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Physics 200 Class #10 Notes
October 10, 2005
Reading Assignment for Wednesday
Text: Chapter 6 pp. 141-160
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Continuation of Chapter 5 with added notes on Energy and Blackbody Radiation
Reminder: The MidTerm Exam is on October 17
We discussed force, and energy...
Now what about a force acting for a given length of time, independent of distance
(great examples at http://www.kungfuscience.org )
Momentum
Descartes and Huygens defined momentum in order to “quantify motion”. Collision problems lead
to the following definition:
Linear Momentum (usually denoted by p) = mass x velocity (note, it has direction)
Momentum  mv
units : kg m / s
Conservation of Linear Momentum Principle:
When two objects collide, momentum (linear momentum) may be transferred from one object to
the other, but the total linear momentum of the system does not change.
(we return to this, but first)
What produces a change in momentum?
Force
But time plays a roll. It is convenient to summarize the effect as follows:
Impulse = force x time interval
Impulse = F t
Use Newton’s second law (for a constant mass)
F  ma
v
t
F t  mv
F m
 means a change in the quantity following it
F t  (mv)
The impulse is equal to the change in momentum of the object.
“Bouncing” – what effect if any?
No Bounce: object moving to the right with momentum mv
F t = momentum after – momentum before
F t = 0 – momentum before = - mv
The minus sign means that the force must act to the left
Phy 200 Fall 2005 Class_10
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With Bounce:
F t = momentum after – momentum before
F t = - mv – mv = - 2mv
For the same time interval, the force must be twice as great
Collisions
Class Demonstrations on Collisions on an Air Track or karate chopping boards
Linear momentum of the system is conserved in collisions! Why?
If we consider the colliding objects to be the system, the force between them is an internal force.
The net external force (and therefore impulse) is zero.
Examples
Suppose we have two masses:
m1  2kg
m2  3kg
Conservati on of momentum requires : m1v1before  m2 v 2before  m1v1after  m2 v 2 after
Case I: The masses stick together after the collision (inelastic collision)
v1before  10m/s v 2before  0
(2kg)(10m / s )  (3kg)(0m / s )  (5kg)(v final )
20kgm / s  (5kg)(v final )
v final  4m / s
Case II: Elastic collision with the 2kg mass moving to the left afterwards, find the speed of the
3kg mass.
(2kg)(10m / s )  (3kg)(0m / s )  (2kg)( 2m / s )  (3kg)(v final )
20kgm / s  4kgm / s  (3kg) v final
v final  8kgm / s
Case III: There is an explosion during the collision and the 2kg mass moves to the left at 95m/s,
find the speed of the 3kg mass afterwards.
(2kg)(10m / s )  (3kg)(0m / s)  (2kg)( 95m / s)  (3kg)(v final )
20kgm / s  190kgm / s  (3kg)(v final )
v final  70m / s
Now with the conservation of energy too!
Total energy is conserved: E = KE + PE = constant
Elastic collisions, kinetic energy is conserved on it's own - things bounce.
Inelastic - energy goes out of the system into the environment, or into smashing things - but
momentum still stays in the system.
1
Remember KE =
mv2
2
Phy 200 Fall 2005 Class_10
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Case I: Inelastic collision of a moving car (1) with a stationary car (2).
m1 = 1000 kg. (2,200 pounds)
m2 = 1000 kg
v1initial = 20 m/s (about 45 miles per hour)
v2initial = 0
Since it's inelastic, let's assume the cars stick together so v1final = v2 final which we just call vfinal
conservation of momentum:
m1v2initial + m2v2initial = m1v2final + m2v2final = m1 vfinal + m2vfinal = (m1 + m2) vfinal
(1000 kg)(20m/s) + (100 kg)(0m/s) = (2000 kg)(vfinal)
(2000 kg m/s) = (2000 kg)(vfinal)
so: vfinal = 10 m/s
Let's look at what happens to the energy. It's inelastic, so some of the energy is put into smashing
the cars:
1
1
1
1
m1v2 1 initial + m2v2 2 initial = m1v2 1 final + m2 v2 2 final + E put into smashing the cars
2
2
2
2
(0.5)(1000 kg)(20 m/s)2 + (0.5)(1000 kg)(0 m/s)2 = (0.5)(2000 kg)(10 m/s)2 + E smash
(0.5)(1000 kg)(400 m2/s2) = (0.5)(2000 kg)(100 m2/s2) + E smash
200,000 J = 100,000 J + E smash
so the energy put into deforming the metal of the cars is
Esmash = 200,000 J - 100,000 J = 100,000 J
That's the same amount of energy it would take you to pick up one of the cars and raise it 10
meters. That's means lifting a 2,200 lb car up to the top of a 3 story house!
Case II: Work that out for an initial velocity of 40 m/s, just double the speed. Really, try it.
Answer below:
Answer: that doubles the total final velocity. vfinal = 20 m/s
The energy equation is
(0.5)(1000 kg)(40 m/s)2 + (0.5)(1000 kg)(0 m/s)2 = (0.5)(2000 kg)(20 m/s)2 + E smash
(0.5)(1000 kg)(1600 m2/s2) = (0.5)(2000 kg)(400 m2/s2) + Esmash
800,000 J = 200,000 J + E smash
so the energy put into deforming the metal of the cars is
Esmash = 800,000 J - 200,000 J = 600,000 J That's a big difference from just doubling the speed!
That's like lifting a 2,200 lb car up to the top of a 18 story house!
Drive carefully.
Phy 200 Fall 2005 Class_10
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