Download Momentum

11/6/2014
Momentum
Momentum and Collisions
• Momentum: p = mv
• Units are kg(m/s): no derived units
• A vector quantity: same direction as velocity
Phy 114
v=2m/s
Linear momentum
p= 3 kg (2m/s)
Important points about linear momentum
• Linear momentum is a vector quantity; it is important to
consider the direction in which the colliding objects are
moving before and after the collision.
• Momentum depends on the velocity of the object, and
the velocity depends on the choice of the reference
frame. Different observers will measure different
momenta for the same object.
• To establish that momentum is a conserved quantity, we
need to ensure that the momentum of a system changes
in a predictable way for systems that are not isolated.
© 2014 Pearson Education, Inc.
Momentum constancy of an isolated system
• For a system with more than two objects, we simply
include a term on each side of the equation for each
object in the system.
© 2014 Pearson Education, Inc.
© 2014 Pearson Education, Inc.
Example 5.1: Two rollerbladers
• Jen (50 kg) and David
(75 kg), both on
rollerblades, push off
each other abruptly.
Each person coasts
backward at
approximately constant
speed. During a certain
time interval, Jen
travels 3.0 m.
• How far does David
travel during that same
time interval?
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From Newton’s 2nd
ΣF = ma
∆v
ΣF = m
∆t
ΣF ( ∆t ) = m(∆v)
ΣF ( ∆t ) = mv2 − mv1
• Impulse: J=ΣF Δt
• Units are same as
momentum
• Impulse causes a
change in momentum
• What happens to ΣF
with
Impulse due to a force exerted on a single
object
• We need a way to account for change in momentum when
the net external force on a system is not zero
• A relationship can be derived from Newton's laws and
kinematics:
– Small Δt?
– Large Δt?
ΣF ( ∆t ) = ∆p
© 2014 Pearson Education, Inc.
• Elastic
– Momentum is Conserved
– Kinetic Energy is Conserved
• Inelastic
– Momentum is Conserved
– Kinetic Energy is NOT Conserved
Impulse: The product of the external force
exerted on an object and the time interval
Graphs
A force is acting in the
x-direction on a 4.00kg particle. The force
varies in time and is
shown on the graph.
Find:
• the impulse
• the final velocity if
initially at rest
• the final velocity if initial
velocity is –2.00 m/s.
Force vs Time for a Moving Cart
4.5
4
3.5
3
Force (N)
Collisions
2.5
2
1.5
1
0.5
0
0
5
10
Time (sec)
15
20
Impulse-momentum equation for a single
object
• If the magnitude of the force changes during the
time interval considered in the process, we use the
average force.
© 2014 Pearson Education, Inc.
© 2014 Pearson Education, Inc.
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Example 5.2: Abrupt stop in a car
• A 60-kg person is traveling in a car that is
moving at 16 m/s with respect to the ground
when the car hits a barrier. The person is not
wearing a seat belt, but is stopped by an air
bag in a time interval of 0.20 s.
• Determine the average force that the air bag
exerts on the person while stopping him.
Problem Solving
Multiple Plans
• Draw a MD
– What new motion
equation can you add?
• Draw a FBD
– What new equation do
we have that relates
force and motion?
• A 0.500-kg football is
thrown toward the east
with a speed of 15.0
m/s. A stationary
receiver catches the ball
and brings it to rest in
0.020 0 s.
• What is the average
force exerted on the
receiver?
Ans: F avg = 375 N
© 2014 Pearson Education, Inc.
Problem Solving: …..this way
Problem Solving: solve this way or…
Multiple Plans
• Draw a MD
Multiple Plans
• Draw a MD
– What new motion
equation can you add?
• Draw a FBD
x=0
t=0
v=15 m/s
x=
t=
0
a=
– What new equation do
we have that relates
force and motion?
– What new motion
equation can you add?
• Draw a FBD
– What new equation do
we have that relates
force and motion?
Impulse --- Momentum
equation
− nhand on football ∆t = m∆v
− nhand on football = ma x
Ans: F avg = 375 N
Ans: F avg = 375 N
Problem in 2-D
Collision
• Momentum always
conserved
– Do this first
– Remember it is a vector
• Kinetic Energy conserved in
elastic collision
– Check this next
A billiard ball moving at 5.00
m/s strikes a stationary
ball of the same mass.
After the collision, the first
ball moves at 4.2 m/s at an
angle of 30° with respect
to the original line of
motion.
• Find the velocity
(magnitude and direction)
of the second ball after
collision.
• Was the collision inelastic or
elastic?
Conservation of Momentum
Motion
Momentum
Momentum (x)
Momentum (y)
Mass 1: Initial
5 m/s
Mass 2: Initial
0 m/s
Mass 1: Final
4.2 m/s at 30°
Mass 2: Final
Conserve
Momentum
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Conservation of Momentum
The generalized impulse-momentum principle
Assume 0.5 kg balls
Motion
Momentum
Momentum (x)
Momentum (y)
Mass 1: Initial
5 m/s
2.5 kgm/s to the
right
+2.5 kgm/s
0
Mass 2: Initial
0 m/s
0
0
0
Mass 1: Final
4.2 m/s at 30°
2.1 kgm/s at 30°
+2.1 cos30° =
1.82 kgm/s
+2.1 sin30° =
1.05 kgm/s
Mass 2: Final
unknown
Conserve
Momentum
px
py
2.5 + 0 = 1.82 + px
Px = +0.68
0 = 1.05 + py
Py=-1.05
Use pythagorean equation to get p=1.25 kgm/s @57 degrees below +x axis.
Use p=mv to get final vel of Mass 2: v=2.5 m/s @57 degrees below +x axis
© 2014 Pearson Education, Inc.
Impulse-momentum bar charts
Example 5.5: Stopping the fall of a movie stunt
diver
• The record for the highest movie stunt fall
without a parachute is 71 m, held by 80-kg
A. J. Bakunas. His fall was stopped by a large
air cushion, into which he sank about 4.0 m.
His speed was approximately 36 m/s when he
reached the top of the air cushion.
• Estimate the average force that the cushion
exerted on this stunt diver's body while
stopping him.
© 2014 Pearson Education, Inc.
© 2014 Pearson Education, Inc.
Determining the stopping time interval from the
stopping distance
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