Ch. 7: Momentum,
Impulse & Collisions
THE COURSE THEME:
NEWTON’S LAWS OF MOTION!
• Chs. 1-5: Motion analysis with Forces.
• Ch. 6: Alternative analysis with Work & Energy.
– Conservation of Energy: NOT a new law!
We’ve seen that this is Newton’s Laws reformulated or translated from Force Language
to Energy Language.
• NOW (Ch. 7): Another alternative
analysis using the concept of (Linear)
Momentum.
• Ch. 7: Another alternative analysis using
the concept of (Linear) Momentum.
• Conservation of (Linear)
Momentum: NOT a new law!
–We’ll see that this is just Newton’s
Laws of Motion re-formulated or reexpressed (translated) from Force
Language to (Linear) Momentum
Language.
• Chs. 1-5: Newton’s Laws of Motion
using the concepts of position,
displacement, velocity, acceleration, force.
• Newton’s Laws with Forces: General!
• In principle, they could be used to solve
any dynamics problem, But, often, they
are very difficult to apply, especially to
very complicated systems. So, alternate
formulations have been developed.
They are often easier to apply.
• In Ch. 6, we expressed Newton’s
Laws using
Work & Energy Language.
• Newton’s Laws with Work &
Energy: These are also general! In
principle, they could be used to
solve any dynamics problem. But,
often, it’s more convenient to use
still another formulation.
• Now, we’ll be discussing another
approach that uses
Momentum
instead of Energy or Force as the
basic physical quantity.
• Newton’s Laws in a different
language (Momentum). Before we
discuss these, we need to learn
vocabulary in
Momentum Language.
Linear Momentum
• Momentum: The momentum of an object
is DEFINED as:
(a vector || v)
SI Units: kgm/s = Ns
• In 3 dimensions, momentum has 3 components:
px = mvx py = mvy pz = mvz
• Newton called mv “quantity of motion”.
• Question: How is the momentum of an object
changed?
• Answer: By the application of a force F!
Momentum: p = mv
Most general statement of Newton’s 2nd Law:
∑F = (p/t)
(1)
The total force acting on an object = the time
rate of change of the object’s momentum.
• (1) is more general than ∑F = ma because it
allows for the mass m to change with time also!
• Example, rocket motion!
• Note: if m is constant, (1) becomes:
∑F = (p/t) = [(mv)/t]
= m(v/t) = ma
Newton’s 2nd Law (General Form!)
∑F = p/t
(1)
• Suppose that, for a short time t,
m = constant
Initial Momentum p0 = mv0
Final Momentum p = mv
• (1) then becomes:
∑F = p/t = m(v-v0)/t
= m(v/t) = ma
(as before)
Example: Force of a Tennis Serve
For a top player, a tennis ball leaves
the racket on the serve with speed
v2 = 55 m/s (~ 120 mi/h).
Assume that v1 = 0 & the ball’s mass is
m = 0.06 kg
If the ball is in contact with the
racket for the short time
t = 4 ms (4 × 10-3 s).
Estimate
The average force Favg
on the ball. Would this be large
enough to lift a 60-kg person?
Solution:
v2 = 55 m/s (~ 120 mi/h) v1 = 0, m = 0.06 kg
t = 4 ms (4 × 10-3 s).
Newton’s 2nd Law:
Favg  (p/t) & p = m(v2 - v1) So,
Favg  800 N
To lift a person with mass M = 60-kg,
requires a force greater than
Mg = (60)(9.8) = 588 N
So, yes the calculated
force is more than enough
to lift such a person!!
• Consider an isolated system with 2
masses: m1 moves at velocity v1 & m2
moves at velocity v2. m1 feels a force F21
exerted on it by m2. m2 feels a force F12
exerted on it by m21.
See figure

NOTE: Misconception! The masses do
NOT have to touch!
Newton’s 3rd Law: F21 = - F12
Or:
F21 + F12 = 0 (1)
Newton’s 2nd Law: (if no other forces act)
F21 = m1a1
F12 = m2a2
• Put (2) & (3) into (1) 
(2).
(3)
m1a1 + m2a2 = 0 (4)
Note
v’s & F’s are
vectors!!
• 2 moving masses interacting.  
N’s 3rd Law: F21 + F12 = 0
N’s 2nd Law: F21 = m1a1. F12 = m2a2
Together:
m1a1 + m2a2 = 0
(4)
• Acceleration definition: time rate of change of
velocity 
a ≡ (v/t). (4) becomes:
m1(v1/t) + m2(v2/t) = 0
Use simple manipulation:
(m1v1)/t + (m2v2)/t = 0
or (m1v1 + m2v2)/t = 0
(5)
• Time rate of change of m1v1 + m2v2 is = 0.
 This means that
m1v1 + m2v2 = constant!
A Vector Equation!
(6)
• So, for 2 moving masses interacting
& isolated from the rest of the world:
m1v1 + m2v2 = constant
(6)
With the definition of momentum:
p1 = m1v1, p2 = m2v2
(6) becomes:
p1 + p2 = constant
(7)
(7) says that, no matter how they
Note: The plural of
interact & what motions they undergo,
the vector sum of the momenta of
otherwise isolated masses is
ALWAYS THE SAME
FOR ALL TIME!
“momentum” is
“momenta”, NOT
“momentums”!!
• Consider now, one mass m & write:
Newton’s 2nd Law: ∑F = ma
• Use the definition of acceleration as the time
rate of change of the velocity: a ≡ (v/t).
• Put this into Newton’s 2nd Law: ∑F = m(v/t)
• If m doesn’t depend on time: ∑F = [(mv)/t]
Put the definition of momentum, p ≡ mv into
Newton’s 2nd Law: ∑F = (p/t)
• Did this for constant m.
• Can be shown it’s more general & is valid
even if m changes with time.
• A general statement of Newton’s 2nd Law is:
∑F = (p/t)
(1)
• The total or net force acting on a mass =
time rate of change in the mass’s momentum.
• (1) is more general than ∑F = ma because it
allows for the mass m to change with time also!
• Example, rocket motion!
• Note: if m is constant, (1) becomes:
∑F = (p/t) = (mv)/t =
m(v/t) = ma
Force and Momentum
 Assume the force and acceleration are constant
 Since momentum is mass times velocity, the force
can be related to the momentum


This is sometimes called the impulse theorem, but
it is nothing more than Newton’s 2nd Law!
• Back to 2 moving masses interacting &
isolated from the rest of the world. We found:
(p1 + p2)/t = 0
or
p1 + p2 = constant
(1)
• This says that the total momentum of the
2 masses ptot = p1 + p2 = constant
• Suppose due to the forces F21 & F12, p1 & p2
change with time. (1) tells us that, no matter
how they change individually,
ptot = constant
So, the total VECTOR momentum of
the 2 masses is conserved!
• If, at some initial time, the 2 momenta are p1i & p2i & if
at some final time they are p1f & p2f, we can write:
ptot = p1i + p2i = p1f + p2f = constant
Example: 2 billiard balls collide (zero external force)
Momentum Before = Momentum After!
m1v1i
m1v1f
m2v2i
m1v1i + m2v2i =
m1v1f + m2v2f
The vector sum
m v is constant!
2 2f
Simple Example:
Initial Momentum = Final Momentum (1D)
v1i = 24 m/s
Masses M
= 10,000 kg
m1v1i+m2v2i = (m1 + m2)vf
v2i = 0
vf = ??
v2i = 0, v1f = v2f = vf
 vf = [(m1v1)/(m1 + m2)] = 12 m/s
Example: Rocket Propulsion
Momentum Before = Momentum After
0 = Procket - Pgas
Example: Rifle Recoil
Momentum Before = Momentum After
mBvBi + mRvRi = mBvB + mRvB
vR
vBi = vRi = 0
vB
pR
pB
mB = 0.02 kg, mR = 5.0 kg, vB = 620 m/s
0 = mB vB + mRvR
 vR = - 2.5 m/s (to left, of course!)
Example: Archer
• An archer, m1 = 60 kg, v1i = 0,
stands on frictionless ice. Arrow
mass, m2 = 0.5 kg, v2i = 0. Archer
shoots arrow horizontally at v2f =
50 m/s to the right. What velocity
v2f does the archer have as a result?
Possible Approaches:
N’s 2nd Law in force form:
No information about F or a!
Energy approach:
No information about work, energy!
Momentum approach:
Easily used!!!
• m1 = 60 kg, v1i = 0, m2 = 0.5 kg
v2i = 0, v2f = 50 m/s, v2f = ?
Momentum
• No external forces in x-direction, so the arrow is
isolated in the x-direction
• Total momentum before shooting is 0
 The total Momentum after shooting
the arrow is also 0!
• Before Shooting the Arrow:
ptot = m1v1i + m2v21 = m1(0) + m2(0) = 0
• Momentum is conserved!
ptot = p1i + p2i = p1f + p2f = constant
• After Shooting Arrow
m1v1f + m2v2f = 0 or, v1f = - (m1/m1)v2f
v1f = - 0.42 m/s (Minus means archer slides to left!)
Collisions & Impulse
Collisions & Impulse
• Briefly consider the details of a collision.
• Assume that a collision lasts a very
small time t. During the collision, the
net force on the object is
Newton’s 2nd Law:
∑F = p/t or p = (∑F)/t
During a collision,
objects can be
deformed due to the
large forces involved.
• p = momentum change of the object due
to the collision. Define: p  Impulse  J
that the collision gives the object (the p
for the object!)
• In the usual case: Either only one force is Example: A tennis
ball is hit by a racket
acting or we replace the left side by an
as in the figure.
average collision force: Fc = ∑F
Impulse:
J = p = Fc t
∑F is time dependent & t is often very small. So, not much error is
made in replacing ∑F with an average force, Fc or Favg. The true
Impulse in the collision is the area under the F vs. t curve. In the
approximation that Fc ≈ ∑F, the Impulse is approximately:
J = p ≈ Fct
This approximation is the same as replacing the green area in the left
figure by the green rectangle in the right figure.
Replace the green area at the left with the green rectangle at the right.
The true impulse
is the area under
the Fc vs. t curve.
t is usually very small & Fc is time dependent
It is often a good approximation to replace the area under
the Fc vs. t curve with the area of the green rectangle.
The approximate impulse is then
J = p ≈ Fct
Example: Crash Test
• Crash test: Car, m = 1500 kg, hits wall. 1
dimensional collision. +x is to the right.
Before crash, v = -15 m/s. After crash, v =
2.6 m/s. Collision lasts Δt = 0.15 s.
• Find: Impulse car receives & average
force on car.
• Assume: Force exerted by wall is large
compared to other forces
• Gravity & normal forces are perpendicular
& don’t effect the horizontal momentum
 Use the impulse approximation
p1 = mv1 = -2.25 kg m/s, p2 = mv2 = 2.64 kg m/s
J = Δp = p2 – p1 = 2.64  104 kg m/s
(∑F)avg = (Δp/Δt) = 1.76  105 N
How Good Are the Bumpers?
In a crash test, a car of mass 1.5103 kg collides with a wall and
rebounds as in figure. The initial and final velocities of the car are
vi=-15 m/s and vf = 2.6 m/s, respectively. If the collision lasts for
0.15 s, find (a) the impulse delivered to the car due to the
collision. (b) the size and direction of the average force
exerted on the car
How Good Are the Bumpers?
In a crash test, a car of mass 1.5103 kg collides with a wall and
rebounds as in figure. The initial and final velocities of the car are
vi=-15 m/s and vf = 2.6 m/s, respectively. If the collision lasts for
0.15 s, find (a) the impulse delivered to the car due to the
collision. (b) the size and direction of the average force
exerted on the car
pi  mvi  (1.5 103 kg)( 15m / s)  2.25 10 4 kg  m / s
p f  mv f  (1.5 103 kg)(2.6m / s)  0.39 104 kg  m / s
I  p f  pi  mv f  mvi
 (0.39 10 4 kg  m / s )  (2.25  10 4 kg  m / s )
 2.64 10 4 kg  m / s
p I
2.64 10 4 kg  m / s
Fav 


 1.76 105 N
t t
0.15s
Example: Karate blow
Estimate the impulse & the average
force delivered by a karate blow
that breaks a board a few cm thick.
Assume the hand moves at roughly
10 m/s when it hits the board.