Physics I
Class 13
Conservative Forces,
Non-Conservative Forces,
and Collisions
Rev. 13-Aug-03 GB
13-1
Conservation of Energy
(Review)
1.
Multi-dimensional
form of work integral:

xf
 
W   F  dx

xi
2.
3.
Conservative force = work doesn’t depend on path.
Potential Energy defined for a conservative force:
 
U(A)    F  dx
A
0
Ug  m g ( y  y0 )  m g h
U s  12 k ( x  x 0 ) 2
4.
Gravity:
5.
6.
Spring:
Conservation of energy if only conservative forces operate:
 K   U or  K   U  0
13-2
Conservative Forces
Non-Conservative Forces
Examples of Conservative Forces:
 Gravity
 Ideal Spring (Hooke’s Law)
 Electrostatic Force (later in Physics 1)
Examples of Non-Conservative Forces:
 Human Pushes and Pulls
 Friction
13-3
Is Mechanical Energy
Always Conserved?
Total Mechanical Energy
EKU
E  K  U  0 if only conservative forces act
When Non-Conservative Forces Act
E  K  U  Wnon cons
This is equivalent to
K f  U f  K i  U i  Wnon cons
Non-conservative forces add (+) or subtract (–) energy.
13-4
Example of Energy Lost to
Friction (Non-Conservative Force)
Kf  0
U f  m g h  70  9.8  1.5  1029J
K i  12 m v 2  12  70  64  2240J
Ui  0
v

d
h
h  d sin( )  3  12  1.5m
A skateboarder with mass = 70 kg starts up a 30º
incline going 8 m/s. He goes 3 m along the
incline and comes to a temporary stop. What was
the force of friction, assuming it was constant?
13-5
Example of Energy Lost to
Friction (Non-Conservative Force)
Kf  0
U f  m g h  70  9.8  1.5  1029J
K i  12 m v 2  12  70  64  2240J
Ui  0
v
d
h

Wfriction  K f  U f  K i  U i  1029  2240  1211J
Wfriction  1211
Ffriction 

 403.7 N
d
3
What do the – signs mean?
13-6
Elastic and Inelastic Collisions
Momentum is conserved in a collision process because only internal
forces are acting. External forces are considered separately.
In many collisions a large percentage of the kinetic energy is lost.
These are known as inelastic collisions. For example, any collision
in which two objects stick together is always inelastic.
If the kinetic energy after a collision is the same as before, then we
have an elastic collision. During the collision process, some of the
kinetic energy can convert to potential energy of various kinds, but
after the collision is over all of the kinetic energy is restored.
13-7
Elastic Collisions in
One Dimension
+X
Initial
v1i
Final
v1f
v2i
v2f
Conservation of Momentum:
m1v1i  m 2 v 2i  m1v1f  m 2 v 2 f
Conservation of Energy:
1
2
m1v 21i  12 m 2 v 2 2i  12 m1v 21f  12 m 2 v 2 2 f
Two equations, two unknowns (final velocities).
13-8
Elastic Collisions in
One Dimension
+X
Initial
Final
v1i
v2i
v1f
v2f
m1  m 2
2 m2
v1f 
v1i 
v 2i
m1  m 2
m1  m 2
2 m1
m 2  m1
v 2f 
v1i 
v 2i
m1  m 2
m1  m 2
13-9
Elastic Collisions in
One Dimension - Example
m1  2 m 2
v1i  0
v 2i  1 m/s
+X
v2i
v1f
2 1
2
2
v1f 
( 0) 
(1)   m/s
2 1
2 1
3
2
1 2
1
v 2f 
( 0) 
(1)   m/s
2 1
2 1
3
v2f
Initial: m2 has all (–1) of the
mom. and KE.
Final: m1 has –4/3 of the mom.
and 8/9 of the KE.
m2 has +1/3 of the mom.
and 1/9 of the KE.
13-10
Class #13
Take-Away Concepts
1.
Modification of energy conservation including nonconservative forces:
 E   K   U  Wnon cons
2.
3.
4.
Non-conservative work adds (+) or subtracts (–)
energy from the system.
Elastic collision preserves KE before and after.
(Don’t assume all collisions are elastic, most are not.)
Special equations for 1D elastic collisions.
m1  m 2
2 m2
v1i 
v 2i
m1  m 2
m1  m 2
2 m1
m  m1
v 2f 
v1i  2
v 2i
m1  m 2
m1  m 2
v1f 
13-11
Class #13
Problems of the Day
___ 1. An elastic collision is one in which:
A. momentum is not conserved but kinetic energy is the same
before and after.
B. a rubbery object bounces off something else.
C. kinetic energy is the same before and after, and momentum
is conserved.
D. momentum is conserved but kinetic energy changes.
E. the total impulse is equal to the change in kinetic energy.
13-12
Answer to Problem 1 for Class #13
The answer is C.
This is explained in the book in Section 10-3, page 198.
13-13
Class #13
Problems of the Day
2. A bicycle rider with combined mass of rider+bicycle = 80 kg
starts down a hill at 10 m/s. After coasting (not pedaling) 100 m,
the rider has dropped 10m below his starting height and now has a
speed of 15 m/s. What was the force of friction plus air resistance
on the bicycle (magnitude) assuming it was constant?
13-14
Answer to Problem 2 for Class #13
This is another difficult problem to solve using F = m a and
kinematics, but easier to solve with energy concepts.
Counting the final height as zero, the initial mechanical energy of
the rider is given by 0.5 * m * vi2 + m g h= 0.5 * 80 * (10)2 + 80 *
9.8 * 10 = 11840 J.
The final mechanical energy is 0.5 * m * vf2 = 0.5 * 80 * (15)2 =
9000 J. The difference is –2840 J.
The force is found by dividing the energy loss by 100 m and
ignoring the minus sign because we want magnitude:
F = 28.4 N.
13-15
Activity #13
Conservation of Energy II
Objective of the Activity:
1.
2.
Further studies of conservation of energy
(same setup as the last activity).
Add friction and see how that affects
conservation of energy.
13-16
Class #13 Optional Material
Elastic Collisions in 2/3D
In order to solve 2D and 3D two-body elastic collision problems
more easily, we need to introduce two key concepts.
1.
2.
The laws of physics are identical in all inertial frames of
reference. (See the optional material on Special Relativity.)
Use a moving coordinate system called the Center of
Momentum System in which the total momentum is zero.
13-17
Center of Momentum
Reference Frame
We introduce a moving reference frame in which the
total momentum of the system is zero.
The X' axis is aligned with
the initial momenta.
p1
Y'
Laboratory frame of reference.
p2
p1'
X'
p2'
13-18
Analyzing the Collision in the
Center of Momentum
 p1' 
Before the collision: p1 '  p 2 '
p2'
p1''

Original direction of momentum.
 depends on the details of the collision.
p2''
After collision, to conserve momentum (zero):


p1"   p 2 "
And to conserve kinetic energy:


| p1"|  | p1 ' |
13-19
Transform Back to the
Laboratory Frame of Reference
p1f
This is just an example, not all elastic collisions turn out this way.
p2f = 0
Apply the inverse coordinate transform to get back to the
laboratory reference frame.
In Special Relativity, this is complicated because time mixes into
the transformation.
The one-dimensional equations shown on previous slides can be
derived using this method by setting  = 180º.
13-20