Download Lecture 17 - Wayne State University Physics and Astronomy

General Physics (PHY 2130)
Lecture 17
•  Momentum
  Momentum and impulse
  Conservation of momentum
http://www.physics.wayne.edu/~apetrov/PHY2130/
Lightning Review
Last lecture:
1.  Work and energy:
  elastic potential energy
  momentum
Review Problem: An 8.0 kg crate, initially at rest, slides down a slope as in the
15 m
picture. The work of the force of friction is 300 J. What is the speed of the crate at
the bottom of the slope?
Review problem
An 8.0 kg crate, initially at rest, slides down a slope as in the picture. The work of
the force of friction is 300 J. What is the speed of the crate at the bottom of the
slope?
Idea: use energy considerations with nonconservative forces present
15 m
Wnc = ΔE = E f − Ei
At the top:
Ei = mgh + 0
mv 2
At the bottom: E f = 0 +
2
Thus:
mv 2
Wnc =
− mgh
2
or
2
v=
( mgh + Wnc ) = 19.2 m / s
m
Recall: Momentum
•  From Newton’s laws: force must be present to change an
object’s velocity (speed and/or direction)
  Wish to consider effects of collisions and corresponding
change in velocity
Golf ball initially at rest, so
some of the KE of club
transferred to provide motion
of golf ball and its change in
velocity
  Method to describe is to use concept of linear momentum
Linear momentum = product of mass
scalar
× velocity
vector
Momentum
p = mv
•  Vector quantity, the direction of the momentum is the
same as the velocity’s
•  Applies to two-dimensional motion as well
p x = mv x and p y = mv y
Size of momentum: depends upon mass
depends upon velocity
Impulse
•  In order to change the momentum of an object
(say, golf ball), a force must be applied
•  The time rate of change of momentum of an object
is equal to the net force acting on it
• 
 


 m(v f − vi ) Δ p
F net = ma =
=
Δt
Δt
or :
 
Δ p = F net Δt
•  Gives an alternative statement of Newton’s second law
•  (F Δt) is called the impulse
•  Impulse is a vector quantity, the direction is the same as
the direction of the force
Graphical Interpretation of Impulse
•  Usually force is not constant,
but time-dependent
impulse = ∑ Fi Δti = area under F (t ) curve
Δti
•  If the force is not constant,
use the average force applied
•  The average force can be
thought of as the constant
force that would give the same
impulse to the object in the
time interval as the actual
time-varying force gives in the
interval
If force is constant: impulse = F Δt
Example: Impulse Applied to Auto Collisions
•  The most important factor is the collision time or
the time it takes the person to come to a rest
•  This will reduce the chance of dying in a car crash
•  Ways to increase the time
•  Seat belts
•  Air bags
  The
air bag increases the time of the collision and
absorbs some of the energy from the body
9
Example: A car of mass 1500 kg collides with a wall and rebounds as shown. If the
collision lasts for 0.150 s, Find (a) the impulse delivered to the car due to the collision
and (b) the magnitude and direction of the average force exerted on the car.
–  Assume force exerted by wall is large
compared with other forces
–  Gravitational and normal forces are
perpendicular and so do not effect the
horizontal momentum
–  Can apply impulse approximation
(a) Impulse delivered to the car
 

= Δp = p f − pi
= (1500kg × 2.60m / s) − (1500kg × (−15.0m / s))
= 2.64 × 10 4 kg .m / s
(b) The average force exerted on the car


Δp 2.64 × 10 4 kg .m / s
Fav =
=
= +1.76 × 10 5 N
Δt
0.150s
ConcepTest
Suppose a ping-pong ball and a bowling ball are rolling toward
you. Both have the same momentum, and you exert the same
force to stop each. How do the time intervals to stop them
compare?
1. It takes less time to stop the ping-pong ball.
2. Both take the same time.
3. It takes more time to stop the ping-pong ball.
ConcepTest
Suppose a ping-pong ball and a bowling ball are rolling toward
you. Both have the same momentum, and you exert the same
force to stop each. How do the time intervals to stop them
compare?
1. It takes less time to stop the ping-pong ball.
2. Both take the same time.  
3. It takes more time to stop the ping-pong ball.
Note: Because force equals the time rate of change of
momentum, the two balls loose momentum at the same
rate. If both balls initially had the same momenta, it
takes the same amount of time to stop them.
Problem: Teeing Off
A 50-g golf ball at rest is hit by “Big
Bertha” club with 500-g mass.
After the collision, golf leaves
with velocity of 50 m/s.
a)  Find impulse imparted to ball
b)  Assuming club in contact with
ball for 0.5 ms, find average force
acting on golf ball
Problem: teeing off
1. Use impulse-momentum relation:
Given:
impulse = Δp = mv f − mvi
= (0.050 kg )(50 m s ) − 0
mass: m=50 g
= 0.050 kg
velocity: v=50 m/s
Find:
impulse=?
Faverage=?
= 2.50 kg ⋅ m s
 
2. Having found impulse, find the average
force from the definition of impulse:
Δp 2.50 kg ⋅ m s
=
Δt
0.5 ×10 −3 s
= 5.00 ×10 3 N
Δp = F ⋅ Δt , thus F =
 
Note: according to Newton’s 3rd law, that is also a reaction force to club hitting the ball:
of club
F ⋅ Δt = − F R ⋅ Δt , or
(
)
mv f − mv i = − M V f − M V i , or
mv f + M V f = mv i + M V i
CONSERVATION OF MOMENTUM
Conservation of Momentum
•  Definition: an isolated system is the one that has
no external forces acting on it
Momentum in an isolated system in which a
collision occurs is conserved (regardless of the
nature of the forces between the objects)
•  A collision may be the result of physical contact
between two objects
•  “Contact” may also arise from the electrostatic
interactions of the electrons in the surface atoms of the
bodies
Conservation of Momentum
The principle of conservation of momentum states when
no external forces act on a system consisting of two
objects that collide with each other, the total momentum of
the system before the collision is equal to the total
momentum of the system after the collision
Conservation of Momentum
•  Mathematically:
m1 v1i + m2 v 2i = m1 v1 f + m2 v 2 f
•  Momentum is conserved for the system of objects
•  The system includes all the objects interacting with each other
•  Assumes only internal forces are acting during the collision
•  Can be generalized to any number of objects
Problem: Teeing Off (cont.)
Let’s go back to our golf ball and club problem:
Ball : Δp = 2.50 kg ⋅ m s , m = 50 gramm
Δv = 50 m s
(
(
)
)
Club : m v f − v i = −2.50 kg ⋅ m s , so
− 2.50 kg ⋅ m s
v f − vi =
= −5m s
0.5 kg
factor of 10 times smaller
18
More on Conservation of Momentum
v1i
m1>m2
v2i
m1
m2
A short time later the masses
collide.
m1
m2
What happens?
19
During the interaction:
N1
N2
y
F21
F12
w1
∑F
∑F
y
= N1 − w1 = 0
x
= − F21 = m1a1
w2
∑F
∑F
y
= N 2 − w2 = 0
x
= F12 = m2 a2
There is no net external
force on either mass.
x
20
Conservation of Momentum


F21 = − F12


F21Δt = − F12 Δt


Δp1 = −Δp 2




p1 f − p1i = p 2 f − p 2i




p1i + p 2i = p1 f + p 2 f
If net external force acting on a system is zero, then the momentum of the
system is conserved



If ∑ Fext = 0, pi = p f
21
Example: A rifle has a mass of 4.5 kg and it fires a bullet of 10.0 grams at a
muzzle speed of 820 m/s. What is the recoil speed of the rifle as the bullet
leaves the barrel?
Given:
mr = 4.5 kg
mb = 10.0 g
=0.01 kg
vb = 820 m/s
vir = vib = 0 m/s
Find:
vr = ?
Idea: as long as the rifle is horizontal, there will
be no net external force acting on the rifle-bullet
system and momentum will be conserved.
pi = p f
0 = mb vb + mr vr
" 0.01 kg %
mb
vr = − vb = − $
' 820 m/s
mr
# 4.5 kg &
= −1.82 m/s
ConcepTest
Suppose a person jumps on the surface of Earth. The Earth
1. will not move at all
2. will recoil in the opposite direction with tiny velocity
3. might recoil, but there is not enough information
provided to see if that could happened
ConcepTest
Suppose a person jumps on the surface of Earth. The Earth
1. will not move at all
2. will recoil in the opposite direction with tiny velocity  
3. might recoil, but there is not enough information
provided to see if that could happened
Note: momentum is conserved. Let’s estimate Earth’s velocity
after a jump by a 80-kg person. Suppose that initial speed of
the jump is 4 m/s, then:
Person : Δp = 320 kg ⋅ m s
Earth : Δp = M EarthVEarth = −320 kg ⋅ m s , so
VEarth =
− 320 kg ⋅ m s
− 23
=
−
5
.
3
×
10
m s
6 ×10 24 kg
tiny negligible velocity, in opposite direction
24
Do you know why police use skid
marks to find velocities of vehicles
before collision???
See page 246
Rocket Propulsion
•  The basic equation for rocket propulsion is:
⎛ Mi
v f − v i = v e ln⎜⎜
⎝ Mf
⎞
⎟⎟
⎠
•  Mi is the initial mass of the rocket plus fuel
•  Mf is the final mass of the rocket plus any remaining fuel
•  The speed of the rocket is proportional to the exhaust speed
Thrust of a Rocket
•  The thrust is the force exerted on the rocket
by the ejected exhaust gases
•  The instantaneous thrust is given by
Δv
ΔM
Ma = M
= ve
Δt
Δt
•  The thrust increases as the exhaust speed
increases and as the burn rate (ΔM/Δt) increases