Download Chapter 7: Impulse, Momentum, and Collisions

Chapter 7: Impulse,
Momentum, and Collisions
 Up to now we have considered forces which have a
constant value throughout the motion and no explicit
time duration
 Now, lets consider a force which has a time duration
(usually short) and with a magnitude that may vary with
time – examples: a bat hitting a baseball, a car crash, a
asteroid or comet striking the Earth, etc.
F
 It is difficult to deal with a
time-varying force, so we
usually take the mean value
F
t0
t
tf
t
 Define a new quantity by multiplying the force

by the time duration 
F  t  I  Impulse
- a vector, points in the same direction as
the force
- has units of N s
 Define another quantity, but which gives a
measure of the motion (similar to KE)
 
mv  p  linear momentum
- a vector, points in same direction as the
velocity
- units of kg m/s = N s
 Linear momentum and KE are related
1 2 2
1 2
KE  mv 
m v 
p
2m
2m
1
2
2
Example
A car of mass 760 kg is traveling east at a speed
of 10.0 m/s. The car hits a wall and rebounds
(moving west) with a speed of 0.100 m/s.
Determine its momentum and KE before and
after the impact. Determine the impulse.
Solution:

Given: m = 750 kg, v 0  10.0 ms east

v f  0.10 ms west  -0.10 ms east


p0  mv 0  (750 kg)(10.0 m/s east)
3
 7.50x10 kg m/s east
2
2
1
1
KE0  2 mv 0  2 (750 kg)(10.0 m/s)
4

3
.
75
x
10
J


p f  mv f  (750 kg)(-0.100 m/s east)
1
 7.50x10 kg m/s east
2
2
1
1
KE f  2 mv f  2 (750 kg)(-0.100 m/s)
 3.75 J
 Now, from the definition of acceleration
and Newton’s 2nd Law:


v f  v0


mv f  mv 0


a
 ma 
t
t


F  t  p f  p0


Impulse-Momentum Theorem
I  p
 The Impulse-Momentum Theorem says that if
an impulse (force*time duration) is applied to an
object, its momentum changes
 In this example, the impact of the car with
the wall applies an impulse to the car  car’s p



changes I  p
 p  (75.0)  7.50x10 3
f
0
 7.58x10
3 kg m
s2
east
Example – Problem 7.13
A 0.500-kg ball is dropped from rest at a point
1.20 m above the floor. The ball rebounds straight
upward to a height of 0.700 m. What are the
magnitude and direction of the impulse of the net
force applied to the ball during the collision with
the floor?
y
Solution:
Given: m = 0.500 kg,
h0=1.20 m, h3=0.7 m,
h1=h2=0
0
3
1
2
Method: need momentum before and after
impact  need velocities  use conservation of
energy
Conservation of mechanical energy is satisfied
between 0 and 1 and between 2 and 3, but not
between 1 and 2
E0  E1
PE0  KE1
2
1
mgh0  2 mv1
2
2 2
1
m gh0  2 m v1
2
2
1
m gh0  2 p1
E3  E2
PE3  KE2
2
1
mgh3  2 mv 2
2
2 2
1
m gh3  2 m v 2
2
2
1
m gh3  2 p2
p1  m 2 gh0

p1  m 2 gh0
p2  m 2 gh3

p2  m 2 gh3

 

I  p  p2  p1
 m 2 gh3  (m 2 gh0 )
 m( 2 gh3  2 gh0 )
 0.5( 2(9.8)(0.7)  2(9.8)(1.2) )
 4.28 N s

I
Collisions
 Involves two (or more) objects which may have
their motion (velocity, momentum) altered by
collisions
 These concepts are applicable to the collisions
of atoms, billiard balls, cars, planetary objects,
galaxies, etc.
 Say, we have a collection of interacting particles
numbered 1, 2, 3, … We can define the Total
Momentum of the system (all the particles) as just
the sum of all the individual momenta

   
P   pi  p1  p2  p3  ...
 Imagine that these particles interact in some
way – collide and scatter
 As long as there are no net external forces
acting on the system (collection of objects), the
Total Linear Momentum does not change
 Which means the Total Linear Momentum is
the same before the collision, during the collision,
and after the collision
 
P0  Pf
 Conservation of Linear Momentum