Transcript Liner Momentum Power Point

Linear Momentum
5-1 Linear Momentum
Linear Momentum, p – defined as mass x
velocity


p  mv
The unit is kgm/s
A quantity used in collisions
So a small object with a large velocity could
have the same momentum as a large object
with a small velocity
9-1 Linear Momentum
5.2 Momentum and Newton’s Second Law
Newton’s Second Law is


F  ma
This is only true for objects with a constant
mass
The original form of the equation was

 p
F 
t
This statement is true even if the mass varies
5.2 Momentum and Newton’s Second Law
5.3 Impulse
A baseball player hits a pitch
Bat delivers an impulse
We actually on consider
average force
Impulse is define as


Ft  p
Impulse Sim
5.3 Impulse
An increase in time produces a decreases in
force
Airbag
A decrease in time produces an increase in
force
5.3 Impulse
5.4 Conservation of Linear Momentum
If no net external force is applied to a system
Then momentum is conserved
p0  p
Explode Sim
5.4 Conservation of Linear Momentum
External Forces will result in a change in
momentum, so no conservation
1.Force added in
Shuttle Launch
2.Force removed
5.4 Conservation of Linear Momentum
5.5 Inelastic Collisions
Inelastic collision – two objects collide and
stick together
Momentum is conserved
p0  p
mAvA  mBvB  (mA  mB )v
Energy is not conserved
5.5 Inelastic Collisions
Example: On a touchdown attempt, a 95 kg
running back runs toward the end zone at 3.75
m/s. A 111kg linebacker moving at 4.10 m/s
meets the runner in a head on collision. If the
two players stick together what is their velocity
immediately after the collision?
mAvA  mBvB  (mA  mB )v
(95)(3.75)  (111)( 4.10)  (95  111)v
v  0.48 ms
5.5 Inelastic Collisions
Example: In a ballistic pendulum, a 100g
bullet is fired at a velocity of 200 m/s at the
bob of a pendulum. The bob has a mass of 10
kg. After the collision, the object and the bob
stick together and swing through an arc. How
high does it get?
This is first a conservation of momentum
problem (how fast does the combination of
bullet and bob go after the collision)
Then it is a conservation of energy problem.
5.5 Inelastic Collisions
Example: In a ballistic pendulum, a 100g bullet is fired at a velocity of 200 m/s at
the bob of a pendulum. The bob has a mass of 10 kg. After the collision, the
object and the bob stick together and swing through an arc. How high does it get?
mAvA  mBvB  (mA  mB )v
(0.1)( 200)  (10)(0)  (.1  10)v
v  1.9 ms
1
2
1
2
mv  mgy
2
(1.9)  (9.8) y
2
y  0.18m
5.5 Inelastic Collisions
If the collision occurs in two dimensions
We need to consider the x
and y axis separately
m Av Ax  mB vBx  (m A  mB )v x
mAvAy  mBvBy  (mA  mB )v y
5.5 Inelastic Collisions
The we use vector addition to calculate the
magnitude and velocity.
v  v v
2
x
2
y
 vy 
  tan  
 vx 
1
5.5 Inelastic Collisions
Example: A 950kg car traveling east at 16m/s
collides with a 1300 kg car traveling north at
21 m/s. If the collision is inelastic, what is the
magnitude and direction of the cars after the
collision?
mAvAyx  mBvByx  (m A  mB )vxy
2
2
)(
16
)

(
1300
)(
0
)

(
950

1300
)
v
v(950

6
.
76
m
/
s
v

6
.
76

12
.
1
 13.9 ms
x
2
2 x
v

v

v
x
y
v(v950
6
.
76
m
/
s
x 
 )(
120.)1m(1300
/ s )( 21)  (950  1300)v y   tan 1  12.1   60.8o
y
v y  12.1m / s
 vy 
  tan  
x 
 vCollisions
5.5 Inelastic
1


 6.76 
5.6 Elastic Collisions
Elastic collision – two objects collide and
bounce apart
Momentum is conserved
m Av A0  mB vB0  m Av A  mB vB
In a perfect elastic collision, energy is
conserved too
1
2
mv  mv  mv  mv
2
A A0
1
2
2
B B0
1
2
2
A A
5.5 Inelastic Collisions
1
2
2
B B
A 10 kg car moving at 2 m/s runs into a 5 kg
car that is parked. What is the velocity of each
car after the collision?
m Av A0  mB vB0  m Av A  mB vB
(10)(2)20
 (510
)(0v)A10
5vvBA  5vB
1
2
1
2
mv  mv  mv  mv
2
A A0
1
2
2
B B0
1
2
2
A A
1
2
2
B B
(10
(10
)()(
2)2)40
(5
)(10
0) v (10
5(10
v)v)v (5)v(5)v
2 2 1
2
2 2 1
A 2
2 2 2
B A A
5.5 Inelastic Collisions
1 2
2 B
2
B
A 10 kg car moving at 2 m/s runs into a 5 kg
car that is parked. What is the velocity of each
car after the collision?
20  10vA  5vB
vA  2  .5vB
40  10v  5v
2
A
2
B
40  10(2  .5vB )  5v
2
2
B
5.5 Inelastic Collisions
A 10 kg car moving at 2 m/s runs into a 5 kg
car that is parked. What is the velocity of each
car after the collision?
vA  2  .5vB
40  10(2  .5vB )  5v
2
2
B
40  10(.25v  2vB  4)  5v
2
B
40  2.5v  20vB  40  5v
2
B
2
B
2
B
0  7.5v  20vB
2
B
5.5 Inelastic Collisions
A 10 kg car moving at 2 m/s runs into a 5 kg
car that is parked. What is the velocity of each
car after the collision?
0  7.5v  20vB
2
B
vB  2.67 ms
m
v

0
.
67
2020
 10

vAvA 5(52s v.67
A 10
B )
Confirm
10(0.67
5(v2.67
40) 10
 5)v  40.1
2
2
A
22
B
5.5 Inelastic Collisions
5.7 Center of Mass
The point where the system can be balanced
in a uniform gravitational field
Center of mass of
Triangle
Uniform objects
center of mass is in the
center
Motion of CM
5.7 Center of Mass
Center of mass is not always in the object
Objects balance if supported at their center of
mass
5.7 Center of Mass
5.8 Systems with Changing Mass: Rockets
When a rocket is launched (or a plane takes
off).
Fuel is used as the rocket launches
This causes the mass to decrease
5.8 Systems with Changing Mass: Rockets
So
P  m 
F

v
t  t 
The force due to the ejected fuel is called
thrust
 m 
thrust  
v
 t 
5.8 Systems with Changing Mass: Rockets
In a Saturn V rocket, fuel is ejected at 13,800
kg/s and at a speed of 2440 m/s
 m 
thrust  
v
 t 
thrust  13,800 2440   33,700,000 N
Since the initial weight of the rocket is
28,500,000N (or mass of 2,850,000 kg)
F  33700000  28500000  5,700,000 N
F 5700000
a 
 2 sm2
m 2850000
5.8 Systems with Changing Mass: Rockets
As the rocket travels its mass drops, so the
acceleration will actually increase
5.8 Systems with Changing Mass: Rockets