Video Transcript
In this video, we’re going to learn
about momentum in two dimensions. We’ll see how momentum is conserved
across multiple dimensions and how to use this principle in two-dimensional examples
to solve problems. To get started, imagine that a
gigantic iron ball is rolling toward a small child. Noticing this imminent disaster,
you take out a gun and begin to fire bullets at the iron ball. Knowing the mass of each bullet as
well as the speed with which they move, you want to know how many bullets you’ll
need to fire in order to deflect the path of the rolling iron ball enough so that it
misses the child.
To figure this out, we’ll want to
know something about momentum in two dimensions. When it comes to this topic, the
first thing we can recognize is that linear momentum is conserved in any collision
or interaction. This means that if we add up all
the momentum in our system at some initial point in time, then that’s equal to the
momentum of our system at some final point. Our system could be a person
swinging back and forth on a swing or two billiard balls colliding or an object in
space being redirected by the gravitational attraction of a large body. The conservation of momentum
applies to all of these scenarios.
Since momentum is a vector
quantity, this conservation of momentum applies to every single dimension of a
scenario independently. Say we have two hockey pucks, puck
number one and puck number two. Puck two starts at rest and number
one runs into number two and bounces off at some angle, we’ve called 𝜃, relative to
the horizontal. Puck number two is likewise put in
motion at an angle we’ve called 𝜙 below the horizontal. Clearly, this is a two-dimensional
scenario. And the fact that momentum is
conserved and is a vector quantity means that we can write two separate, two
independent, momentum conservation expressions.
Focusing just on the 𝑥-direction,
we might write something like 𝑚 one times 𝑣 one initial in the 𝑥-direction plus
𝑚 two times 𝑣 two initial in the 𝑥-direction equals 𝑚 one 𝑣 one 𝑓 in the
𝑥-direction plus 𝑚 two 𝑣 two 𝑓 in the 𝑥-direction. We can do the same thing in the
𝑦-direction, this time using the 𝑦-component of our speeds instead of the
𝑥-component. This all means we now have two
independent equations of motion we can use to solve for information in this
scenario.
If our scenario involved three
dimensions instead of two, we would have three separate equations. Working with momentum in multiple
dimensions, it’s important to be careful about subscripts — how we identify
different symbols we use. For example, notice that, for each
of our speed values, we have three subscripts, one which identifies the object it
refers to, the second which identifies whether it’s in the initial or final state,
and the third which identifies its dimension of motion. That can be a lot to keep track
of. So as we work through examples, we
want to be careful to keep track of our notation. Let’s get some practice now with an
example exercise.
A rocket of mass 200.0 kilograms in
deep space moves with a velocity of 121𝑖 plus 38.0𝑗 metres per second. Suddenly, it explodes into three
pieces. The first piece has a mass of 78.0
kilograms and moves at negative 321𝑖 plus 228𝑗 metres per second. The second piece has a mass of 56.0
kilograms and moves at 16.0𝑖 minus 88.0𝑗 metres per second. Find the velocity of the third
piece.
We can label this velocity 𝑣 sub
three. And we know it will have two
components, an 𝑖 or 𝑥- and 𝑗 or 𝑦-component. Considering the information we’ve
been given, we’ll label the mass of the rocket as capital 𝑀. And we’ll call its velocity 𝑣 sub
𝑀. We’ll label the first piece of
exploded mass 𝑀 sub one and the velocity of that mass 𝑣 sub one. And we’ll call the second piece
that exploded 𝑀 sub two. And the velocity of that piece
we’ll label 𝑣 sub two.
If we drew a sketch of the initial
and final states of this scenario, initially, we have our large rocket mass capital
𝑀 moving at a velocity 𝑣 sub 𝑀. And finally, after the explosion,
our big mass 𝑀 is broken up into three smaller pieces each moving with their own
velocity. To solve for the velocity of the
third unknown piece, we’ll use the principle of the conservation of momentum. This principle tells us that if we
add up all the momentum in our system initially, that’s equal to all the momentum in
our system finally. And because momentum is a vector,
that means that this conservation applies independently to every dimension of our
situation.
In our case, we have a
two-dimensional scenario, whose dimensions we can call 𝑥 and 𝑦. We can consider these dimensions
separately one at a time. We’ll start in the 𝑥-direction and
write out the conservation of linear momentum in that dimension. We can say that the initial
momentum in the 𝑥-direction is equal to the final momentum of our system in that
same direction. Knowing that the momentum of an
object is equal to its mass times its velocity, we can write that the initial
momentum of our system in the 𝑥-direction is the mass capital 𝑀 times the speed of
that mass in the 𝑥-direction.
When we consider the momentum in
the 𝑥-direction finally after a rocket has broken into three separate pieces,
that’s equal to 𝑚 one times the 𝑥-component of 𝑚 one’s velocity plus 𝑚 two times
𝑣 two 𝑥 plus 𝑚 three times 𝑣 three 𝑥, that is, the sum of the momenta of each
of our three separate masses in the 𝑥-direction. In the problem statement, we’re
told the value for capital 𝑀, 𝑚 sub one, and 𝑚 sub two. We can use these three values to
solve for 𝑚 sub three. 𝑚 sub three is equal to capital 𝑀
minus 𝑚 sub one minus 𝑚 sub two or plugging in for the given values 200.0
kilograms minus 78.0 kilograms minus 56.0 kilograms. This equals 66.0 kilograms which
we’ll keep track of as the mass of our third particle.
We now have all the information
needed in order to solve for the 𝑥-component of our third piece 𝑣 sub three
𝑥. We rearrange this equation to solve
for 𝑣 sub three 𝑥. When we plug in for these many
values, we’re inserting the values of the masses of our original and final three
pieces as well as the 𝑥-component of the velocities of our original piece and two
of our final pieces. In terms of the units involved,
notice that the unit of kilograms cancel out and we’ll be left with a result in
metres per second. When we calculate this fraction to
three significant figures, we find a result of 732 metres per second. So we now know the 𝑥-component of
the velocity of our third piece, 732𝑖 metres per second.
We next wanna solve for the 𝑗
component of that velocity. To do this, we’ll follow a similar
process. In the 𝑦-direction, we know that
the initial momentum is equal to the final momentum. Initially, the momentum in this
direction is capital 𝑀 times 𝑣 sub 𝑀 in the 𝑦-dimension. And finally, the 𝑦 momentum is 𝑚
one times 𝑣 one 𝑦 plus 𝑚 two times 𝑣 two 𝑦 plus 𝑚 three times 𝑣 three 𝑦. Once again, we rearrange this
expression to solve for the variable we’re interested in. In this case, the 𝑦-component of
particle three is velocity. When we plug all these values in
this time using the 𝑦-component of our velocities is for our original mass and two
of our final masses, we find that to three significant figures 𝑣 sub three 𝑦 is
negative 79.6 metres per second. We now have an expression for the
complete two-dimensional velocity of the third piece of exploded mass.
Let’s summarize what we’ve learned
about momentum in two dimensions. First, we’ve seen that linear
momentum is conserved in any collision or interaction whether elastic or inelastic
under no matter how many dimensions are involved. Written as an equation, we can say
that 𝑝 sub 𝑖, the initial momentum in our system, is equal to 𝑝 sub 𝑓, the final
momentum. And since momentum is a vector,
this conservation of momentum applies to every single dimension of the vector. In other words, an 𝑁-dimension
scenario yields 𝑁 independent linear momentum equations. When we’re careful to keep track of
all of our subscripts in two-dimensional scenarios, the conservation of linear
momentum is a powerful tool for understanding object motion.
