Momentum and Impulse


So far we’ve studied
the properties of a
single object; i.e. its
motion and energy
How do we analyze the
motion of two or more
objects that interact
with each other??
Momentum

It’s the product of mass times velocity
p = mv


Units: (kg) x (m/s) = kg·m/s
Momentum is a technical term for something we
already know!
–
If a train and a car are going the same speed, it’s harder to
stop the train.

–
The greater mass of the train gives it more momentum than the
car.
A bullet fired from a gun has more penetrating power than one
thrown by hand.

Even though they have the same mass, the first bullet has more
momentum due to its higher velocity.
Give it a try!!

Which vehicle has more momentum, a
1500kg truck moving at 0.3m/s or a 105kg
go-cart moving at 5 m/s?
– Truck
p = mv  (1500kg) x (0.3m/s) = 450 kg·m/s
–
Gocart
p = mv  (105kg) x (5m/s) = 525 kg·m/s
Interaction


When two objects interact they exert forces on each
other.
Newton’s Third Law states that these forces must be
equal and opposite.
Change in Momentum




For one of these objects, Newton’s First Law gives
F = ma  a = F/m
If we use the average acceleration for this case
aavg = v/t
then
v/t = Favg/m  mv = Favgt
Looking at the left side of this equation
mv = m(vf – vi) = mvf – mvi
= pf – pi = p
Impulse-Momentum Theorem

Therefore, the average force times the time
interval over which it acts is equal to the
change in momentum.
p = Favgt
–

We call Favgt the impulse that acts on an object.
Units: (force) x (time) = N·s


Are N·s equal to kg·m/s?
Do dimensional analysis to find out.
Impulse, Day-to-Day

Like momentum, you already understand how this
idea works, now you have a scientific name for it!

If a constant net force acts on an object (say a box is
pulled along a slippery floor); the longer you apply
the force, the greater will be the change in the
object’s speed.

Similarly, if you apply a force to an object for a
specific amount of time (say a push on a swing), the
greater the force, the greater will be the change in
the object’s speed.
Give it a try!!

A golf club strikes a 46g
golf ball for 0.5ms, the
ball leaves the face of
the club at 70 m/s.
Find the average force
that the club exerts on
the ball during the
impact. (the ball is
initially at rest)
Solution

mv = Favgt  Favg= mv/ t

Favg= (.046kg)(70m/s – 0m/s)/(5x10-4s)
Favg= 6440 N = 1448 lbs

What then is the average force that the ball
exerts on the club?

Newton’s Third Law says -6440 N
Conservation of Momentum

Whenever two objects interact, it has been
found that the sum of their momentum is the
same before and after the interaction.
ptot,i = ptot,f  m1v1,i + m2v2,i = m1v1,f + m2v2,f

This is called the Law of Conservation of Momentum
–
only true if there are no net external forces acting on the
objects during their interaction.
Types of Interaction
(Explosions)

One object splitting into
two or more parts

Two objects being
separated by some
force

Rockets!!!
Types of Interaction
(Collisions)

whether it’s a collision
in the day-to-day sense
(car accidents)

or not (catching a
baseball)
Explosive Interactions
(One-Dimensional)



Imagine an object of mass m at rest exploding into two parts of
mass m1 and m2.
Conservation of momentum tells us that the initial momentum of
the system must equal the final momentum of the system.
but, since

then
–

pi = pf or mv = m1v1,f + m2v2,f
mv = 0
m1v1,f + m2v2,f = 0
therefore
m1v1,f = -m2v2,f
The minus sign tells us that the two parts must be moving in the
opposite direction (makes sense right!!)
Give it a try!!

a)
Neil is a 150kg astronaut floating at
rest in outer space. He decides he
wants his picture taken with the Earth
behind him, so he throws his camera
to Buzz, another astronaut floating
near-by.
Is Neil still at rest after throwing the
camera?
No, conservation of momentum says that he must
move in the opposite direction in order to have
equal and opposite momentum from the camera.
Give it a try!! (continued)
b)
If the camera has a mass of 0.80kg and it moves away
with a velocity of 12m/s to the left, what is Neil’s
velocity after he throws it?
(m1 = 0.80kg, v1,f = 12m/s, m2 = 150kg, v2,f = ?)
pi = pf = 0, so m1v1,f + m2v2,f = 0 or m1v1,f = -m2v2,f
which gives us
c)
-(m1v1,f)/m2 = v2,f so, v2,f = -0.064m/s
How far will Neil be from the spot where he threw the
camera after 1 hour?
d = v2,ft
given t = 3600s, then d = 230m
Collision Types

Elastic Collisions
–
–
Momentum is conserved
Kinetic energy is conserved too!


No permanent deformation, no sound, no friction
Inelastic Collisions
–
–
Momentum is conserved
Kinetic energy is not conserved


Possible permanent deformation, sound, or friction between
objects
Work done by non-conservative forces
Give it a try!!

Two balls are rolling along a table with negligible friction.
One ball, with a mass of 0.250kg, has a velocity of
0.200m/s eastward. The other ball, with a mass of
0.100kg, has a velocity of 0.100m/s eastward. The first
ball hits the second from directly behind. If the final
velocity of the first ball is 0.143m/s eastward, what is the
final velocity of the second ball?

Is this an elastic collision?
Solution (part 1)

Using conservation of momentum
m1v1,i + m2v2,i = m1v1,f + m2v2,f

solving for v2,f
v2,f = (m1v1,i + m2v2,i - m1v1,f)/m2
v2,f = 0.243m/s eastward
(Remember, velocity is a vector quantity!!)
Solution (part 2)

Elastic collisions mean kinetic energy is also conserved
KEi = KEf
– or
½m1(v1,i)2 + ½m2(v2.i)2 = ½m1(v1,f)2 + ½m2(v2,f)2
–
–

Before the collision
KEi = 0.00550J
Afterward
KEf = 0.00551J
Yes! It is an elastic collision.
Inelastic collisions
Two possible outcomes
The objects bounce
apart afterwards
The objects stick
together afterwards
(perfectly inelastic)
Perfectly Inelastic Collisions

What is conserved?

How do you think the final velocities of the objects
after the collision will be related to each other?

Remember, velocity is a vector. You have to take
direction into account!!
Give it a try!!

A 5kg lump of clay traveling at 10m/s to the left
strikes a 6kg lump of clay moving at 12m/s to the
right. Find the final velocity of the resulting object
if they stick together.

How much kinetic energy is lost in the collision?
Solution (part 1)

Given:

m1v1,i + m2v2,i = (m1 + m2)vf

(m1v1,i + m2v2,i)/ (m1 + m2) = vf

vf = +2m/s or 2m/s to the right
m1 = 5kg m2 = 6kg v1,i = -10m/s v2,i = 12 m/s
Solution (part 2)


Kinetic energy is not conserved so let’s find KE.
KE = KEf – KEi

KEi = ½m1(v1,i)2 + ½m2(v2,i)2 = 682J
KEf = ½(m1 + m2)(vf)2 = 22J

KE = 22J – 682J = -660J

The energy did not disappear! It was converted to heat energy
in the clay and probably into sound energy.
Collisions
(Two Dimensions)

Momentum is a vector like velocity.

Use conservation of momentum along
each axis separately when solving
problems.
–
Pick an orientation for your coordinate system
that simplifies the problem
Give it a try!!

A 60kg man is sliding east
at 0.5m/s on a frozen pond
man
(assume it’s frictionless). A
friend throws him a 5kg bag
of salt to help him stop. If the
bag’s velocity is 5m/s to the
north, what is the man’s velocity
after catching the bag?
bag
Solution




Given:
1: m1 = 60kg, v1,i,x = 0.5m/s, v1,i,y = 0
2: m2 = 5 kg, v2,i,x = 0, v2,i,y = 5m/s
Use conservation of momentum for each component
X-axis
– m1v1,i,x + m2v2,i,x = (m1 + m2)vf,x
– (m1v1,i,x + 0)/ (m1 + m2) = vf,x
– vf,x = 0.46m/s
Y-axis
– m1v1,i,y + m2v2,i,y = (m1 + m2)vf,y
– (0 + m2v2,i,y)/ (m1 + m2) = vf,y
– vf,y = 0.38m/s
Solution


Find the magnitude and direction of the
velocity
Magnitude
–
–

Direction
–
–

Use the Pythagorean Theorem
v = 0.60m/s
 = tan-1(vy /vx)
 = 400
So, v = 0.60m/s @ 400 N of E
Impulse
vs
Conservation Of Momentum

Impulse: only correct if applied to one of the objects
in an interaction
p = Favg t
Conservation of momentum: only correct if
–

applied to all the objects in an interaction
–
ptot,i = ptot,f
m1v1,i + m2v2,i = m1v1,f + m2v2,f
continued

putting the object 2 terms on the left, and object 1
terms on the right:
m2v2,i - m2v2,f = m1v1,f - m1v1,i
p2,i – p2,f = p1,f – p1,i
-p2 = p1
-Favg,2 t2 = Favg,1 t1
- therefore, from Newton’s third law, the magnitude
of the impulse on each object in an interaction is
the same
The Ballistic Pendulum



One projectile, one hanging
object
Perfectly Inelastic Collision
Collision:
–

Pendulum Swing:
–

conservation of momentum!!!
before
conservation of energy!!!
Different initial and final
points for each part.
after
Give it a try!!

A 100g bullet is fired into a 1.35kg block of
wood. If the block rises 25cm, how fast was
the bullet going when it hit the block?
25cm
Solution

There are two parts

The perfectly inelastic collision
– The upswing of the block/bullet
Knowns: m1 = .100kg, m2 = 1.35kg, y = .25cm
–
–

MEi = MEf
–


Need velocities to analyze the collision so let’s examine the
up swing (i.e. conservation of energy)
(be careful choosing initial and final points)
what forms of energy are present??
Ki = Uf or ½mvi2 = mghf
so vi2 = 2ghf
vi = 2.21m/s
Solution

Now we know the speed of the block/bullet
after the collision. So

vf = 2.21m/s, v2 = 0, v1 = ?, m = m1 + m2 = 1.45kg

we get

m1v1 + m2v2 = mvf
v1 = mvf/m1
or
v1 = 32m/s