Chapter 6
Linear Momentum and Collisions
In everyday language, the term
“momentum” refers to a tendency
to “keep moving”. The same is
true in classical mechanics, where
momentum (and the conservation
thereof) is often used to analyze
situations where two or more
objects interact with each other.
In the photo, a runaway train (with
lots of momentum) interacted with
the wall of the station, with a
predictable outcome.
PC141 Intersession 2013
Day 12 – June 4 – WBL 6.1-6.3
Slide 1
6.1 Linear Momentum
Linear momentum, given the symbol p, is a property of mass
that is in motion. By definition,
the linear momentum of an object is the
product of its mass and velocity
Or, in mathematical terms,
p = 𝑚v
Since it is the product of a mass and a velocity, linear momentum
has SI units of kg·m/s. Unlike the Joule or Watt, there is no
common name for this unit.
A couple of notes on momentum: (1) although it involves both
m and v, linear momentum isn’t the same as kinetic energy,
which depends on the square of the speed. (2) Unlike kinetic
energy, linear momentum is a vector property of motion. It has
the same direction as velocity.
PC141 Intersession 2013
Day 12 – June 4 – WBL 6.1-6.3
Slide 2
6.1 Linear Momentum
Since p = 𝑚v is a vector equation, we can split it into equivalent
scalar equations in each direction, just as we did with Newton’s
second law:
𝑝𝑥 = 𝑚𝑣𝑥 ,
𝑝𝑦 = 𝑚𝑣𝑦 ,
𝑝𝑧 = 𝑚𝑣𝑧
It’s worth noting that linear momentum is often referred to
simply as “momentum”. We will try to avoid this lazy
terminology, since there is another type of momentum (angular
momentum), which we will cover in the next chapter.
We will often need to find the total linear momentum P of a
system of several masses. This is simply the vector sum of each
of the individual linear momenta:
P = p1 + p2 + p3 + ⋯
PC141 Intersession 2013
Day 12 – June 4 – WBL 6.1-6.3
Slide 3
6.1 Linear Momentum
Many physical problems can be analyzed by considering changes
in linear momentum. A non-zero ∆p can arise from a change in
speed or a change in direction (or a change in mass…this is rare,
but in some circumstances it may be relevant).
Two examples of changes in momentum (due to an object
rebounding from a wall) are shown here.
PC141 Intersession 2013
Day 12 – June 4 – WBL 6.1-6.3
Slide 4
6.1 Linear Momentum
Force and Momentum
Let’s return to Newton’s 2nd law. Assuming that mass remains
constant, we know that – at least over a short time –
acceleration is a = (v − v0 )/∆𝑡. Therefore,
F𝑛𝑒𝑡
𝑚 v − v0
𝑚v − 𝑚v0 𝐩 − 𝐩0 ∆𝐩
= 𝑚a =
=
=
=
∆𝑡
∆𝑡
∆𝑡
∆𝑡
In other words, the net external force acting on an object is
equal to the time rate of change of the object’s momentum.
The preceding equation represents an equivalent form of
Newton’s second law – in fact, this is the form in which Newton
first presented the law. It’s also applicable in more situations, for
reasons that we won’t need in PC141.
PC141 Intersession 2013
Day 12 – June 4 – WBL 6.1-6.3
Slide 5
6.1 Linear Momentum
Force and Momentum
The preceding equation tells us that the following statements are
equivalent:
1. A non-zero net force results in a non-zero acceleration
2. A non-zero net force results in a change in momentum.
Projectile motion provides an
example. As we know, the
horizontal velocity is constant.
Therefore, the horizontal linear
momentum is constant, which is
required, since there is no
horizontal force in this case.
PC141 Intersession 2013
Day 12 – June 4 – WBL 6.1-6.3
Slide 6
Problem #1: Net Force
WBL LP 6.3
A net force on an object can cause…
PC141 Intersession 2013
A
…an acceleration
B
…a change in momentum
C
…a change in velocity
D
…all of the preceding
Day 12 – June 4 – WBL 6.1-6.3
Slide 7
Problem #2: Baseball
WBL Ex 6.5
A 0.150-kg baseball traveling with a horizontal speed of 4.50 m/s is hit by a bat
and then moves with a speed of 34.7 m/s in the opposite direction. What is the
change in the ball’s momentum?
Note: It’s always good to get a sense for “reasonable” numbers…if a baseball is
traveling toward home plate at 4.50 m/s, it certainly wasn’t pitched…a pitched
baseball would be traveling closer to 40 m/s. Perhaps this one was lobbed by a
nearby coach.
Solution: In class
PC141 Intersession 2013
Day 12 – June 4 – WBL 6.1-6.3
Slide 8
Problem #3: Bad Truck Driver
WBL Ex 6.13
A loaded tractor-trailer with a total mass of 5000 kg traveling at 3.0 km/h hits a
loading dock and comes to a stop in 0.64 s. What is the magnitude of the average
force exerted on the truck by the dock?
Solution: In class
PC141 Intersession 2013
Day 12 – June 4 – WBL 6.1-6.3
Slide 9
6.2 Impulse
Much of this chapter is concerned with collisions, in which two
objects (at least one of which is initially in motion), are in contact
for a period of time†. Examples are the contact between a
baseball and bat, or between two balls on a snooker table. In
most of these cases, the force is not constant as a function of
time – it increases from zero to some maximum value, then
decreases back to zero. The total time
during which the force is non-zero is usually
on the order of milliseconds.
The figure shows one example of a force – vs
– time graph.
†technically
speaking, a “collision” doesn’t have to actually involve physical
contact between objects. As we know from chapter 4, there exist “noncontact” forces as well. The math is the same, but we’ll avoid these
situations in PC141.
PC141 Intersession 2013
Day 12 – June 4 – WBL 6.1-6.3
Slide 10
6.2 Impulse
Even though the force may vary with time, we can still use the
“new” version of Newton’s 2nd law from a few slides ago, if we
use average values: F𝑎𝑣𝑔 =
∆𝐩
.
∆𝑡
This can be rearranged as
F𝑎𝑣𝑔 ∆𝑡 = ∆𝐩
The term F𝑎𝑣𝑔 ∆𝑡 is known as the impulse of the force. It is a
vector property, with the symbol 𝐈. As the product of a force and
a time, it has units of N·s.
The preceding equation indicates that during a collision, the
impulse exerted on an object is equal to the change in the
object’s linear momentum. This is referred to as the impulsemomentum theorem.
PC141 Intersession 2013
Day 12 – June 4 – WBL 6.1-6.3
Slide 11
6.2 Impulse
Recall from chapter 5 that the area under a graph of force vs.
position is equal to the net work done on an object (which,
furthermore, is equal to the change in the object’s kinetic
energy). Here, the area under a graph of force vs. time is seen to
be equal to the impulse on the object, which is furthermore
equal to the change in its linear momentum.
Although in most cases the actual
relation between force and time is
complicated, using the average force
facilitates finding the area under the
curve.
PC141 Intersession 2013
Day 12 – June 4 – WBL 6.1-6.3
Slide 12
Problem #4: Astronaut
WBL Ex 6.25
An astronaut (mass of 100 kg, with equipment) is headed back to her space
station at a speed of 0.750 m/s but at the wrong angle. To correct her direction,
she fires rockets from her backpack at right angles to her motion for a brief time.
These directional rockets exert a constant force of 100.0 N for only 0.200 s.
a) What is the magnitude of the impulse delivered to the astronaut?
b) What is her new direction (relative to the initial direction)?
c) What is her new speed?
Solution: In class
PC141 Intersession 2013
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Slide 13
Problem #5: Pop Fly
WBL Ex 6.36
A baseball player pops a pitch straight up. The ball (mass 200 g) was traveling
horizontally at 35.0 m/s just before contact with the bat, and 20.0 m/s just after
contact. Determine the direction and magnitude of the impulse delivered to the
ball by the bat.
Solution: In class
PC141 Intersession 2013
Day 12 – June 4 – WBL 6.1-6.3
Slide 14
6.3 Conservation of Linear Momentum
In chapter 5, we learned that energy is a conserved quantity (in
that the total energy of the universe is constant). For systems
smaller than the entire universe, there are conditions to this
conservation – it only holds true for systems that are closed, and
that contain no nonconservative forces. We also learned that
when nonconservative forces are present, the total mechanical
energy changes only by the extent that the nonconservative
forces do net work on the system.
Likewise, under certain conditions, the total linear momentum
of a system is a conserved quantity. This fact allows us to
analyze the dynamics of objects that undergo collisions.
PC141 Intersession 2013
Day 12 – June 4 – WBL 6.1-6.3
Slide 15
6.3 Conservation of Linear Momentum
If the linear momentum of an object is to be conserved, we
simply require that the net force acting on it be zero. In other
words, if
∆𝐩
F𝑛𝑒𝑡 =
=0
∆𝑡
then ∆p = p𝑓 − p𝑖 = 0.
This conservation law also holds true for systems of objects, in
which we simply write that ∆P = P𝑓 − P𝑖 = 0 if F𝑛𝑒𝑡 = 0.
The condition of zero net force is met by closed systems. Note
that such systems may have internal forces, but these always act
as Newton’s 3rd-law force pairs. Since they are equal and
opposite to each other, they always sum to zero when calculating
the net force. Examples will be discussed in class.
PC141 Intersession 2013
Day 12 – June 4 – WBL 6.1-6.3
Slide 16
Problem #6: Conserved Linear Momentum
WBL LP 6.9
The linear momentum of an object is conserved if
PC141 Intersession 2013
A
the force acting on the object is conservative
B
a single, unbalanced internal force is acting on the
object
C
the mechanical energy is conserved
D
none of the preceding
Day 12 – June 4 – WBL 6.1-6.3
Slide 17
Problem #7: Really Bad Ice Skaters
WBL Ex 6.43
Two ice skaters not paying attention collide in a completely inelastic collision.
Prior to the collision, skater 1, with a mass of 60 kg, has a velocity of 50 km/h
eastward, and moves at a right angle to skater 2, who has a mass of 75 kg and a
velocity of 7.5 km/h southward. What is the velocity of the skaters after the
collision?
Solution: In class
PC141 Intersession 2013
Day 12 – June 4 – WBL 6.1-6.3
Slide 18
Problem #8: Exploding Bomb
WBL Ex 6.47
An explosion of a 10.0-kg bomb releases only two separate pieces. The bomb
was initially at rest and a 4.00-kg piece travels westward at 100 m/s immediately
after the explosion.
a) What are the speed and direction of the other piece immediately after the
explosion?
b) How much kinetic energy was released in this explosion?
Solution: In class
PC141 Intersession 2013
Day 12 – June 4 – WBL 6.1-6.3
Slide 19
Problem #9: Ballistic Pendulum
WBL Ex 6.55
A ballistic pendulum is a device used to measure the velocity of a projectile – for
example, the muzzle velocity of a rifle bullet. The projectile is shot horizontally
into, and becomes embedded in, the bob of a pendulum. The pendulum swings
upward to some height h, which is measured. The masses of the block and bullet
are known. Show that the initial velocity of the projectile is given by
𝒎+𝑴
𝒗𝟎 =
𝟐𝒈𝒉
𝒎
Solution: In class
PC141 Intersession 2013
Day 12 – June 4 – WBL 6.1-6.3
Slide 20