Lesson: Chapter - 9
Conservation of Momentum
If we combine Newton’s Third Law with what we know about impulse, we can derive
the important and extremely useful law of conservation of momentum.
Newton’s Third Law tells us that, to every action, there is an equal and
opposite reaction. If object A
exerts a force F on object
B, then object
B exerts a force –F on object
A. The net force exerted between
objects A and
B is zero.
The impulse equation,
J = F?t = ?p,
tells us that if the net force acting on a system is zero, then the impulse, and
hence the change in momentum, is zero. Because the net force between the objects
A and
B that we discussed above is zero, the momentum of the system
consisting of objects A and
B does not change.
Suppose object A is a cue ball and
object B is an eight ball on a pool
table. If the cue ball strikes the eight ball, the cue ball exerts a force on
the eight ball that sends it rolling toward the pocket. At the same time, the
eight ball exerts an equal and opposite force on the cue ball that brings it to
a stop. Note that both the cue ball and the eight ball each experience a change
in momentum. However, the sum of the momentum of the cue ball and the momentum
of the eight ball remains constant throughout. While the initial momentum of the
cue ball,
PA,
is not the same as its final momentum,
PÁ,
and the initial momentum of the eight ball,
PB,
is not the same as its final momentum,
P,
the initial momentum of the two balls combined is equal to the final momentum of
the two balls combined:
The conservation of momentum only applies to systems that have no external
forces acting upon them. We call such a system a closed or isolated system:
objects within the system may exert forces on other objects within the system
(e.g., the cue ball can exert a force on the eight ball and vice versa), but no
force can be exerted between an object outside the system and an object within
the system. As a result, conservation of momentum does not apply to systems
where friction is a factor.
Conservation of Momentum on subject test Physics
The conservation of momentum may be tested both quantitatively and qualitatively
on subject test Physics. It is quite possible, for instance, that subject test Physics will
contain a question or two that involves a calculation based on the law of
conservation of momentum. In such a question, “conservation of momentum” will
not be mentioned explicitly, and even “momentum” might not be mentioned. Most
likely, you will be asked to calculate the velocity of a moving object after a
collision of some sort, a calculation that demands that you apply the law of
conservation of momentum.
Alternately, you may be asked a question that simply demands that you identify
the law of conservation of momentum and know how it is applied. The first
example we will look at is of this qualitative type, and the second example is
of a quantitative conservation of momentum question.
Example 1
|
|
|
An apple of mass m falls into the bed of a moving toy truck of mass M.
Before the apple lands in the car, the car is moving at constant velocity
v on a frictionless track.
Which of the following laws would you use to find the speed of the toy truck
after the apple has landed?
|
|
(A) |
Newton’s First Law |
|
(B) |
Newton’s Second Law |
|
(C) |
Kinematic equations for constant acceleration |
|
(D) |
Conservation of mechanical energy |
|
(E) |
Conservation of linear momentum |
Although the title of the section probably gave the solution away, we phrase the
problem in this way because you’ll find questions of this sort quite a lot on
subject test Physics. You can tell a question will rely on the law of conservation of
momentum for its solution if you are given the initial velocity of an object and
are asked to determine its final velocity after a change in mass or a collision
with another object.
Some Supplemental Calculations
But how would we use conservation of momentum to find the speed of the toy truck
after the apple has landed?
First, note that the net force acting in the x
direction upon the apple and the toy truck is zero. Consequently,
linear momentum in the x
direction is conserved. The initial momentum of the system in the
x direction is the momentum of the
toy truck,Pi = Mv.
Once the apple is in the truck, both the apple and the truck are traveling at
the same speed,
Vf.
Therefore,
Vf = mvf + Mvf = (m + M)vf.
Equating
Pi
and
Pf,
we find:
As we might expect, the final velocity of the toy truck is less than its initial
velocity. As the toy truck gains the apple as cargo, its mass increases and it
slows down. Because momentum is conserved and is directly proportional to mass
and velocity, any increase in mass must be accompanied by a corresponding
decrease in velocity.
Example 2
A cannon of mass 1000 kg launches a cannonball of mass 10 kg at a velocity of
100 m/s. At what speed does the cannon recoil?
Questions involving firearms recoil are a common way in which subject test Physics may
test your knowledge of conservation of momentum. Before we dive into the math,
let’s get a clear picture of what’s going on here. Initially the cannon and
cannonball are at rest, so the total momentum of the system is zero. No external
forces act on the system in the horizontal direction, so the system’s linear
momentum in this direction is constant. Therefore the momentum of the system
both before and after the cannon fires must be zero.
Now let’s make some calculations. When the cannon is fired, the cannonball
shoots forward with momentum (10 kg)(100 m/s) = 1000
kg · m/s. To keep the total momentum of the system at zero, the cannon must then
recoil with an equal momentum:
Pcannon = mvcannon
= 1000 kg.ms = (1000 kg)vcannon
vcannon = 1 m/s
Any time a gun, cannon, or an artillery piece releases a projectile, it
experiences a “kick” and moves in the opposite direction of the projectile. The
more massive the firearm, the slower it moves.
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Practice Questions
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