Video Transcript
In this video, we will learn how to
distinguish between the effects of collisions which do and which do not conserve
kinetic energy. Before we get into the specifics of
the elasticity of collisions, let’s review more generally collisions between two
objects.
There are three general cases when
two objects collide together. First, they can collide and bounce
off of each other. Second is an explosion when the two
objects are linked together and then explode apart. And the third case is when the two
objects come together and stay together. In each of these three situations,
we have conservation of momentum, which we should recall means that the initial
momentum of the system is equal to the final momentum of the system. Expanding that relationship out
farther for our two objects in each case, we could say that the 𝑚 one 𝑣 one
initial plus the 𝑚 two 𝑣 two initial is equal to 𝑚 one 𝑣 one final plus 𝑚 two
𝑣 two final. As we recall, the momentum of an
object is equal to its mass times its velocity.
Let’s look back at our three
scenarios again, but this time through the lens of elasticity. To differentiate our three
scenarios, we can call them bounce, explosion, and stick together, where bounce
refers to two objects that stay separate after they collide, explosion refers to two
objects that start together and then separate into two separate ones, and stick
together refers to objects that start separate and then, after the collision, stay
together. When we talk about the elasticity
of a collision, we usually talk about it in two ways: whether the collision was
elastic or inelastic.
Elastic collisions are one in which
the initial kinetic energy of the system before the collision is equal to the final
kinetic energy of the system after the collision, whereas in an inelastic collision,
the initial kinetic energy of the system before the collision does not equal the
final kinetic energy of the system after the collision. To determine whether a collision is
elastic or inelastic, we must compare the kinetic energies before the collision to
the kinetic energies after the collision. Of the three scenarios, number two,
the explosion, and number three, stick together, must always be inelastic, meaning
that their kinetic energies will not be the same before as after. In fact, the scenario where they
stick together is often times referred to as being perfectly inelastic.
However, in the first scenario,
when the two objects stay separate, this can be either elastic or inelastic,
depending on whether or not the kinetic energy is conserved or is not conserved. Once again, the momentum will be
conserved during the collision, but we’d have to look at the kinetic energies to
determine whether or not it is elastic or inelastic. Let’s apply the definitions of
elastic and inelastic collisions to a few scenarios. If we have two objects traveling
towards each other, the blue object having a kinetic energy of 10 joules and the
pink object having a kinetic energy of 20 joules, how much kinetic energy do we have
before the collision?
To find the initial kinetic energy,
we need to add the energy of the blue object plus the energy of the pink object
before the collision, in this case, 10 joules plus 20 joules, giving our system a
total initial kinetic energy of 30 joules. We’ve seen that earlier that
elastic collisions are one in which the kinetic energy is conserved. This means that the final kinetic
energy of our system must be 30 joules, but what might they look like? It could mean that the blue object
has an energy of 20 joules and the pink object has an energy of 10 joules, which add
together to give us 30. Or maybe the blue object has 15
joules and the pink object has 15 joules. Perhaps they even have the same
energy as initially, with 10 joules for the blue and 20 joules for the pink.
Without knowing the objects’
initial masses or initial velocities, it’s impossible to tell which combination we
will have. However, the one thing we do know
is that since the collision is elastic, our final kinetic energy must be the same as
the initial, in this case, 30 joules. Let’s analyze the situation where
we do know the kinetic energies of all the objects before and after the
collision.
Before the collision, the blue
objects still has 10 joules of kinetic energy and the pink object has 20 joules of
kinetic energy. Adding our two energies together,
10 joules plus 20 joules, we have the initial kinetic energy of 30 joules once
again. After the collision, we know that
the blue object has a kinetic energy of six joules and the pink object has a kinetic
energy of 21 joules. Adding these two energies together,
we get six joules plus 21 joules is 27 joules for the final kinetic energy of the
system. We now know that this is an
inelastic collision as the kinetic energy final does not equal the kinetic energy
initial.
In fact, we can determine the
percent loss of kinetic energy during the collision. We do this by subtracting the final
kinetic energy from the initial kinetic energy, dividing by the initial kinetic
energy, and multiplying by 100. Subtracting 27 joules from 30
joules, we get three joules. When we divide three joules by 30
joules and multiply it by 100, we get 10 percent, which means that our system has
lost 10 percent of its overall kinetic energy.
What if instead of knowing the
kinetic energies of our objects, we knew the masses and velocities? If our blue object had a mass of 10
kilograms and was traveling at two meters per second to the right of our screen
before the collision and came to rest directly after the collision and our pink
object had a mass of two kilograms and was traveling at an initial speed of four
meters per second to the left of our screen before the collision, what would be the
final velocity of the pink object after the collision? If we are told that our collision
is elastic, then we know that the initial kinetic energy of the system is equal to
the final kinetic energy of the system. And recalling that the kinetic
energy of an object is equal to one-half 𝑚𝑣 squared, we can combine these two
equations to solve for our unknown velocity.
Plugging in our values, our
equation becomes one-half times 10 kilograms times two meters per second squared for
the initial kinetic energy of the blue object plus one-half times two kilograms
times four meters per second squared for the initial kinetic energy of the pink
object is equal to one-half times 10 kilograms times zero meters per second squared
for the final kinetic energy of the blue object plus one-half times two kilograms
times the final velocity 𝑣 f of the pink object squared.
Multiplying out the values for the
initial kinetic energy of the blue object, we get 20 joules. And multiplying out the values for
the initial kinetic energy of the pink object, we get 16 joules. Multiplying out the final kinetic
energy of the blue object, we get zero joules. And multiplying out the final
kinetic energy for the pink object, we get one kilogram times 𝑣 final squared. Adding our two initial kinetic
energies of our objects, we get a total initial kinetic energy of the system of 36
joules, with the right side of our equation becoming one kilograms times 𝑣 final
squared. To isolate the final velocity, we
divide both sides by one kilogram.
To solve for our final velocity, we
need to square root both sides of the equation, giving us a final speed for our pink
object of six meters per second. Now that we have gone over the
difference between elastic and inelastic collisions and solved for an unknown
variable, let’s do a few practice problems.
The following diagram shows the
changes in the positions of two objects from time 𝑡 zero to their positions at time
𝑡 one. The objects have the same mass as
each other. The arrows above the objects
indicate the direction of their velocities, but the magnitudes of the velocities are
not known; they may or may not be equal to each other. A point 𝑃 is a distance 𝐷 from
both of the objects at 𝑡 zero, but at 𝑡 one, the objects are at distances 𝑑 one
and 𝑑 two from 𝑃, where 𝑑 one is less than 𝑑 two. The directions of the velocities at
𝑡 one are shown in the diagram, but the magnitudes are not known and may be
zero. No external forces act on either
object.
In our diagram, we can see our two
objects, the green and blue object, which are initially at time 𝑡 zero along their
path. Sometime later, at time 𝑡 one, we
can see our same objects, now at new positions along their path. We are told that the objects have
the same mass, but that their velocities are unknown. The arrows above the objects
indicate the direction that the objects are moving but not the magnitude of their
velocities. We’re gonna make room on our screen
for our questions.
Which of the following is the least
distance from the point 𝑃 to the point at which the objects could have come into
contact with each other? (A) 𝑑 two minus 𝑑 one, (B) 𝑑
two, (C) zero, (D) 𝑑 one, (E) 𝑑 one minus 𝑑 two.
The question asked us to find the
least distance from a comparison point 𝑃 right here on our diagram. We can see from the diagram that
both the green ball and the blue ball are moving towards each other. We have three options for the
relative motion of our objects. The blue one could be moving faster
than the green, the blue could be moving slower than the green, or the blue and the
green could be moving at the same speed. Because point 𝑃 is directly in the
middle between blue and green, they would reach point 𝑃 at the same time if they
were going at the same speed. As this is one of our options of
motion, we can say this is a possibility.
If the blue object and the green
object are moving at the same speed towards each other before the collision, then
the distance from point 𝑃, at which the objects could have come in contact with
each other, would be (C) zero.
Which of the following is the
greatest distance from the point 𝑃 to the point in which the objects could have
come into contact with each other? (A) 𝑑 two, (B) 𝑑 two minus 𝑑
one, (C) 𝑑 one, (D) 𝑑 one minus 𝑑 two, (E) zero.
Since the problem wants to know the
greatest distance from point 𝑃, let’s start with the extreme of our answer
choices. The greatest distance from point 𝑃
of our answer choices would be 𝑑 two, remembering that in the original problem, it
told us that 𝑑 one was less than 𝑑 two. Looking at the motion to the
objects at 𝑡 one, this must occur after the collision as we can see that that
motion arrows are in the opposite directions to what they were at 𝑡 zero.
What scenario would allow our
greatest distance from the diagram 𝑑 two to be correct? 𝑑 two could be the point where the
objects come in contact if the green ball after the collision comes immediately to
rest and the blue object would have to move a total distance of 𝑑 one plus 𝑑 two
from the collision point. In the original blurb, it told us
that the objects have the same mass but that the velocities were unknown. And that even though the arrow show
the direction of velocity, the magnitudes may be zero. It is plausible then that the green
object stops immediately after the collision and the blue object moves through a
distance of 𝑑 one plus 𝑑 two. The greatest distance from the
point 𝑃 to the point at which the objects could have come in contact with each
other is 𝑑 two, answer choice (A).
Which of the following quantities
is the value the same at 𝑡 zero and at 𝑡 one? (A) The speed of either object, (B)
the momentum of either object, (C) total momentum, (D) the kinetic energy of either
object, (E) total kinetic energy.
In the original blurb, it said we
do not know the speeds of either of the objects. Therefore, we can eliminate answer
choice (A) as this is not necessarily have to be the same at both 𝑡 zero and 𝑡
one. And since the speeds of either
object are not the same, then the kinetic energy of either object does not have to
be the same. Answer choice (D) can also be
eliminated. And the problem does not tell us
that it’s an elastic collision, which means that the total kinetic energy does not
need to be conserved, answer choice (E). This leaves us with the momentum,
either the total momentum or the momentum of either object.
The problem states that there are
no external forces, and during a collision, we have conservation of momentum
assuming there’s no external forces. Conservation of momentum tells us
that the initial momentum is equal to the final momentum. This applies to the momentum of the
system not the individual objects. The total momentum, answer choice
(C), would be the same at both 𝑡 zero and at 𝑡 one.
Key Points
Kinetic energy is conserved for
elastic collisions. The initial kinetic energy of the
system is equal to the final kinetic energy of the system. Kinetic energy is not conserved for
inelastic collisions. The initial kinetic energy of the
system does not equal the final kinetic energy of the system.
