Transcript Momentum 2013

nfl football momentum
Momentum is a commonly used term in sports.
A team that has the momentum is on the move
and is going to take some effort to stop.
Momentum as a physics term; refers to the
quantity of motion that an object has. A sports
team which is "on the move" has the
momentum. If an object is in motion (on the
move) then it has momentum.
An object’s momentum will change
if its mass and/or velocity changes.
Most common… a change in velocity.
What is a change in velocity called?
Acceleration
a = Vf -Vo
t
According to Newton’s laws,
a net force causes an object to accelerate,
or change its velocity.
IMPULSE
F
Dt
Impulse J is a force F
acting for a small time
interval Dt.
Impulse:
J = F Dt
Example 1: The face of a golf club exerts an
average force of 4000 N for 0.002 s. What is the
impulse imparted to the ball?
Impulse:
J = F Dt
F
J = (4000 N)(0.002 s)
J = 8.00 Ns
Dt
The unit for impulse is the Newton-second (N·s)
Impulse from a Varying Force
Normally, a force acting for a short interval is
not constant. It may be large initially and then
play off to zero as shown in the graph.
F
In the absence of calculus, we
use the average force Favg.
time, t
J  Favg Dt
Unless told otherwise, treat forces as average forces
Impulse Changes Velocity
Consider a mallet hitting a ball:
F  ma; a 
v f  vo
 v f  v0 
F  m

 Dt 
F
Dt
F Dt  mv f  mvo
Impulse = Change in “mv”
Momentum Defined
Momentum ρ is defined as the product of mass
and velocity, mv. Units: kg m/s
p = mv
m = 1000 kg
v = 16 m/s
ρ = (1000 kg)(16 m/s)
ρ = 16,000 kg m/s
Impulse & Momentum
Impulse = Change in momentum
F Dt = mvf - mvo
F
Dt
mv
A force F acting on a ball
for a time Dt increases its
momentum mv.
Example 2: A 50-g golf ball leaves the face of
the club at 20 m/s. If the club is in contact for
0.002 s, what average force acted on the ball?
Given: m = 0.05 kg; vo = 0;
+
F
Dt
mv
Dt = 0.002 s; vf = 20 m/s
Choose right as positive.
0
F Dt = mvf - mvo
F (0.002 s) = (0.05 kg)(20 m/s)
Average Force:
F = 500 N
A 1000 kg car moving at 30 m/s (p = 30,000 kg m/s)
can be stopped by
30,000 N of force acting for 1.0 s (a crash!)
or
by 3000 N of force acting for 10.0 s (normal stop)
http://www.youtube.com/watch?v=vtj6
Th6wb8g
http://www.youtube.com/watch?v=Z5BlObat1pQ
Another applications of impulse
Contact time is reduced if arm's
deceleration is kept as small as
possible.
This is done by using "followthrough", which means to
continue to push during the
contact period.
So to summarize…
To minimize the effect of the force on an object involved
in a collision, the time must be increased.
To maximize the effect of the force on an object
involved in a collision, the time must be decreased.
BUT the change in momentum is the SAME either way!
Vector Nature of Momentum
Consider the change in momentum of a
ball that is dropped onto a rigid plate:
+
vf
vo
A 2-kg ball strikes the plate with
a speed of 20 m/s and rebounds
with a speed of 15 m/s. What is
the change in momentum?
Dp = mvf - mvo = (2 kg)(15 m/s) - (2 kg)(-20 m/s)
Dp = 30 kg m/s + 40 kg m/s
Dp = 70 kg m/s
Example 3: A 500-g baseball moves to the left at 20
m/s striking a bat. The bat is in contact with the ball
for 0.002 s, and it leaves in the opposite direction at 40
m/s. What was average force on ball?
+
m = 0.5 kg
-20 m/s
F Dt = mvf - mvo
F
+ 40 m/s
Dt
vo = -20 m/s; vf = 40 m/s
F(0.002 s) = (0.5 kg)(40 m/s) - (0.5 kg)(-20 m/s)
Continued . . .
Example Continued:
+
m = 0.5 kg
+
F
- 20 m/s
40 m/s
Dt
F Dt = mvf - mvo
F(0.002 s) = (0.5 kg)(40 m/s) - (0.5 kg)(-20 m/s)
F(0.002 s) = (20 kg m/s) + (10 kg m/s)
F(0.002 s) = 30 kg m/s
F = 15,000 N
Summary of Formulas:
Impulse
Momentum
J = FavgDt
ρ = mv
Impulse = Change in momentum
F Dt = mvf - mvo
Conservation of Momentum
According to the law of conservation of linear momentum,
when the vector sum of the external forces that act on a
system of bodies equals zero, the total linear momentum of
the system remains constant no matter what momentum
changes occur within the system
For two objects interacting with one another, the
conservation of momentum can be expressed as:
m1v1  m2v2  m v  m v
'
1 1
v1 and v2 are initial velocities,
velocities
v1'
'
2 2
and
'
are final
2
v
ELASTIC AND INELASTIC COLLISIONS
Elastic Collision: A collision in which objects collide and
bounce apart with no energy loss.
Inelastic Collision: A collision in which objects collide and
some mechanical energy is transformed into heat energy.
The animation below portrays the elastic collision
between a 1000-kg car and a 3000-kg truck. The
before- and after-collision velocities and
momentum are shown in the data tables.
The animation below portrays the elastic collision
between a 3000-kg truck and a 1000-kg car.
The before- and after-collision velocities and
momentum are shown in the data tables.
Before the collision, the momentum of the truck is
60 000 Ns and the momentum of the car is 0 Ns;
the total system momentum is 60 000 Ns.
After the collision, the momentum of the truck
is 30 000 Ns and the momentum of the car is
30 000 Ns; the total system momentum is
60 000 Ns.
The animation below portrays the inelastic collision
between a very massive diesel and a less massive flatcar.
The diesel has four times the mass of the freight car.
After the collision, both the diesel and the flatcar move
together with the same velocity.
In elastic collisions no permanent deformation
occurs; objects elastically rebound from each other.
In head-on elastic collisions between equal masses,
velocities are exchanged.
Example 4: A 0.50-kg ball traveling at 6.0 m/s collides headon with a 1.00-kg ball moving in the opposite direction at a
velocity of -12.0 m/s. The 0.50-kg ball moves away at -14 m/s
after the collision. Find the velocity of the second ball.
M1 = 0.50 kg
V1 = 6.0 m/s
M2 = 1.00 kg
V2 = -12.0 m/s
p before = p
Vf1 = -14 m/s
after
m1V1 + m2V2 = m1Vf1 + m2V2f
(.5kg)(6m/s) + (1kg)(-12m/s) = (.5kg)(-14m/s) + (1kg)(V2f)
V2f = - 2 m/s
Inelastic collisions are characterized by objects
sticking together and permanent deformation.
Example 5: A 3000-kg truck moving rightward with a speed
of 5 km/hr collides head-on with a 1000-kg car moving
leftward with a speed of 10 km/hr. The two vehicles stick
together and move with the same velocity after the collision.
Determine the post-collision velocity of the car and truck.
p before = p
M1 = 3000 kg
V1 = 5.0 km/hr
M2 = 1000 kg
V2 = -10 km/hr
after
m1V1 + m2V2 = (m1+ m2 )V
m1v1  m2v2
V
m1  m2
(3000kg)(5km/hr) + (1000kg)(-10km/hr)
(3000kg + 1000kg)
V = 1.25 km/hr, right
Explosions
• When an object separates suddenly, as in
an explosion, all forces are internal.
• Momentum is therefore conserved in an
explosion.
• There is also an increase in kinetic energy
in an explosion. This comes from a
potential energy decrease due to chemical
combustion.
Recoil
• Guns and cannons “recoil” when fired.
• This means the gun or cannon must move
backward as it propels the projectile
forward.
• The recoil is the result of action-reaction
force pairs, and is entirely due to internal
forces. As the gases from the gunpowder
explosion expand, they push the projectile
forwards and the gun or cannon
backwards.
Summary of Formulas:
Impulse
J = FavgDt
Momentum p
= mv
Impulse = Change in momentum
F Dt = mvf - mvo
Conservation of Momentum
m1v1  m2v2  m v  m v
'
1 1
'
2 2
Example 6 A 7500-kg truck traveling at 5 m/s east
collides with a 1500-kg car moving at 20 m/s from a
direction of 210. After the collision, the two vehicles
remain tangled together. With what speed and in what
direction does the wreckage begin to move?
m1 = 7500 kg
m2 = 1500 kg
v1 = 5 m/s, 0º
v2 = 20 m/s, 210º
m1 v1+ m2 v2 =( m1 +m2)V
m1 v1+ m2 v2 =( m1 +m2)V
x-comp
7500 kg (5 m/s)
1500kg (20m/s cos 210º)
Σx = 11,519 kg m/s
y-comp
0
1500 kg (20m/s sin 210º)
Σy = - 15,000 kg m/s
Initial Momentum  (11519) 2  (15000) 2 = 18,912.7 kg m/s
initial momentum = final momentum
= (m1 + m2) V
initial
18912.7
V

= 2.1 m/s
m1  m2 7500  1500
15000
  tan
= 52.5º
11519
-1
I quadrant
V = (2.1 m/s,52.5º)
Example 7: Suppose a 5.0-kg projectile launcher
shoots a 209 gram projectile at 350 m/s. What is the
recoil velocity of the projectile launcher?
2 m/s
y
2 kg
y
3 m/s
50o
x
2 kg
x
8 kg
0 m/s
8 kg
v
Before
After
Sample problem
• Calculate velocity of 8-kg ball after the collision.
2 m/s
y
2 kg
y
3 m/s
50o
x
2 kg
x
8 kg
0 m/s
8 kg
v
Before
After
2D-Collisions
• Momentum in the x-direction is conserved.
▫ SPx (before) = SPx (after)
• Momentum in the y-direction is conserved.
▫ SPy (before) = SPy (after)
• Treat x and y coordinates independently.
▫ Ignore x when calculating y
▫ Ignore y when calculating x
• Let’s look at a simulation:
▫ http://surendranath.tripod.com/Applets.html
Center of Mass
Center of Mass
• The center of mass of a system is the point at
which all the mass can be assumed to reside.
• Sometimes the system is an assortment of
particles and sometimes it is a solid object.
• Mathematically, you can think of the center of
mass as a “weighted average”.
Center of Mass – system of points
xcm
xm


ycm 
zcm 
i
i
M
 yi mi
• Analogous equations
exist for velocity and
acceleration, for
example
M
 zi mi
M
vx ,cm
v


x ,i
M
mi
• Determine the Center of Mass.
y
2
2 kg
x
0
3 kg
1 kg
-2
-2
0
2
4
Certain problems involve the use of
conservation of energy and the conservation of
momentum
These problems require a special kind of
Physics aptitude and are only attempted by the
most savvy of Physics students…. You are those
students.
The Ballistic Pendulum is used in the
lab setting to find the velocity of fast
moving projectiles
How fast was the bullet traveling before
it struck the catcher? Is mechanical
energy conserved before and after the
collision?
821.3 m/s
Conservation of Momentum and Energy
Example 2
A 4 kg tennis racket moving with a velocity of 21 m/s strikes a 50
g tennis ball initially at rest. Find the velocity of the tennis ball
and tennis racket after the collision.