1
For better understanding we’ll do another example:
Momentum and Impulse
h=2m
 Momentum is mass times velocity:
vector!
p=mv
unit: (p) = kg m/s
Let’s go back to Newton’s second law: F = ma.
Actually, Newton formulated his second law as:

m = 30 g = 0.03 kg
h
h
Force = time rate of change of momentum
F=
Δp
Δt
Δp is the change in momentum produced
by the force F in time Δt.
If the mass doesn’t change, then
F=
Δ(mv)
Δv
=m
= ma
Δt
Δt
F = ma is actually a special case of Newton’s second law and it
can not be applied to the situations in which mass can change.
Another very useful form of Newton’s 2. law:
 Impulse F∆t will produce change in momentum Δp
F∆t = ∆p
Δp = mv - mu
F∆t is called the impulse of the force F.
action of a force F over time Δt
units: (F∆t) = Ns
Ns = kg m/s
REMEMBER: Although we write F for simplicity, we actually
mean Fnet , because only Fnet and not individual forces can
change momentum (by producing an acceleration)
Achieving the same change in momentum over
a long time requires smaller force and over a
short time requires greater force.
both eggs fall the same distance, so the velocity
of both eggs just before impact is:
v = u2 +2gh = 2gh = 6m/s
Impact: before impact u = 6 m/s
just after impact is: v = 0
∆p = mv – mu = – 0.18 kg m/s
In both cases momentum is reduced to zero during
impact/interaction with the floor. But the time of interaction
is different. In the case of concrete, time is small while in the
case of pillow, the stopping time is greatly increased.
If you look at the impulse -momentum relation F∆ t = ∆p, you
see that for the same change in momentum (– 0.18 kg m/s in
this case), if the time is smaller the ground must have exerted
greater force on the egg. And vice versa. The pillow will exert
smaller force over greater period of time.
 Getting smart and smarter by knowing physics:
Often you want to reduce the momentum of an object to zero
but with minimal impact force (or injury). How to do it? Try to
maximize the time of interaction; this way you’ll decrease the
stopping force.
• Car crash on a highway, where there’s either a concrete wall
or a barbed-wire fence to crash into. Which to choose?
Naturally, the wire fence – your momentum will be decreased
by the same amount, so the impulse to stop you is the same,
but with the wire fence, you extend the time of impact, so
decrease the force.
• Bend your knees when you jump down from high! Try keeping
your knees stiff while landing – it hurts! (only try for a small
jump, otherwise you could get injured…) Bending the knees
extends the time for momentum to go to zero, by about 10-20
times, so forces are 10-20 times less.
• Safety net used by acrobats, increases impact time,
decreases the forces.
• Catching a ball – let your hand move backward
with the ball after contact…
• bungee jumping
• Wearing the gloves when boxing versus boxing
with bare fists.
2
Sometimes you want to increase the force over a short time
• This is how in karate,
an expert can break a stack
of bricks with a blow of a hand:
Bring in arm with tremendous
speed (large momentum), that
is quickly reduced on impact
with the bricks.
The shorter the time, the larger
the force on the bricks.
Such a system is called an “isolated system”.
mathematically:
p is momentum of the system
pafter = pbefore
 Impulse due to a time varying force
Formulas we had are for constant force.
What if the force changes over time ∆t?
The graph shows the variation with time of the
force on the football of mass 0.5 kg.
ball was given an impulse
of approximately
100 x 0.01 = 1 Ns
during this 0.01s
area under graph is the total
impulse given to the ball
≈ 2x(100x0.05)/2 = 5 Ns
Change in momentum, Δp, in time Δt, is the area
under the graph force vs. time.
F∆t = ∆p
The total momentum of a system of interacting
particles is conserved - remains constant, provided
there is no resultant external force.
→ ∆p = 5 kg m/s
∆p = m∆v → ∆v = 10 m/s
v = u + ∆v
p = p1 + p2
m1v1 + m2v2 = m1u1 + m2u2
Certain situations (collisions, explosions, ejections) do not allow
detailed knowledge of forces. The problem is that it is difficult to
see exactly how to apply Newton’s laws . One cannot easily
measure neither forces involved in the collision nor acceleration
(velocity appears to be instantaneously acquired).
The law of conservation of momentum gives us an easy and
elegant way to predict the outcome without knowing forces
involved in process. It is much easier to measure velocities and
masses before and after strong interaction.
WE CAN APPLY THE LAW OF CONSERVATION OF
MOMENTUM TO COLLISIONS AND EXPLOSIONS
(EJECTIONS) IF DURING INTERACTION THE NET EXTERNAL
FORCE IS EITHER ZERO OR CAN BE NEGLECTED.
Example: baseball is struck with a bat – duration of the collision is
about 0.01 s, and the average force the bat exerts on the ball is
several thousand Newtons what is much greater than the force of
gravity, so you can ignore it.
 Examples how to use law of conservation of
momentum in the case of ejection or explosion.
• A 60.0-kg astronaut is on a space walk when her tether line
breaks. She is able to throw her 10.0-kg oxygen tank away from
the shuttle with a speed of 12.0 m/s to propel herself back to
the shuttle. What is her velocity?
Collisions – conservation of momentum
• collisions can be very complicated
• two objects bang into each other and exert strong
forces which are very hard to measure over short time intervals
• fortunately, we can predict the future without going
into pesky details of force.
• What will help us is the law of conservation of momentum:
+
pafter = pbefore
0 = m1 v 1 + m 2 v 2
 Law of Conservation of Momentum
particle 1 and particle 2 collide with one another.
velocities just before
collision, before they
touch each other
velocities just after
interaction (collision)
0 = 60.0 v1 + 10.0 (12.0) = 0
v1 = − 2.0 m/s
moving toward shuttle
• Very similar case is spaceship propulsion which is actually
example of conservation of momentum. The momentum
gained by fuel ejected in the backward direction must be
balanced by forward momentum gained by the spaceship.
hot gas ejected at
very high speed
• the same as
untied balloon.
3
• Similar examples are: recoil of the firing gun, recoil of the firing
cannon, ice-skater’s recoil, throwing of the package from the boat
etc.
The negative momentum of the small fish is very
effective in slowing the large fish.
c. Small fish swims toward the large fish at 3 m/s.
p before lunch = p after lunch
(6) (1 ) + (2) (—3 ) = (6 + 2 ) v
6 —6 = 8v
Two stationary ice skaters push off
Both skaters exert equal forces on each other however,
the smaller skater acquires a larger speed (due to
larger acceleration) than the larger skater.
Momentum is conserved!
+
pafter = pbefore

v=0
fish have equal and opposite momenta. Zero momentum before
lunch would equal zero momentum after lunch, and both fish
would come to a halt.
d. Small fish swims toward the large fish at 4 m/s.
0 = p1 + p2
p1 = – p2
 Example how to use law of conservation of
momentum in the case of collisions.
(6) (1 ) + (2) (—4 ) = (6 + 2 ) v
6 —8 = 8v

v = — 0.25 m/s
There are two fish in the sea. A 6 kg fish and a 2 kg fish.
The big fish swallows the small one. What is its velocity
immediately after lunch
The minus sign tells us that after lunch the two-fish system moves
in a direction opposite to the large fish’s direction before lunch.
a. the big fish swims at 1 m/s toward and swallows
the small fish that is at rest.
• A red ball traveling with a speed of 2 m/s along the x-axis hits
the eight ball. After the collision, the red ball travels with a speed
of 1.6 m/s in a direction 37o below the positive x-axis. The two
balls have equal mass. At what angle will the eight ball fall in the
side pocket? What is the speed of the blue (8th) ball after collision.
+ direction
p before lunch = p after lunch
Mu1 + mu2 = (M+m)v
(6 kg)(1 m/s) + (2 kg)(0 m/s) = (6kg + 2 kg)v
6 kg.m/s = (8 kg) v

v = 0.75 m/s
in the direction of the large fish before lunch
b. Suppose the small fish is not at rest but is swimming toward
the large fish at 2 m/s.
in x – direction
in y – direction
m u1 + 0 = m v1 cos 370 + m v2 cos q2
v2 cos q2 = u1 - v1 cos 370 = 0.72 m/s
(1)
0 = - m v1 sin 370 + m v2 sin θ2
v2 sin θ2 = v1 sin 370 = 0.96 m/s
(2)
direction of v2 ; (2)/(1)
p before lunch = p after lunch
(6) (1 ) + (2) (—2 ) = (6 + 2 ) v
6 —4 = 8v

v = 0.25 m/s
in the direction of the large fish before lunch
(2) →
v2 = 0.96 / sin 530
tan θ2 = 1.33
θ2 = 530
v2 = 1.2 m/s