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### Course: Physics library > Unit 19

Lesson 1: AP Physics 1 concept review- AP Physics 1 review of 1D motion
- AP Physics 1 review of 2D motion and vectors
- AP Physics 1 review of Forces and Newton's Laws
- AP Physics 1 review of Centripetal Forces
- AP Physics 1 review of Energy and Work
- AP Physics 1 review of Momentum and Impulse
- AP Physics 1 review of Torque and Angular momentum
- AP Physics 1 review of Waves and Harmonic motion
- AP Physics 1 Review of Charge and Circuits

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# AP Physics 1 review of Momentum and Impulse

In this video David quickly reviews the momentum and impulse topics on the AP Physics 1 exam and solves an example problem for each concept. Created by David SantoPietro.

## Want to join the conversation?

- At5:42, I dont see why the answer is D. The problem asks for the velocity not the momentum. Since p=mv, shouldnt the velocity equal to p/m which is -20?? Thank you!(91 votes)
- Yes, you're right. I just totally forgot to divide by the mass at the end. My bad! I'll add an annotation.(2 votes)

- Where's the dropbox link?(17 votes)
- I found the link (but it's a google doc)! I know it's too late for some people but still... trying to help other people out!! Good Luck ;)

Momentum and Impulse:

https://docs.google.com/document/d/1rbYFE2fcrJ8M7r9tpiCfKBtE-N3-_AH4AgUYJg4byGg/edit?usp=sharing

Torque and Angular Momentum:

https://docs.google.com/document/d/16y9raqwIqzYdwwicajFwahIoWyQsWhYTfuOUnO3seUw/edit?usp=sharing

Waves and Harmonic Motion:

https://docs.google.com/document/d/1R7OrdVjavRMSsRy08iDIofPpOJSkQV9A9lO0Ojzee-c/edit?usp=sharing

Charge and Circuits:

https://docs.google.com/document/d/1nLZuDnSJEgi0errdAcxRIfLJVKuAktvkgelsjPukrRY/edit?usp=sharing(13 votes)

- For the 2D collisions problem, how do we know that there are no net forces in the system?(1 vote)
- That is an excellent question, Shawn.

You may consider the phrase "net force" as "the external force", or "the force left over after subtracting all individual forces in a system", according to Giancoli AP Physics. My suggestion is that you decide whether there is a third party force exerted, say, by an experimenter.

Consider the circumstance: A rocket and its additional tanks are fired up to the sky and after a while the tanks exert a repulsive force in the opposite direction of the rocket so they may separate with the rocket and fall. Will we consider the forces the tanks exert "net forces" to the whole system? Apparently no. They are simply regarded as "inner-forces" existed in the whole rocket-tank system. However, if we now expand a little bit and consider air resistance, will air resistance be a net force? Well, yes, since that is indeed a third party force that is not part of the system.

To reiterate, what to consider is merely the net third party force that does not belong to the original whole system you have been asked to consider. The sum of forces that are not exerted BY the system itself but are exerted ON the system.

Again, thank you for asking. And sorry for writing this long. A rookie here.

Adrian Li

AP Center, High School Affiliated to RUC, Beijing, China(6 votes)

- can you link the document(3 votes)
- So then, where does the energy go in an inelastic collision?(2 votes)
- At5:25The annotation needs to be changed. David solved for the "Final Momentum" instead of the final velocity. To say that he solved the problem for the change in momentum is the purpose of the of the example.(1 vote)
- Is the answer of the question at4:30wrong? Shouldn't we choose C instead of D? We need to consider about the unit of the velocity, right?(1 vote)
- Yes, the correct answer is C since you would divide that -40 kg m/s by the mass (2kg) to get the velocity of -20 m/s.(1 vote)

- At8:16should the answer not be C because the Ke is conserved in that the momentum is the same before and after?(1 vote)
- Energy is NOT in fact conserved in the collision: Some energy is transferred outside of the system and turned into thermal energy.(1 vote)

- At13:09, I would think that the answer is C, not D. I understand that point D. is correct in that IMMEDIATELY after the car arrives at rest, that would be a possible solution. However, the block is on ice so why wouldn´t the block slide up to position c? the momentum causes slight forward movement correct?(1 vote)
- The question asks for the positions at the instant that the car's tires stop moving, not later on. Plus there should be no change in center of mass. In answer choice C, the center of mass is at 5, not 4, which was the original center of mass.(1 vote)

- For the question at8:12, is the momentum conserved?(1 vote)
- Momentum, both linear and angular, is
**always**conserved (unless, of course, there were external unbalanced forces).(1 vote)

## Video transcript

- [Instructor] What does momentum mean? The definition of momentum is
the mass times the velocity. So the formula is simple,
it's just m times v. And why do we care about momentum? We care about momentum
because if there's no net external force on a system, the momentum of that
system will be conserved. In other words, the total
initial momentum of that system would equal the total final
momentum of that system. So momentum will be conserved if there's no net external force. And momentum is a vector,
that means it has components. The total momentum will
point in the direction of the total velocity, and
the momentum in each direction can be conserved independently. In other words, if there's no
net force in the y direction, then the momentum in the y
direction will be conserved, and if there's no net
force in the x direction, the momentum in the x
direction will be conserved. Since the momentum is m
times v, the units are kilograms times meters per second. So what's an example problem
involving momentum look like? Let's say two blocks of mass 3M and M head toward each other, sliding
over a frictionless surface with speeds 2V and 5V, respectively, and after the collision
they stick together. Which direction will the two masses head after the collision? To figure this out, we can
just ask what direction is the total momentum of the system initially. Since momentum's gonna be
conserved, that'll have to be the direction of the momentum finally. So the momentum of the 3M mass is going to be the mass, which is 3M, times the velocity, which is 2v, so we get a momentum of 6Mv. And the momentum of the mass
M is gonna be the mass M, times the velocity, which is negative 5v. Momentum is a vector, so you can't forget the negative signs. Which gives a momentum of negative 5Mv. So the total initial momentum
of the system would be 6Mv plus -5Mv, which is one
Mv, and that's positive, which means the total momentum,
initially, is to the right. That means, after the collision, the total momentum will also
have to be to the right, and the only way that could be the case if these two masses joined
together, is for the total combined mass to also move to the right. What does impulse mean? The impulse is the amount of
force exerted on an object or system multiplied by
the time during which that force was acting. So in equation form, that means
J, the impulse, is equal to the force multiplied by how
long that force was acting. And the net impulse is gonna
be equal to the net force times the time during which
that net force was acting. And this is also going
to be equal to the change in momentum of that system or object. In other words, if a mass
had some initial momentum and ends with some final momentum, the change in momentum of that mass, p final minus p initial, is
gonna equal the net impulse, and that net impulse is
gonna equal the net force on that object multiplied by the time during which
that force was acting. And since impulse is a change in momentum, and momentum is a vector,
that means impulse is also a vector, so it can be
positive and negative. And the units are the same as momentum, which is kilograms
times meters per second. Or, since it's also force
times time, you could write the units as Newtons times seconds. So what's an example problem
involving impulse look like? Let's say a bouncy ball
of mass M is initially moving to the right with a speed 2v. And it recoils off a wall with a speed v. We want to know, what's the
magnitude of the impulse on the ball from the wall? So the impulse, J, is going to be equal to the change in momentum. The change in momentum is
p final minus p initial, so the final momentum is gonna
be the mass times the final velocity, but this
velocity's heading leftwards, so you can't forget the negative sign, minus the initial momentum,
which would be M times 2v, which gives a net impulse of -3Mv. This makes sense. The net impulse has to point in the same direction as the net force. This wall exerted a force to the left, that means the impulse also points left, and has a magnitude of 3Mv. If you get a force versus time graph, the first thing you
should think about is that the area under that
graph is going to equal the impulse on the object. So if you graph the force on
some object as a function of time, the area under that
curve is equal to the impulse. Just be careful, since
area above this time axis is going to count as positive
impulse, and area underneath the time axis would count
as negative impulse, since those forces would be negative. Why do we care that the
area's equal to the impulse? Well, if we can find the area,
that would equal the impulse, and if that's the net
impulse on an object, that would also equal the change
in momentum of that object. Which means we could figure out the change in velocity of an object. So what's an example
problem involving impulse as the area under a graph look like? Let's say a toy rocket
of mass two kilograms was initially heading to the right with a speed of 10 meters per second, and a force in the horizontal direction is exerted on the rocket,
as shown in this graph, and we want to know, what's
the velocity of the rocket at the time t equals 10 seconds? To figure that out, we'll figure out the area under the curve. This triangle would
count as positive area. This triangle would
count as negative area. And since this triangle
is just as positive as this triangle is negative,
these areas cancel completely. And the only area we'd have
to worry about is the area between eight seconds and 10 seconds. This is going to end up
being a negative area, since the height of the rectangle is -30, and the width of the rectangle
is going to be two seconds. This gives an impulse
of -60 Newton seconds. So if the impulse on this
object is -60 Newton seconds, that's going to equal the change
in momentum of that object. How much momentum did
this object start with? The initial momentum of this object is going to be two kilograms
times the initial velocity, which was 10 meters per
second to the right, which is positive 20
kilogram meters per second. So if the initial momentum
of the rocket is positive 20 and there was a change in momentum of -60, the final momentum just has to be -40. Or in other words, since the
change in momentum would have to be the final momentum
minus the initial momentum, which was positive 20, we could
find the final momentum by adding 20 to both sides,
which would give us -60 plus 20, which is -40. What's the difference between an elastic and an inelastic collision? What we mean by an
elastic collision is that the total kinetic energy of that system is conserved during the collision. In other words, if a
sphere and a cube collide, for that collision to be elastic, the total kinetic energy
of the sphere plus the kinetic energy of the
cube before the collision would have to equal the
kinetic energy of the sphere plus the kinetic energy of
the cube after the collision. If the total kinetic
energy before the collision is equal to total kinetic
energy after the collision, then that collision is elastic. It's not enough for the system
to just bounce of each other. If two objects bounce,
the total kinetic energy might not be conserved. Only when the total
kinetic energy is conserved can you say the collision is elastic. For an inelastic collision,
the kinetic energy is not conserved during the collision. In other words, the total
initial kinetic energy of the sphere plus cube would not equal the total final kinetic energy
of the sphere plus cube. Where does this kinetic energy go? Typically, in an inelastic collision, some of that kinetic energy is transformed into thermal energy during the collision. While masses could bounce
during an inelastic collision, if they stick together, the
collision is typically called a perfectly inelastic collision, since in this collision
you'll transform the most kinetic energy into thermal energy. And when two objects stick
together, it's a surefire sign that that collision is
definitely inelastic. So what's an example problem
that involves elastic and inelastic collisions look like? Let's say two blocks of mass
2M and M head toward each other with speeds 4v and 6v, respectively. After they collide,
the 2M mass is at rest, and the mass M has a
velocity of 2v to the right. And we want to know, was this collision elastic or inelastic? Now you might want to say that, since these objects
bounced off of each other, the collision has to be
elastic, but that's not true. If the collision is elastic,
then the objects must bounce, but just because the objects bounce does not mean the collision is elastic. In other words, bouncing
is a necessary condition for the collision to be elastic,
but it isn't sufficient. If you really want to know
whether a collision was elastic, you have to determine whether
the total kinetic energy was conserved or not. And we can figure that
out for this collision without even calculating anything. Since the speed of the 2M mass decreased, the kinetic energy of
the 2M mass went down. And since the speed of
the M mass also decreased after the collision, the kinetic energy of the
mass M went down, as well. So if the kinetic energy
of both masses go down, then the final kinetic
energy after the collision has to be less than the
initial kinetic energy. Which means kinetic
energy was not conserved, and this collision had to be inelastic. One final note, even though kinetic energy wasn't conserved during this process, the momentum was conserved. The momentum will be conserved for both elastic and inelastic collisions. It's just kinetic energy
that's not conserved for an inelastic collision. How do you deal with
collisions in two dimensions? Well the momentum will be
conserved for each direction in which there's no net impulse. If there's no net impulse
in both directions, then the momentum in both directions will be conserved independently. In other words, if there's no
net force in the x direction, the total x momentum has to be constant, and if there's no net
force in the y direction, the total momentum in the y
direction has to be constant. So in other words, if two spheres collide in a glancing collision,
the total momentum in the x direction initially should equal the total momentum in
the x direction finally, if there's no net impulse
in that x direction. And the total momentum in
the y direction initially, of which there is none in this case, would have to equal the total
momentum in the y direction finally, if there's no net
impulse in the y direction. So what's an example involving collisions in two dimensions look like? Let's say a metal sphere
of mass M is traveling horizontally with five meters per second when it collides with an
identical sphere of mass M that was at rest. After the collision,
the original sphere has velocity components of
four meters per second and three meters per second
in the x and y directions. And we want to know, what
are the velocity components of the other sphere right
after the collision? So assuming there were no net forces in the x or y direction in this case, then the momentum will be
conserved for each direction, and since the mass of
each sphere's the same, we can simply look at
the velocity components. So if we started with
five units of momentum in the x direction, we have to end with five units of momentum in the x direction. So the x component of the second sphere has to be one meter per second. And since we started
with no momentum in the vertical direction initially, we have to end with no
momentum vertically. So if the first sphere has
three units of momentum vertically after the collision,
then the second sphere has to have three units of
momentum vertically downward after the collision, which
gives us an answer of D. What's the center of mass mean? The center of mass of an
object or a system is the point where that object or system would balance. And the center of mass is also the point where you can treat the entire
force of gravity as acting. The way you can solve
for the center of mass is by using this formula. You multiply each mass
by how far that mass is from the reference point. If there's no reference point specified, you get to choose the
arbitrary reference point, which would designate where x equals zero. You continue adding each
mass times its position. For positions to the left
of the reference point, those would count as negative positions. And when you're done
accounting for every mass in your system, you
divide by the total mass, which would be all the masses
added up, and the number you get would be the position
of the center of mass. The center of mass is going
to have units of meters, since it's a location. The location where the system
or object would balance, and the location where you can treat the entire force of gravity as acting. Something else that's
extremely important to remember is that the center of mass of
a system will not accelerate unless there's an external
force on that system. In other words, the
center of mass of a system follows Newton's first law. If the center of mass
of a system is at rest, then even if the masses in that system exert forces on each
other and move around, the center of mass of
that system will stay put until there's a net external
force on the system. And if the center of mass was initially moving to the right at some speed, that center of mass will continue moving to the right at that speed, even if the masses are moving
in different directions, until there's a net force on that system. So what's an example problem involving center of mass look like? Lets say a remote control car of mass M is sitting at rest on a
wooden plank, also of mass M, in the position seen here. There is friction between
the wheels of the car and the plank, but there's
no friction between the plank and the ice upon which
the plank is sitting. The remote control car
is turned on and off. What would be a possible final position of the car and the plank? Now because the car's at rest
and the plank is at rest, that means the center
of mass of this system is also at rest. And since there's no net
force on this system, the center of mass is going
to have to remain at rest. Where is the center of mass? Well the car's mass is at three, the plank's center is at five, so the center of mass
between the car and the plank would be at the location of four. So to find the correct solution, we just need to figure
out which one of these also has the center of mass at four. Option A has the car at three and the center of the plank at three. That'd put the center
of mass at three meters, but that can't be right, the
center of mass can't move, there were no external
forces on our system, and the center of mass started at rest, so it's got to remain at rest. For option B, the center
of the car is at four, the center of the plank is at three, this would put the center
of mass somewhere between three and four, but again,
that can't be right. We need our center of mass
to be at the location four. Option C has the car at six and the center of the plank at four. This would put the center of
mass of the system at five, that can't be right. We need our center of mass at four. Option D has the car at five, and the center of the plank at three. That puts the center of
mass at location four, just like it was before. So D is a possible solution.