Download 07-1 Note 07 Impulse and Momentum ∑ = ∑ =

Note 07
Impulse and Momentum
Sections Covered in the Text: Chapter 9, except 9.6 & 9.7
In this note we consider the physics of particles in
collision. Collisions form a class of problems in
physics that involve two or more particles interacting
in such a way that their states of motion are changed.
We shall simplify our description by defining another
kinematic quantity: linear momentum. 1
Isaac Newton called momentum an object’s
“quantity of motion”. To appreciate the idea of
momentum, suppose two objects, a car and a pingpong ball, are moving towards you both at a speed of
2.00 m.s–1. You have the option of avoiding being hit
by only one of the objects. Which do you choose?
Which object has the lesser quantity of motion?
Even if you haven’t studied physics, you would
surely choose the ping-pong ball. Being hit by a pingpong ball is likely to hurt less than being hit by a car
moving at the same speed. The car has the greater
mass of the two and therefore the greater “quantity of
motion” or momentum.
We begin by defining linear momentum. We then
develop the law of conservation of linear momentum
for an isolated system. We then describe a few cases of
one-dimensional collisions. We conclude with the
concept of the center of mass.
This statement allows for a mass m that can vary
(which our earlier statement of the law in Note 04 did
not). Eq[7-2] reduces to eq[4-1] in the special case
where m is constant. To see this we substitute eq[7-1]
into eq[7-2]. Simplification yields
∑F =
=m
Momentum and Isolated Systems
The instantaneous state of an isolated system of two
interacting particles may be depicted as in Figure 7-1.
By isolated system we mean one in which the particles
interact with one another but not with their environment. This figure depicts two particles not in contact
(for example interacting via the force of gravity or the
electrostatic force), though we could easily modify it
to describe two particles in contact.
Let the linear momentum of particles 1 and 2 be p1
and p 2 respectively. Applying the second law, eq[72], to each particle yields:
Linear Momentum and its Conservation
…[7-1]
Linear momentum, being the product of a scalar and a
vector, is a vector. It has dimensions M.L.T–1 and units
kg.m.s–1. The direction of the momentum vector is the
same as the velocity vector.
We can express Newton’s second law in terms of
linear momentum in this way:
dp
∑ F = dt .
dv
= ma .
dt
The second line follows if m is constant.3 T h u s
the
resultant force on an object (system) equals the time
rate of change of linear momentum of the object
(system). The definition of linear momentum enables
us to put the second law into a more general powerful
form. If, in addition, the system is an isolated one then
we can formulate a law of conservation of linear
momentum as we shall see next.
The linear momentum of a particle of mass m moving
with velocity v is defined2
p = mv .
d p d(mv)
=
dt
dt
F 21 =
d p1
dt
and
F12 =
d p2
.
dt
r
Here F12 is the force that particle 1 exerts on particle 2.
The third law requires that the forces be equal and
oppositely directed. Their sum is therefore:
…[7-2]
€
1
The use of collisions as a tool in physics research has been of
immense value. What is known about the structure of solids has
been largely discovered with collision techniques. The structure of
elementary particles is studied by means of high-energy collisions.
2
This is the non-relativistic definition, the definition that applies
at speeds much less than the speed of light. The relativistic definition that applies at any speed will be given in PHYA21H3S.
F 21 + F12 = 0 ,
3
The first line holds whether m is constant or not. A system in
which m might not be constant is a relativistic one. This topic is
discussed in PHYA21H3S.
07-1
Note 07
eq[7-4] are known, then the fourth can be calculated.
Eq[7-4] applies in a space of any dimension and can
be extended to apply to any number of particles. Let
us consider an example.
Example Problem 7-1
Finding an Unknown Momentum
An isolated system consists of two particles that
interact in some manner. A sophisticated instrument
enables us to measure the momentum of the particles
before the interaction; they are found to be,
respectively,
)
)
)
p1i = 1.00 i + 2.00 j + 3.00 k ,
…[7-5a]
Figure 7-1. Two particles making up an isolated system are
shown interacting via non-contact forces.
and
or using eq[7-2],
After the interaction, we measure the momentum of
particle 1 and obtain
)
)
)
p1 f = 3.00i + 4.00 j + 5.00k . …[7-5c]
d p1 d p 2 d
+
= (p1 + p 2 ) = 0 .
dt
dt
dt
Momentum is a vector and the sum p1 + p 2 , the total
momentum of the system, is also a vector.4 Since the
time derivative of the total linear momentum is zero,
then it follows that the total linear momentum is
constant:
r
ptot = const .
…[7-3]
p1i + p2i = p1 f + p2 f ,
…[7-4]
where i and f denote initial and final values of the
linear momentum for the elapsed time over which the
particles interact. Eq[7-4] states in so many words that
the total momentum of the final state of the system
equals the total momentum of the initial state of the
system.
Eq[7-3] is a mathematical statement of the law of
conservation of linear momentum. In words, in any
interaction the total linear momentum of any isolated
system is constant, or remains unchanged, or is
conserved. Obviously, if three of the four quantities in
4
We shall shorten the phrase linear momentum to just momentum
in those cases when no confusion with angular momentum is likely
to occur.
07-2
Calculate the momentum of particle 2 after the
interaction.
Solution:
The system is an isolated one so the law of conservation of linear momentum applies. From eq[7-4] we
solve for p 2 f :
p 2 f = p1i + p 2i – p1 f .
Specifically, for a system of two particles,
€
)
)
)
p 2i = 2.00i + 3.00 j + 4.00k . …[7-5b]
Thus by substituting eqs[7-5] into the above expression we have what we seek:
)
)
)
p 2 f = (0.00)i + (1.00) j + (2.00)k .
You can see that the x-component of the final
momentum of particle 2 is zero. This means that after
the interaction particle 2 moves in the yz-plane.
The law of conservation of linear momentum is very
powerful and enables us to solve certain collision
problems more easily than would otherwise be the
case. Can you think of an easier way of solving this
problem by applying Newton’s laws or the kinematic
equations developed in earlier notes? Clearly, we
must add the law of conservation of linear
momentum to our collection of physics tools.
Note 07
Let us consider a second, more real world example.
Example Problem 7-2
Conservation of Linear Momentum
A 60.0 kg boy and a 40.0 kg girl, both wearing ice
skates on an ice surface face each other initially at rest.
The girl pushes the boy, sending him eastward with a
speed of 4.0 m.s–1. Describe the subsequent motion of
the girl. (Ignore friction.)
Solution:
We consider the system of boy and girl as an isolated
one since the boy and girl interact only with each
other and not with their environment. Applying the
law of conservation of momentum, the total momentum of the system before the push equals the total
momentum of the system after the push. Let us give
the momentum vector pointing east a positive sign,
the momentum vector pointing west a negative sign.
Thus we have
can be quite large. Because the elapsed time is usually
short, it is useful to define a quantity called impulse,
which we do next.
Impulse and Momentum
Impulse is another word used loosely in everyday
speech. In physics, the word impulse is defined as
simply change in momentum. In a collision between two
particles (and especially a contact collision) the force
of interaction might vary with time as is shown in
Figure 7-2. The force is relatively short-lived, being
zero before clock time ti and zero after clock time t f
and having a relatively large value at maximum. The
elapsed time for the interaction is to a good
approximation ∆t = tf – ti.
p before = 0 = p boy + pgirl
p girl = – p boy .
so
The girl’s momentum vector points in a direction
opposite the boy’s, so as he moves eastward she
moves westward. Her speed is
v=
mboy vboy (60.0kg)(4.0m.s–1 )
=
= 6.0 m.s –1.
mgirl
(40.0kg)
A point to appreciate here is that the boy is heavier
than the girl and both boy and girl are subject to the
same magnitude of force (the push—the third law).
But because momentum is conserved here not speed,
the girl moves away after the interaction at a higher
speed than does the boy.
Figure 7-2. A force that varies over a relatively short elapsed
time. The area under the force curve is equal to the
magnitude of the impulse.
Recall that Figure 7-1 depicts two particles interacting
without making apparent “contact”. In the event that
the particles do make contact then the elapsed time
over which the forces act with appreciable magnitude
might be quite short. (A good example is the collision
between two billiard balls).5 During this time the force
of interaction can vary widely, and at its maximum
d p = ∑ F(t)dt .
5
A somewhat less ideal example is the collision of a glider with
the elastic band at the end of an air track.
Rearranging eq[7-2] we have for an infinitesmal
change in momentum:
Notice that the sum of forces is a function of time. To
find the change in momentum we must integrate over
the elapsed time for the interaction:
tf
Δ p = p f – pi =
∫ ∑ F(t)dt .
ti
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Note 07
The change in momentum
is defined as the impulse
r
and given the symbol J . Thus
€
r
J=
tf
r
r
∫ ∑ F (t)dt = Δp .
…[7-5]
ti
Impulse has the same dimensions and units as
momentum and is also a vector. You should be able to
spot from
€ eq[7-5] and Figure 7-2a that the impulse has
a magnitude equal to the area under the force curve
between the two clock times, that is, over the elapsed
time of the collision. The direction of the impulse
vector is the same as the direction of the change in
momentum vector.
Collisions
As implied above, two particles colliding may or may
not make contact in the conventional sense. Figure 7-3
shows how contact and non-contact collisions might
be depicted when the force of interaction is repulsive.
contact force as shown in Figure 7-3a is, in essence, a
collision involving a non-contact force such as the
electromagnetic force.6
Collisions are generally classified as being inelastic,
perfectly inelastic or elastic. An inelastic collision is the
most general and common type of collision, with
elastic and perfectly inelastic collisions being special
cases. We consider the perfectly inelastic collision first
in the next section because in many respects it is the
simplest kind of collision to describe mathematically.
One-Dimensional Perfectly Inelastic Collision
A perfectly inelastic collision is defined as a collision
of two objects in which the objects stick together and
move as one object after the collision. Let the objects
have mass m1 and m2 and initial velocity components
v1i and v 2i along a straight line (Figure 7-4). As before
we assume that the system is isolated.
Figure 7-4. Velocity vectors for a perfectly inelastic head-on
collision between two particles in an isolated system before
the collision (a) and after the collision (b).
Figure 7-3. A collision may or may not involve the conventional notion of contact. In classical physics, collisions (a)
and (b) are described by the same mathematics (allowing for
the fact that (a) is a 1D collision and (b) is a 2D collision) if
the force of interaction is repulsive.
After the collision the objects stick together and move
as one object with some common velocity component
vf. Because the total linear momentum of the twoobject isolated system before the collision equals the
total linear momentum of the combined-object system
after the collision we have
m1 v1i + m2 v 2i = (m1 + m2 )v f ,
Whether two particles actually make contact in a
collision or not is irrelevant to the mathematical
description of the problem. An argument can be made
that if we were to zero in on smaller and smaller
distances (using some kind of super microscope) then
even a collision involving what appears to be a
07-4
6
…[7-6]
The visualization of this assertion requires a stretch of the
imagination. Its complete description is a part of most higher-level
courses in classical mechanics.
Note 07
and therefore
vf =
m1v1i + m2 v2i
.
m1 + m2
…[7-7]
Thus if the masses and their initial velocities are
known, then the final velocity of the combination can
be calculated. Let us consider a numerical example.
Example Problem 7-3
A Perfectly Inelastic Collision
Two spherical lumps of plasticine of masses 0.50 kg
and 1.00 kg approach each other with velocities 1.00
m.s –1 and 0.50 m.s–1, respectively, along the same line.
The first lump is moving to the right, the second to the
left. They collide in a perfectly inelastic collision. Find
the velocity of the combined lump after the collision.
Solution:
Applying eq[7-7] and being careful to give the
velocity of the second lump a negative sign, since it is
moving to the left, we have
vf =
(0.50kg)(1.00m.s –1 ) + (1.00kg)(–0.50m.s –1)
0.50kg + 1.00kg
=0!
Figure 7-5. The velocity vectors for two objects in an isolated system undergoing a head-on elastic collision.
Using the fact that kinetic energy is also conserved we
must also have
1
2
1
2
1
2
1
2
m1v1i2 + m2 v 22i = m1v12 f + m2 v22 f . …[7-9]
Collecting over m1 and m2, eq[7-9] reduces to
m1 (v1i2 – v12 f ) = m2 (v22 f – v 22i ) .
When factored, this equation becomes
After the collision the combined lump has a speed of
zero and is therefore at rest.
Question: Because the velocity of the combination
after the collision is zero, the total momentum after
the collision is also zero. But does this mean that the
collision violates the law of conservation of linear
momentum?
…[7-10]
Now consider eq[7-8]. Collecting over terms in m 1 and
m2 in eq[7-8] gives:
m1 (v1i – v1 f ) = m2 (v2 f – v 2i ) .
…[7-11]
Dividing eq[7-10] by eq[7-11] yields
One-Dimensional Elastic Collision
An elastic collision is defined as a collision in which
kinetic energy is conserved as well as linear momentum. Consider the head-on elastic collision of two
objects (Figure 7-5). Let the particles have mass m1 and
m2, respectively, and initial velocity v1i and v 2i ,
respectively, with m 1 initially moving to the right and
m2 initially moving to the left. Let the particles collide
head-on, and after the collision have velocity v1 f and
v 2 f respectively. Applying the law of conservation of
linear momentum we have
m1 v1i + m2 v 2i = m1 v1f + m2 v2 f .
m1 (v1i – v1 f )(v1i + v1 f ) = m2 (v2 f – v 2i )(v2 f + v2i ) .
…[7-8]
v1i + v1f = v 2 f + v2i ,
which, when rearranged, is
v1i – v2i = –(v1 f – v2 f ) .
…[7-12]
Eq[7-12] states in so many words that the relative
velocity of the two particles after the collision is equal
to the negative of the relative velocity of the two
particles before the collision—or the relative velocity
of the two particles undergoes a reversal in the
collision.
07-5
Note 07
If the masses and the initial velocities of the two
particles are known then the final velocities can be
calculated from eqs[7-8] and [7-12]. After some
manipulation the results are:
 m – m2 
 2m2 
v1 f =  1
 v1i + 
v
 m1 + m2 
 m1 + m2  2i
…[7-13]
and
The Center of Mass
In any object or collection of objects a special point
exists called the center of mass. The idea of the center of
mass can be understood with the help of the object
drawn in Figure 7-6, two balls connected by a rigid
rod. The center of mass (denoted CM) of the object is,
as we shall see, a point located somewhere on the rod
between the balls.
 2m1 
 m2 – m1 
v2 f = 
 v1i + 
v .
 m1 + m2 
 m1 + m2  2i
When using eqs[7-13], you must take care to assign
the correct signs to the velocities. For example, if
particle 2 is initially moving to the left then v2i must be
given a negative sign.
Eqs[7-13] are general expressions in that they
describe two simpler special cases:
1 If m 1 = m2 then vif = v 2i and v2f = v 1i. In other words,
if the particles have the same mass then they
exchange speeds in the collision—particle 1 moves
off with the initial speed of particle 2 and vice
versa.
2 If m 2 is initially at rest then v2i = 0 and eqs[7-13]
reduce to
 m – m2 
v1 f =  1
v
 m1 + m2  1i
…[7-14]
and
 2m1 
v2 f = 
v .
 m1 + m2  1i
If m 1 is very large in comparison to m 2 then eqs[7-14]
show that v 1f ≈ v1i and v2f ≈ 2v1i. In other words, if a
very large mass collides head-on with a very small
mass then the large mass continues on afterwards
with the same speed whereas the small mass takes on
twice the initial speed of the large mass.
Thus far we have studied head-on collisions only, in
other words, collisions in 1D space. But a collision can
be a “glancing” one, in 2D or 3D space, as was shown
in Figure 7-3b. Because of the complexity of the
algebra and lack of time we shall leave 2D and 3D
collisions to a higher-level course in classical
mechanics. We have established the physics, the rest
is algebra.
Figure 7-6. Illustration of the meaning of the center of mass.
To get the idea of the center of mass suppose we
apply an external force horizontally from the left to a
point in the object. If the point is above the center of
mass, then the object will tend to rotate in a clockwise
direction. If the point is below the center of mass, then
the object will tend to rotate in a counter-clockwise
direction. If the point is at the center of mass, then the
object will tend to move to the right with translational
motion only. It will not rotate at all.
Without resorting to a lot of mathematics we can
state a number of attributes of the center of mass of
certain objects. One attribute, the most important one,
is that if the object is a uniform rigid object then its
center of mass is located at its geometric center.7
7
Isaac Newton refused to publish his theory of gravitation in
his “Principia Mathematica” until he had proved this attribute
geometrically. The task took him several years.
07-6
Note 07
To Be Mastered
•
•
•
•
•
Definitions: linear momentum, isolated system
Definition: impulse
Definition: center of mass
Definitions: elastic collision, inelastic collision, perfectly inelastic collision
Laws: Conservation of Linear Momentum
Typical Quiz/Test/Exam Questions
1.
Describe one example of each of the following:
(a) an elastic collision,
(b) a perfectly inelastic collision,
(c) a collision in which linear momentum is not conserved.
2.
A 15.0 g bullet is fired horizontally into a 3.00 kg block of wood suspended by a long cord and initially at rest
(see the figure). The bullet sticks in the block and the block swings 10.0 cm above its initial level. Answer the
following questions.
10.0 cm
bullet
(a) What kind of collision is involved here?
(b) What was the speed of the bullet in m.s–1 at the time of impact?
07-7
Note 07
3.
A ball of mass 0.500 kg is released from rest a height h = 30.0 cm above a hard smooth surface. Assume the
collision the ball makes with the surface is elastic. Answer the following questions:
m = 0.500 kg
h = 30.0 cm
h'
(a) Is h’ greater than h, less than h or equal to h? Explain your answer.
(b) How long does the ball take to reach the surface?
(c) What is the speed of the ball when it collides with the surface?
4.
An apparatus consists of two sections of straight metal track A and B as shown. The track is grooved so that a
hard rubber ball of mass 200.0 g placed near one end of the track rolls down without falling off. The ball
rebounds from a bumper at C. On one run, the ball is released from a height h = 30.0 cm in section A; it
subsequently rolls through sections A and B and bounces back from C. Kinetic friction is negligible. Answer
the following questions:
h
45˚
A
C
B
(a) What is the acceleration of the ball in section A?
(b) What is the acceleration of the ball in section B?
(c) What is the speed of the ball when it hits the bumper at C?
(d) If the ball bounces back from C in an elastic collision, to what vertical height does the ball return to in
section A?
07-8