The Law of the Conservation of

Momentum
• Newton’s third law tells us that
when two intersecting objects
collide, they exert force on each
other. If F1, the force that mass
m1 exerts on mass m2 and F2 the
force that mass m2 exerts on
mass m1, then F1 = -F2
• This can be written as
m1a1 = -m2a2
where a = v2-v1
t
Thus, m1 v1’-v1 = -m2 v2’ – v2
t t

m1v1 – m1v1’ = -m2v2 + m2v2’

m1v1 + m2v2 = m1v1’ + m2v2’
where
m1 = mass of the first object
m2 = mass of the second object
V1 = velocity before collision of the first object
v1’ = velocity after collision of the first object
v2 = velocity before collision of the second object
v2’= velocity after collision of the second body
Law of Conservation of
Momentum
-the total momentum of an
isolated system before
collision is equal to the total
momentum after collision.
 Conservation of Momentum
• The law of conservation of momentum states
when a system of interacting objects is not
influenced by outside forces (like friction), the
total momentum of the system cannot
change.

• If you throw a rock forward from a

skateboard, you will move backward
in response.
 The Law of the Conservation of Momentum
states…
The total momentum of all objects interacting
with one another remains constant regardless
of the nature of the forces between the
objects.

The total momentum of a system will stay the

same before and after a collision.
 Why does Mr Stickfigure on the ice
move
Before Collision
backwards?
After Collision

-p +p

p = -5 kgm/s p = 5 kgm/s

Momentum = zero (not moving) Momentum = zero

ptotal = 0kgm/s ptotal = 5kgm/s + (-5kgm/s)
ptotal = 0kgm/s
Why does Mr Stickfigure move backwards?
•Answer the Question
–Mr Stickfigure moves backwards because he threw
the ball forward.
•Explain the Relationship
–The Law of the Conservation of Momentum states
that the total momentum of a system must stay the
same before and after. Before the ball and Mr
Stickfigure had a total momentum of zero, so after the
total momentum needed to stay zero.
•Support with Data/Observations
–When the ball moved forward it had a positive
momentum, so Mr Stickfigure needed a negative
momentum to cancel it out. This is why he moved
backwards.
 “recoil”

Recoil is a term that refers to moment when a gun

moves backwards after it is shot.
Recoil happens because everything must follow “The
Law of the Conservation of Momentum”!!!
 Why a gun “recoils”?
The total momentum before was
zero
So the total momentum after has to
be zero
The gun moves with a negative momentum
because the bullet moves with a positive
momentum and they cancel out, the total
momentum stays zero.
 3 Types of Collisions
• Elastic

• Inelastic

• Perfectly Inelastic
Elastic
• A collision in which two
objects move separately with
different velocities, but not
permanent deformation
Inelastic
• A collision in which two objects
deforms so that the objects move in
the same direction but with different
final velocities after colliding.
Perfectly Inelastic
• A collision in which two
objects stick together and
move with the same velocity
after colliding.
For each of the following
examples, identify the
type of collision…
Perfectly Inelastic
Elastic
Inelastic
 Explain why the final velocity of the moving object
“makes sense” in order to conserve total momentum.

After the collision the 1st ball transferred its momentum

to the 2nd ball. Since the balls have the same mass, the
velocity of the second ball should be the same as the
first in order to conserve momentum.
 Explain why the final velocity of the moving object
“makes sense” in order to conserve total momentum.

After the collision the cart

and McDonald are
moving.
Since there is more mass,
there should be less
velocity to conserve
momentum.
Sample Problems:
• A 50.0kg girl jumps into a 100kg raft at rest on the water. If the
velocity of the girl is 4.00m/s as she jumps, what is the final
velocity of the girl and the raft?
Before After
G
G
R R
m1 = 50.0kg m2 = 100kg m1 = 50.0kg M2’ = 100kg
V1 = 4.00m/s V2 = 0m/s Vf = ?

m1V1 + m2V2 = (m1 + m2)Vf

(50.0kg)(4.00m/s)+(100.0kg)(0m/s) = (50.0kg +100kg)(Vf)
200kgm/s = (150kg)Vf
1.33m/s = Vf
• A 63.0kg astronaut throws a 5.0kg hammer in a direction
away from the shuttle with a speed of 18.0m/s, pushing the
astronaut back to the shuttle. Assuming that the astronaut
and hammer start from rest, find the final speed of the
astronaut after throwing the hammer. +90kgm/s
-90kgm/s
Before After
A H A H

mA = 63.0kg mH = 5.0kg mA = 63.0kg mH = 5.0kg

VA = 0m/s VH = 0m/s VA = ? VH = 18m/s

mAVA + mHVH = mAVA + mHVH

(63.0kg)(0m/s)+(5.0kg)(0m/s) = (63.0kg)VA + (5.0kg)(18.0m/s)
0kgm/s = (63.0kg)VA + 90.0kgm/s
-90.0kgm/s = (63.0kg)VA
-1.43m/s = VA
• A 15.0kg cart moving to the right with a speed of 4.0m/s collides with a 6.5kg cart moving to
the left with a speed of 2.0m/s. If the carts stick together, find the final speed of the two
carts.

Before After
1 2 A H
m1 = 15.0kg m2 = 6.5kg m1 = 15.0kg m2 = 6.5kg
V1 = 4.0m/s V2 = -2.0m/s Vf = ?

m1V1 + m2V2 = (m1 + m2)Vf

(15.0kg)(4.0m/s)+(6.5kg)(-2.0m/s) = (15.0kg + 6.5kg)(Vf)
60kgm/s + (-13.0kgm/s) = (21.5kg)Vf
60kgm/s -13.0kgm/s = (21.5kg)Vf
47.0kgm/s = (21.5kg)VA

2.19m/s = VA
Your Turn:
1.A 2750 kg van runs into the
back of an 850 kg compact car
at rest. They move off
together at 8.5 m/s. Assuming
that the friction with the road
is negligible, calculate the
initial speed of the van.
Your Turn:
2. Two ice skaters with masses
of 50 kg and 60 kg respectively
are initially at rest and facing
each other over a surface of ice.
Find the velocity of 60 kg skater
if the 50 kg skater moves with a
velocity of 5 m/s.
Your Turn:
3. A 1400 kg vehicle is moving at 30
m/s westward collides with a 1100 kg
vehicle moving at a certain velocity in
the opposite direction. After the
collision, the 1400 kg vehicle attains a
velocity of 35 m/s westward while the
1100 kg vehicle attains a velocity of 25
m/s eastward. Find the initial velocity
of the 1100 kg vehicle.
Your Turn:
4. Two cars traveling in the same direction
skid on a path of smooth level wet road.
The 1200 kg car skids straight into the back
of 1000 kg car. The two cars become
entangled after the impact. If the 1000 kg
car is moving at 8 m/s before the impact
and both cars are moving at 15 m/s after
the impact, calculate the velocity of the
1200 kg car just before the collision takes
place.
Your Turn:
5. A 0.0005 kg bullet is fired
horizontally and hits an 8 kg block
of wood initially at rest which can
move freely. The wood and the
bullet move with a velocity of 0.50
m/s after impact. What is the initial
velocity of the bullet?
 Key Concepts
Elastic “Starts from rest” Vi = 0m/s
“Stops” Vf = 0m/s

m1v1i + m2v2i = m1v1f + m2v2f Perfectly Inelastic

Inelastic
m1v1i + m2v2i = (m1+ m2)vf

m1v1i + m2v2i = m1v1f + m2v2f

 Impulse
• Another factor to be considered in collisions is
how much energy is transferred from one object
to another. Energy transfer occurs when they are
in contact with each other (or”collision
duration”).
• Where F is the force vector (in newtons), t is the
time (in seconds) that the constant force is
pushing, m is the mass (in kilograms) of the ball
and v is the change in the velocity vector (in m/s)
 Impulse
• Thus the impulse is the force needed to
produce a change in the body’s momentum
through a combination of changes in its mass
and/or velocity. In the case of the baseball
and bat, an impulse , such as the force of a bat
on the ball, causes the ball’s momentum to
change. The above equation not only defines
what impulse is, but it also shows that impulse
equals momentum.
Life Lesson:
Life is a race. Most of the
time, how well you finish the
race depends on how
prepared are you for the
journey. Success is often
measured by how fast you
reached your destination.
 But travel sometimes
involves dangers and risk. And
life in its perspectives needs to
be conserved like energy and
momentum. Thus, it requires
precaution or safety. Since life is
unpredictable, you must be
ready to the challenges it will
bring you at every twists and
turns.