# Physics ch. 6

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```Physics ch. 6
Momentum and Collisions
Vocab Words
 Elastic collision
 Impulse
 Momentum
 Perfectly inelastic collision
6-1Momentum and Impulse
 In this chapter, we will examine how the
force and duration of a collision between
two objects affects the motion of the
objects
 The linear momentum of an object is
defined as the product of the mass and
the velocity of the object.
 P=mv
 Momentum is a vector quantity, direction
matches the direction of the velocity
 More massive object has more
momentum than a less massive object at
the same velocity
 However, small objects with a large
velocity also have a large momentum
 Ex: hail
6A Sample Problem
 A 2250 kg pickup truck has a velocity of
25 m/s to the east. What is the
momentum of the truck?
A change in momentum takes
force and time
 Ex: It takes more force to stop a fast
moving ball than a slower moving ball
 Ex: It takes more force to stop a real
dump truck moving at the same speed at
a toy dump truck
 Momentum is closely related to force
 F = Δp/Δt
 Impulse-Momentum Theorem
 FΔt = Δp or FΔp = mvf – mvi
 Where FΔt is the impulse
 A small force acting for a long time can
produce the same change in momentum
as a large force acting for a short time.
 In this book, all forces exerted on an
object are assumed to be constant
unless otherwise stated.
 This equation helps describe some
common sports examples:
Ex: baseball-the ball will experience a
greater momentum if the force of the bat
is kept in contact with the ball for a
longer time period (FΔt)
6B Sample Problem
 A 1400 kg car moving westward with a
velocity of 15 m/s collides with a utility
pole and is brought to rest in 0.30 s.
Find the magnitude of the force exerted
on the car during the collision.
Stopping times and distances
depend on impulse-momentum
theorem
 Highway engineers use the theorem to
determine safe stopping distances and
safe following distances
 Fig. 6-3
 The stopping distance and time is 2
times greater for the loaded truck than
the unloaded truck.
6C Sample Problem
 A 2250 kg car traveling to the west slows
down uniformly from 20.0 m/s to 5.00
m/s. How long does it take the car to
decelerate if the force on the car is 8450
N to the east? How far does the car
travel during the deceleration?
 Hint: use x = ½ (Vi + Vf)Δt from ch. 2
A change in momentum over a
longer time requires less force
 This theorem helps design safety
equipment that exerts a less force during
a collision
 Ex: nets and air mattresses
 The change in the momentum of falling
is the same, just the net extends the time
of the collision so that the change in the
person’s momentum occurs over an
extended time period, thus creating a
smaller force on the person
 Ex: fig. 6-4 and 6-5
6-2 Conservation of
Momentum
 Now, we’ll examine the momentum of
two or more objects interacting together
Pg. 215 soccer balls
The momentum that ball B gains is exactly
the same amount of momentum that ball
A loses during the collision.
Table 6-1-Mass, Velocity, and Momentum
Law of Conservation of
Momentum
 M1v1i + m2v2i = m1v1f + m2v2f
 The total momentum of all objects
interacting with one another remains
constant regardless of the nature of the
forces between the objects.
Momentum is Conserved in
Collisions
 Notice that the total momentum of all
objects interacting in a system are
conserved but the momentum of each
object is not conserved.
 Frictional forces will be disregarded in
most problems in this book so that most
problems only have two objects
interacting.
Momentum is conserved for
objects pushing away from each
other
 Ex: when you jump up, your momentum
is not conserved but when you factor in
Earth, the total momentum is conserved
 Fig. 6-7 Skaters Total Momentum before
is zero and total momentum after is also
zero
6D Sample Problem
 A 76 kg boater, initially at rest in a
stationary 45 kg boat, steps out of the
boat and onto the dock. If the boater
moves out of the boat with a velocity of
2.5 m/s to the right, what is the final
velocity of the boat?
Newton’s third law leads to
conservation of momentum
 Remember that the force exerted by one
body on another is equal in magnitude
and opposite in direction to the force
exerted on the first body by the second
body.
 F1 = -F2
 The two forces act in the same time
interval so that:
 F1Δt = -F2Δt
 The impulse on 1 is equal and opposite
in magnitude to the impulse on 2
 This is true in every collision between
two objects
 The change in momentum in the first object is
equal to and opposite the change in momentum
of the second object
 M1v1f - m1v1i = -(m2v2f – m2v2i)
 This mean if the momentum in one object
increases, the momentum in the second object
must decrease by that amount
Formula Rearranged
 M1v1i + m2v2i = m1v1f + m2v2f
Forces in real collisions are
not constant
 Fig. 6-9 Although forces vary in a real
collision, they are always equal but
opposite in magnitude to each other
 For problems, we’ll use the average
force during the collision, which is equal
to the constant force required to cause
some change in momentum as the real,
changing force.
6-3 Elastic and Inelastic
Collisions
 Some objects stick together and move
with the momentum equal to their
combined momentum before the
collision
 Some objects collide and bounce so that
they move away with two different
velocities
 Total momentum is conserved in a
collision but kinetic energy is generally
not conserved
 Some is transferred to internal energy
Perfectly Inelastic Collisions
 When two objects collide and move
together as one mass
 The final mass is equal to the combined
mass of the objects and they move with
the same velocity after the collision
Formula-Inelastic Collisions
m1v1i + m2v2i = (m1 + m2)vf
Pay close attention to -/+ signs that
indicate direction
+ To the right
- To the left
6E Sample Problem
 A 1850 kg luxury sedan stopped at a
traffic light is struck from the rear by a
compact car with a mass of 975 kg. The
two cars become entangled as a result
of the collision. If the compact car was
moving at a velocity of 22.0 m/s to the
north before the collision, what is the
velocity of the entangled mass after the
collision?
Kinetic energy is not constant
in inelastic collisions
 Some of the kinetic energy is converted
to sound energy and internal energy as
objects deform
 Elastic-means it keeps its shape
 Inelastic-means it is deformed and loses
some kinetic energy
6F Sample Problem
 Two clay balls collide head-on in a
perfectly inelastic collision. The first ball
has a mass of 0.5 kg and an initial
velocity of 4.00 m/s to the right. The
mass of the second ball is 0.250 kg, and
it has an initial velocity of 3.00 m/s to the
left. What is the final velocity of the
composite ball of clay after the collision?
What is the decrease in kinetic energy
during the collision?
Elastic Collisions
 Two objects collide and return to their
original shapes with no change in total
kinetic energy
 After the collision, the two objects move
separately
 In an elastic collision, the total
momentum and total kinetic energy
remain constant
Most collisions are neither
elastic nor perfectly inelastic
 Most objects do not stick together and move as
one but also, most collisions result in some
decrease in kinetic energy
Ex: football being kicked is deformed a little which
converts kinetic energy into internal elastic
potential energy
OR the formation of sound in any collision
represents a decrease in kinetic energy so the
collision cannot be elastic
Inelastic Collisions
 Most collisions fall into this type
 The colliding objects bounce and move
separately after the collision, but the total
kinetic energy decreases in the collision
 For this book, we will assume the all
collisions in which the objects do not
stick will be elastic (p and KE will be
constant)
KE is conserved in elastic
collisions
 The total momentum and total kinetic
energy remain constant throughout the
collision if it is a perfectly elastic collision
M1v1i + m2v2i = m1v1f + m2v2f
½ m1v1i2 + ½ m2v2i2 = ½ m1v1f2 + ½
m2v2f2
Sample Problem 6G
 A 0.015 kg marble moving to the right at 0.225
m/s makes an elastic head-on collision with a
0.030 kg shooter marble moving to the left at
0.180 m/s. After the collision, the smaller
marble moves to the left at 0.315 m/s. Assume
that neither marble rotates before or after the
collision and that both marbles are moving on a
frictionless surface. What is the velocity of the
0.030 kg marble after the collision?
 (remember that direction to the right is + and
direction to the left is -)
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