Video Transcript
In this video, we’ll learn how to
apply the law of conservation of momentum to study collisions in one dimension and
differentiate between elastic and inelastic collisions.
We’ll begin by recalling the
relationship between the impulse produced by some force and its change of
momentum. Specifically, let’s think about a
body of constant mass. The impulse vector 𝐉 produced by
the action of some force vector 𝐓 over an interval of time 𝑡 sub one to 𝑡 sub two
is equal to the change in momentum of the body. Let’s demonstrate in particular
what that looks like.
Let’s consider two particles one
and two with momenta vector 𝐏 sub one 𝑖 and vector 𝐏 sub two 𝑖,
respectively. During the moment at which they
collide, particle one exerts a vector force 𝐅 sub one two on particle two, whilst
particle two exerts a vector force of 𝐅 sub two one on particle one. These two forces result purely from
the interaction between the two particles. Specifically, Newton’s third law of
motion tells us that these forces are equal in magnitude but opposite in
direction. So we can say that the vector 𝐅
sub one two is equal to the negative vector 𝐅 sub two one.
Due to the force that is exerted by
particle two on particle one during the collision, particle one experiences a change
in momentum. This change in momentum, the change
in the vector 𝐏 sub one, is equal to the difference between the vector 𝐏 sub one
𝑓, that’s the final momentum, and the vector 𝐏 sub one 𝑖, that’s the initial
momentum. And that of course is equal to the
impulse vector 𝐉 sub one, which can in turn be considered as the definite integral
between 𝑡 sub one and 𝑡 sub two. That’s the time of the collision of
the vector 𝐅 sub two one with respect to 𝑡.
In a similar way, we can calculate
the change in momentum of particle two. It’s equal to the difference
between the vector 𝐏 sub two 𝑓, that’s the final momentum of this particle, and
the vector 𝐏 sub two 𝑖, the initial momentum of this particle. But remember, we said that the
vector 𝐅 sub one two is equal to the negative of the vector 𝐅 sub two one. So in turn, we can say that the
impulse 𝐉 sub one is equal to the negative impulse 𝐉 sub two. This then allows us to identify
that the change in momentum of the first particle is equal to the negative change in
momentum of the second. We can then rearrange this formula
to show a really important fact. By adding the change in momentum of
the second particle to both sides, we find that the change in momentum of particle
one plus the change in momentum of particle two is equal to zero.
So what does this final statement
actually tell us? It says that there is no change in
the total momentum of the two particles during the collision. In other words, the total momentum
is constant. And this allows us to say that
momentum is a conserved quantity.
Let’s define this a little more
formally. The law of the conservation of
momentum tells us that if we take two or more bodies in an isolated system, in other
words, a system which means that there are no external forces acting on it and these
bodies are acting upon each other, their total momentum remains constant. In particular, the vector sum of
the momentum is equal to zero. So we know that for any collision
between two particles, the total momentum is conserved if we assume there are no
other interactions than the one between the colliding particles. And note that whilst we’ve provided
these definitions using vector quantities, in practice, we tend to work with scalar
values. And the conservation of momentum
law allows us to do so with minimal concern.
So with this in mind, let’s look at
solving a problem which involves finding the impulse during a collision between two
objects that are moving towards one another.
Two spheres, 𝐴 and 𝐵, of
equal mass were projected toward each other along a horizontal straight line at
19 centimeters per second and 29 centimeters per second, respectively. As a result of the impact,
sphere 𝐵 rebounded at 10 centimeters per second. Find the velocity of sphere 𝐴
after the collision given that its initial direction is the positive
direction.
We first see that we’re told
that the initial direction of sphere 𝐴 is positive. So we might begin by drawing a
sketch of the scenario. We have sphere 𝐴 and 𝐵 moving
towards one another at 19 centimeters per second and 29 centimeters per second,
respectively. Sphere 𝐴 is moving in the
positive direction, which means sphere 𝐵 is moving in the negative
direction. We’re also told that each
sphere has an equal mass. So let’s define the mass of
each of our spheres to be 𝑚 grams.
Now, in practice, we tend to
work with kilograms and meters per second. But here we’re working with
centimeters per second. And so it’s more usual to be
consistent and use grams. We’re going to begin by
calculating the initial momentum of this system. We know that momentum is
conserved. So this will allow us to
compare the initial momentum with the final momentum, which should then in turn
allow us to calculate the final velocity of sphere 𝐴. We also know that we often
calculate momentum by using vector quantities for momentum and velocity. Since scalar momentum and
scalar velocity can be thought of as vectors in one direction, this means we can
generalize this for scalar quantities also.
So, we know that we can define
the initial momentum of sphere 𝐴 as 𝑝 sub 𝐴, 𝑖. And it’s its mass times
velocity. That’s 𝑚 times 19 or 19𝑚. In a similar way, we can
calculate the initial momentum for sphere 𝐵. 𝑝 sub 𝐵, 𝑖, that’s the
initial momentum, is mass times velocity. So that’s 𝑚 times negative 29
or negative 29𝑚. Then, we can calculate the
total momentum in the system — let’s call that 𝑝 sub net 𝑖, the initial net
momentum — by finding the sum of the initial momentum of 𝐴 and 𝐵. That’s 19𝑚 plus negative 29𝑚,
which is negative 10𝑚.
Now that we’ve identified
what’s happening before the collision, let’s think about what’s happening
immediately after. We’re told that the spheres
rebound. In other words, they move
immediately after the collision in opposite directions. We’re told sphere 𝐵 rebounds
at 10 centimeters per second, so it travels in the opposite direction at this
velocity. We’re trying to find the
velocity of sphere 𝐴. So let’s define that to be 𝑣
centimeters per second. And we’re assuming it’s moving
in the negative direction.
Then, we calculate the net
momentum by considering the final momentum of each sphere. The final momentum of sphere 𝐴
is going to be negative 𝑚𝑣, whilst the final momentum of sphere 𝐵, let’s call
that 𝑝 sub 𝐵, 𝑓, is mass times velocity, that’s 10𝑚. Then, we find their sum to find
the final net momentum. That’s negative 𝑚𝑣 plus
10𝑚.
Now, according to the principle
of conservation of momentum, the net momentum before the collision must be equal
to the net momentum after the collision. In other words, negative 10𝑚
must be equal to negative 𝑚𝑣 plus 10𝑚. Let’s clear a little bit of
space and solve this equation. To do so, we might first begin
by noticing that each expression in this equation contains a factor of 𝑚. We also defined 𝑚 to be the
mass of each object, so it cannot be equal to zero, meaning that we can divide
our entire equation by 𝑚. When we do, we get negative 10
equals negative 𝑣 plus 10. Subtracting 10 from both sides,
and our equation becomes negative 20 equals negative 𝑣.
And so the velocity of the
sphere after the collision is 20 centimeters per second. But of course, we define this
motion to be to the left. Since velocity is directional
and we define the direction to the right to be positive, we have to say that the
final velocity of sphere 𝐴 is negative 20 centimeters per second.
Now, at this stage, it’s worth
noting that we could have alternatively modeled the motion after the collision
slightly differently. In other words, we could have
defined the arrow 𝑣 centimeters per second to be moving to the right. If we’ve done that at this
stage, we would have ended up with 𝑣 equals negative 20 centimeters per second,
thereby still indicating that the object is moving to the left. Either way, the velocity is
negative 20 centimeters per second.
We’ll now demonstrate how this
process can determine the impulse of the force of the collision of bodies moving in
opposite directions as well as the velocity of a body after the collision.
Two spheres of masses 200 grams
and 350 grams were moving toward each other along the same horizontal straight
line. The first was moving at 14
meters per second and the second at three meters per second. The two spheres collided. As a result, the first sphere
rebounded at seven meters per second in the opposite direction. Given that the positive
direction is the direction of motion of the first sphere before the impact,
determine the impulse 𝐼 the second sphere exerted on the first one and the
speed 𝑣 of the second sphere after impact.
Now before we go any further,
we might observe that with our speeds, we’re working in meters per second,
whilst with the masses, we’re working in grams. So we’re going to convert their
masses to SI base units of mass, the kilogram. To do so, we divide each by
1000. We get 0.2 kilograms and 0.35
kilograms, respectively.
So how do we calculate the
impulse 𝐼 that the second sphere exerted on the first? Well, first, we know that the
magnitudes of the impulses on the spheres are equal. We also have information about
the velocity of the first sphere before and after the collision. This means we can calculate the
change in velocity of the first sphere due to the collision. If we call that the change in
𝑣 sub 𝑎, where 𝑎 is the name of the first sphere, then we can say that this
is equal to its final velocity minus its initial velocity.
Now, since it was initially
moving in the positive direction, it will rebound and move in the negative
direction. This means if it’s for
traveling at a speed of seven meters per second, its velocity is negative
seven. So the change in velocity is
negative seven minus 14. And that’s negative 21 meters
per second. Then, we can calculate the
impulse on the first sphere by finding its change in momentum. Momentum is of course mass
times velocity. So we could calculate the
momentum before and after and subtract them. Or alternatively, we can
multiply the momentum by the change in velocity. For our first sphere, we can
then say that its mass is 0.2 and its change in velocity is negative 21. So the impulse on this sphere
is 0.2 times negative 21, and that’s negative 4.2. And the units we use are newton
seconds.
Next, we know that an impulse
of equal magnitude, but opposite in sign, acts on the second sphere. In other words, the impulse on
the second sphere must be 4.2 newton seconds. But of course, we can link that
to its change in momentum. This means it’s equal to its
mass times its change in velocity. Now, since it was initially
moving in the negative direction at a speed of three meters per second, its
change in velocity is 𝑣 minus negative three. So its change in momentum is
0.35 times 𝑣 minus negative three. This right-hand side is
equivalent to 0.35 times 𝑣 plus three. And we calculated the impulse
on the second sphere was 4.2. So we can say 4.2 equals 0.35
times 𝑣 plus three.
Next, we divide through by
0.35. Now, 4.2 divided by 0.35 is
simply 12. Then, we subtract three from
both sides. 12 minus three is nine. So we’ve calculated that 𝑣 is
equal to nine meters per second. And so we’ve answered the
question. The impulse that the second
sphere exerted on the first, which we defined earlier to be 𝐽 sub 𝑎, is in
fact 𝐼 equals negative 4.2 newton seconds. Similarly, the speed 𝑣 of the
second sphere is the magnitude of nine meters per second, which is simply nine
meters per second.
In the two examples we’ve seen so
far, we demonstrated how to use conservation of momentum to calculate various
quantities after a collision. It’s worth noting at this stage
though that whilst momentum is conserved, this doesn’t necessarily mean kinetic
energy is conserved. In fact, there are three
scenarios. The first is said to be an elastic
collision. This is a collision during which
the kinetic energy is conserved. In an inelastic collision, part of
the kinetic energy is dissipated in friction between the particles. Then finally, we have something
called a perfectly inelastic collision. Now, in this collision, kinetic
energy is dissipated in friction as in the previous definition. But the particles are also bonded
together, and this results in them having the same velocity after collision. This might sometimes be called
coalescence.
In our final example, we’ll look at
the motion of two particles after an inelastic collision.
Two spheres are moving along a
straight line. One has a mass of 𝑚 and is
moving at speed 𝑣, whereas the other one has a mass of 10 g and is moving at 36
centimeters per second. If the two spheres were moving
in the same direction when they collided, they would coalesce into one body and
move at 30 centimeters per second in the same direction. However, if they were moving in
opposite directions, they would coalesce into one body, which would move at six
centimeters per second in the direction the first sphere had been traveling. Find 𝑚 and 𝑣.
With the information we’ve been
given, we’re going to set up a pair of simultaneous equations in 𝑚 and 𝑣 using
conservation of momentum. In particular, we’re going to
use the formula momentum is equal to mass times velocity.
Let’s begin by identifying the
direction to the right to be positive. Then, we can say the momentum
of the first sphere is 𝑚 times 𝑣. The momentum of the second
sphere is 10 times 36. That’s 360, giving us a total
or net momentum before the collision of 𝑚𝑣 plus 360. By the conservation of
momentum, we know this must be equal to the net momentum after the
collision. The new mass of the body is 10
plus 𝑚, whilst its velocity is 30. Since it’s in the same
direction, it’s still positive, meaning the momentum after the collision is 30
times 10 plus 𝑚. By distributing parentheses and
rearranging, we can form an equation, 60 equals 30𝑚 minus 𝑚𝑣.
Now that we have this equation,
let’s complete the process for the second scenario. In this scenario, these spheres
are moving in opposite directions. This means the velocity of the
second sphere must be negative 36. So the net momentum is 𝑚𝑣
minus 360 before the collision. After the collision, they
coalesce, giving us a total mass of 10 plus 𝑚. And they move at six
centimeters per second, still in the positive direction. This time, if we distribute the
parentheses and rearrange, we get the equation negative 420 equals six 𝑚 minus
𝑚𝑣.
And now we might observe that
we can quite quickly calculate the value of 𝑚 here. We have a system of equations
in which we have the same term negative 𝑚𝑣. Let’s define these equations to
be one and two, respectively. Then, if we subtract one
equation from the other, and it doesn’t matter which way round we do it, that
negative 𝑚𝑣 term will disappear. 60 minus negative 420 is
480. Then, 30𝑚 minus six 𝑚 is
24𝑚, and negative 𝑚𝑣 minus negative 𝑚𝑣 is zero. If we then divide through by
24, we find that 𝑚 is equal to 20 or 20 grams.
Then, we can find the value of
𝑣 by substituting it into either of our earlier equations. When we do, we get 60 equals 30
times 20 minus 20𝑣, which gives us negative 540 equals negative 20𝑣. Finally, we divide through by
negative 20, and we find 𝑣 is equal to 27 centimeters per second. 𝑚 is 20 grams and 𝑣 is 27
centimeters per second.
We’re now going to recap the key
points from this lesson. In this lesson, we saw that the
momentum of a body vector 𝐩 is given by 𝑚 times the velocity vector 𝐯 but that we
generally work with the scalar form 𝑝 equals 𝑚𝑣. We also saw that if we’re working
in a closed system, the total or net momentum is conserved. In other words, the total momentum
before the collision is equal to the total momentum after. We saw that impulse is equal to the
change in momentum. And we learned what it means for a
collision to be elastic, inelastic, and perfectly inelastic.
