LESSON 36 – MOMENTUM &
IMPULSE
By: Fernando Morales
September 18, 2013
LEARNING GOALS
Familiarize with
Cornell Style of Note
Taking
 Define Momentum
and Impulse
 Derive Impulse from
Newton’s Third Law
 Solve Problems
related to momentum,
impulse, and area
under the curve of a
Force vs. time graph

THE CORNELL NOTE-TAKING SYSTEM
MOMENTUM
Linear momentum is the product of
an object’s mass and velocity
p = mv
Linear momentum is a vector
direction is that of the velocity
& magnitude is mv
The value of the momentum depends on reference frame
(since velocity depends on the reference frame)
The units are kgm/s
MOMENTUM & NEWTON’S 2ND LAW: NET FORCE
Mathematical
derivation:
 F = Fnet = ma
2nd Law
v
=m
t
vf - vi
=m
t
pf - pi
= t
p
=
t
conclusion:
Def’n of acceleration
pf = mvf; pi = mvi
Force is needed to change momentum
The rate of change of momentum
equals the net force applied
p
 F = t
EXAMPLE: A RAIN STORM
Assuming the rain
comes to rest upon
striking the car
(vf=0m/s), find the
average force exerted
by the raindrop.
6
Rain comes straight down with velocity of v0=15m/s and hits the roof of a car perpendicularly.
Mass of rain per second that strikes the car roof
is 0.06kg/s.
[A]
F
m vf  m v0
t
m
 ( ) v 0
t
F = -(0.06kg/s)(15m/s)=0.9 N
According to action-reaction law, the
force exerted on the roof also has a
magnitude of 0.9 N points downward:
-0.9N
7
IMPULSE AND 3RD LAW FORCES
Forces of Interaction (collisions)
cause deformations of objects –
the forces vary over time
Forces of Interaction rise from 0 to
a maximum and then fall to 0 again
IMPULSE & MOMENTUM
Fnet
p
= t
Fnet t = p
The Impulse of a force is Ft and
equals the change in momentum p
The area under a Force-time
curve is the Impulse
X2
-3
EXAMPLE
What is the value of airbags in a car? Explain in terms of
impulse and momentum.
[T, C]
By slowing the passengers down gradually as they move into the
windscreen or dashboard, airbags increase the time of contact. The
momentum to be dissipated is fixed, whether the airbags are there
or not. Momentum change = impulse = force x time, so if the time
of impact increases then the average force will decrease.
A smaller average force means less injury.
X2
-3
EXAMPLE
In the early days cars were made very rigid, so that damage to
the vehicle would be minimized. Now cars are built with
“crumple zones”: the drive train folds up, and engine
compartment bends like an accordion.
What is the advantage?
[T, C]
The folding of the crumple zone increases the amount of time of
the impact, which decrease the average force. A lower average
force means that passengers are more likely to survive, although
the car will be more severely damage. A good trade-off.
EXAMPLE
X7
-5
Impulse and Momentum
Find the impulse, velocity change and average acceleration when a
force as graphed is applied to a 33 kg object.
[K]
9N
Solution
(a) Impulse = Area under the curve
=  (9 N x 3 s) + (9 N x 8 s) +  (9 N x 3 s)
3s
= 99 Ns, the impulse, momentum change is 99 Ns
(b) Velocity change = total momentum change divided by mass
= 99 Ns / 33 kg = 3 m/s, in the direction of the force.
(c) Acceleration = velocity change divided by time
= 3 m/s / 14 s = 0.21 m/s² , in the direction of the force.
11 s
14 s
RUNNING SHOE ANALYSIS
Which causes a greater impulse, heel running or forefoot running?
Is it safer for you to run with shoes that cause high impulse or low
impulse?
[T]


Impulse can be calculated from Force vs. Time graphs
Impulse is the area under the F-T graph
EXAMPLE
Bend your knees when landing!
[A]
(a) find the impulse felt by a 70 kg man landing on hard ground after
jumping from 3.0 m;
estimate the average force exerted on him if
(b) he lands stiff-legged and
(c) he lands with bent legs.
In (b) assume the body moves 1.0 cm on landing, and in (c) about 50 cm
Solution
(a) Impulse = Change in Momentum; Ft = mv
We need the speed change
from Conservation of energy: KE = PE
v = (2gh) = (2 · 9.8 ms² · 3.0 m) = 7.67 ms
Ft = mv
= 70 kg (0 - 7.67 ms ) = -539 Ns
Is the Ns a unit of momentum?
kgm
1 Ns = 1 kgm
·
1s
=
1
s²
s
EXAMPLE
X7
-5
Bend your knees when landing!
(cont)
Solution
(b) stiff-legged – comes to rest in 1 cm
v
vavg =
= 3.84 ms
2
d
10-2 m
The collision lasts t =
=
= 2.60 (10-3) s
vavg
3.84 ms
F=
p
t
=
539 Ns
2.60 (10-3) s
5
= 2.07 (10 ) N
F is the net force on the man; it is the sum of the normal force, Fgrd, the
ground exerts upward and the weight of the man,  690 N downward:
F = Fgrd - mg so Fgrd = F + mg = 2.1(105) N + 0.0069 (105) N  2.1 (105) N
EXAMPLE
X7 - 5
(cont)
Bend your knees when landing!
Solution
(c) springy – comes to rest in 50 cm
v
vavg =
= 3.8 ms
2
d
0.5 m
The collision lasts t =
=
vavg
3.8 ms
p
540 Ns
F=
=
= 4.2 (103) N
t
0.13 s
= 130 (10-3) s
F is the net force on the man; it is the sum of the normal force, Fgrd, the
ground exerts upward and the weight of the man,  690 N downward:
F = Fgrd - mg so Fgrd = F + mg = 4.2 (103) N + 0.69 (103) N  5 (103) N
So bending the legs reduces the force considerably from about
210 (103) N to  5 (103) N – a factor of 40, thus saving broken
legs and hips – the bones could not stand the force in (b)
EXAMPLE
Washing a car – momentum and force
[A]
X7 - 1
Water leaves a hose at 1.5 kg/s with a speed of 20 m/s, and hits the side of a
car which stops it (we ignore splashing). Find the force exerted by the car on
the water
Solution
Take x +ve right.
Each second 1.5 kg hits the car with a speed of
20 m/s; the momentum change is -30 kgm/s.
The force required is
p
t
This is -30 N, since t = 1 s
Washing a car – momentum and force
EXAMPLE
X7 - 1
water splashes back? Is the force greater?
What if the
Solution
The change in momentum is now increased
The force required is p but p is larger so the force must be larger
t
REQUIRED BEFORE NEXT CLASS
Read Section 5.2 from Nelson Textbook
 Pg. 227 # 1, 6, 7, 9, 12
