Physics 231
11
12
Wed., 11/13 10.1-.4 Introducing Collisions Quiz 9
Lab
L9 Multi-particle Systems
Fri., 11/15
10.6-.8 Scattering
Mon., 11/18
Tues. 11/19
Wed.,11/20
Lab
Fri., 11/22
2013 1
Ch 10 Day 1
RE 10.a bring laptop, smartphone, pad,…
RE 10.b
10.9-.10 Collision Complications: Inelastic, Relativistic, & Quantized RE 10.c
EP9
10.5, .11 Different Reference Frames
L10 Collisions (ballistic pendulum?)
(Rep of Washinton’s 3/2 visits Thurs night 7pm)
11.1 Translational Angular Momentum Quiz 10
RE 11.a; HW10: Ch 10 Pr’s 13*, 21, 30, “39”,
Equipment
Lab carts and track (not air track) & cart weights (black bars)
Bouncy ball
Ch. 10 Collisions: Exploring the Nucleus
In this chapter, we are going to apply the laws of physics to describe and model
collisions.
Collision: A relatively brief and strong interaction.
Why focus on these? For a start, they’re all around us. Bats hit balls; protons hit
other protons, etc. Secondly, such collisions can be used to probe the inner
structure of atoms – while an atom is too small to see into, we can see the
ricochets of particles that collide with it and thus infer the structure.
What’s special about a collision? That it’s relatively brief and strong allows us
to a) think of the before and after without focusing on the rather messy during,
and b) ignore most weak, background interactions.
o Baseball and Bat: Take for instance a ball hitting a bat. This happens so
strongly and so briefly that during that split second of the interaction, the
hit, by far, is the dominant influence on the ball’s change in motion comparatively, gravitation has a negligible affect during this period. Also,
think about the collision itself – how the ball momentarily deforms. The
interaction itself can be pretty messy; but we can look around this brief
interval and just focus on the affect of the collision, not the collision itself.
o Ex. Say that the ball has mass 1 kg. It is in contact with the bat for, say, 5
ms. During that time, it switches from going forward at 100mph = 44.7
m/s to going 80 mph backward = 35.8 m/s.
 So what net force must it be subject to?

p 1kg
35 .8m / s 1kg 44 .7 m / s
Fnet
1.61 10 4 N
t
0.005 s

Meanwhile, what’s the gravitational force on the ball?
Fb E mg 1kg 9.8m / s 2 9.8N
 Conclusion: during the split second of the bat-ball collision, the
Earth’s pull and air’s resistance, for that matter are negligible.
o Demo: Carts on track (from lab)
Today’s Goal
o Our goal today is to build a mathematical model for the simplest case:
collisions of two objects.
o One could use such a model to answer any number of questions about
such a collision, but a fairly natural way to structure it is so that, given the
Ch 10 Day 1
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initial conditions (masses, velocities), we can use the laws of physics to
predict the final conditions (velocities).
o Two unknowns need two equations. Note that, with two objects
colliding, we have two final velocities to figure out. So we have, at least,
two unknowns (in 1-D). Algebraically that will require, at least, two
equations.
Momentum & Impulse
Conservation of Momentum

dp system 
Fnet.ext , so, to the
The first tool we’ll use is momentum. Recall that
dt
extent that there is no net external force, or at least that the interval we’re
tf


considering is quite brief, p system
Fnet.ext dt 0 . In such a case, we say that
Physics 231
8.1
8.2
ti
momentum is (approximately) conserved during the brief collision.
The conservation of momentum, like the conservation of energy, is a very
powerful tool.
System = carts.
o Let’s take our system to be the two carts on the air track, and our time
interval to be just before to just after they collide.
o




o
p1a
p 1 .b
p 2 .b
p 2. a
o Then


p system Fsystem net.ext t







if t is
p1
p 2 ( F1 air F1 table . friction F1 table .normal F1 Earth. gravl F2 air ...) t


p1
p2 0
small enough.
o Note: it’s true that there is friction, and if you watch a little while, it will
have a considerable affect, but just for the time during the collision, the
affect is quite small.
o To invoke this relationship, you don’t need to know the exact nature of the
collision, just that there are no significant external interactions, so the total
momentum in the system can’t change.
o There are three ways of phrasing this relationship
 Momentum is conserved, there is not change in momentum:

p system 0


p1
p2 0
 Momentum is conserved, the total amount we had initially equals
the total amount we have Finally:


p sys.b p sys.a




p1b p 2.b p1.a p 2.a
 Momentum is exchanged, what one member of the system gives
up, the other gains:


p1
p2




p1a p1.ib
p 2.a p 2.b
Physics 231
Ch 10 Day 1
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However you phrase it, conservation of momentum gives us one of the two
required equations.
o Demo: Maximally (Perfectly) Inelastic. (collide carts and stick) In the
case that the two carts hit and stick together, you also have the equation


v1.a v 2.a .
o The carts are clearly going nowhere near the speed of light, so we can


approximate the momenta as p mv .
o We can now predict the final velocities based on the initial conditions and
laws of physics:



m1 v1.i m small v small .i mbig m small v f


mbig v big .i m small v small .i
 
va
mbig m small
Qualitatevly, what does this say?
Q: In the limit that mbig>>msmall, the final velocity is
approximately what?
A: Vbig.i
Q: In the limit that mbig=msmall, the final velocity is what?
A: The average of the two velocities: (Vbig +Vsmall)/2 .
CW: 9.X.1 A 1kg cart traveling at speed 2m/s strikes a stationary 0.5
kg cart head-on, and they two masses stick together. (a) what was the
initial total kinetic energy? (b) What is the final speed? (c) what is the
final total kinetic energy? (d) what is the increase in internal energy of
the two masses?
Sometimes a Collision is just part of the story.
Example: Imagine this scenario, a 60kg kid on a swing, say she’s gotten herself
going so that at her highest her center of mass is 4m above the ground and when
she swoops down her center of mass is just 0.5 m above the ground. On her
down-sweep, she reaches down and picks up her 3kg backpack that’s sitting on
the ground (maybe her cell phone started ringing).
a) How high will she get on her upswing with the pack in her lap?
b) If she drops the pack the next time she comes down, how high will she rise?
Clearly there’s a “collision” when she picks up the bag, and another one (of
sorts) when she drops it. But those are imbedded in the context of smoother
motion. The collision itself looks like one of these “maximally inelastic” ones
we just modeled. If we knew how fast she was going just before the collision,
we could say how fast she should be going just after, and if we know how fast
she’s going just after, we might be able to say how far up she’ll swing. Let’s
build our model in the same way it plays out.
i)
She swings from ro to r1, meanwhile her speed goes from 0 to v1 which is
the speed just before the collision.
Q: What principle is good at relating a change in position to a change
in speed?
Physics 231
Ch 10 Day 1
A: Energy Principle.
System: Girl & Earth
Surroundings: Swing/Chain
E system Wsur system


K trans
U Girl Earth
Fswing drswing . force
2013 4
0
A nice thing about the swing’s force is that it’s exerted
perpendicular to the motion at any given instant, so it does no work!
That’s not so say it doesn’t have any effect on her motion, it redirects
her motion, but it doesn’t speed her up or slow her down.
K trans
U Girl Earth 0
1
2
mG v12.a
1
2
mG vo2
mG g y1
yo
0
v1.a
2 g y o y1
ii) She Collides with the backpack. This plays out just as we predicted for
the colliding carts, except it’s even simpler since the second object, the
backpack is initially stationary.

mG v1.a

v1.b
mG mbp
iii) She swings up again. This plays out just like the down-swing, but in
reverse.
K trans
U Girl Earth 0
1
2
mbp v 22
mG
y2
y1
1
2
1
2
mG
mbp v12b
mG
mbp g y 2
y1
0
v12b
g
iv) She swings down again. This is just the same process in reverse, and she’ll have
the same speed just before dropping the bag as she did just after on the
upswing. Now here’s a tricky point:
a. if she simply lets go of the bag, then she exerts no force on it one
way or another, and both she and it will continue moving in the
horizontal direction at the same speed as just before she let go
(meanwhile it will arc down to the Earth while she arcs up.) So
simply letting go, her speed’s unchanged, her kinetic energy’s
unchanged, and she’s going to climb to the exact same height y2.
b. Alternatively, if she throws the backpack, away from her, then we
have a maximally inelastic collision in reverse.



mG v1c mbp vbp.ic mG mbpl v1b


mG mbpl v1b mbp v bp.ic

v1c
mG
Then this is the speed that comes into our energy consideration
v2
y 3 y1 12 1c
g
Physics 231
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o Demo: Elastic. (collide carts and bounce magnetically). Now, in this


case, v1. f v 2. f , we still need a second equation if we want to solve for
two unknowns. What other relation holds through this collision?
Conservation of Energy.
rf


E system
Fnet.ext dr 0
ri

o
o
o
K1.cm
E1. int
K 2.cm
E 2. int
U 1, 2
0
K1.cm
E1. int
K 2.cm
E 2. int 0
In general, both the center of mass and the internal energy of the system
can change. This is rather unfortunate for the internal energy is rather
difficult to track. In the special case that the internal energy is the same
after the collision as before (though it may change during the collision, as
in the case of a compressing squash ball), we say the collision is elastic,
for it bounces back with just as much translational kinetic energy as it had
to begin with.
Let’s see how things work out in this case.
Kt
Kp 0
Rephrase all this terms of 1 and
2
2
p t2.a p t2.b p p.a p p.b
2 instead of small and big
0
2m t
2m p




p t .b p pb p.ta p p.a
o
o between these two equations, a general solution can be found for the two
final momenta.
o Let’s consider the simple case in which one of the carts is initially
stationary.

p p .b
o
p t2.a
2mt

p p.b

p p .a
p 2p.a
p 2p.b
2m p


p t .a p p .a

p t .a
0 (1)



p t .a p pl .b p p.a
o
o being 1-D, we’re just dealing with x-components, so we can drop the
vector notation. If a term is positive, the motion is to the right, if it is
negative, the motion is to the left.
p t .a p p.b p p.a (2)

o Target: p p .a
o Predictions:
 Q: Would you expect the light cart to gain or loose momentum
through the collision?
Loose, since the other cart gains momentum, so
|psmall.a|<|psmall.b|
 Collision not built into equations: Note that, while you and I are
thinking about there being a collision between the “before” and
“after” shots, we’re not directly modeling the collision, there are
Ch 10 Day 1
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no terms in our math for the force of one object on the other or the
work done by one on the other. Instead, we’re invoking the fact
that, whether there’s a collision or not momentum and energy are
conserved.
With that in mind, I suggest to you that the math will lead
to two solutions: one for if there is a collision and one for if
there isn’t.
What should be the no-collision answer?
o p p .a = p p .b . So we better find this in our math.
Physics 231
o Algebra
(2) -> (1):
2
p p.b p p.a
p 2p.a
2mt
p 2p.b
p 2p.b
0
2m p
p 2p.a
2 p p.b p p.a
p 2p.a
mt
p 2p.b
mp
(3)
0
Digress for Equal Mass Special Case:
Now, in the special case that the two objects have the same mass, mbig =
msmall and our equation reduces to
2
2
2
2
p small
p small
2 p small .b p small .a p small
p small
.b
.a
.a
.b
0
2
2 p small
.a
2 p small .b p small .a
0
p small .a p small .a p small .b 0
Which means that either
p small .a 0 or p small .a p small .b
In the first case, the cart that was initially moving came to a stop; in the
second case, it rolled on by – no collision happened.
Returning to the conservation of momentum relation:
p big .a p small .b p small .a
If it did come to a stop, then the second cart acquired the other’s
momentum. If it didn’t, the other cart remained stationary.
Demo: Equal mass carts collide.
Let’s return to the more general case, where they don’t have the same mass. The
equation we’d developed (3) could be rephrased as
2
p small
.a
1
1
mbig
m small
p small .a
2 p small .b
mbig
2
p small
.b
1
mbig
1
m small
0
This is of the mathematical form
x 2 A xB C 0
Which, from some distant math class you must have learned, is solved by
x
B
B2
2A
4 AC
Physics 231
Ch 10 Day 1
Not putting you through the tedium, I’ll just quote the result:
msmall mbig
p small .a p small .b
msmall mbig
2013 7
Compare with predictions.
o Indeed, there are two solutions (following from the +/- sign in the
quadratic equation).
o One is simply that the small cart has the same momentum after the
collision as before.
msmall mbig
o The other is p small .a p small .b
(4)
msmall mbig
p p. f
p p.i
mp
mt
mp
mt
o Q: How does varying the relationship between the two masses
effect the ratio of new to old momentum of the projectile?
 Say mt > mp
p p. f
mt m p
Pf/Pi
p p.i
mp
mt

-1
As long as the target mass is bigger than the projectile
mass, the bigger it is, the bigger (all be it, more negative),
the final momentum.
mtarget =
 mLimit
target of much lighter mass.
mprojectile
Q: What do you expect? Say you bounce ball off a
wall (a very heavy object indeed). How do the
ball’s initial and final momenta compare?
 Notice that, in the limit that the big mass is
much greater than the small mass, we have
p p. f
p p.i .
 In terms of the change in the small cart’s
2 p p.i . This is
momentum: p p p p. f p p.i
like bouncing a ball off a wall.
Demo: Collide with wall.
Demo: Bounce two carts off each other without any additional masses, to the
extent that they have equal masses, see one stop and the other continue on.
Magnitude of Momentum: Note that the more massive the target, the less
the magnitude of the projectile’s momentum changes.
CW
Final Momentum of Target. As for the final velocity of the big cart, plug
this solution back into Eq’n 2 and get
Physics 231
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Ch 10 Day 1
pt. f
p p.i
p p. f
pt. f
p p.i
p p.i
pt. f
p p.i
mp
mt
mp
mt
p p.i 1
mp
mt
mp
mt
2mt
m p mt
8.2.1 Heavy projectile, light target
If the target were lighter than the projectile, then, it would make sense to write
m p mt
p p. f p p.i
m p mt
If the big cart is significantly more massive, its final momentum is just slightly
less than its original momentum.
Pf/Pi
projectile
1
1
mtarget
/mprojectile
-1
Again, in the limit that the more massive object is quite more massive,
2m pl
p t . f p p.i
mt
mt v t . f
vt . f
m p v p.i
2mtl
mp
2v p.i
o Rather surprisingly, the small object undergoes a change in velocity of
2vp.b. and bounces off the more massive one at about twice the speed of
the more massive one.
o Change Frames to understand
 Imagine being in the frame that is moving along with the massive
cart initially, at speed vbig.b. Then you see the small cart coming at
you with velocity -vbig.b. Upon colliding, the massive cart hardly
changes course, but the small one, as we say in the previous case,
bounces off with nearly equal and opposite momentum, so now it
has velocity vbig.b relative to us, i.e., undergoes a change in speed
of 2vbig.b.
Physics 231
Ch 10 Day 1
2013 9
Class Work.
A ball of mass m1 hits a stationary target of mass m2 head-on. The total initial and final
kinetic energies are the same. Which of the following statements is false?
(1) If m1 = m2, the ball comes to rest after the collision.
(2) If m1 << m2, the momentum of the ball hardly changes.
(3) If m1 < m2, the ball bounces straight back.
(4) If m1 < m2, the ball bounces straight back with less kinetic energy than it had originally.
(5) If m1 >> m2, the ball keeps going without change of direction.
Which of the following is a property of all “elastic” collisions?
(1) The colliding objects interact through springs.
(2) The kinetic energy of one of the objects doesn’t change.
(3) The total kinetic energy is constant at all times -- before, during, and after the
collision.
(4) The total kinetic energy after the collision is equal to the total kinetic energy before
the collision.
(5) The elastic spring energy after the collision is greater than the elastic spring energy
before the collision.
Equipment
RW22
Lab carts and track (not air track) & cart weights (black bars)
Hover discs (make sure they’re charged up)
Laptop with 08_Rutherford_dist.py, and Scattering.exe
Ch. 8 Collisions: Exploring the Nucleus
Collision: A relatively brief and strong interaction.
Why focus on these? Ubiquitous and useful
What’s special about a collision? can ignore other interactions
8.3
Conservation of Momentum
Physics 231
Ch 10 Day 1
201310
o Demo: Perfectly Inelastic, call our carts, “p” and “t” for “projectile”
and “target.”







mt vt .i m p v p.i mt vt . f m p v p. f , vt . f v p. f v f



 mt vt .i m p v p.i mt m p v f


mt vt .b m p v p.b

vf
mt m p
8.3.1 Light projectile, heavy target
o Demo: Elastic. (collide carts and bounce magnetically). Now, in this
case, we still need a second equation if we want to solve for two
unknowns. What other relation holds through this collision?
Conservation of Energy.
rf


E system
Fnet.ext dr 0
ri

K t .cm
E t . int
K p.cm
E p. int
K t .cm
E t . int
K p.cm
E p. int
U t, p
0
0
o In general, both the center of mass and the internal energy of the system
can change. This is rather unfortunate for the internal energy is rather
difficult to track. In the special case that the internal energy is the same
after the collision as before (though it may change during the collision, as
in the case of a compressing squash ball), we say the collision is elastic,
for it bounces back with just as much kinetic energy as it had to begin
with.
o Let’s see how things work out in this case.
Kt
Kp 0
o
p t2.i
p t2. f
p 2p.i
2m t


p t .i p p.i
p 2p. f
2m p


p t . f p p. f
0




p t . f p t .i p p.i p p. f
o
o between these two equations, a general solution can be found for the two
final momenta.
o Let’s consider the simple case in which one of the carts is initially
stationary.
initial
final
o

p t .i

p p .i

p p. f
p t2. f
2mt
p 2p. f
2m p
0

pt. f
p 2p.i
0 (1)
Physics 231

p p.i

pt. f
201311
Ch 10 Day 1



p tf p p.i p p. f

p p. f
o
o being 1-D, we’re just dealing with x-components, so we can drop the
vector notation. If a term is positive, the motion is to the right, if it is
negative, the motion is to the left.
 xˆ : p t . f p p.i p p. f (2)
o Target: p p . f
o Predictions:
 Q: Would you expect the light cart to gain or loose forward
momentum through the collision?
Loose, since the other cart gains forward momentum, so
p p. f p p.i .
 Collision not built into equations: Note that, while you and I are
thinking about there being a collision between the “before” and
“after” shots, we’re not directly modeling the collision, there are
no terms in our math for the force of one object on the other or the
work done by one on the other. Instead, we’re invoking the fact
that, whether there’s a collision or not momentum and energy are
conserved.
With that in mind, I suggest to you that the math will lead
to two solutions: one for if there is a collision and one for if
there isn’t.
What should be the no-collision answer?
o p p . f = p p .i . So we better find this in our math.
o Algebra
CW (1 of 2): Plug Eq’n 2 into Eq’n 1 and get it into the form of A ps.f2 + B ps.f + C = 0
(2) -> (1):
p p.i
2
p p. f
p 2p. f
2mt
p 2p.i
p 2p.i
0
2m p
p 2p. f
2 p p.i p p. f
p 2p. f
mt
p 2p.i
p 2p. f
1
mp
0
mp
2 p p.i p p. f
p 2p. f
mt
p 2p. f
p 2p.i
1
mt
p 2p.i
mp
p p. f 2 p p.i
p 2p.i
0
1
mt
1
mt
0
Plug into the quadratic equation and simplify
m p mt
p p. f p p.i
m p mt
Compare with predictions.
o Indeed, there are two solutions (following from the +/- sign in the
quadratic equation).
Physics 231
Ch 10 Day 1
201312
o Choosing the + sign, gives simply that the small cart has the same
momentum after the collision as before. p p. f p p.i
o The other is p p. f
p p.i
mp
mt
mp
mt
which does indeed have the
final momentum being less than the initial momentum.
o Q: How does varying the relationship between the two masses
effect the ratio of new to old momentum of the projectile?
 Say mt > mp
p p. f
mt m p
p p.i
mp
mt

Pf/Pi
mtarget =
mprojectile
-1
As long as the target mass is bigger than the projectile
mass, the bigger it is, the bigger (all be it, more negative),
the final momentum.
 Limit of much lighter mass.
Q: What do you expect? Say you bounce ball off a
wall (a very heavy object indeed). How do the
ball’s initial and final momenta compare?
mtarget
 Notice that, in the limit that the big mass is
much greater than the small mass, we have
p p. f
p p.i .
 In terms of the change in the small cart’s
2 p p.i . This is
momentum: p p p p. f p p.i
like bouncing a ball off a wall.
Demo: Bounce two carts off each other without any additional masses, to the
extent that they have equal masses, see one stop and the other continue on.
Magnitude of Momentum: Note that the more massive the target, the less
the magnitude of the projectile’s momentum changes.
CW
Final Momentum of Target. As for the final velocity of the big cart, plug
this solution back into Eq’n 2 and get
pt. f
p p.i
p p. f
pt. f
p p.i
p p.i
pt. f
p p.i
mp
mt
mp
mt
2mt
m p mt
8.3.2 Heavy projectile, light target
p p.i 1
mp
mt
mp
mt
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Ch 10 Day 1
If the target were lighter than the projectile, then, it would make sense to write
m p mt
p p. f p p.i
m p mt
If the big cart is significantly more massive, its final momentum is just slightly
less than its original momentum.
Pf/Pi
projectile
1
1
mtarget
/mprojectile
-1
Again, in the limit that the more massive object is quite more massive,
2m pl
p t . f p p.i
mt
mt v t . f
vt . f
m p v p.i
2mtl
mp
2v p.i
o Rather surprisingly, the small object undergoes a change in velocity of
2vp.b. and bounces off the more massive one at about twice the speed of
the more massive one.