Download Momentum and Impulse

Momentum
Momentum is defined as the product of the
mass × velocity of an object.
p=m×v
where m is the mass in kg, v is the velocity
in m/s, and the momentum, p, in kg•m/s.
Momentum is a vector quantity meaning that it
has magnitude and direction.
The direction of momentum is that of the velocity
of the object.
From Newton’s 2nd Law:
 Fnet = ma
Δv
a =
Δt
mΔv
Therefore, Fnet =
Δt
FnetΔt = mΔv
FnetΔt is called impulse where Fnet is measured in
N and Δt measured in s.
 FnetΔt = N•s
 Impulse is a vector quantity and its direction
is that of Fnet.
Impulse and Momentum
Momentum can be thought of as “inertia in
motion”.
When a car on Thurber’s Avenue crashes into a
guard rail, a large force is exerted on the guard
rail.
 The force of impact comes from a change
in velocity or deceleration of the car.
 The force of impact comes from a change
in velocity or deceleration of the car.
 The force of impact is directly proportional
to the change in velocity of the moving
car and to its mass.
Fnet α m × v
To change the momentum of an object, you
must consider the impulse.
 Impulse depends on the magnitude of the
force and the time of impact (time the
objects are in physical contact).
Hitting a golf ball for distance (teeing off),
throwing a baseball or football for distance
requires a “follow through”.
 These situations call for the largest
change in momentum (mΔv) which is
achieved by using a large force for as
long as possible (FΔt).
When receiving a punch, it is best to “roll with
the punch”.
 By rolling with the punch, you extend the
time of impact which minimizes Fnet.
 Fnet = mΔv/Δt
 The same change in momentum will occur
in either case but it is best to maximize the
value of Δt rather then the Fnet.
When throwing a punch, you try to pull your fist
back as quickly as possible.
 By pulling back your fist as quickly as
possible, it minimizes the Δt which
maximizes the Fnet.
Didn’t you use this idea as a youngster jumping
from a high height?
Momentum Problems
A 0.144 kg baseball with a velocity of 7.5 m/s
strikes a wall perpendicularly and rebounds at
8.9 m/s.
m = 0.144 kg
v1 = 7.5 m/s
v2 = -8.9 m/s
(a) What is the change in momentum of the
baseball?
Δmv = mΔv = 0.144 kg × (7.5 m/s –
-8.9 m/s) = 2.4 kg•m/s2
(b) If the force of impact acted 0.045 s, what
was the average force on the wall?
Fnet
mΔv
=
Δt
Fnet
2.4 kg•m/s2
=
= 53 N
0.045 s
(c) What average force is exerted on the
baseball?
By Newton’s 3rd Law, 53 N.
A 26 kg boy is riding on a wagon going at
3.0 m/s N. The boy jumps off the back of a
wagon in such a way that he has no velocity
when he lands on the ground. The mass of the
wagon is 10. kg.
mb = 26 kg
vb = 3.0 m/s N
vb’ = 0 m/s
mw = 10. kg
(a) What was the boy's change in momentum?
Δpb = mΔv = 26 kg × -3.0 m/s = -78 kg•m/s S
(b) What was the wagon's change in
momentum?
Δpsystem = 0, therefore the change in
momentum of the wagon is 78 kg•m/s N.
(c) How fast does the wagon go after the boy
jumps off?
Δpw = mΔv
Δv = Δpw/m = 78 kg•m/s/10. kg = 7.8 m/s N
vf = vi + Δv = 3.0 m/s + 7.8 m/s = 10.8 m/s N
Conservation of Momentum
When studying the momentum changes in
collisions, you must focus on a closed and
isolated system.
 A system is defined as a collection of
objects.
 A system is closed if no objects enter or
leave the system.
 A system is isolated if no external forces
are exerted on it.
Consider a game of pool when there is only the
8-ball and the cue ball.
 The system is closed because both balls
only interact with each other.
 The system is isolated because rolling
friction (µk = 0) can be ignored.

A
B
8
pB
pA
where mA = mB, vA > vB, and pA > pB
Assume neither ball is spinning (no english
or angular momentum) because that would
complicate matters.
 At the instant of impact.
8
FB
FA
After impact.
8
vB > 0
vA = 0
The Conservation of Momentum states that in a
closed and isolated system, the momentum
before the collision equals the momentum after
the collision.
 Δp = 0
pi = p f
 Momentum may be redistributed in a
collision but Δpsystem = 0.
 Internal forces are forces between
objects within the system.
 External forces are exerted by objects
outside a system.
 Momentum is conserved only when
internal forces are present.
Conservation of Momentum Problems
A 7450 kg truck traveling at 4.9 m/s 090°
collides with a 1575 kg car traveling at 21 m/s
240°. After colliding, the truck and car move
together. What is their velocity after the
collision?
m1 = 7450 kg
v1 = 4.9 m/s 090°
m2 = 1575 kg
v2 = 21 m/s 240°
m1v1
θ2
.
θ1
mRvR
θ1 = 30°
θ2 = ?
mRvR = ((m1v1)2 + (m2v2)2 - 2×m1v1×m2v2cosθ)½
mRvR = ((7450 kg × 4.9 m/s)2 + (1575 kg × 21 m/s)2
- 2 × 7450 kg × 4.9 m/s ×
1575 kg × 21 m/s × cos30°)½
mRvR = 1.8 x 104 kg•m/s
.
vR = 2.0 m/s
sinθ1
mRvR
=
sinθ2
m2v2
sinθ2
sin30°
=
4
1575 kg × 21 m/s
1.8 x 10 kg•m/s
θ2 = 67°
vR = 2.0 m/s, 157°
Two angry hockey players push each other.
The mass of the first hockey player is 70. kg
and the second has a mass of 50. kg. If the first
hockey player has a velocity of 6.2 m/s N, how
far apart are they after 3.0 s? Assume friction is
negligible.
m1 = 70.kg
v1 = 6.2 m/s N
Δt = 3.0 s
m2 = 50. kg
v2 = ?
µ=0
Δp = 0
pi = p f
m1v1 + m2v2 = m1v1’ + m2v2’
0 = m1v1’ + m2v2’
0 = 70. kg × 6.2 m/s + 50. kg × v2’
V2’ = -8.7 m/s = 8.7 m/s S
ΔxT = Δx1 + Δx2 = v1 × Δt1 + v2 × Δt2
ΔxT = 6.2 m/s × 3.0 s + 8.7 m/s × 3.0 s
ΔxT = 45 m
Conservation of Kinetic Energy
Kinetic energy is the energy due to the motion
of an object.
 KE = ½mv2
where m is the mass in kg, v is the
velocity in m/s, and KE is the kinetic
energy in J (joules).
 Kinetic energy is a scalar quantity.
The Law of Conservation of Energy states that
within a closed and isolated system, energy can
change form but the total amount of energy is
constant.
 ΔE = 0
When two objects collide, the force each object
exerts on the other slightly change the shape of
the object.
 KE
PE
Momentum And Energy Together
These two conservation laws, momentum and
energy, are both needed to give a “complete
picture” of what happens in any collision.
 The conservation of momentum gives the
final momentum after the collision.
What it does not tell us is the distribution
between mass and velocity.
The conservation of kinetic energy tells us how
the distribution between mass and velocity by
considering the elastic properties of the colliding
objects.
 The conservation of energy puts an upper
limit of the velocity of the impacted object
and how close you get to this upper limit
depends on the elasticity.
Remember the three types of collisions:
 Elastic Collisions
After the collision, the objects separate.
Both momentum and kinetic energy are
conserved.
 Perfectly Inelastic Collisions
After the collision, the objects stick
together and share a common velocity.
Only momentum is conserved.
 Inelastic Collisions
The objects deform during the collision so
the total kinetic energy decreases but the
objects separate after the collision.
Only momentum is conserved.
A 600. mg popcorn kernel feeling the heat is
moving around at 7.4 cm/s before it pops and
breaks into two pieces of equal mass. If one
piece comes to an abrupt halt, determine the
change in kinetic energy.
m1 = 600. mg
v1 = 7.4 cm/s
m1’ = 300. mg
m2’ = 300. mg
v 1’ = v 1’
v 2’ = 0
Δp = 0
pi = p f
m1v1 = m1v1’ + m2v2’
600. mg × 7.4 m/s = 300. mg × v1’
v1’ = 14.8 m/s
ΔKE = 0
KEi = KEf
½m1v12 = ½m1v1’2 + ½m2v2’2
1g
½ × (600. mg ×
cm
(7.4 s
×
103 mg
1m
102 cm
×
1 kg
103 g
×
)2 = 1.6 x 10-6 J
½ × (300. mg ×
1g
103 mg
×
1 kg
103 g
×
cm
1m
2 = 3.3 x 10-6 J
(14.8 s ×
)
102 cm
ΔKE = KEf – KEi = 3.3 x 10-6 J - 1.6 x 10-6 J
ΔKE = KEf – KEi = 1.6 x 10-6 J
A Toyota with a mass of 575 kg moving at
15 m/s collides into the rear of a Ford pick-up
which has a mass of 1575 kg and is moving at
5.0 m/s. The two vehicles lock together and
continue sliding forward.
(a) What is their final velocity?
m1 = 575 kg
v1 = 15 m/s
v 1’ = ?
m2 = 1575 kg
v2 = 5.0 m/s
v 2’ = ?
Δp = 0
pi = p f
m1v1 + m2v2 = (m1+m2)v’
575 kg × 15 m/s + 1575 kg × 5.0 m/s =
(575 kg + 1575 kg) × v’
v’ = 7.7 m/s
(b) How much kinetic energy was “lost” in the
collision?
ΔKE = 0
KEi = KEf
½m1v12 + ½m2v22 = ½(m1 + m2)v’
½ × 575 kg × (15 m/s)2 + ½ × 1575 kg ×
(5.0 m/s)2 = ½ × (575 kg + 1575 kg) ×
(7.7 m/s)2
84000 J ≠ 64000 J
ΔKE = 84000 J – 64000 J = 2.0 x 104 J
Therefore, 2.0 x 104 J were converted into heat
and sound which represent a wasted form of
energy.
A 15 g bullet is shot into a wooden block which
has a mass of 5085 g. The block is at rest on a
horizontal surface. After the bullet becomes
imbedded in the block, the two move at 1.0 m/s.
Determine the velocity of the bullet before
striking the block.
mb = 15 g
vb = ?
vb’ = 1.0 m/s
mw = 5085 g
vw = 0
vw’ 1.0 m/s
Wrap Up Questions
If two particles have equal kinetic energies, are
their momenta always equal?
No, the momenta would only be equal if their
masses were equal.
Does a large force always produce a larger
impulse than a smaller force? Justify your
answer.
The impulse, FnetΔt, is dependent on both the
net force and the time interval the net force
acts.
The statement would be true only if the time
intervals were equal.
The cue ball collides with the 8-ball which is
initially at rest. Is it possible for both the cue
ball and 8-ball to be at rest immediately after
the collision?
The conservation of momentum (Δp = 0)
prohibits this from happening. If the system
(the cue ball and the 8-ball) had momentum
before the collision, then there has to be the
same momentum before the collision.
Would it be possible for one ball to be at rest
after the collision?
Yes, whenever two equal masses experience a
head-on elastic collision, the stationary ball will
have all of the momentum after the collision.
Is momentum conserved when a basketball
bounces off the floor?
Yes, if the system is considered to be both the
basketball and the earth. However, due to the
mass of the earth, its momentum would be
undetectable.
Too understand why the earth must be
included, take a look at the next slide.
Assume the ball moving downward to have a
positive velocity, then the ball moving upward
would have a negative velocity.
m1 v 1
Δp = 0
pi = pf
m1v1 = - m1v2
How can v1 = - v2?
-m1v1’
Can a collision occur such that all of the kinetic
energy is lost?
Yes, provided that equal masses are
approaching each other with the same speeds,
such that the momentum before the collision is
equal to zero.
If the collision is perfectly inelastic, then the
kinetic energy is converted into elastic potential
energy which deforms the objects.