Momentum

Impulse
Conservation of Linear Momentum
 MOMENTUM
3/12/24 2
Momentum
Product of mass and velocity
! !
p = mv m is the mass of the object
v is the velocity of the object

Standard unit: 1 kg m/s

3/12/24 3
Momentum
Product of mass and velocity
! !
p = mv m is the mass of the object
v is the velocity of the object

Note: VELOCITY, not speed, is a vector

Can be split into components (cannot be added)

px î = mv x î p y ĵ = mv y ĵ pz k̂ = mv z k̂
3/12/24 4
Momentum
Product of mass and velocity
! !
p = mv m is the mass of the object
v is the velocity of the object

Sign convention:
Positive if upward or rightward
Negative if downward or leftward

3/12/24 5
Impulse-Momentum Theorem
Change in the momentum of the system
! !
J = Δp Pf is the final momentum
Pi is the initial momentum

! ! ! ! !
J = p f − pi = m f v f − mi vi
3/12/24 6
Impulse-Momentum Theorem
Change in the momentum of the system
! !
J = Δp Pf is the final momentum
Pi is the initial momentum

! ! ! ! !
J = p f − pi = mv f − mvi
If the mass of the object remains constant

3/12/24 7
Impulse-Momentum Theorem
Comparison of Impulse-momentum and Work-
kinetic energy (for constant mass)

! !
Impulse-Momentum theorem:

! ! !
J = p f − pi = mv f − mvi
Work-Energy theorem:

1 1
W = K f − K i = mv f − mvi
2 2

2 2
3/12/24 8
Impulse-Average Force
Change in the momentum of the system

! !
Δp
Fnet =
Δt
! !
If the mass of the object remains constant

! Δ( mv ) Δv
Fnet = =m = ma
Δt Δt
3/12/24 9
Impulse-Average Force
Change in momentum in differential form

! !
dp
Fnet =
dt
How to change momentum: Apply a force
For a constant change in momentum:
Large force in short time
Weak force sustained for a longer period of time
3/12/24 10
Impulse-Average Force
Change in momentum in differential form

! !
dp
Fnet =
dt
Physical interpretation
Less momentum: Easier to change state of motion
Greater momentum: Harder to change state of motion

3/12/24 11
Impulse-Average Force
Change in the momentum of the system

! !
Δp
Fnet =
Δt
! ! !
J = Δp = Fnet Δt
Momentum changes if there is a net force
that acts for a certain amount of time.

3/12/24 12
Graphical Analysis – Impulse
Impulse is the AREA
under the Force
versus Time graph.

13
3/12/24
Graphical Analysis – Impulse
Impulse is the AREA Similar to work:
under the Force area under the
F vs. x curve J
versus Time graph.

For an average force:

! !
J = Fav ( t 2 − t1 )

14
3/12/24
Graphical Analysis – Impulse
Impulse is the AREA Similar to work:
under the Force area under the
F vs. x curve J
versus Time graph.

If function of Force is
given:
t2
!
J = ∫ F dt
t1
15
3/12/24
Problem
Which is easier to catch (i.e. brought to rest)?
a. 0.50-kg ball moving at 4.0 m/s or
b. 0.10-kg ball moving at 20 m/s
How much impulse should be imparted to the balls
in each of the previous cases?

16
3/12/24
Solution
1. Either ball is easy to catch.
a. 0.50-kg ball moving at 4.0 m/s
𝒎 𝒎
𝑷𝒊 = 𝟎. 𝟓 𝒌𝒈 𝟒 = 𝟐 𝒌𝒈
𝒔 𝒔
𝒎 𝒎
𝑷𝒇 = 𝟎. 𝟓 𝒌𝒈 𝟎 = 𝟎 𝒌𝒈 Brought to a stop
𝒔 𝒔
𝒎
𝑱 = 𝑷𝒇 − 𝑷𝒊 = −𝟐 𝒌𝒈
𝒔

b. 0.10-kg ball moving at 20 m/s

𝒎 𝒎
𝑷𝒊 = 𝟎. 𝟏 𝒌𝒈 𝟐𝟎 = 𝟐 𝒌𝒈
𝒔 𝒔
𝒎 𝒎
𝑷𝒇 = 𝟎. 𝟏 𝒌𝒈 𝟎 = 𝟎 𝒌𝒈 Brought to a stop
𝒔 𝒔
𝒎
𝑱 = 𝑷𝒇 − 𝑷𝒊 = −𝟐 𝒌𝒈
𝒔
17
3/12/24
Solution
2. How much impulse should be imparted to the balls in
each of the previous cases?
a. 0.50-kg ball moving at 4.0 m/s
𝒎
𝑱 = 𝑷𝒇 − 𝑷𝒊 = −𝟐 𝒌𝒈
𝒔
b. 0.10-kg ball moving at 20 m/s
𝒎
𝑱 = 𝑷𝒇 − 𝑷𝒊 = −𝟐 𝒌𝒈
𝒔

18
3/12/24
Problem
A 0.400-kg ball hits a wall with an initial
velocity of 30 m/s, to the left. It
rebounds with a final speed of 20 m/s,
to the right. Take the right to the the
positive direction.

1. What is the initial momentum of the ball?

2. What is the final momentum of the ball?
3. What is the impulse imparted on the ball?
4. If the ball stays in contact with the wall for
0.0080 s, what is the average force
experienced by the ball?

3/12/24 19
Solution
1. Initial momentum
𝒎 𝒎
𝑷𝒊 = 𝒎𝒗𝒊 = 𝟎. 𝟒 𝒌𝒈 −𝟑𝟎 = −𝟏𝟐 𝒌𝒈
𝒔 𝒔
2. Final momentum
𝒎 𝒎
𝑷𝒇 = 𝒎𝒗𝒇 = 𝟎. 𝟒 𝒌𝒈 +𝟐𝟎 = +𝟖 𝒌𝒈
𝒔 𝒔
3. Impulse
𝒎
𝑱 = 𝑷𝒇 − 𝑷𝒊 = 𝟖 − −𝟏𝟐 = +𝟐𝟎 𝒌𝒈
𝒔
4. Net force (magnitude)
𝑱 𝟐𝟎
𝑭𝒏𝒆𝒕 = =+ = +𝟐𝟓𝟎𝟎 𝑵
𝚫𝐭 𝟎. 𝟎𝟎𝟖𝟎

3/12/24 20
Momentum and Impulse
To break hollow block:
Same momentum change
from hand, requires short
time of contact.
®
 
Dp = J =
F Dt

Resulting large average force

causes block to break. J

21
3/12/24
Large force for short time interval

Resulting large
average force causes
block to break. J

22
3/12/24
Momentum and KE Compared

Impulse-Momentum Theorem Work-KE Theorem

Changes in momentum Kinetic energy changes
are due to impulse when work is done

→ → →
J = p f − pi W = K f − Ki

23
3/12/24
Momentum and KE Compared

Impulse-Momentum Theorem Work-KE Theorem

Changes in mom Kork is done
Changes in momentum Total work depends on
are due to time over distance over which
which the force acts force acts

 t2  x2
 
J = ò å F dt W = ò å F × dx
t1 x1

24
3/12/24
Conservation of Energy and Momentum
Conservation of mechanical energy
Mechanical energy is
ΔEmech = Wother conserved if Wother=0
No non-conservative forces
acting on the system.

Conservation of Linear momentum

! !
Δp = J ext
Momentum is conserved
if Jext =0
there are no external forces acting
on the system.

3/12/24 25
Conservation of Momentum as Observed Everyday

3/12/24 26
Conservation of Momentum as Observed Everyday

3/12/24 27
Conservation of Momentum as Observed Everyday

3/12/24 28
Conservation of Momentum as Observed Everyday

3/12/24 29
 COLLISIONS

3/12/24 30
Perfectly elastic collisions
Completely inelastic collisions
Momentum conservation in non-collision
situations
Remarks on conservation of momentum
Interacting bodies are ISOLATED from outside
Internal forces ARE ZERO
Action-Reaction forces cancel each other
External forces HAVE NO EFFECT
Not along the collision direction (perpendicular)
Gravity acts along the vertical, does not affect horizontal

3/12/24 32
Collisions
Systems where momentum is conserved
Our scale: Billiard balls, colliding vehicles, etc.

3/12/24 33
Collisions
Consider two colliding objects A and B with masses
mA and mB. If there are no external forces acting on
them, then

! ! ! !
Δp = J ext = 0 pi,total = p f ,total
3/12/24 34
Collisions
Consider two colliding objects A and B with masses
mA and mB. If there are no external forces acting on
them, then

The Total Momentum of the

! !
pi,total = p f ,total
Objects BEFORE AND AFTER
COLLISION MUST BE CONSTANT.
Pi = Initial total momentum BEFORE COLLISION
Pf = Final total momentum AFTER COLLISION

3/12/24 35
Coliisions
Consider two colliding objects A and B with masses
mA and mB. If there are no external forces acting on
them, then

! ! ! !
m Ai v Ai + m Bi v Bi = m Af v Af + m Bf v Bf
Initial momentum Initial momentum Final momentum Final momentum
of mA of mB of mA of mB

3/12/24 36
Coliisions
Consider two colliding objects A and B with masses
mA and mB. If there are no external forces acting on
them, then

If the masses of the objects remain the same after collision:

! ! ! !
mAv Ai + mBv Bi = mAv Af + mBv Bf
3/12/24 37
Collision problems we will consider
Perfectly elastic collisions

Completely inelastic collisions

3/12/24 38
 PERFECTLY ELASTIC COLLISIONS
3/12/24 39
Perfectly elastic collisions
Perfectly elastic collisions
Ideal case!
After collision, the interacting
objects will NOT stick together

Total Momentum is Total Kinetic Energy is

CONSERVED. CONSERVED.
!
Δptotal = 0 ΔK total = 0
3/12/24 40
Two objects in perfectly elastic collision
Total momentum: Contribution from entire system
Vector sum of momenta of two bodies
Let: A & B be the colliding bodies:
If the masses of the two interacting objects remain the same after collision:

! ! ! ! !
Δptot = 0 m Av Af + m B v Bf = m Av Ai + m B v Bi

ΔK tot = 0 1 1 1 1
m Av Af + m B v Bf = m Av Ai + m B v Bi
2 2 2 2

2 2 2 2
3/12/24 41
Two objects in perfectly elastic collision
Total momentum: Contribution from entire system
Rearrange terms:

!
Δptot = 0
!
( ! !
) !
m A v Af − v Ai = −m B v Bf − v Bi ( )
ΔK tot = 0 1⎡
2⎣
m A (
v Af
− v )(
Ai
v Af
+ v )
Ai
⎤
⎦ = −
1⎡
2⎣
m B (
v Bf
− v Bi)(
v Bf
+ v Bi )
⎤
⎦

3/12/24 42
Two objects in perfectly elastic collision
If moving in 1D,

m A (v Af - v Ai ) = -mB (vBf - vBi )

1
2
[ ] [
m A (v Af - v Ai )(v Af + v Ai ) = - mB (vBf - vBi )(vBf + vBi )
1
2
]

(vBi − v Ai ) = − (vBf − v Af )
Relative velocity of approach equal to negative
relative velocity of recession in perfectly elastic collisions. J
3/12/24 43
General case: Both moving initially, different masses

Use the two equations to find the two unknowns

In general, you must use:
    1 1 1 1
m Av Af + mB vBf = m Av Ai + mB vBi 2
m Av Af + mB vBf2 = m Av Ai
2
+ mB vBi2
2 2 2 2
Note: One is quadratic

For 1D motion, you can use:

- (vBf - v Af ) = +(vBi - v Ai )
   
m Av Af + mB vBf = m Av Ai + mB vBi

Both are linear

3/12/24 44
Special case: One object stationary
Before:
Relative to big, small ball approaches
After:
Relative to big, small ball recedes
At the same rate

Or the other way around

i.e. Relative to small, big ball
approaches then recedes at same rate

3/12/24 45
Special case: One object stationary
Before:
Relative to big, small ball approaches After collision:
If bigger object did not
After:
move, smaller object
Relative to big, small ball recedes gets same speed to the
At the same rate other direction.

Or the other way around After collision:

i.e. Relative to small, big ball If bigger object did
approaches then recedes at same rate stops, smaller object
gets thrown forward at
a very fast speed.
3/12/24 46
Special special case: One stationary, same masses

Change state of motion

One ball is initially approaching then Stops
Other ball is initially stationary then Moves with same velocity

3/12/24 47
Special special case: One stationary, same masses

Multiple uniform masses

Middle masses will not significantly move at all

3/12/24 48
Summary: Elastic collision
Elastic collision:
Negligible energy losses due to nonconservative forces
Perfectly elastic collision conserves the ff:
Momentum    
m Av Af + mB vBf = m Av Ai + mB vBi

Kinetic energy 1 m Av Af2 + 1 m B v Bf2 = 1 m Av Ai2 + 1 m B v Bi2

2 2 2 2

Speed of approach/recession

( vBi − vAi ) = − ( vBf − vAf )

3/12/24 49
Quiz: An Elastic Straight-Line Collision
Consider the figure below. The two gliders have
ideal spring bumpers to make collision elastic.
What are the velocities of A and B after collision?

   
m Av Af + mB vBf = m Av Ai + mB vBi

1 2
m Av Af
1
+ m B v Bf
2 1
= m Av Ai
2 1
+ m B v Bi
2
( vBi − vAi ) = − ( vBf − vAf )
2 2 2 2
3/12/24 50
Quiz: An Elastic Straight-Line Collision
Consider the figure below. The two gliders have
ideal spring bumpers to make collision elastic.
What are the velocities of A and B after collision?
! !
pi,total = p f ,total EQUATION 1
! ! ! !
mAv Ai + mBv Bi = mAv Af + mBv Bf
! ! !
( !
− ( v Bi − v Ai ) = v Af − v Bf ) EQUATION 2

3/12/24 51
Quiz: An Elastic Straight-Line Collision
Consider the figure below. The two gliders have
ideal spring bumpers to make collision elastic.
What are the velocities of A and B after collision?

EQUATION 1
! !
0.5 ( 2 ) + 0.3( −2 ) = 0.5v Af + 0.3v Bf

! !
(
− (( −2 ) − ( 2 )) = v Af − v Bf ) EQUATION 2

3/12/24 52
Quiz: An Elastic Straight-Line Collision
Consider the figure below. The two gliders have
ideal spring bumpers to make collision elastic.
What are the velocities of A and B after collision?

!
v Af = −1m/s
!
v Bf = +3m/s AFTER COLLISION,
A moves to the left, while B moves to the right

3/12/24 53
 COMPLETELY INELASTIC COLLISIONS

3/12/24 54
Completely inelastic collisions
Completely inelastic collisions Total Momentum is
Momentum is conserved CONSERVED.
! !
Energy is NOT conserved
Objects stick to each other!
pi,total = p f ,total
SAME Final velocity for
the two objects.
v Af = v Bf
.

3/12/24 55
Two objects in completely inelastic collision
Easier to solve: Involves only one equation

! ! !
( mA + mB ) v f = mAv Ai + mBv Bi
SAME final velocity for objects in a completely inelastic
collision. J

3/12/24 56
The Ballistic Pendulum
A ballistic pendulum is a system for measuring the speed
of a bullet. It consists of a wooden block of mass M
suspended from vertical strings. When a bullet strikes the
block, it embeds itself in the block. If the bullet has mass
m, and if the block rises a maximum height h after the
collision, find the speed of the bullet.

3/12/24 57
Quiz: The Ballistic Pendulum


Dp = 0
 
pi = p f
The pendulum is at rest before collision
   
mb vbi + m p v pi = mb vbf + m p v pf

3/12/24 58
Quiz: The Ballistic Pendulum


Dp = 0
 
pi = p f
The bullet and the pendulum stick
together after collision

mb vbi + m p v pi = (mb + m p )v pf
  

3/12/24 59
Quiz: The Ballistic Pendulum

Result from Conservation of

momentum

mb vbi = (mb + m p )v pf
 

Kinetic energy upon collision

K = (mb + m p )v pf
1 2

2
3/12/24 60
Quiz: The Ballistic Pendulum

Result from Conservation of

momentum

mb vbi = (mb + m p )v pf
 

Kinetic energy upon collision The bullet-pendulum system rises

K = (mb + m p )v pf U G = (mb + m p )gy

1 2

2
3/12/24 61
Quiz: The Ballistic Pendulum

Result from Conservation of

momentum

mb vbi = (mb + m p )v pf
 

Since only gravity is present, mechanical energy is conserved at all times

1
(mb + m p )v pf = (mb + m p )gy
2

2
3/12/24 62
Quiz: The Ballistic Pendulum

Result from Conservation of

momentum

mb vbi = (mb + m p )v pf
 

And the initial velocity of the bullet can be solved:

vbi =
( m b + mp ) 2gy
mb
3/12/24 63
Quiz: The Ballistic Pendulum

Result from Conservation of

momentum

mb vbi = (mb + m p )v pf
 

Using projectile motion KINEMATICS:

d g From Projectile motion:

vbi = = d d = range or horizontal distance
t 2y y = vertical distance

3/12/24 64
Quiz: The Ballistic Pendulum

Result from Conservation of

momentum

mb vbi = (mb + m p )v pf
 

These two equations must be equal:

vbi =
( m b + mp ) 2gy = d
g
mb 2y
3/12/24 65
Quiz: Inelastic Straight-Line Collision
Consider the figure below. The two gliders stick
together after collision. What are the velocities of
A and B after collision?

! !
pi,total = p f ,total v Af = v Bf

3/12/24 66
Quiz: Inelastic Straight-Line Collision
Consider the figure below. The two gliders stick
together after collision. What are the velocities of
A and B after collision?
! !
pi,total = p f ,total EQUATION 1
! ! ! !
mAv Ai + mBv Bi = mAv Af + mBv Bf
! !
v Af = v Bf EQUATION 2

3/12/24 67
Quiz: Inelastic Straight-Line Collision
Consider the figure below. The two gliders stick
together after collision. What are the velocities of
A and B after collision?

EQUATION 1
!
0.5 ( 2 ) + 0.3( −2 ) = ( 0.5 + 0.3) v Bf
! !
v Af = v Bf EQUATION 2

3/12/24 68
Quiz: Inelastic Straight-Line Collision
Consider the figure below. The two gliders stick
together after collision. What are the velocities of
A and B after collision?

!
v Af = +0.5m/s
!
v Bf = +0.5m/s AFTER COLLISION,
A and B move together with the same velocity

3/12/24 69
Problem
A 5.00 g bullet is fired horizontally into a 1.20 kg
wooden block resting on a horizontal surface. The
coefficient of kinetic friction between the block
and the surface is 0.20. The bullet remains
embedded in the block, which is observed to slide
0.230 m along the surface before stopping. What is
the initial speed of the bullet?

Hint: You can think of this case as a TWO-STAGE

process

3/12/24 70
 MOMENTUM CONSERVATION
IN 2D/3D

3/12/24 71
Momentum in 2D/3D
Momentum is conserved for each dimension
where there are no external forces
IDEA: We cannot combine different components

pxi = pxf
p yi = p yf
pzi = pzf
3/12/24 72
Momentum in 2D/3D
For two interacting objects A and B, the
conservation relation is:
! ! ! !
pAi + pBi = pAf + pBf
IDEA: We cannot combine different components
pxi = pxf Only if
Fxnet = 0
p yi = p yf Only if Fynet = 0
pzi = pzf Only if Fznet = 0
3/12/24 73
Momentum in 2D/3D
For two interacting objects A and B, the
conservation relation is:
! ! ! !
pAi + pBi = pAf + pBf
IDEA: We cannot combine different components
mAv Axi + mB vBxi = mAv Axf + mB vBxf
mAv Ayi + mB vByi = mAv Ayf + mB vByf
mAv Azi + mB vBzi = mAv Azf + mB vBzf
3/12/24 74
Momentum in 2D/3D
For two interacting objects A and B, the
conservation relation is:
Get the components of the velocities
!
v
+y
vy
vx = v cos θ
v y = v sin θ vx
+x

The angle is measured counterclockwise from the positive x-axis

3/12/24 75
Momentum in 2D/3D
For two interacting objects A and B, the
conservation relation is:
For completely inelastic collision
The objects stick together!

! ! v Axf = v Bxf
v Af = v Bf v Ayf = v Byf
v Azf = v Bzf

3/12/24 76
Momentum in 2D/3D
For two interacting objects A and B, the
conservation relation is:
For perfectly elastic collision
The objects DO NOT stick together!
RELATIVE SPEED OF APPROACH:

(v Bfx )
- v Afx = - (v Bix - v Aix )
(v Bfy - v Afy ) = -(v Biy - v Aiy )
(v Bfz - v Afz ) = -(v Biz - v Aiz )
3/12/24 77
Momentum in 2D/3D
For two interacting objects A and B, the
conservation relation is:
For gravity that acts perpendicular to the ground
The ground is set as xy plane

pAxi + pBxi = pAxf + pBxf

pAyi + pByi = pAyf + pByf
Momentum along z-axis is NOT
CONSERVED due to gravity

3/12/24 78
Example: Banggaan…
A small 1000-kg Altis car is traveling north of Shaw Boulevard at 15
m/s. At the intersection of Shaw and EDSA, it collides with a Land
Cruiser with a mass of 2000 kg, traveling east on EDSA at 10 m/s.
The cars get tangled and move away from impact point as one
mass.
Treating each car as a particle, find the total momentum just before collision
and the velocity of the wreckage after collision.

3/12/24 79
Example: Banggaan…
A small 1000-kg Altis car is traveling north of Shaw Boulevard at 15
m/s. At the intersection of Shaw and EDSA, it collides with a Land
Cruiser with a mass of 2000 kg, traveling east on EDSA at 10 m/s.
The cars get tangled and move away from impact point as one
mass.
Treating each car as a particle, find the total momentum just before collision
and the velocity of the wreckage after collision.
!
pi,total = pix,total iˆ + piy,total ĵ
! 0 ˆ
ALTIS
(moving upward) v Ai = 15m/s cos 90 i +15m/s sin 90 ĵ
0

! 0ˆ
(moving rightward) v Bi = 10m/s cos 0 i + 10m/s sin 0 ĵ
CRUISER 0

3/12/24 80
Example: Banggaan…
A small 1000-kg Altis car is traveling north of Shaw Boulevard at 15
m/s. At the intersection of Shaw and EDSA, it collides with a Land
Cruiser with a mass of 2000 kg, traveling east on EDSA at 10 m/s.
The cars get tangled and move away from impact point as one
mass.
Treating each car as a particle, find the total momentum just before collision
and the velocity of the wreckage after collision.
!
pi,total = pix,total iˆ + piy,total ĵ
pix,tot = 1000kgv Aix + 2000kgv Bix
piy,tot = 1000kgv Aiy + 2000kgv Biy

3/12/24 81
Example: Banggaan…
A small 1000-kg Altis car is traveling north of Shaw Boulevard at 15
m/s. At the intersection of Shaw and EDSA, it collides with a Land
Cruiser with a mass of 2000 kg, traveling east on EDSA at 10 m/s.
The cars get tangled and move away from impact point as one
mass.
Treating each car as a particle, find the total momentum just before collision
and the velocity of the wreckage after collision.
!
pi,total = pix,total iˆ + piy,total ĵ
pix,tot = 1000kg ( 0 ) + 2000kg (10m/s) = 20000kg.m/s
piy,tot = 1000kg (15m/s ) + 2000kg ( 0 ) = 15000kg.m/s

3/12/24 82
Example: Banggaan…
A small 1000-kg Altis car is traveling north of Shaw Boulevard at 15
m/s. At the intersection of Shaw and EDSA, it collides with a Land
Cruiser with a mass of 2000 kg, traveling east on EDSA at 10 m/s.
The cars get tangled and move away from impact point as one
mass.
Treating each car as a particle, find the total momentum just before collision
and the velocity of the wreckage after collision.
! !
pi,total = p f ,total
pix,tot = p fx,tot = 20000kg.m/s
piy,tot = p fy,tot = 15000kg.m/s

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Example: Banggaan…
A small 1000-kg Altis car is traveling north of Shaw Boulevard at 15
m/s. At the intersection of Shaw and EDSA, it collides with a Land
Cruiser with a mass of 2000 kg, traveling east on EDSA at 10 m/s.
The cars get tangled and move away from impact point as one
mass.
Treating each car as a particle, find the total momentum just before collision
and the velocity of the wreckage after collision.
! !
pi,total = p f ,total
p fx,tot = 20000kg.m/s = 1000kgv Afx + 2000kgv Bfx
p fy,tot = 15000kg.m/s = 1000kgv Afy + 2000kgv Bfy

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Example: Banggaan…
A small 1000-kg Altis car is traveling north of Shaw Boulevard at 15
m/s. At the intersection of Shaw and EDSA, it collides with a Land
Cruiser with a mass of 2000 kg, traveling east on EDSA at 10 m/s.
The cars get tangled and move away from impact point as one
mass.
Treating each car as a particle, find the total momentum just before collision
and the velocity of the wreckage after collision.

p fx,tot = 20000kg.m/s = (1000 + 2000 ) kgv Afx

p fy,tot = 15000kg.m/s = (1000 + 2000 ) kgv Afy

v Afx = v Bfx v Afy = v Bfy For INELASTIC

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Example: Banggaan…
A small 1000-kg Altis car is traveling north of Shaw Boulevard at 15
m/s. At the intersection of Shaw and EDSA, it collides with a Land
Cruiser with a mass of 2000 kg, traveling east on EDSA at 10 m/s.
The cars get tangled and move away from impact point as one
mass.
Treating each car as a particle, find the total momentum just before collision
and the velocity of the wreckage after collision.

v Afx = v Bfx = 6.67m/s

v Afy = v Bfy = 5.00m/s
The Altis and the Cruiser move together with the same velocity

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 MOMENTUM CONSERVATION IN
NON-COLLISION SITUATIONS

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Recoil velocity of a Rifle

A marksman holds 3.00-kg rifle. He fires a 5.00 g

bullet horizontally with a velocity relative to the
ground of 300 m/s.
What is the recoil velocity of the rifle?

88
Recoil velocity of a Rifle
A marksman holds 3.00-kg rifle. He fires a 5.00 g bullet horizontally with a
velocity relative to the ground of 300 m/s.
What is the recoil velocity of the rifle?

Initial (before firing) Final (after firing)

bullet 𝑣!" = 0 𝑚/𝑠 (stationary) 𝑣!# = 300 𝑚/𝑠 (to the right)
rifle 𝑣$" = 0 𝑚/𝑠 (stationary) 𝑣$# = ?

𝒎𝒃𝒗𝒃𝒊 + 𝒎𝒓𝒗𝒓𝒊 = 𝒎𝒃𝒗𝒃𝒇 + 𝒎𝒓𝒗𝒓𝒇

𝒎𝒃𝒗𝒃𝒇 𝒎
𝒗𝒓𝒇 =− = −𝟎. 𝟓
𝒎𝒓 𝒔
89
Flowchart
Is it a collision? Or just an object that Use momentum conservation
Collision (the net force along this collision direction must be zero)
moves alone or changes displacement?
1st equation:
PF = PI or MAVAF + MBVBF = MAVAI + MBVBI
Just one object
Will they stick together after collision?
Use energy conservation. No Yes
Are the forces involved all conservative Perfectly elastic Completely inelastic
or not all?
2nd equation: 2nd equation:
- ( VBF – VAF )= ( VBI – VAI ) VBF = VAF
All Conservative For momentum, initial and final points are before and after collision
All conservative (gravity or spring) Not all
then EmechF – EmechI = 0
Non-conservative (friction, drag) Constant Two ways
KEF + UGF + UEF - KEI - UGI - UEI = 0
Then EmechF – EmechI = Wother. 1. W = F . D
For energy, you have to set the initial and Wother is work done by non-conservative force 2. W = FD cos θ
final positions of the objects.
Is the force constant θ is the angle bet F and D
KE = 0.5 mv2, UG = mgh, UE = 0.5 kx2 initial or varying?

Varying
final hF= 0 hI = h Table as reference
Two ways:
d hF= d hI = d+h Ground as reference (1) Integrate the function of Force with x
(2) Get the area under the force plot
Height is measured from a reference which you can set arbitrarily

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