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In this video, we're going
to take a Closer Look at
the Conservation of Momentum.
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Recall that momentum
is a property of a body
that is in motion.
It's equal to the mass of
the body times its velocity.
We've seen that when objects
interact with each other,
the total momentum of the object
remains constant, regardless of
the nature of the forces
acting between the objects.
In other words,
momentum is conserved.
As an equation, the conservation
of momentum says that
the initial momentum of a system
is equal to its final momentum.
Here's an example of a problem
in which the conservation
of momentum can be used to
calculate the final velocity of
an object, after an interaction
with another object.
Suppose a boater has reached
the shore, and is stepping out
of his boat onto the land.
The boater has a mass
of 81 kilograms, and was
in a 65 kilogram boat.
If he steps out of the boat
with a velocity of 2.3 meters
per second towards the shore,
what is the final velocity
of the boat?
The first thing we want to do is
set up a data table that
clearly shows the information
that is given to us, and what
we're trying to solve for.
M sub 1, which will refer
to the mass of the boater,
is equal to 81 kilograms.
V sub 1i, which is the initial
velocity of the man, is going to
be zero, because he is initially
at rest in the boat.
The final velocity of the man,
v sub 1f, is going to be equal
to 2.3 meters per second towards
the shore, because that is
the final velocity of the boater
as he steps onto the shore.
The mass of the boat, m sub 2,
is equal to 65 kilograms.
The boat's initial velocity,
v sub 2i, is equal to zero,
since the boat is also
initially at rest.
We want to solve for the boat's
final velocity, v sub 2f.
Since the man and the boat
are both initially at rest,
both initial velocities
will be zero.
Therefore, the total initial
momentum of the system
is also equal to zero.
Since momentum has to be
conserved, the final momentum
of the system must also
be equal to zero.
We can write that out as m sub 1
times v sub 1f, plus m sub 2
times v sub 2f is equal to zero.
We can move m sub 2 times
v sub 2f to the other
side of the equation,
and see that m sub 1 times
v sub 1f equals negative
m sub 2 times v sub 2f.
Now all we need to do is
rearrange the equation to
solve for the final velocity
of the boat.
We divide both sides by m2,
and see that it is equal
to negative m sub 1
times v sub 1f,
divided by m sub 2.
We plug in 81 kilograms
for m sub 1,
2.5 meters per second
for v sub 1f, and 65 kilograms
for m sub 2,
and find that the final
velocity of the boat is negative
2.9 meters per second,
or 2.9 meters per second
away from the shore.
Here's another example
of a problem in which
conservation momentum is
used to find the answer.
If a 41 kilogram student
is on ice skates, holding
a 23 kilogram exercise ball,
and then the student
throws the ball horizontally,
causing the student
to move backwards
at 3.4 meters per second,
what would be the final
velocity of the ball?
The first thing we want to do
is create a data table
that lists everything
that we're given,
along with what
we're solving for.
The mass of the student,
which we will call m sub 1,
is 41 kilograms.
The mass of the exercise ball,
which we will call m sub 2,
is 23 kilograms.
Initially, both the student
and the ball have a velocity of
zero, before the ball is thrown.
So this will be a similar setup
to the last problem,
in which we know
the final momentum must
be zero, since the initial
momentum is zero.
We can write that out as m sub 1
times v sub 1f, plus m sub 2
times v sub 2f equals zero.
We want to solve for v sub 2f,
the final velocity of the ball.
To do that, we'll rearrange this
equation by subtracting m sub 1,
v sub 1f from both sides,
and then dividing
both sides by m sub 2,
the mass of the ball.
Plugging our values into
the equation and solving,
we find that the final velocity
of the ball must be 6.1 meters
per second, after rounding
to two significant figures.
The direction of the velocity
vector would be in the opposite
direction of the student that is
gliding after throwing the ball.
Now here's one for you to try.
A soccer player places
a 74 kilogram automatic
soccer ball throwing machine
on a frictionless surface,
and gives it a push,
so it moves at 1 meter
per second to the left.
While it is sliding,
the machine fires
a 0.43 kilogram
soccer ball horizontally,
with a velocity of 28 meters
per second to the right.
What is the final velocity
of the machine?
Take a moment, press pause,
and try it for yourself.
Did you get negative 1.2 meters
per second?
If so, great work.
If not, let's take a look
at how we got that answer.
As always, we want to construct
a data table showing
all the data that's given,
along with what
we're solving for.
The mass of the soccer ball
machine, m sub m,
is 74 kilograms.
The initial velocity of
the machine and the soccer ball
is one meter
per second to the left,
in the negative x direction.
We're solving for the final
velocity of the machine,
which we will designate
as v sub mf.
The mass of the soccer ball
we will designate as m sub s,
which is 0.43 kilograms.
The final velocity of
the soccer ball, v sub sf,
is 28 meters per second.
Starting with the law
of momentum conservation,
in this case,
the initial momentum of
our system is not zero.
Both the machine and soccer ball
are initially moving
at -1 meter per second.
We want to solve for the final
velocity of the machine.
To do that,
let's rearrange this equation.
First, we subtract the final
momentum of the soccer ball
from both sides.
And then we divide both sides
by the machine's mass.
We now see that the final
velocity of the machine equals
the mass of the machine
times its initial velocity,
plus the mass of the soccer ball
times its initial velocity,
minus the soccer ball's mass,
times its final velocity,
all divided by
the mass of the machine.
Plugging in all our numbers
from the data table,
we find that the final
velocity of the machine
is -1.2 meters per second,
or 1.2 meters per second
to the left,
in the negative direction.
That's a Closer Look at
Conservation of Momentum.
For more "Physics in Motion",
go to our homepage,
where the entire series
is available to you.