Physics 207: Lecture 21
Today’s Agenda
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Momentum Conservation
Inelastic collisions in one dimension
Inelastic collisions in two dimensions
Explosions
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Comment on energy conservation
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Ballistic pendulum
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Elastic collisions in one dimension
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Impulse and average force
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1
Momentum Conservation
FEXT =
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z
z
dP
dt
dP
=0
dt
FEXT = 0
The concept of momentum conservation is one of the most
fundamental principles in physics.
This is a component (vector) equation.
ÍWe can apply it to any direction in which there is no
external force applied.
You will see that we often have momentum conservation
even when energy is not conserved.
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Page 1
Elastic vs. Inelastic Collisions
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A collision is said to be elastic when kinetic energy as well as
momentum is conserved before and after the collision.
Kbefore = Kafter
ÍCarts colliding with a spring in between, billiard balls, etc.
vi
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A collision is said to be inelastic when kinetic energy is not
conserved before and after the collision, but momentum is
conserved.
Kbefore ≠ Kafter
ÍCar crashes, collisions where objects stick together, etc.
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Inelastic Collision in 1-D: Example 2
M
m
v=0
V
ice
(no friction)
M+m
v=?
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Page 2
Example 2...
Use conservation of momentum to find v after the collision.
After the collision:
Before the collision:
Pf = ( M + m )v
Pi = M V + m ( 0 )
Pi = Pf
Conservation of momentum:
M V = ( M + m )v
v =
M
V
(M + m)
vector equation
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⎛M + m⎞
v =⎜
⎟V
⎝ m ⎠
Example 2...
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Now consider the K.E. of the system before and after:
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Before:
E BUS =
z
After:
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So
1
1
M + m⎞ 2 2 1 ⎛M + m⎞
MV 2 = M ⎛⎜
⎟ v = ⎜
⎟ ( M + m )v 2
2
2 ⎝ M ⎠
2⎝ M ⎠
EA =
1
( M + m )v 2
2
M ⎞
E A = ⎛⎜
⎟E
⎝M + m⎠ B
Kinetic energy is NOT conserved
in an inelastic collision!
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Page 3
Lecture 21, Act 1
Momentum Conservation
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Two balls of equal mass are thrown horizontally with the same
initial velocity. They hit identical stationary boxes resting on a
frictionless horizontal surface.
The ball hitting box 1 bounces back, while the ball hitting box 2
gets stuck.
ÍWhich box ends up moving faster?
(a) Box 1
(b) Box 2
(c) same
2
1
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Lecture 21, Act 1
Momentum Conservation
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z
z
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Since the total external force in the x-direction is zero,
momentum is conserved along the x-axis.
In both cases the initial momentum is the same (mv of ball).
In case 1 the ball has negative momentum after the collision,
hence the box must have more positive momentum if the total is
to be conserved.
The speed of the box in case 1 is biggest!
x
V1
2
1
V2
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Page 4
Lecture 21, Act 1
Momentum Conservation
mvinit = MV1 - mvfin
mvinit = (M+m)V2
V1 = (mvinit + mvfin) / M
V2 = mvinit / (M+m)
V1 numerator is bigger and its denominator is smaller
than that of V2.
V1 > V2
V1
x
2
1
V2
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Inelastic collision in 2-D
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Consider a collision in 2-D (cars crashing at a slippery
intersection...no friction).
V
v1
m1 + m2
m1
m2
v2
before
after
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Page 5
Inelastic collision in 2-D...
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There are no net external forces acting.
ÍUse momentum conservation for both components.
m1
Px ,i = Px ,f
m1v1 = (m1 + m 2 )V x
Vx =
v
(m1 + m 2 ) 1
m 2 v 2 = (m1 + m 2 )V y
Py ,i = Py ,f
Vy =
v1
m2
(m1 + m 2 )
v2
V = (Vx,Vy)
m1 + m2
m1
m2
v2
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Inelastic collision in 2-D...
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So we know all about the motion after the collision!
Vx =
V = (Vx,Vy)
θ
Vx
Vy
Vy =
tan θ =
m1
(m1 + m 2 )
m2
(m1 + m 2 )
Vy
Vx
=
v1
v2
m2v 2 p2
=
m1v1 p1
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Page 6
Inelastic collision in 2-D...
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We can see the same thing using vectors:
P
P
p1
p2
p2
θ
p1
tan θ =
p2
p1
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Lecture 21, Act 2
Center of Mass
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A bomb explodes into 3 identical pieces. Which of the following
configurations of velocities is possible?
(a) 1
(b) 2
v
v
m
m
v
V
m
(c) both
v
m
m
m
(1)
v
(2)
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Page 7
Lecture 21, Act 2
Center of Mass
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z
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No external forces, so P must be conserved.
Initially: P = 0
In explosion (1) there is nothing to balance the upward
momentum of the top piece so Pfinal ≠ 0.
v
mv
m
mv
mv
v
v
m
m
(1)
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Lecture 21, Act 2
Center of Mass
z
z
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No external forces, so P must be conserved.
All the momenta cancel out.
Pfinal = 0.
mv
v
mv
mv
m
v
m
m
(2)
Page 8
v
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Comment on Energy Conservation
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We have seen that the total kinetic energy of a system
undergoing an inelastic collision is not conserved.
ÍEnergy is lost:
» Heat (bomb)
» Bending of metal (crashing cars)
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Kinetic energy is not conserved since work is done during
the collision!
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Momentum along a certain direction is conserved when
there are no external forces acting in this direction.
ÍIn general, momentum conservation is easier to satisfy
than energy conservation.
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Lecture 21, Act 3
Collisions
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A box sliding on a frictionless surface collides and sticks to a
second identical box which is initially at rest.
What is the ratio of initial to final kinetic energy of the system?
(a) 1
(b) 2
(c) 2
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Page 9
Lecture 21, Act 3
Solution
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No external forces in the x direction, so PX is constant.
v
K I = mv
m
m
⎛v ⎞
PF = (2m )⎜ ⎟
⎝2⎠
m
m
v/2
x
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Lecture 21, Act 3
Solution
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Compute kinetic energies:
KI =
1
mv 2
2
v
m
m
2
KF =
1
(2 m )⎛⎜ v ⎞⎟ = 1 K I
2
2
⎝2⎠
KI
KF
m
m
v/2
=2
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Page 10
Lecture 21, Act 3
Another solution
1
P2
mv 2 =
2
2m
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We can write K =
z
P is the same before and after the collision.
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The mass of the moving object has doubled, hence the
kinetic energy must be half.
m
KI
KF
m
=2
m
m
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Lecture 21, Act 3
Another Question:
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Is it possible for two blocks to collide inelastically in such a
way that the kinetic energy after the collision is zero?
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Page 11
Lecture 21, Act 3
Another Question
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Is it possible for two blocks to collide inelastically in such a
way that the kinetic energy after the collision is zero?
YES: If the CM is not moving!
CM
CM
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Ballistic Pendulum
L
L
m v
V=0
H
M+m
V
M
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L
L
A projectile of mass m moving horizontally with speed v
strikes a stationary mass M suspended by strings of length
L. Subsequently, m + M rise to a height of H.
Given H, what is the initial speed v of the projectile?
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Page 12
Ballistic Pendulum...
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Two stage process:
1. m collides with M, inelastically. Both M and
m then move together with a velocity V
(before having risen significantly).
2. M and m rise a height H, conserving K+U energy E.
(no non-conservative forces acting after collision)
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Ballistic Pendulum...
Stage 1: Momentum is conserved
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in x-direction: mv = ( m + M )V
z
m ⎞
V = ⎛⎜
⎟v
⎝m + M ⎠
Stage 2: K+U Energy is conserved
( EI = EF )
1
2
V 2 = 2 gH
( m + M )V 2 = ( m + M ) gH
Eliminating V gives:
M⎞
⎛
v = ⎜1 + ⎟ 2 gH
m⎠
⎝
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Page 13
Elastic Collisions
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Elastic means that kinetic energy is conserved as well as
momentum.
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This gives us more constraints
ÍWe can solve more complicated problems!!
ÍBilliards (2-D collision)
ÍThe colliding objects
have separate motions
after the collision as
well as before.
Initial
Final
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Start with a simpler 1-D problem
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Elastic Collision in 1-D
m2
m1
initial
v1,i
v2,i
x
m2
m1
final
v2,f
v1,f
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Page 14
Elastic Collision in 1-D
before
Conserve PX:
m2
m1
v1,i v2,i
x
m1v1,i + m2v2,i = m1v1,f + m2v2,f
after
v1,f
Conserve Kinetic Energy:
1/
2
v2,f
m1v21,i + 1/2 m2v22,i = 1/2 m1v21,f + 1/2 m2v22,f
Suppose we know v1,i and v2,i
We need to solve for v1,f and v2,f
Should be no problem
2 equations & 2 unknowns!
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Elastic Collision in 1-D
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However, solving this can sometimes get a little
bit tedious since it involves a quadratic
equation!!
m1v1,i + m2v2,i = m1v1,f + m2v2,f
1/
2
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m1v21,i + 1/2 m2v22,i = 1/2 m1v21,f + 1/2 m2v22,f
A simpler approach is to introduce the
Center of Mass Reference Frame
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Page 15
Aside: Kinetic Energy
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We know that
K = 1/2mv2
Kinetic energy can also be expressed in
terms of momentum:
K = 1/2mv2
=
=
m 2v 2
2m
( m v )2
2m
K=
p2
2m
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Collision “timescales”
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Collisions typically involve interactions that happen quickly.
vf
vi
Vf
F
initial
final
The balls are in contact
for a very short time.
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Page 16
Collision “timescales”
During this brief time, the forces involved can be quite large
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∆t
t1
t2
p1
p2
t5
t4
t3
p3 = 0
p5
p4
F2
F4
F3
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Force and Impulse
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The diagram shows the force vs. time for a typical collision.
The impulse, I, of the force is a vector defined as the
integral of the force during the collision.
F
tf
I = ∫t F dt
i
Impulse I = area under this curve !
t
∆t
ti
Impulse has units of Ns.
tf
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Page 17
Force and Impulse
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F=
Using
dP
dt
the impulse becomes:
dP
dt
dt
= ∫tt dP = Pf − Pi ≡ ∆P
tf
tf
i
i
I = ∫t F dt = ∫t
F
f
i
I = ∆P
t
∆t
impulse = change in momentum!
ti
tf
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Force and Impulse
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Two different collisions can have
the same impulse since I depends
only on the change in momentum,
not the nature of the collision.
F
F
ti
same area
∆t
∆t big, F small
tf
t
∆t
t
tf
ti
∆t small, F big
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Page 18
Force and Impulse
soft spring
F
F
stiff spring
ti
∆t
∆t big, F small
tf
t
∆t
t
tf
ti
∆t small, F big
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Lecture 21, Act 4
Force & Impulse
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Two boxes, one heavier than the other, are initially at rest on a
horizontal frictionless surface. The same constant force F acts
on each box for exactly 1 second.
ÍWhich box has the most momentum after the force acts?
(a) heavier
F
light
(b)
lighter
F
(c) same
heavy
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Page 19
Lecture 21, Act 4
Solution
We know Fav =
∆p
∆t
∆ p = Fav ∆ t
so
In this problem F and ∆t are the same for both boxes!
The boxes will have the same final momentum.
F
F
light
heavy
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Force and Impulse
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We can use the notion of impulse to define “average force”,
which is a useful concept.
F
The time average of a force for the
time interval ∆t = tf - ti is:
Fav =
1 t
I
∫ F dt =
∆t t
∆t
f
i
or:
Fav
Fav
t
∆t
∆P
=
∆t
ti
tf
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Page 20
Force and Impulse
soft spring
F
Fav
F
stiff spring
Fav
ti
∆t
∆t big, Fav small
tf
t
∆t
t
tf
ti
∆t small, Fav big
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Force and Impulse:
Baseball Example
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A pitcher pitches the ball (m = .7 kg)
at 145 km/hr (about 90 mph).
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The batter makes contact with the
ball for .001 s causing the ball to
leave the bat going 190 km/hr
(about 120 mph).
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Find the average force on the ball,
disregarding gravity.
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Page 21
Baseball Example
First convert everything to m/s:
145 km/hr = 40.28 m/s
190 km/hr = 52.78 m/s
Next find the change in momentum ( = the impulse):
Pf - Pi = (.7 kg)(52.78 m/s) - (.7 kg)(-40.28 m/s)
Pf - Pi = 65.14 kg-m/s
Finally find the average force:
F av =
∆p
65 . 14 Ns
=
= 65142 N
∆t
.001s
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Page 22