PHYS 218 sec. 517-520
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Transcript of PHYS 218 sec. 517-520
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PHYS 218sec. 517-520
ReviewChap. 8
Momentum, Impulse, and Collisions
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What you have to know
• Momentum and impulse• Momentum conservation• Various collision problems
– The typical examples where momentum conservation is used are collisions!
• Center of mass
• Section 8.6 is not in the curriculum.
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Newton’s 2nd law of motion
: Newton's 2nd law when the mass is constant.F ma m=å r r
This cannot be applied to the problems where mass m is changing.E.g.Rocket propulsion problem. (See Sec 9.6)
Original form of the 2nd law of motion written by Newton
( )dF mvdt
=å r r ( )when is constant, it becomes d dvm F mv m madt dt
= = =årr r r
This defines the (linear) momentum ( . momenta)plr rp = mv Newton's 2nd law can be written as
dpFdt
=årr
Momentum is a vector quantity.
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The impulse-momentum theorem
Impulse ( )J F t= Dår rImpulse is a vector quantity.
2 1
2 1
time interval
if the force is acting
between and
t t
F
t t
-å
2 1
2 1
If the net force is constant,p pFt t
-= -år r
Note that Newton's 2nd law is dpFdt
=årr
( )( )2 1 2 1F t t p p- = -å r r2 1J p p= -
r r r ( )since J F t= Dår r
Impulse-momentum theoremThe change in momentum of a particle
during a time interval equals the impulse of the net force.
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Impulse
2 2 2
1 1 12 1
If forces are changing with time,t t t
t t t
dpFdt dt dpJ p pdt
J= - == = =åò ò ò r rrr r rr
General definition Impulse-momentum theorem
Impulse: area under the curve of net force in F-t graph
Få
tSame area Same impulse
Large force for a short time
Smalle force for a longer time
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Momentum vs. Kinetic Energy
2 1
Impulse-momentum theorem
change in a particle's momentum is due to impulse
depends on
J p p= -®
Þ time
2 1
Work-energy theorem
change in a particle's kinetic energy is due to work
depends on
W K K= -®
Þ distance
Integral form of Newton’s 2nd law
Differential form of Newton’s 2nd lawdpFdt
=årr
identical
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Ex 8.2
x
A ball hits a wall
before
after
1v
2v
1 20.4 kg, 30 m/s, 20 m/s, 0.01 simpulse? average force that the wall exerts on the ball?m v v t= =- = D =
( )( ) ( ) ( ){ }
1 1
2 2
2 1 2 1 2 1
Initial momentum: Final momentum: Impulse
0.4 kg 20 m/s 30 m/s
20 N s
p mvp mv
J p p mv mv m v v
==
= = - = - = -= ´ - -= ×
average force:
20 N s 2000 N0.01 s
av
av
J F tJFt
= D×Þ = = =D
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Ex 8.31 2kicking a soccer ball: 0.4 kg, 20 m/s, 30 m/s,
45 , 0.01 simpulse of the net force? average net force?
m v vtq
= = == ° D =
q
Before
After
1v
2v
x
y
This is a 2-dim. motion.First decompose the momentum into
x and y directions
( )
( )
1 1 1
2 22 2 2 2
22 1 2 1 1
22 1 2 1
20 m/s, 0
cos 45 , sin 452 2
The impulse is
16.5 kg m/s2
0 8.5 kg m/s2
x y
x y
x x x x x
y y y y y
v v v
v vv v v v
vJ p p m v v m v
vJ p p m v v m
=- =- =
= °= = °=
Þæ ö÷ç= - = - = + = ×÷ç ÷÷çè øæ ö÷ç= - = - = - = ×÷ç ÷÷çè ø
, ,
,2 2 3, ,
,
1650 N, 850 N
8501.9 10 N, tan 271650
yxav x av y
av yav av x av y
av x
JJF Ft t
FF F F
Fq q
= = = =D DÞ = + = ´ = = Þ = °
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Conservation of Momentum
Internal forces
External force
Isolated system: External force = 0
A B
B on AFr
A on BFr
B on A A on B,A Bdp dpF Fdt dt
= =r rr r
Newton’s 2nd law
Newton’s 3rd law
B on A A on B 0F F+ =rr r
( )B on A A on B 0A BA B
dp dp dF F p pdt dt dt
+ = + = + =r r rr r r r
You can add these forces to obtain the net (internal) force for
the system The sum of these forces cannot be applied to a single object.
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Conservation of Momentum
( )B on A A on B 0
If we define , 0
A B
A B
dF F p pdt
dPP p pdt
+ = + =
= + =
rr r r rr rr r r
Total momentum
Total momentum of an isolated system is conserved.
When the system contains many particles,the total momentum is
A B C A A B B C CP p p p m v m v m v= + + + = + + +r r r r r r rL L
constantP =r
Momentum conservation gives a vector equation.
Therefore, each component of the total momentum is conserved!constantconstant
constant
x Ax Bx Cx
y Ay By Cy
z Az Bz Cz
P p p pP p p p
P p p p
= + + + == + + + == + + + =
LLL
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Ex 8.4 Recoil of a rifle
Before
After
x
x
Rifle + bullet
Rifle bullet
Rv Bv
Note that momentum conservation is valid for an
isolated system.So it is useful when the system
changes. Therefore, draw diagrams for “before” &
“after” the event
mass of the rifle 3 kg, mass of the bullet 5 g300 m/s. Then (recoil speed of the rifle)?
Momentum and kinetic energy of the rifle? Of the bullet?
R B
B R
m mv v
= ==
( )
Use momentum conservationInitial momentum: 0 ( the rifle+bullet is at rest)Final momentum:
gives 0
0.005 kg 300 m/s 03 kg
initial
final R R B B
initial final R R B B
BR B
R
PP m v m v
P P m v m v
mv vm
== +
= + =æ ö÷çÞ =- =- ÷́ =-ç ÷ç ÷çè ø
Q
.5 m/s
change the unit so that the both masses have the same unit
The negative sign means that the rifle is moving back!
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Ex 8.4 (cont’d)
( ) ( )
( ) ( )
( ) ( )22
momentum of the bullet0.005 kg 300 m/s 1.5 kg m/s
momentum of the rifle3 kg 0.5 m/s 1.5 kg m/s
kinetic energy of the bullet1 1 0.005 kg 300 m/s 225 J2 2
kinetic energy of the rif
B B B
R R R
B B B
p m v
p m v
E m v
= = ´ = ×
= = ´ - =- ×
= = ´ =
( ) ( )22
le1 1 3 kg 0.5 m/s 0.375 J2 2R R RE m v= = ´ - =
The sum of these two momenta vanishes as it should be.
The kinetic energy is NOT conserved.
The kinetic energy before the event was 0.
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Ex 8.5 Collision along a straight line
A B
A B
Before
After1Av 1Bv
2Av 2Bv
1 1
2 2
0.5 kg, 0.3 kginitial speeds: 2.0 m/s, 2.0 m/sfinal speeds: 2.0 m/s, ?
A B
A B
B A
m mv v
v v
= == =-
= =
If you have a collision problem, always use momentum conservation.
initial 1 1
final 2 2
initial final
1 1 2 2
1 12
total momentum before the collision: total momentum after the collision: momentum conservation gives
A A B B
A A B B
A A B B A A B B
A A B B B BA
P m v m vP m v m v
P Pm v m v m v m v
m v m v m vv
= += +
=+ = +
+ -Þ = ( ) ( ) ( ) ( ) ( ) ( )2 0.5 kg 2.0 m/s 0.3 kg 2.0 m/s 0.3 kg 2.0 m/s0.5 kg
0.4 m/sAm
´ + ´ - - ´=
=-
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Ex 8.6 Collision in a horizontal plane
Before
After1Av 1 0Bv =
2Av
2Bv
x
x
y
ab
1
1
2 2
20 kg, 12 kginitial speeds: 2.0 m/s in the (+) -direction, 0 m/sfinal speeds: 1.0 m/s with 30 , ?
A B
A
B
A A
m mv xv
v va
= ==== = ° =r
Again, we use momentum conservation.
1 1 2 2
1 1
by momentum conservation
A A B B A A B B
A A x B B x
m v m v m v m v
m v m v
+ = +Þ +
r r r r
2 2
1
in -directionA A x B B x
A A y
m v m v x
m v
= +
1B B ym v+ 2 2
1 2 1 22
2 22
in -direction
cos 1.89 m/s
sin 0.83 m/s
A A y B B y
A A x A A x A A A AB x
B B
A A y A AB y
B B
m v m v y
m v m v m v m vvm m
m v m vvm m
a
a
= +- -Þ = = =
=- =- =-
Also note that momentum is a vector quantity.
2 22 2 2
2
2
2.1 m/s,
arctan 24
B B x B y
B y
B x
v v v
vv
b
= + =æ ö÷ç ÷= =- °ç ÷ç ÷çè ø
Use the numbers above
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Momentum conservation and collisions
The key to solve collision problems is Momentum conservation.
1. At the instant of collision, very complex and very large forces are acting on the bodies.
2. Then the other forces can be neglected and the system is almost an isolated system. momentum conservation can be applied
3. The forces are very complex and we do not know much about the forces. momentum conservation can give useful information for the motions of the colliding bodies
Elastic collision
If the forces between the colliding bodies are conservative, no mechanical energy is lost. The total kinetic energy of the system is conserved.
For elastic collisions, the total momentum and the total kinetic energy of the system are conserved.
initial final initial finaland K K P P= =r r
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Inelastic collision
In this collision, the total kinetic energy of the system after the collision is less than before the collision.
initial final initial finalbut K K P P> =r r
The forces during the collision are non-conservative.
Completely inelastic collision
This is a special case of inelastic collisions.
This is where the colliding bodies stick together and move as one body after the collision.
1,final 2,final
initial final
,
the mass of the combined body after the collision is
But we still have A B
v v
m m
P P
=+
=
r r
r r
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Completely inelastic collisions
What happens to momentum and kinetic energy in a completely inelastic collision of two bodies
A B
A B
Before
After1Av
r1Bv
r
2 2 2A Bv v v= =r r r
( ),1 ,1 2
momentum conservation gives
A A B B A Bm v m v m m v+ = +r r r
Consider a 1-dim collision where vB1 = 0
,1 ,1
momentum conservation:
A A B Bm v m v+ ( ) 2
2 ,1
A B
AA
A B
m m v
mv vm m
= +
Þ = +( )
( ) ( )
21 ,1 ,1
22 2
2 ,1 ,1
2
1
Kinetic energies1 02
1 12 2
< 1
A A B
AA B A A B A
A B
A
A B
K m v v
mK m m v m m vm m
K mK m m
= =
æ ö÷ç ÷= + = + ç ÷ç ÷ç +è ø
Þ = +
Q
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Ex 8.8 The ballistic pendulum
This is a device which can measure the speed of a bullet.
y
Before collision
Immediatelyafter collision
Swing of the bodyafter collision
Analyze this event in two stages !
Collision process One-body motion
This gives a relation between v1 and v2.
This gives a relation between v2 and the height y.
(A) (B)1v
2v
mass of the bullet , mass of the wood blockB Wm m= =
0v =
2v
So, if you know y, you can know v2 and v1
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Ex 8.8 (cont’d)
The first stage: (A) collision( )1 2
1 2
momentum conservation
completely inelastic scattering & is at rest initially
So we have
B B W
W
B W
B
m v m m v
m
m mv vm
Þ = +
æ ö+ ÷ç ÷=ç ÷ç ÷çè ø
Q
The second stage: (B) one-body motion
1 1Energy conservation K U+ 2K=
( ) ( )2
22 2
1 22 B W B W
U
m m v m m gy v gy
+
+ = + Þ =
Combining the two
1 2B W
B
m mv gy
m+= By measuring the height y, you can know the
speed of the bullet
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Ex 8.9 Automobile collision
x
y
Tvr
Cvr
Cm
Tm
Before
x
y
T Cm m+
vr
After
Completely inelastic collision
( ) ( )ˆ ˆ2000 kg, 10 m/s ; 1000 kg, 15 m/sT T C Cm v m v= = = =i jr r
initial
initial, ,
Total momentum:
-component: C C T T
x C C x
P m v m v
x P m v
= +=
r r r
, , initial,y , ,; -component: T T x T T x C C y T T ym v m v y P m v m v+ = = +
( ) ( )
( )
,
final initial , ,
22 4final , ,
final, ,
final, ,
final final
ˆ ˆ
2.5 10 kg m/s
tan 0.75 37
8.3 m/s
C C y
T T x C C y
T T x C C y
y C C y
x T T x
C T
m v
P P m v m v
P m v m v
P m vP m v
V P m m
q q
== = +
Þ = + = ´ ×
Þ = = = Þ = °
= + =
i jr r
q
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Elastic collisions
initial final initial finalIn an elastic collision, and K K P P= =r r
Consider 1-dim elastic collision.
A B
A B
Before
After1Av 1Bv
2Av 2Bv
1 1 2 2
2 2 2 21 1 2 2
Momentum conservation
Kinetic energy is conserved for elastic collision1 1 1 12 2 2 2
A A B B A A B B
A A B B A A B B
m v m v m v m v
m v m v m v m v
+ = +
+ = +Am Bm
1 2 1 2
If the masses are known, the unknown quantities are , , ,A A B Bv v v v
These give TWO equations.
1 1You can set up the initial condition, then you can know the initial velocities and A Bv v
You now have TWO unknowns (the final velocities).
Therefore, you can completely determine the unknowns.
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Elastic collisions: one body initially at rest
1
1 2 2
2 2 21 2 2
0. Then we have
1 1 12 2 2
B
A A A A B B
A A A A B B
vm v m v m v
m v m v m v
== +
= +
rearrange( )( ) ( )( )
2 1 2
2 2 22 1 2 1 2 1 2
B B A A A
B B A A A A A A A A
m v m v v
m v m v v m v v v v
= -= - = - +rearrange
( ) ( )
2 1 2
1 2 1 2
2 1 2 1
Using the first equation, the second equation gives
This relation is used to rewrite the first equation as
2 and
B A A
B A A A A A
A B AA A B A
A B A B
v v v
m v v m v v
m m mv v v vm m m m
= +
+ = --Þ = =+ + A B
A B
Before
After1Av 1 0Bv =
2 0Av =2 1B Av v=
2 2 1
If ,0 and
A B
A B A
m mv v v
=Þ = =
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AB
A
Before
After1Av 1 0Bv =
2 1A Av v» - 2 0Bv »
2 1 2
If , and 0
A B
A A B
m mv v vÞ » - »
=
B
AB
B
Before
After1Av 1 0Bv =
2 1A Av v» 2 12B Av v»
2 1 2 1
If , and 2
A B
A A B A
m mv v v vÞ » »
?
A
1. The motion of the heavy object doesn’t change.
2. The light object moves to the opposite direction but with same speed.
1. The motion of the heavy object doesn’t change.
2. The light object moves in the direction of the heavy one but (two times) faster
than the heavy one.
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Elastic collisions & relative velocity
/ / / /
/ / /
relative velocity: , where is the velocity of measured by .Therefore,
A P A B B P X Y
A B A P B P
v v v v X Yv v v
= += -
2 1 2
1 2 2
From the slide (p.22), B A A
A B A
v v vv v v
= +Þ = - The velocity of B relative to A after the
collision
( ) ( )1 1 1A B Av v v=- - =- - The negative of the velocity of B relative to A before the collision
1
1
we have used that 0, but this is true for 0
B
B
vv
=¹ ( )2 2 1 1
In general,
for elastic collisions.B A B Av v v v- =- -r r r r
If you watch the motion of A sitting on B, then you will see:A is approaching to B with speed v before the collision.
After the collision, A is moving away from B with the same speed
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Ex 8.12 Two-dim elastic collision (If this is a head-on collision, it is the same as the 1-dim collision.)
Before
After1Av 1 0Bv =
2Av
2Bv
x
x
y
ab
FOUR unknowns after collision (two velocity vectors)
Momentum conservation gives two eqs.Kinetic energy conservation gives one eq.
So totally THREE equations.
To uniquely determine the final velocities, we need ONE more information for the
final velocities
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Ex 8.12 (cont’d) ( )1 1
2
2
0.5 kg, 0.3 kg, 4.0 m/s , 0
2.0 m/sThen what are , , and ?
A B A B
A
B
m m v v
vv a b
= = = ==
ir rr r
2 22 2 2 2 1 2
1 2 2 2 2
This is an elastic collision, so
1 1 1 4.47 m/s2 2 2
i f
A A A AA A A A B B B B
B
K K
m v m vm v m v m v v vm
=-= + Þ = Þ =
1 2 2
2 2
2 2
momentum conservation-direction: 2 cos 1.34cos-direction: 0 0 sin 1.34sin
By solving the coupled equations (use sin cos 1)36.9 , 26.6
A A A A x B B x
A A y B B y
x m v m v m vy m v m v
a ba bq q
a b
= + Þ = += + Þ = -
+ == ° = °
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Center of mass
We now introduce the concept of center of mass (CM)
( ), ,i i i ir x y z=r
i-th particle
( )1 1 2 2 3 3 4 4
1 2 3 4
1 1 2 2 3 3 4 4
1 2 3 4
The center of the mass of the system has the position , , defined as
where is the t
cm cm cm
i i i icm
i
i i i icm
i
x y z
m x m xm x m x m x m xx
m m m m m M
m y m ym y m y m y m yym m m m m M
M
+ + + += = =+ + + ++ + + += = =+ + + +
å ååå åå
LL
LL
otal mass of the system,
iM m=åThen,
1 : the postion of the center of mass of the systemcm i ir m rM
= år r
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Motion of the center of mass
( ) ( )
1Since ,cm i i
icm cm i i i i i i
r m rM
drd dM v r m r m m v p Pdt dt dt
=
= = = = = =
åå å å å
r rr rr r r r r
Total momentum of the systemMomentum of the CM
You can represent an extended object as a point-like particle; the total mass of the
object is at its Center of Mass (CM)
( )
If there is no external force, is constant, so is the velocity of the CM
const.
If there is net external force,
cm
cmcmext cm
PPvM
d Mvdv dPF M a Mdt dt dt
= =
= = = =å
r
r
rrrr r
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Motion of the CM
`
When a shell explodes
This is the trajectory of the shell if it doesn’t explode.