmvmKKK
afterbefore
after
before
from conservation of momentum:
Since collision is elastic, kinetic energy is also conserved:
3'2 v0'1 v
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 1.0 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?
vHe= 0vP = 20,000
HeP
beforebefore
HePHeHePPafter
HeHePPbefore
vvpp
vvvmvmp
vmvmp
'4'1000,20
'4'1''
000,200)20000(1
Example. A nuclear collision.
vHe= 0vP = 20,000
HeP vv '4'1000,20
22
228
2222
122
1
822
122
1
'2)'420000(5.0'2'5.0102
'2'5.0''
102
HeHe
HeP
afterbefore
HePHeHePPafter
HeHePPbefore
vvvvx
KKvvvmvmK
xvmvmK
from conservation of momentum:
Since collision is elastic, kinetic energy is also conserved:
8000' Hev000,12' Pv
Example. Propulsion in space: explosion.
Ion Thruster
tFJ
Nt
pF fuel
5000
Rocket
Rocket thrust: LARGE force over short time
tFJ
Nt
pF ion
092.0
Ion Thruster: 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?
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
vmvmK
K
GGAAafter
before
404,13
''
02
212
21
v’G = 875
Conservation of Momentum can also be applied in 2 or 3
dimensionsFor 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.
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
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.
'870025.0
'2175.0
')(')(0
vv
v
m
vmmv
vmmvmppp
pp
b
b
Pbb
Pbbb
PbPb
afterbefore
85.42'
'221
ghv
ghmvmEE
BB
TB
smvvb /422'87
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)?