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Final
Momentum – chapter 9
Finish momentum & review for exam
Exam 2 (5 – 8)
Rotation
Gravity
Waves & Sound
Temperature
Finish up & Review
Review (last day of classes 12/02)
12/06 @ 2:00 – 4:30 for section 2
1
Racing Balls
2
When the two balls are launched from one end of the track with the
same initial velocity, what will happen:
1) Ball #1 on the straight track arrives at the other end first
2) Ball #2 on the track with the dip arrives at the other end first
3) the race is a tie - both balls reach the other end at the same time?
Chapter 9
Linear Momentum and Collisions
Linear momentum is defined as:
p = mv
Momentum is given by mass times velocity.
Momentum is a vector.
The units of momentum are (no special unit):
[p] = kg·m/s
Momentum
Momentum = p = mv (kg.m/s)
Since p is a vector, we can also consider the
components of momentum:
px = mvx
py = mvy
Note: momentum is “large” if m and/or v is large.
• Name an object with large momentum but small
velocity.
• Name an object with large momentum but small mass
Is there a relationship between F and P?
Another way of writing Newton’s Second Law is
F = Dp/Dt= rate of change of momentum
This form is valid even if the mass is changing.
This form is valid even in Relativity and Quantum
Mechanics.
Impulse
We can rewrite F = Dp/Dt as:
FDt = Dp
I = FDt is known as the impulse.
The impulse of the force acting on an object equals the
change in the momentum of that object.
Exercise: Show that impulse and momentum have the
same units.
Example
A 0.3-kg hockey puck moves on frictionless ice at 8 m/s toward
the wall. It bounces back away from the wall at 5 m/s. The
puck is in contact with the wall for 0.2 s.
(a) What is the change in momentum of the hockey puck during
the bounce? [3.9 kg.m/s]
(b) What is the impulse on the hockey puck during the bounce?
[3.9 kg.m/s]
(c) What is the average force of the wall on the hockey puck
during the bounce? [19.5 N]
If there are no external forces on a system,
then the total momentum of that system is
constant. This is known as:
The Principle of
Conservation
of
Momentum
In that case, pi = pf.
Internal vs. External Forces
Here the system is just the box and table.
Any forces between those two objects are
internal. Example: The normal forces
between the table and the box are internal
forces. Internal forces on the system sum
to zero.
system
External forces do not necessarily sum to zero.
Something outside the circle is pushing or pulling
something inside the circle.
Example
Two skaters are standing on frictionless ice. Skater A has a mass of 50
kg and skater B has a mass of 80 kg. Skater A pushes Skater B for
0.25 s, causing Skater B to move away at 10 m/s.
(a) What is the velocity of Skater A after he pushes Skater B? [16 m/s]
(b) What is the change in momentum of Skater B? [800 kg.m/s]
(c) What is the average force exerted on Skater B by Skater A during
the 0.2 s push? [4000 N]
(d) What is the change in momentum of Skater A? [800 kg.m/s]
(e) What is the average force exerted on Skater A by Skater B during
the 0.2 s? [4000 N]
Problem #4 (HW 10)
A honeybee with a mass of 0.175 g lands on one end of a
popsicle stick. After sitting at rest for a moment, the bee
runs toward the other end with a velocity 1.55 cm/s relative
to the still water. What is the speed of the 4.65 g stick
relative to the water? (Assume the bee's motion is in the
negative direction.)
[0.583e-3] m/s = 0.583 mm/s
Collisions
In general, a “collision” is an interaction in which
• two objects strike one another
• the net external impulse is zero or negligibly
small (momentum is conserved)
Examples: car crash; billiard balls
Collisions can involve more than 2 objects
From the conservation of momentum:
v1,i
v1,f
v2,i
v2,f
pi = pf
m1v1,i + m2v2,i = m1v1,f + m2v2,f
What about conservation of energy?
We said earlier that the total energy of an
isolated system is conserved, but the total
kinetic energy may change.
• elastic collisions: K is conserved
• inelastic collisions: K is not conserved
• perfectly inelastic: objects stick together
after colliding
Perfectly Inelastic Collisions
After a perfectly inelastic collision the two
objects stick together and move with the
same final velocity:
pi = pf
m1 v1,i + m2 v2,i = (m1+ m2)vf
This gives the maximum possible loss of kinetic energy.
In non-relativistic collisions, the total mass is conserved
Problem #3 (HW 10)
A car with a mass of 950 kg and an initial speed of v1 = 17.8 m/s approaches an
intersection, as shown in the figure. A 1300 kg minivan traveling northward is
heading for the same intersection. The car and minivan collide and stick together.
If the direction of the wreckage after the collision is 37.0° above the x axis, what
is the initial speed of the minivan and the final speed of the wreckage?
[9.8] m/s , [9.41] m/s
Elastic Collisions
Kinetic energy is conserved in addition to
momentum:
m1v1,i  m2 v 2,i  m1v1, f  m2 v 2, f
pi = pf
Ki = Kf
1
1
1
1
2
2
2
m1v1,i  m2v2,i  m1v1, f  m2v22, f
2
2
2
2
Lots of variables, keep track!
Elastic Collisions in 1-dimension
Kinetic energy is conserved in addition to
momentum:
m v m v m v m v
m v  v   m v  v 
p i = pf
1 1,i
1
Ki = Kf

2 2 ,i
1,i
1, f
1 1, f
2 2, f
2
2 ,i
2, f
1
1
1
1
m1v12,i  m2 v22,i  m1v12, f  m2 v22, f
2
2
2
2
m1 v12,i  v12, f  m2 v22, f  v22,i
m1 v1,i  v1, f

v
1,i
v
1,i
 v1, f
 
  m v
2
2, f
 v2 ,i
Divide :
 
 v1, f  v2 , f  v2 ,i

v
2, f
 v2 , i


v1,i  v2 ,i  v2 , f  v1, f
v1,i  v1, f  v2 ,i  v2 , f
Relative velocity of approach before collision = relative velocity of separation after collision
Center of Mass
The center of mass (CM) of an object or a group of
objects (system) is the “average” location of the mass in
the system. The system behaves as if all of its mass were
concentrated at the center of mass.
The center of mass is not always located on
the object.
Where is the CM for
this object?
Calculating the Center of Mass
For two objects:
X cm
m1x1  m2 x2 m1x1  m2 x2


m1  m2
M
In general, X coordinate of center of mass:
X cm
m1x1  m2 x2  ... mx


m1  m2  ...
M
In general, Y coordinate of center of mass:
Ycm
m1 y1  m2 y2  ... my


m1  m2  ...
M
Exercise
Find the x coordinate of the center of mass of
the bricks shown in the Figure.
x = 0.92 L