If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

What is conservation of momentum?

Learn what conservation of momentum means and how to use it.

What is the principle of conservation of momentum?

In physics, the term conservation refers to something which doesn't change. This means that the variable in an equation which represents a conserved quantity is constant over time. It has the same value both before and after an event.
There are many conserved quantities in physics. They are often remarkably useful for making predictions in what would otherwise be very complicated situations. In mechanics, there are three fundamental quantities which are conserved. These are momentum, energy, and angular momentum. Conservation of momentum is mostly used for describing collisions between objects.
Just as with the other conservation principles, there is a catch: conservation of momentum applies only to an isolated system of objects. In this case an isolated system is one that is not acted on by force external to the system—i.e., there is no external impulse. What this means in the practical example of a collision between two objects is that we need to include both objects and anything else that applies a force to any of the objects for any length of time in the system.
If the subscripts i and f denote the initial and final momenta of objects in a system, then the principle of conservation of momentum says

Why is momentum conserved?

Conservation of momentum is actually a direct consequence of Newton's third law.
Consider a collision between two objects, object A and object B. When the two objects collide, there is a force on A due to B—FAB—but because of Newton's third law, there is an equal force in the opposite direction, on B due to A—FBA.
The forces act between the objects when they are in contact. The length of time for which the objects are in contact—tAB and tBA—depends on the specifics of the situation. For example, it would be longer for two squishy balls than for two billiard balls. However, the time must be equal for both balls.
Consequently, the impulse experienced by objects A and B must be equal in magnitude and opposite in direction.
If we recall that impulse is equivalent to change in momentum, it follows that the change in momenta of the objects is equal but in the opposite directions. This can be equivalently expressed as the sum of the change in momenta being zero.

What is interesting about conservation of momentum?

There are at least four things that are interesting—and sometimes counter-intuitive—about momentum conservation:
  • Momentum is a vector quantity, and therefore we need to use vector addition when summing together the momenta of the multiple bodies which make up a system. Consider a system of two similar objects moving away from each other in opposite directions with equal speed. What is interesting is that the oppositely-directed vectors cancel out, so the momentum of the system as a whole is zero, even though both objects are moving.
  • Collisions are particularly interesting to analyze using conservation of momentum. This is because collisions typically happen fast, so the time colliding objects spend interacting is short. A short interaction time means that the impulse, FΔt, due to external forces such as friction during the collision is very small.
  • It is often easy to measure and keep track of momentum, even with complicated systems of many objects. Consider a collision between two ice hockey pucks. The collision is so forceful that it breaks one of the pucks into two pieces. Kinetic energy is likely not conserved in the collision, but momentum will be conserved.
    Provided we know the masses and velocities of all the pieces just after the collision, we can still use conservation of momentum to understand the situation. This is interesting because by contrast, it would be virtually impossible to use conservation of energy in this situation. It would be very difficult to work out exactly how much work was done in breaking the puck.
  • Collisions with "immovable" objects are interesting. Of course, no object is really immovable, but some are so heavy that they appear to be. Consider the case of a bouncy ball of mass m traveling at velocity v towards a brick wall. It hits the wall and bounces back with velocity v. The wall is well attached to the earth and doesn't move, yet the momentum of the ball has changed by 2mv since velocity went from positive to negative.
If momentum is conserved, then the momentum of the earth and wall also must have changed by 2mv. We just don't notice this because the earth is so much heavier than the bouncy ball.

What kind of problems can we solve using momentum conservation?

Exercise 1a: The recoil of a cannon is probably familiar to anyone who has watched pirate movies. This is a classic problem in momentum conservation. Consider a wheeled, 500 kg cannon firing a 2 kg cannonball horizontally from a ship. The ball leaves the cannon traveling at 200 m/s. At what speed does the cannon recoil as a result?
Exercise 1b: Suppose the cannon was raised to fire at an angle α=30 to the horizontal. What is the recoil speed in this case? Where did the additional momentum go?
Exercise 2a: A golf club head of mass mc=0.25 kg is swung and collides with a stationary golf ball of mass mb=0.05 kg. High speed video shows that the club is traveling at vc=40 m/s when it touches the ball. It remains in contact with the ball for t=0.5 ms; after that, the ball is traveling at a speed of vb=40 m/s. How fast is the club traveling after it has hit the ball?
Exercise 2b: What is the average force on the club due to the golf ball in the previous problem?
Exercise 3: Suppose a 100 kg football player is at rest on an ice rink. A friend throws a 0.4 kg football towards him at a speed of 25 m/s. In a smooth motion he receives the ball and throws it back in the same direction at a speed of 20 m/s. What is the speed of the player after the throw?

Want to join the conversation?

  • leafers ultimate style avatar for user Eka Terina
    In the exercise 2b when we calculate the force, why is it the mass of the ball in the equation? Shouldn't it be the mass of the golf club?
    (22 votes)
    Default Khan Academy avatar avatar for user
    • piceratops tree style avatar for user 1 Prime
      Here's what I found:
      When calculating the force exerted during a collision, the focus is on the change in momentum of the object experiencing the force. In this case, it's the golf club that undergoes the change in velocity due to its collision with the golf ball. The force exerted on the golf club during the collision is determined by its mass, as it reflects the club's resistance to changes in motion. The mass of the golf ball, while influencing the overall dynamics of the collision, does not directly impact the force experienced by the golf club. Therefore, when applying the impulse-momentum theorem to calculate the force, it's appropriate to use the mass of the golf club.
      (6 votes)
  • leafers ultimate style avatar for user Robert Ireland
    Hello. I believe in 2b acceleration is negative because the final velocity of the club minus the initial velocity is 32-40=-8, which would make the force negative, thus the answer to 2b should be -4kN. Which make sense since the ball applies an impulse force opposite to the swing of the club (Newton's 3rd).
    (15 votes)
    Default Khan Academy avatar avatar for user
  • starky ultimate style avatar for user aca92bg00
    in exercise 2a: can we do it like this:

    mc*deltaVc= - mb*deltaVb
    mc*(Vcf-Vci)= - mb*(Vbf-Vbi)
    0.25kg*(Vcf - 40m/s) = - 0.05kg*(40m/s - 0m/s)
    0.25kg*Vcf - 0.25kg*40m/s = - 0.05*40 kg*m/s
    0.25Vcf = -0.05*40 + 0.25*40
    0.25Vcf = -2 + 10
    Vcf= 8/0.25
    Vcf= 32m/s
    (15 votes)
    Default Khan Academy avatar avatar for user
  • hopper cool style avatar for user JPhilip
    So I understand the concept of conservation of momentum now, but I have a question: If an object such as a car runs into a very strong wall, the car would completely stop (without bouncing back) but the wall would not move either. Where does the momentum go?
    (12 votes)
    Default Khan Academy avatar avatar for user
  • starky tree style avatar for user Milen Paul
    Why did we not subtract 20m/s from 25m/s in exercise 3?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • male robot hal style avatar for user Charles LaCour
      In the problem the ball is thrown from person A to person B at 25 m/s then the ball is thrown from person B to person A at 20 m/s.

      The direction from A to B is the opposite direction than B to A so if you consider the velocity of the ball from A to B to be positive then the velocity from B to A is negative giving you Vi = 25 m/s and Vf = -20 m/s.

      When you put them into the Vi - Vf part of the equation you have (25 m/s) - (-20 m/s) = 25 m/s + 20 m/s = 45 m/s.
      (11 votes)
  • blobby green style avatar for user Tierney Donahue
    what types of situations does conservation of momentum apply to?
    (6 votes)
    Default Khan Academy avatar avatar for user
  • duskpin ultimate style avatar for user Jons9
    In question 1a in bold, it says Pc + Pb = 0. Shouldn't the change in momentum be equal to Pc - Pb = 0? I thought a change is a difference between two magnitudes instead of a sum?
    (4 votes)
    Default Khan Academy avatar avatar for user
  • primosaur seedling style avatar for user Pranav Nayak
    In the facts about conservation of momentum in third point you said KE is likely not conserved, can you please explain that in detail and state some examples where KE cannot be conserved
    (0 votes)
    Default Khan Academy avatar avatar for user
    • male robot hal style avatar for user Charles LaCour
      Here is a video of an experiment where KE is not conserved but momentum is: https://www.youtube.com/watch?v=N8HrMZB6_dU

      The KE and momentum of the two bullets should equivalent to each other. The two blocks travel to the same height because of the conservation of liner momentum but the KE of the bullet gets split between heat, sound, block and bullet deformation as well as rotation in one of the blocks..

      Since both blocks attain the same height so the KE from vertical motion but the block that was shot off center also has KE from rotation so it has retained more the the bullets KE as KE. In the block that is shot in the center there is more of the KE transferred to heat, sound, block and bullet deformation than in the one shot off center.
      (9 votes)
  • blobby green style avatar for user alona maria
    Is conservation of momentum only applied for two objects that are in motion and are colliding? Is there a conservation of momentum for a ball hitting a wall? How is conservation of momentum differ from Newton's third law of motion?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • male robot hal style avatar for user Charles LaCour
      Conservation of momentum is separate from Newton's third law and it is applicable in any isolated system. So a ball hitting a wall is subject to conservation of momentum.

      Even when you have a system that seems to violate conservation of momentum you can almost always increase the system to include enough to have momentum conserved.
      (2 votes)
  • duskpin sapling style avatar for user Violet
    I thought conservation means 'to save'
    (2 votes)
    Default Khan Academy avatar avatar for user