Transcript Momentum&ItsConservation

AP Physics C
 Wahl,
Michael. SparkNote on Conservation of
Momentum. 25 Oct. 2008
<http://www.sparknotes.com/physics/linear
momentum/conservationofmomentum>.
 Wahl, Michael. SparkNote on Collisions. 25
Oct. 2008
<http://www.sparknotes.com/physics/linear
momentum/collisions>.

Center of Mass - The point at which a given net
force acting on the system will produce the same
acceleration as if all the mass were
concentrated at that point.

Impulse - A force applied over a period of time.

Momentum - The product of an object's mass
and velocity.

Conservation of Momentum - The principle
stating that for any system with no external
forces acting on it, momentum will be
conserved.
 We
have been studying the mechanics of
single particles.
 We will now expand our study to systems of
several particles.
 The concept of the center of mass allows us
to describe the movement of a system of
particles by the movement of a single point.
 X-coordinate
of the center of mass:
xcm
mx


m
i i
i
 Y-coordinate
of the center of mass:
ycm
my


m
i
i
i
 Calculate
the center of mass of the following
system: A mass of 5 kg lies at x = 1, a mass of
3 kg lies at x = 4 and a mass of 2 kg lies at x
= 0.
 Calculate
the center of mass of the following
system: A mass of 10 kg lies at the point
(1,0), a mass of 2 kg lies at the point (2,1)
and a mass of 5 kg lies at the point (0,1).
 Velocity
of the center of mass:
vcm
mv


m
i i
i
 Acceleration
of the center of mass:
acm
ma


m
i i
i
 External
Net Force
Fext  Macm

Consider the system
from problem 2, but
now with forces acting
upon the system. On
the 10 kg mass, there is
a force of 10 N in the
positive x direction. On
the 2 kg mass, there is
a force of 5 N inclined
45o above horizontal.
Finally, on the 5 kg
mass, there is a force of
2 N in the negative y
direction. Find the
resultant acceleration
of the system.
 Two
masses, m1 and m2, m1 being larger, are
connected by a spring. They are placed on a
frictionless surface and separated so as to
stretch the spring. They are then released
from rest. In what direction does the system
travel?
A
50 kg man stands at the edge of a raft of
mass 10 kg that is 10 meters long. The edge
of the raft is against the shore of the lake.
The man walks toward the shore, the entire
length of the raft. How far from the shore
does the raft move?
 We
shall define
this concept, force
applied over a
time period, as
impulse. Impulse
can be defined
mathematically,
and is denoted by
J:
J = FΔt
 It
is a vector
quantity.
 Can we predict the
motion of an
object?
J = FΔt = (ma)Δt
v
J = m t Δt
J = mΔv = Δ(mv) =
mvf - mvo
 What
is the impulse of a force of 10 N acting
on a ball for 2 seconds?
 The
ball has a mass of 2 kg and is initially at
rest. What is the velocity of the ball after
the force has acted on it?
 From
our equation relating impulse and
velocity, it is logical to define the momentum
of a single particle, denoted by the vector p,
as such:
p = mv
 Again,
momentum is a vector quantity,
pointing in the direction of the velocity of
the object. From this definition we can
generate two very important equations, the
first relating force and acceleration, the
second relating impulse and momentum.
A
particle has linear momentum of 10 kgm/s, and a kinetic energy of 25 J. What is
the mass of the particle?
 If
we take a time derivative of our
momentum expression we get the following
equation:
dp
dv
m
 ma  F
dt
dt
dp
F 
dt
 Newton’s
Second Law of Motion
J  F t
J  mat
J  m (v f  v i )
J  mv f  mvi
J  p f  pi
J  p
A
2 kg bouncy ball is dropped from a height
of 10 meters, hits the floor and returns to its
original height. What was the change in
momentum of the ball upon impact with the
floor? What was the impulse provided by the
floor?
A
ball of 2 kg is thrown straight up into the
air with an initial velocity of 10 m/s. Using
the impulse-momentum theorem, calculate
the time of flight of the ball.
 Recall:


Impulse is a change in momentum.
Work done is a change in kinetic energy.
p  mv
1
2
K  mv
2
1
1 p2
 K  pv 
2
2 m
 Suppose
we have a system of N particles,
with masses m1, m2,…, mn. Assuming no mass
enters or leaves the system, we define the
total momentum of the system as the vector
sum of the individual momentum of the
particles:
P
= p1 + p2 +
...
+ pn = m1v1 + m2v2 +
...
+ mnvn
 Recall:
1
vcm =
(m1v1 + m2v2 + ... + mnvn)
M
 P  Mvcm
P  Mvcm
dvcm
dP
M
 Macm  Fext
dt
dt
dP
 Fext 
dt
 If
the net external force is zero, then the
total momentum of the system is constant.
 That
is,
p p
i
 Remember
quantity.
f
that momentum is a vector
A
60 kg man standing on a stationary 40 kg
boat throws a .2 kg baseball with a velocity
of 50 m/s. With what speed does the boat
move after the man throws the ball?
A
.05 kg bullet is fired at a velocity of 500
m/s, and embeds itself in a block of mass 4
kg, initially at rest and on a frictionless
surface. What is the final velocity of the
block?

An object at rest explodes into
three pieces. Two, each of the
same mass, fly off in different
directions with velocity 50 m/s
and 100 m/s, respectively. A
third piece goes off in the
negative y-direction is also
formed in the explosion, and
has twice the mass of the first
two pieces. Determine the
direction of the second particle
and the speed of the third
particle. Let θ1 = 65o.
θ2
Θ1 = 65o
v3
A
spaceship moving at 1000 m/s fires a
missile of mass 1000 kg at a speed of 10000
m/s. What is the mass of the spaceship it
slows down to a velocity of 910 m/s?

Collision - The brief direct contact between
two bodies that results in a net impulse on each
body.

Elastic Collision - Any collision in which kinetic
energy is conserved.

Inelastic Collision - Any collision in which
kinetic energy is not conserved.

Completely Inelastic Collision - Any collision in
which the two bodies stick together.
 Why
are these collisions special? We know
with all collisions that momentum is
conserved. If two particles collide we can
use the following equation:
m1v1o + m2v2o = m1v1f + m2v2f
 However, we also know that, because the
collision is elastic, kinetic energy is
conserved. For the same situation we can use
the following equation:
1 m1v1o2 + 1 m2v2o2 = 1 m1v1f2 + 1 m2v2f2
2
2
2
2
 Two
balls, each with mass 2 kg, and
velocities of 2 m/s and 3 m/s collide head
on. Their final velocities are 2 m/s and 1
m/s, respectively. Is this collision elastic or
inelastic?
 Two
balls of mass m1 and m2, with velocities
v1 and v2 collide head on. Is there any way
for both balls to have zero velocity after the
collision? If so, find the conditions under
which this can occur.
 Two
balls with equal masses, m, and equal
speed, v, engage in a head on elastic
collision. What is the final velocity of each
ball, in terms of m and v?
 One
pool ball traveling with a velocity of 5
m/s hits another ball of the same mass,
which is stationary. The collision is head on
and elastic. Find the final velocities of both
balls.
 So
what if kinetic energy is not conserved?
Our knowledge of such situations is more
limited, since we no longer know what the
kinetic energy is after the collision. However,
even though kinetic energy is not conserved,
momentum will always be conserved.
Consider the case in which two particles collide,
and actually physically stick together. In this
case, called a completely inelastic collision we
only need to solve for one final velocity, and the
conservation of momentum equation is enough
to predict the outcome of the collision. The two
particles in a completely inelastic collision must
move at the same final velocity, so our linear
momentum equation becomes:
m1v1o + m2v2o = m1vf + m2vf
Thus
m1v1o + m2v2o = Mvf

A
car of 500 kg, traveling at 30 m/s rear ends
another car of 600 kg, traveling at 20 m/s. in
the same direction The collision is great
enough that the two cars stick together after
they collide. How fast will both cars be going
after the collision?
 Two
balls of equal masses move toward each
other on the x-axis. When they collide, each
ball ricochets 90 degrees, such that both
balls are moving away from each other on
the y-axis. What can be said about the final
velocity of each ball?
 Two
pool balls traveling in opposite
directions collide. One ball travels off at an
angle θ to its original velocity, as shown
below. Is there any possible way for the
second ball to be completely stopped by this
collision? If so state the conditions under
which this could occur.
 Two
objects are traveling perpendicular to
each other, one moving at 2 m/s with a mass
of 5 kg, and one moving at 3 m/s with a mass
of 10 kg, as shown below. They collide and
stick together. What is the magnitude and
direction of the velocity of both objects?

A common pool shot
involves hitting a ball into
a pocket from an angle.
Shown below, the cue ball
hits a stationary ball at an
angle of 45o, such that it
goes into the corner
pocket with a speed of 2
m/s. Both balls have a
mass of .5 kg, and the cue
ball is traveling at 4 m/s
before the collision.
Calculate the angle with
which the cue is deflected
by the collision.