Conservation of Momentum: Definition, Equation & Examples

Anyone who has ever played a game of pool is familiar with the law of conservation of momentum, whether they realize it or not.

The law of conservation of momentum is fundamental in understanding and predicting what happens when objects interact or collide. This law predicts the motions of billiard balls and is what decides whether that eight ball makes it into the corner pocket or not.

What Is Momentum?

Momentum is defined as the product of an object’s mass and velocity. In equation form, this is often written as p = mv.

It is a vector quantity, which means it has a direction associated with it. The direction of an object’s momentum vector is the same direction as its velocity vector.

The momentum of an isolated system is the sum of the momenta of each individual object in that system. An isolated system is a system of interacting objects that are not interacting in any net way with anything else. In other words, there is no net external force acting on the system.

Studying total momentum in an isolated system is important because it allows you to make predictions of what will happen to the objects in the system during collisions and interactions.

What Are Conservation Laws?

Before embarking on an understanding of the law of conservation of momentum, it is important to understand what is meant by a “conserved quantity.”

To conserve something means to prevent the waste or loss of it in some way. In physics, a quantity is said to be conserved if it remains constant. You might have heard the expression as it relates to the conservation of energy, which is the notion that energy can neither be created nor destroyed, but only changes form. Hence the total amount of it remains constant.

When we talk about conservation of momentum, we are talking about the total amount of momentum staying constant. This momentum can transfer from one object to another within an isolated system and still be considered conserved if the total momentum in that system does not change.

Newton's Second Law of Motion and the Law of Conservation of Momentum

The law of conservation of momentum can be derived from Newton’s second law of motion. Recall that this law related net force, mass and acceleration of an object as Fnet = ma.

The trick here is to think about this net force as acting on a system as a whole. The law of conservation of momentum applies when the net force on the system is 0. This means that, for each object in the system, the only forces that may be exerted on it must come from other objects within the system, or else be canceled out somehow.

External forces may be friction, gravity or air resistance. These need to either not be acting, or they must be counteracted, in order to make the net force on the system 0.

You can begin the derivation with the statement Fnet = ma = 0.

The m in this case is the mass of the entire system. The acceleration in question is the net acceleration of the system, which refers to the acceleration of the center of mass of the system (the center of mass is the average location of the total system mass.)

In order for the net force to be 0, then the acceleration must also be 0. Since acceleration is the change in velocity over time, this implies that velocity must not be changing. In other words, velocity is constant. Hence we get the statement that mvcm = constant.

Where vcm is the velocity of the center of mass, given by the formula:

v_{cm} = \frac{m_1v_1 + m_2v_2 + ...}{m_1 + m_2 + ...}

So now the statement reduces to:

m_1v_1 + m_2v_2 + ... = \text{constant}

This is the equation that describes the conservation of momentum. Each term is the momentum of one of the objects in the system, and the sum of all momenta must be constant. Another way to express this is by stating:

m_1v_{1i} + m_2v_{2i} + ... = m_1v_{1f} + m_2v_{2f} + ...

Where the subscript i refers initial values and f to final values, usually occurring before and then after some sort of interaction, such as a collision between objects in a system.

Elastic and Inelastic Collisions

The reason the law of conservation of momentum is important is that it can allow you to solve for an unknown final velocity or the like for objects in an isolated system that might collide with each other.

There are two main ways in which such a collision can occur: elastically or inelastically.

A perfectly elastic collision is one in which colliding objects bounce off of each other. This type of collision is characterized by the conservation of kinetic energy. The kinetic energy of an object is given by the formula:

KE = \frac{1}{2}mv^2

If kinetic energy is conserved, then the sum of the kinetic energies of all of the objects in the system must remain constant both before and after any collisions. Using conservation of kinetic energy together with conservation of momentum can allow you to solve for more than one final or initial velocity in a colliding system.

A perfectly inelastic collision is one in which when two objects collide, stick to each other and move as a singular mass afterwards. This can simplify a problem as well because you only need to determine one final velocity instead of two.

While momentum is conserved in both types of collisions, kinetic energy is only conserved in an elastic collision. Most real-life collisions are neither perfectly elastic or perfectly inelastic, but lie somewhere in between.

Conservation of Angular Momentum

What was described in the previous section is conservation of linear momentum. There is another type of momentum that applies to rotational motion that is called angular momentum.

Just as with linear momentum, angular momentum is also conserved. Angular momentum depends on an object’s mass as well as how far that mass is from a rotational axis.

When a figure skater spins, you will see them rotate faster as they bring their arms closer to their body. This is because their angular momentum is only conserved if their rotational speed increases in proportion with how close they bring their arms to their center.

Examples of Momentum Conservation Problems

Example 1: Two billiard balls of equal mass roll toward each other. One is traveling with an initial speed of 2 m/s and the other is traveling with a speed of 4 m/s. If their collision is perfectly elastic, what is the final velocity of each ball?

Solution 1: It’s important when solving this problem to choose a coordinate system. Since everything is happening in a straight line, you might decide that motion to the right is positive and motion to the left is negative. Assume the first ball is traveling to the right at 2m/s. The velocity of the second ball is then -4m/s.

Write an expression for the total momentum of the system before the collision, as well as the total kinetic energy of the system before the collision:

m_1v_{1i} + m_2v_{2i} \\ \frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2

Plug in values to get an expression for each:

m_1v_{1i} + m_2v_{2i} = 2m - 4m = -2m \\ \frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m(2)^2 + \frac{1}{2}m(-4)^2 = 10m

Note that since you weren’t given values for the masses, they remain unknown, though both masses were the same, which allowed for some simplification.

After the collision, the expressions for momentum and kinetic energy are:

mv_{1f} + mv_{2f} \\ \frac{1}{2}mv_{1f}^2 + \frac{1}{2}mv_{2f}^2

By setting the initial values equal to the final values of each, you can cancel out the masses. You are then left with a system of two equations and two unknown quantities:

mv_{1f} + mv_{2f} = -2m \implies v_{1f} + v{2f} = -2 \\ \frac{1}{2}mv_{1f}^2 + \frac{1}{2}mv_{2f}^2 = 10m \implies v_{1f}^2 + v{2f}^2 = 20

Solving the system algebraically gives the following solutions:

v_{if} = -4 \text{ m/s} v_{2f} = 2 \text{ m/s}

You'll note that because the two balls had the same mass, they essentially exchanged velocities.

Example 2: A 1,200-kg car traveling east at 20 miles per hour collides head on with a 3,000-kg truck traveling west at 15 miles per hour. The two vehicles stick together when they collide. With what final velocity do they move?

Solution 2: One thing to note about this particular problem are the units. The SI units for momentum are kg⋅m/s. However, you are given mass in kg and speeds in miles per hour. Note that as long as all speeds are in consistent units, there is no need for conversion. When you solve for the final velocity, your answer will be in miles per hour.

The initial momentum of the system can be expressed as:

m_cv_{ci} + m_tv_{ti} = 1200 \times 20 - 3000 \times 15 = -21,000 \text{ kg}\times\text{mph}

The final momentum of the system can be expressed as :

(m_c + m_t)v_f = 4200v_f

The law of conservation of momentum tells you that these initial and final values should be equal. You can solve for the final velocity by setting the initial momentum equal to the final momentum, solving for final velocity as follows:

4200v_f = -21,000 \implies v_f = \frac{-21000}{4200} = -5 \text{ mph}

Example 3: Show that kinetic energy was not conserved in the previous question involving the inelastic collision between the car and the truck.

Solution 3: The initial kinetic energy of that system was:

\frac{1}{2}m_cv_{ci}^2 + \frac{1}{2}m_tv_{ti}^2 = \frac{1}{2}(1200)(20)^2 + \frac{1}{2}(3000)(15)^2 = 557,500 \text{ kg(mph)}^2

The final kinetic energy of the system was:

\frac{1}{2}(m_c + m_t)v_f^2 = \frac{1}{2}(1200 + 3000)5^2 = 52,500 \text{ kg(mph)}^2

Because the initial total kinetic energy and the total final kinetic energy are not equal, then you can conclude that kinetic energy was not conserved.

References

About the Author

Gayle Towell is a freelance writer and editor living in Oregon. She earned masters degrees in both mathematics and physics from the University of Oregon after completing a double major at Smith College, and has spent over a decade teaching these subjects to college students. Also a prolific writer of fiction, and founder of Microfiction Monday Magazine, you can learn more about Gayle at gtowell.com.