Download Impulse and Momentum

Impulse and Momentum
AP Physics
Which do you think has more momentum?
Change in momentum
 Like any change, change in momentum is calculated by
looking at final and initial momentums.
 Dp = pf – pi
 Dp: change in momentum
 pf: final momentum
 pi: initial momentum
Impulse = ∆Momentum
Consider Newton’s 2nd Law and
the definition of acceleration
Units of Impulse:
Units of Momentum: Ns
Kg x m/s
Momentum is defined as “Inertia in Motion”
Impulse is the Area
Since J=Ft, Impulse is the AREA of a Force vs. Time graph.
Impulsive Forces
 Usually high magnitude,
short duration.
 Suppose the ball hits the
bat at 90 mph and leaves
the bat at 90 mph, what
is the magnitude of the
momentum change?
 What is the change in the
magnitude of the
momentum?
Example
A 100 g ball is dropped from a height of h = 2.00 m above the floor. It
rebounds vertically to a height of h'= 1.50 m after colliding with the
floor. (a) Find the momentum of the ball immediately before it collides
with the floor and immediately after it rebounds, (b) Determine the
average force exerted by the floor on the ball. Assume that the time
interval of the collision is 0.01 seconds.
How about a collision?
Consider 2 objects speeding toward
each other. When they collide......
Due to Newton’s 3rd Law the
FORCE they exert on each other
are EQUAL and OPPOSITE.
The TIMES of impact are also equal.
F1   F2
t1  t 2
( Ft )1  ( Ft ) 2
J1   J 2
Therefore, the IMPULSES of the 2
objects colliding are also EQUAL
How about a collision?
If the Impulses are equal then
the change in MOMENTUMS
are also equal!
Momentum is conserved!
The Law of Conservation of Momentum: “In the absence of
an external force (gravity, friction), the total momentum
before the collision is equal to the total momentum after
the collision.”
Law of Conservation of Momentum
 If the resultant external force on a system is zero, then the
vector sum of the momentums of the objects will remain
constant.

SPbefore = SPafter
Sample problem
 A 75-kg man sits in the back of a 120-kg canoe that is at rest in a
still pond. If the man begins to move forward in the canoe at 0.50
m/s relative to the shore, what happens to the canoe?
Collision Types
 Elastic collisions
 Also called “hard” collisions
 No deformation occurs, no kinetic energy lost
 Inelastic collisions
 Deformation occurs, kinetic energy is lost
 Perfectly Inelastic (stick together)
 Objects stick together and become one object
 Deformation occurs, kinetic energy is lost
Sample Problem
 A fish moving at 2 m/s
swallows a stationary fish
which is 1/3 its mass. What is
the velocity of the big fish after
dinner?
Example
Granny (m=80 kg) whizzes around
the rink with a velocity of 6 m/s.
She suddenly collides with
Ambrose (m=40 kg) who is at rest
directly in her path. Rather than
knock him over, she picks him up
and continues in motion without
"braking." Determine the velocity
of Granny and Ambrose.
Elastic Collision
 In elastic collisions, there is no deformation of colliding
objects, and no change in kinetic energy of the system.
Therefore, two basic equations must hold for all elastic
collisions

Spb = Spa (momentum conservation)

SKb = SKa (kinetic energy conservation)
What happens if we have two
unknowns?
 At an amusement park, a 96.0 kg bumper car moving with a
speed of 1.24 m/s bounces elastically off a 135 kg bumper
car at rest. Find the final velocities of the cars.
Let subscript 1 refer to the 96.0kg car and subscript 2 refer to the 135 kg car.
Use momentum conservation.
m1v1  m2v2  m1v1f  m2v2f
Use conservation of kinetic energy.
1
1
2 1
2
2 1
m1v1  m2 v2  m1v1f  m2 v2f 2 .
2
2
2
2
Rearranging the first equation gives
m1 (v1  v1f )
 1.
m2 (v2f  v2 )
Rearranging the second equation gives
m1 (v12  v1f 2 )
m2 (v2f 2  v22 )
1
m1 (v1  v1f )(v1  v1f )
.
m2 (v2f  v2 )(v2f  v2 )
Comparing these two equations implies that
(v1  v1f )
 1,
(v2f  v2 )
or v2f  v1  v1f  v2 .
Substitute for
v2f
in the first equation and solve for
v1f .
m1v1  m2 v2  m1v1f  m2 (v1  v1f  v2 )
(m1  m2 )v1  2m2 v2  (m1  m2 )v1f
 m1  m2 
 2m2 
v1f  
 v1  
 v2
 m1  m2 
 m1  m2 
Since v2f  v1  v1f  v2 , v1f  v2f  v2  v1.
Substitute for
v1f
in the first equation and solve for
m1v1  m2v2  m1 (v2f  v2  v1 )  m2v2f
2m1v1  (m2  m1 )v2  (m1  m2 )v2f
v2f .
Sample Problem
 A 500-g cart moving at 2.0 m/s on an air track elastically strikes a
1,000-g cart at rest. What are the resulting velocities of the two carts?
2D Inelastic Collisions must rely on the
Conservation of Momentum:
 Example: A car with a mass of 950
kg and a speed of 16 m/s
approaches an intersection, as
shown. A 1300 kg minivan
traveling at 21 m/s is heading for
the same intersection. The car and
minivan collide and stick together.
Find the speed and direction of the
wrecked vehicles just after the
collision, assuming external forces
can be ignored.
Sample problem
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