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Collisions
Students
•  Midterm #2 answer key and full solutions are on D2L, as are
your scores.
•  Clicker scores have also been updated.
•  Your estimated class grade will be added by 5pm today..
•  Midterm #2 was
clearly challenging. 80
The average score
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was 59 with a
standard deviation 60
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of 16.5. I try to
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have a mix of
easier and harder
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questions. It
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seems like the mix 10
was tilted toward
0
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the difficult side.
Exam scores
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One of the challenging problems
•  33% got this problem right:
•  Be sure you understand what is going on here. Feel free
to discuss with me after class or set up another
2
appointment.
One of the challenging problems
•  25% got this problem right:
•  Be sure you understand what is going on here. Feel free
to discuss with me after class or set up another
appointment.
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Thinking about momentum
What happens when you drop a tennis ball?
While falling, gravity is the only force acting so velocity
increases in the downward direction (ay = -g).
When it hits the floor there is a normal force. What do we know
about the normal force? It’s much larger than gravity and upward
After bouncing, the velocity is in the opposite direction.
What happens to the momentum of the tennis ball?
Since momentum is mass times velocity, it is similar to the
velocity. Gravity increases momentum and then the normal
force changes it from down to up in a very short time.
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What happens during the collision?
So exactly what happens when
the tennis ball hits the ground
(or a tennis racket)?
The collision time is very short
The force increases rapidly and
then decreases back to zero
The force versus time graph might
look something like what is shown:
F
Reality is likely more complicated.
t
Using momentum ideas we can avoid
the details of the forces involved.
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Force versus time graphs for bouncing balls
For tennis ball,
collision takes 6/1000
of a second with a
maximum force of 85
N compared to a
weight of 0.55 N.
For golf ball, collision
takes 1/1000 of a
second with a
maximum force of
220 N compared to a
weight of 0.45 N
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Impulse-momentum theorem
For a constant force, the
impulse is defined as
! !
!
J = Fnet ⋅ Δt = Fnet ⋅ (t2 − t1 )
If we consider a constant
interval
then
!force over a finite time
!
!
!
dp
Δp
nd
Newton’s 2 law Fnet =
can be written Fnet =
dt
Δt
! !
and therefore Δp = Fnet ⋅ Δt
This gives us the impulse–momentum theorem
!
! ! !
J = Δp = p2 − p1
The change in momentum of a particle over a time interval is
given by the impulse of the net force over the time interval.
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Clicker question 1
Set frequency to BA
Consider two carts, of masses m and 2m, at rest on an air track
(no friction). You push each cart for 3 s, exerting equal force on
each. At the end, the momentum of the light cart is…
A.  ¼ the momentum of the heavy cart
B.  ½ the momentum of the heavy cart
C.  equal to the momentum of the heavy cart
D.  twice the momentum of the heavy cart
E.  four times the momentum of the heavy cart
By the impulse-momentum theorem,
! the
! change in momentum
is given by the impulse which is J = Fnet ⋅ Δt . The same force
is applied over the same time, so the same momentum
change must occur.
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Clicker question 2
Set frequency to BA
A ball bounces off the !floor as shown. The direction of the
impulse of the ball Δp is ...
A.  Straight up ↑
B.  Straight down ↓
C.  Straight right →
D.  Straight left ←
!
E.  At an angle
−p
( )
!
! ! !
Vector subtraction: Δp = p2 − p1 = Δ p !
The horizontal component of momentum (px)
does not change, while the vertical component
goes from down to up so the change is up.
1
p2
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Getting impulse from force vs time graph
Can get impulse from the area under the force vs time curve.
! !
J = F ⋅ Δt
One rectangle is a small force over F
a long time and the other rectangle
is a large force over a small time.
Both have the same area and so
give the same impulse
t
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Conservation of momentum
Consider two objects, A and B, which interact with
each other but there are no other forces acting
By Newton’s 3rd law, the force A exerts on B is
equal and opposite to the force B exerts on A.
Since the forces are equal and opposite and
the time over which the forces act must be
equal, the impulse is also equal and opposite.
Therefore, the change in momentum of B by the
force A exerts on B is equal and opposite to the
change in momentum of A by the force B exerts on A
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Conservation of momentum
We can think of the two objects A and B as a system.
Forces between A and B can change the momentum
of A and B but the net change of momentum of the
system (due to these forces) is 0
Conservation of momentum: If there is no net (external)
force on a system, the total momentum is constant.
So if we know the total momentum of A and B at any point,
then we know the total momentum of A and B at any other
point even if the individual A and B momenta have changed!
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Clicker question 3
Set frequency to BA
Consider two carts, of masses m and 2m, at rest on an air track
and connected by a compressed spring. After releasing the two
carts, the spring expands to its equilibrium point and drops straight
down. What is the total momentum of the system composed of
the two carts after they are released?
BEFORE
A. 0
B. mv to the left
C. mv to the right
D. 2mv to the left
E. Impossible to tell
AFTER
The spring simply allows the
two carts to act on each other
so there is no outside force.
Therefore momentum is conserved and is still 0
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