### Video Transcript

A rocket of mass 200.0 kilograms in
deep space moves with a velocity of 121π plus 38.0π metres per second. Suddenly, it explodes into three
pieces. The first piece has a mass of 78.0
kilograms and moves at negative 321π plus 228π metres per second. The second piece has a mass of 56.0
kilograms and moves at 16.0π minus 88.0π metres per second. Find the velocity of the third
piece.

We can label this velocity π£ sub
three. And we know it will have two
components, an π or π₯- and π or π¦-component. Considering the information weβve
been given, weβll label the mass of the rocket as capital π. And weβll call its velocity π£ sub
π. Weβll label the first piece of
exploded mass π sub one and the velocity of that mass π£ sub one. And weβll call the second piece
that exploded π sub two. And the velocity of that piece
weβll label π£ sub two.

If we drew a sketch of the initial
and final states of this scenario, initially, we have our large rocket mass capital
π moving at a velocity π£ sub π. And finally, after the explosion,
our big mass π is broken up into three smaller pieces each moving with their own
velocity. To solve for the velocity of the
third unknown piece, weβll use the principle of the conservation of momentum. This principle tells us that if we
add up all the momentum in our system initially, thatβs 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 π₯ and π¦. We can consider these dimensions
separately one at a time. Weβll start in the π₯-direction and
write out the conservation of linear momentum in that dimension. We can say that the initial
momentum in the π₯-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 π₯-direction is the mass capital π times the speed of
that mass in the π₯-direction.

When we consider the momentum in
the π₯-direction finally after a rocket has broken into three separate pieces,
thatβs equal to π one times the π₯-component of π oneβs velocity plus π two times
π£ two π₯ plus π three times π£ three π₯, that is, the sum of the momenta of each
of our three separate masses in the π₯-direction. In the problem statement, weβre
told the value for capital π, π sub one, and π sub two. We can use these three values to
solve for π sub three. π sub three is equal to capital π
minus π sub one minus π 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
weβll keep track of as the mass of our third particle.

We now have all the information
needed in order to solve for the π₯-component of our third piece π£ sub three
π₯. We rearrange this equation to solve
for π£ sub three π₯. When we plug in for these many
values, weβre inserting the values of the masses of our original and final three
pieces as well as the π₯-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 weβll 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 π₯-component of
the velocity of our third piece, 732π metres per second.

We next wanna solve for the π
component of that velocity. To do this, weβll follow a similar
process. In the π¦-direction, we know that
the initial momentum is equal to the final momentum. Initially, the momentum in this
direction is capital π times π£ sub π in the π¦-dimension. And finally, the π¦ momentum is π
one times π£ one π¦ plus π two times π£ two π¦ plus π three times π£ three π¦. Once again, we rearrange this
expression to solve for the variable weβre interested in. In this case, the π¦-component of
particle three is velocity. When we plug all these values in
this time using the π¦-component of our velocities is for our original mass and two
of our final masses, we find that to three significant figures π£ sub three π¦ 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.