Transcript N e w t o n` s L a w s

Linear Momentum
5-1 Linear Momentum
Linear Momentum, p – defined as mass x
velocity


p  mv
The unit is kg·m/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 only
consider average force
Impulse is define as


Ft  p
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
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 – momentum is conserved,
but energy is lost
Momentum is conserved


p0  p
mAv A0  mB vB 0  mAv A  mB vB
5.5 Inelastic Collisions
Completely (or perfectly) Inelastic collision –
two objects collide and stick together
mAv A0  mB vB 0  (mA  mB )v
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?
mAv A0  mB vB 0  (mA  mB )v
(95)(3.75)  (111)( 4.10)  (95  111)v
v  0.48 ms
5.5 Inelastic Collisions
If the collision occurs in two dimensions
We need to consider the x
and y axis separately
m Av A0 x  mB vB0 x  (m A  mB )v x
mAvA0 y  mBvB0 y  (mA  mB )v y
5.5 Inelastic Collisions
Then 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 completely inelastic,
what is the magnitude and direction of the
cars’ velocity after the collision?
mAvA0 xy  mB vB0 xy  (m A  mB )v xy
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
Elastic Collisions
Momentum is conserved
m Av A0  mB vB0  m Av A  mB vB
Kinetic 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 Elastic 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 Elastic 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 Elastic 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
0  7.5v  20vB
2
B
5.5 Elastic Collisions
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?
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 Elastic 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