Transcript Physics 103: Lecture Notes

Physics 103: Lecture 10
Impulse and Momentum

02/24/2003
Today’s lecture will cover the following new
Concepts:
Impulse
Momentum
Impulse-Momentum Theorem
Momentum Conservation
Physics 103, Spring 2003, U. Wisconsin
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Impulse and Momentum

Impulse = average force times time
I = Fave t
(nothing new….. Newton’s second law)
v p
F  ma  m

t t
since
p  p f  pi
QuickTime™ and a Animation decompressor are needed to see this picture.
I  Favet
Assumption:

mass m of the object is constant
not necessary, but otherwise there are complications.
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Physics 103, Spring 2003, U. Wisconsin
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You drop an egg onto
a) the floor
Questions
b) a thick piece of foam rubber
In both cases, the egg does not bounce.
In which case is the impulse greater?
A) case 1
B) case 2
C) the same
correct
In which case is the average force greater
A) case 1
correct
B) case 2
C) the same
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Preflight Lecture 10 No.1

The impulse delivered to a body by a force is
1. defined only for interactions of short duration.
2. equal to the change in momentum of the body.
3. equal to the area under an F versus x graph.
4. defined only for elastic collisions.
• Often useful in this situation,
but not ONLY
70
60
50
A
B
C
D
40
30
20
10
0
Preflight 10.1
• equal to the area under an
F versus t graph!
• it is defined for all collisions
elastic and inelastic!
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Preflight Questions 2 & 3
Two identical balls are dropped from the same height onto the
floor. In case 1 the ball bounces back up, and in case 2 the ball
sticks to the floor without bouncing. In which case is the
impulse given to the ball by the floor the biggest?
1. Case 1
correct
2. Case 2
3. The same
45
40
35
30
25
20
15
10
5
0
A
B
C
Preflight 10.2
Pretty Sure
Not Quite Sure
Just Guessing
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the impulse is greater for case one because ...
the change in momentum of the object ... is
proportional to the change in velocity which is
greater in case one because it has a greater
final velocity (down then up) than case 2 (which
is only from down to zero). Impulse must be
greater for case 1.
Example: suppose m=1 kg, v(initial)=-1 m/s
mv(initial)= -1 kg-m/s
Case 1
mv(final)= +1 kg-m/s Impulse = 1- (-1)=2
Case 2 mv(final)= 0
Physics 103, Spring 2003, U. Wisconsin
Impulse = 1 - 0 = 1
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Preflight Question 4 & 5
In both cases of the above question, the direction of the
impulse given to the ball by the floor is the same. What is this
direction?
1. Upward
correct
2. Downward
75%
time
25%
0%
20%
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40%
60%
80%
Pretty Sure
Not Quite Sure
Just Guessing
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Preflight Questions 6, 7 & 8
Is it possible for a system of two objects to have zero total
momentum while having a non-zero total kinetic energy?
1. YES
correct
2. NO
53%
in an isolated system, two ice skaters starting at
rest and pushing on one another will move in
opposite directions thus the momenta of the two
are equal and opposite and total momentum is
zero. but they are moving apart after the push
and therefore the KE is non-zero.
47%
0%
20%
40%
60%
two hockey pucks moving towards each other with
the same speed on a collision course have zero
total momentum, but a non zero total kinetic
energy
Pretty Sure
Not Quite Sure
Just Guessing
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Physics 103, Spring 2003, U. Wisconsin
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Impulse and Momentum

Momentum-Impulse Theorem
Favet  I = pf - pi = p

For single object….
F = 0  momentum conserved (p = 0)
•
For a collection of objects …
o
Fext = 0  total momentum conserved
 p   p   p  0
f
i
o Fext = mtotal a

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“External” is a VERY important qualifier
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Some Terminology
• Elastic Collisions:
collisions that conserve kinetic energy
• Inelastic Collisions:
collisions that do not conserve kinetic energy
* Completely Inelastic Collisons:
objects stick together
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Elastic Collision in 1-Dimension
Quic kTime™ and a Cinepak decompress or are needed to see this picture.
m1v1i  m2v 2i  m1v1 f  m2v 2 f
1
1
1
1
m1v1i2  m2v 2i2  m1v12f  m2v 22 f
2
2
2
2
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Elastic Collision
m1v1i  m2v 2i  m1v1 f  m2v 2 f
1
1
1
1
2
2
2
m1v1i  m2v 2i  m1v1 f  m2v 22 f
2
2
2
2
m1 (v1i2  v12f )  m2 (v 22 f  v 2i2 )
m1(v1i  v1 f )(v1i  v1 f )  m2 (v 2 f  v 2i )(v 2 f  v 2i )
m1 (v1i  v1 f )  m2 (v 2 f  v 2i )
v1i  v1 f  v 2i  v 2 f
v1i  v 2i  (v1 f  v 2 f )
Magnitude of relative velocity is conserved.
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Preflight Lecture 10 No.9

In an elastic collision
1. kinetic energy is conserved.
2. momentum is conserved.
3. the magnitude of the relative velocity is conserved.
4. all of the above are correct.
• True by definition of elastic
• True by definition of collision
10%
12%
3%
75%
0%
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20%
40%
60%
• Total momentum is unchanged
80%
Physics 103, Spring 2003, U. Wisconsin
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Preflight Lecture 10 No.10

In an inelastic collision
1. both kinetic energy and momentum are conserved.
2. only kinetic energy is conserved.
3. only momentum is conserved.
4. neither kinetic energy nor momentum are conserved.
9%
14%
56%
21%
0%
20%
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40%
• False by definition of inelastic collision
• False by definition of inelastic collision
• False by definition of collision
60%
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Collisions
m2
m1
m2
m1
“before”
“after”
Procedure
• Draw “before”, “after”
• Define system so that Fext = 0
• Set up axes
• Compute Ptotal “before”
• Compute Ptotal “after”
• Set them equal to each other
Explosions
“before”
M
m1
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m2
“after”
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Conceptual Example
Consider two blocks (mass m and M) sliding toward one
another (with velocities vm and vM) on a frictionless plane.
m
vm
vM
M
What happens after the collision?
Pm + PM = Pm + PM
What does your result depend on?
Is the collision elastic or?
How would the answer change if there was friction
between the blocks and the plane?
Energy would be lost during the motion. We would
need initial distances to calculate!
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Totally Inelastic Head-on Collision
Draw “before” and “after”
m
v
v
“before”
m
System = two blocks
Axis: postive to right
2m
vf
“after”
• Before: Ptotal,before = mv + (-mv) = 0 !
• After: Ptotal,after = (2m)vf
• Ptotal,before = Ptotal,after
• 0 = (2m)vf
• vf = 0 !
• Therefore KEafter = 0
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Preflight Question 11 & 12
Movies often show someone firing a gun loaded with blanks. In a
blank cartridge the lead bullet is removed and the end of the
shell casing is crimped shut to prevent the gunpowder from
spilling out. When a gun fires a blank, is the recoil greater than,
the same as, or less than when the gun fires a standard bullet?
1. greater than
2. same as
correct
3. less than
Impulse is the same
in the two cases
18%
Pretty Sure
Not Quite Sure
Just Guessing
36%
45%
0%
10%
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20%
30%
40%
50%
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Explosions
“before”
M
v1
m2
m1
v2
“after”
• Example: m1 = M/3 m2 = 2M/3
• Which block has larger momentum?
* Each has same momentum
• Which block has larger velocity?
* mv
same for each
 smaller mass has larger velocity
• Which block has larger kinetic energy?
* KE = mv2/2 = m2v2/2m = p2/2m  smaller mass has larger KE
• Is kinetic energy conserved?
*NO!!
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Example

A rocket of mass 200 kg is fired at with the vertical
component of the initial velocity 1000 m/s and the horizontal
component of 300 m/s. When it reaches the highest point on
the trajectory an explosion occurs. The rocket is split in half
and part A develops a vertical component to its velocity of
100 m/s. How far from the launching point does part A hit
the ground? (Assume the ground is flat and use g = 10
m/s2.)
a)
60.0 km
b)
65.8 km
c)
63.15 km
t2
tdown
t1
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Solution
v 02 1000 2
y max  
 50,000m
2g
20
v 1000
t 0 
100s
g 10
after the explosion part A has 10% of the velocity
t 10s
Y 500m
so t up 110s and y 50,500m
1 2
1
ay t
050,500 10 t 2
2
2
t down  10,100  100.5s
y  y 0  v 0y t 
x  300 110100.5 63.15km
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Reprise
1a) Impulse is average force times time of action
1b) Momentum is mass times velocity.
2) Conservation of Momentum is useful. Depends (as always)
on the situation.
3) Kinetic energy may be conserved (elastic) or may not
(inelastic)
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