Momentum and Collisions
Unit 5, Presentation 1
Momentum

The linear momentum p of an object
of mass m moving with a velocity
is defined as the product of the
mass and the velocity v



p  mv
SI Units are kg m / s
Vector quantity, the direction of the
momentum is the same as the
velocity’s
More About Momentum

Momentum components

px = m vx and py = m vy


Applies to two-dimensional motion
Momentum is related to kinetic
energy
KE 
p
2
2m
Impulse


In order to change the momentum of an
object, a force must be applied
The time rate of change of momentum of
an object is equal to the net force acting
on it


p

m(vf  vi )
 Fnet
t
t
Gives an alternative statement of Newton’s
second law
Impulse cont.

When a single, constant force acts
on the object, there is an impulse
delivered to the object
I  Ft
 I is defined as the impulse
 Vector quantity, the direction is the
same as the direction of the force

Impulse-Momentum Theorem

The theorem states that the impulse
acting on the object is equal to the
change in momentum of the object


I  Ft  p  m v f  m v i
If the force is not constant, use the
average force applied
Average Force in Impulse

The average force can
be thought of as the
constant force that
would give the same
impulse to the object
in the time interval as
the actual timevarying force gives in
the interval
Average Force cont.


The impulse imparted by a force
during the time interval Δt is equal
to the area under the force-time
graph from the beginning to the end
of the time interval
Or, the impulse is equal to the
average force multiplied by the time
interval, F a v  t   p
Impulse Example Problem
A golf ball with mass 5.0 x 10-2 kg is struck with a club.
The force on the ball varies from zero when contact
is made up to some maximum value (when the ball
is maximally deformed) and then back to zero when
the ball leaves the club. Assume that the ball leaves
the club face with a velocity of 44 m/s.
(a) Find the magnitude of the impulse due to the
collision.
(b) Estimate the duration of the collision and the
average force acting on the ball, assuming that the
distance the ball travels on the face of the club is
approximately 2.0 cm.
Example Problem (cntd)
m = 5.0 x 10-2 kg
v = 44 m/s
I=?
d=0.02 m
t=?
v
I  P  mv
I  5 . 0  10
2
d
t
t
d

v avg
kg ( 44 m / s )
0 . 02 m
 9 . 1  10
22 m / s
I  2 . 2 kg  m / s
F avg 
P
t

2 . 2 kg  m / s
9 . 1  10
4
sec
 2400 N
4
sec
Impulse Applied to Auto Collisions

The most important factor is the
collision time or the time it takes
the person to come to a rest


This will reduce the chance of dying in
a car crash
Ways to increase the time


Seat belts
Air bags
Typical Collision Values

Calculate the force and acceleration
acting on a 75 kg person traveling at 27
m/s and coming to stop in 0.010 s
m  75 kg
I   P  m  v  75 kg (  27 m / s )   2025 kg  m / s
t  0 . 010 sec
v i  27 . 0 m / s
v f  0m / s
I  Ft
F 
F ?
I ?
a?
a
I
t
F
m


2025 kg  m / s
 202500 N
0 . 010 sec
202500 N
75 kg
 2700 m / s  275 g
Survival


Increase time
Seat belt


Restrain people so it takes more time
for them to stop
New time is about 0.15 seconds
Air Bags



The air bag increases the time of the
collision
It will also absorb some of the energy
from the body
It will spread out the area of contact


Decreases the pressure
Helps prevent penetration wounds