In this video, we’re going to learn
about the conservation of linear momentum. We’ll learn why it is that linear
momentum is conserved. And how we can use this fact to
help us solve problems.
To start out, imagine that you are
the manager of a fireworks production company. Before every show, once the
fireworks launchers are put in place, one of your roles is to mark off a perimeter
around these launchers. This perimeter forms the boundary
of the closest people are allowed to get to the firework show itself. To figure out where that perimeter
should be, you want to know the furthest possible point from the launchers that a
piece of the exploded fireworks might travel. To get an idea for that maximum
possible distance, it will be helpful to know about the conservation of linear
We know that the linear momentum of
an object is equal to its mass times its velocity. Written as an equation, lower case
𝑝, the representation of momentum, is equal to 𝑚 times 𝑣. Knowing that, let’s imagine a
scenario where we have two masses. One we’ll call mass 𝐴, and the
other we’ll call mass 𝐵. These masses move towards one
another and eventually collide. When this happens, Newton’s third
Law of motion tells us something about this collision.
This law, which says that for every
action there’s an equal and opposite reaction, tells us that the force mass 𝐴
exerts on 𝐵 in the collision is equal and opposite the force mass 𝐵 exerts on
𝐴. We know that, practically, this
collision will not occur instantaneously, but rather over some span of time we can
call Δ𝑡. If we multiply both sides of this
equation by that time interval, then we see that each side now represents an
impulse, on the left-hand side the impulse of 𝐴 on 𝐵, and on the right-hand side
the impulse of 𝐵 on 𝐴.
If we recall the impulse–momentum
theorem, which says simply that the impulse an object experiences is equal to its
change in momentum, then we can rewrite our expression for impulse to involve the
masses and change in speeds of each of our objects. In particular, the mass of object
𝐴 multiplied by its change in speed is equal to negative the mass of object 𝐵
multiplied by its change in speed.
Say that as a next step, we add 𝑚
sub 𝐵 times Δ𝑣 sub 𝐵 to both sides of this equation, cancelling that term out on
the right-hand side. We now have an equation that tells
us that the change in momentum of object 𝐵 over this collision plus the change in
momentum of object 𝐴 is equal to zero. That is, overall, there is no
momentum change over this interaction.
Another way to express this is to
say that momentum is conserved. That is the momentum at the
beginning of an interaction is the same as the overall momentum after the
interaction. This is what the conservation of
linear momentum means. It means if we count up all the
momentum in a scenario before an interaction and then count up all the momentum in
the scenario afterwards, those two amounts will be the same.
Something important to recall about
the conservation of linear momentum is that since momentum is a vector, it’s
conserved in every independent direction. It’s conserved in the 𝑥-direction
as well as in the 𝑦 as well as the 𝑧 and so on. Let’s consider for a moment how to
practically use this principle of the conservation of linear momentum.
Imagine that we had a firework
which when launched up in the air explodes into two different pieces. Say we want to use the fact that
linear momentum is conserved to solve for the final speed of one of those two
exploded pieces. If we set up a coordinate system,
an origin, we can begin by calculating the initial as well as the final momentum
involved. Since momentum is a vector, we can
apply this conservation principle in both the 𝑥- and the 𝑦-directions.
So, we’ll calculate the initial
momentum in the 𝑥-direction as well as in the 𝑦 and the final momentum in those
two directions separately too. If the firework is initially at
rest, then that means it’s initial momentum in both the 𝑥- and 𝑦-directions is
zero. And if we call the two pieces of
the exploded firework 𝑚 one and 𝑚 two, respectively, then we can write out the
final momentum in the 𝑥- and 𝑦-directions, 𝑚 one 𝑣 sub one 𝑥 plus 𝑚 two 𝑣 sub
two 𝑥. And for 𝑦, 𝑚 one 𝑣 sub one 𝑦
plus 𝑚 two 𝑣 sub two 𝑦.
Once all these terms are written
out, we use the fact that linear momentum is conserved to set the initial and final
momenta in each direction equal to one another. Then, if we’re given enough initial
information, we’re able to solve the information we want. Let’s get some practice with the
conservation of linear momentum through a couple of examples.
A man of mass 62 kilograms stands
at rest on an icy surface that has negligible friction. He throws a ball of mass 670 grams,
giving the ball a horizontal velocity of 8.2 meters per second. What horizontal velocity does the
man have after throwing the ball?
We can label this horizontal
velocity of the man 𝑣 sub 𝑚 and start off by drawing a sketch of the
situation. In this situation, initially, we
have a man holding a ball and standing at rest on a frictionless surface. The man then throws the ball,
moving in what we’ll call the positive direction with the speed 𝑣 sub 𝑏 of 8.2
meters per second. Because momentum is conserved, we
know that as a result of throwing the ball, the man himself will move to the left
with some velocity we’ve called 𝑣 sub 𝑚. Let’s use this conservation
principle, that initial momentum is equal to final momentum, to solve for 𝑣 sub
As we consider the initial
condition of our system, since the man and the ball are both at rest, that means
their speed is zero and, therefore, the overall initial momentum of our system is
also zero. When we consider the final momentum
of our system, after the ball has been thrown, that’s equal to the man’s mass times
his velocity plus the ball’s mass times its velocity. Because linear momentum is
conserved, we can equate our initial and final momentum and write that zero is equal
to 𝑚 sub 𝑚𝑣 sub 𝑚 plus 𝑚 sub 𝑏𝑣 sub 𝑏.
We want to solve for the man’s
velocity 𝑣 sub 𝑚. And that’s equal to negative 𝑚 sub
𝑏𝑣 sub 𝑏 over 𝑚 sub 𝑚. We know the mass of the man and the
ball as well as the velocity of the ball. So, we’re ready to plug in and
solve for 𝑣 sub 𝑚. When we do, we’re careful to
convert the mass of the ball into units of kilograms to agree with the units of the
man’s mass. When we calculate this value, we
find it’s equal to negative 0.089 meters per second. That’s the velocity of the man
sliding across the ice after he has thrown the ball.
Now let’s consider an example
involving two masses which initially are in motion.
Two cars of equal masses approach a
frictionless perpendicular intersection. Car A travels north at 22 meters
per second and car B travels east. The cars collide and stick
together, traveling at 17 degrees north of east. What was the initial speed of car
We can label car B’s initial speed
𝑣 sub B and begin our solution by drawing a sketch of the situation. In this scenario, initially, we
have two cars approaching one another at an intersection. Car A is traveling north at a speed
given as 22 meters per second, and car B is traveling to the east at an unknown
speed. At the center of the intersection,
the cars collide and, finally, they move together as one mass in a direction given
as 17 degrees north of east. Based on this information, we want
to solve for the initial speed of car B.
We’ll use the principle that the
total initial momentum in a system is equal to the total final momentum in that
system. In our scenario, initially and
finally, we have momenta pointing in two different directions. In the northern direction, we have
𝑚 sub A times 𝑣 sub A. And in the eastern direction,
orthogonal to the north–south axis, we have 𝑚 sub B times 𝑣 sub B.
If we then consider the final
momentum in our system, again both in the north and eastern directions separately,
to the north we have our combined mass 𝑚 sub A plus 𝑚 sub B, since the cars have
stuck together, times their final speed, whatever that is, we’ll call it 𝑣 sub 𝑓,
multiplied by the sine of the angle 𝛳. And to the east, we have the same
expression except now we use the cosine of that angle to point in the direction east
Applying our conservation
principle, we can write in the northern direction that 𝑚 sub A𝑣 sub A equals 𝑚
sub A plus 𝑚 sub B times 𝑣 sub 𝑓 times the sine of 𝛳. And separately, in the eastern
direction, 𝑚 sub B𝑣 sub B is equal to the sum of the two masses times their final
velocity times the cosine of 𝛳. We recall that we want to solve for
𝑣 sub B, the initial speed of car B. And very importantly, we’re told in
the problem statement that the masses of the two cars are equal. 𝑚 sub A is equal to 𝑚 sub B. So, if we replace every 𝑚 sub A
and 𝑚 sub B we see in our two equations simply with the mass 𝑚, we see then that
that mass 𝑚 cancels out from both sides of both equations.
Looking at the equation in the
northern direction, we can use the fact that 𝑣 sub A and 𝛳 are given information
to solve for 𝑣 sub 𝑓, the final speed of the combined car mass. 𝑣 sub 𝑓 is equal to 𝑣 sub A over
two times the sine of 𝛳, or 22 meters per second divided by two times the sine of
17 degrees. If we substitute this expression
for 𝑣 sub 𝑓 into 𝑣 sub 𝑓 in our equation for 𝑣 sub B and enter our value of 17
degrees for 𝛳. Then we see our expression
simplifies a bit and that factors of two cancel and the cosine of one angle divided
by the sine of the same angle is equal to the cotangent of that angle.
So, entering these values on our
calculator, we find a result for 𝑣 sub B which, to two significant figures, is 72
meters per second. That’s the initial speed of car B
in this scenario.
Let’s summarize what we’ve learned
about the conservation of linear momentum. We’ve seen that Newton’s third law
of motion implies that a system’s overall change in momentum is zero. That is the initial momentum in the
system is equal to the final momentum in that system. This is what it means that linear
momentum is conserved.
We’ve also seen that since momentum
is a vector quantity, it is conserved in each orthogonal direction, or each
perpendicular direction, of a scenario space, for example the 𝑥-direction, the 𝑦-
direction, the 𝑧-direction, separately. And finally, we’ve seen that a
problem-solving strategy for linear momentum involves considering the initial and
final states of a scenario as well as each component direction.