In the right circumstances, the total amount of Momentum of a system never changes. This might not sound very exciting at first, but this principle has multiple applications. For example, we can determine the velocity of a bullet by just using the conservation of Momentum and a woodblock. Take a large wooden block and suspend it with a chord and viola! We have a ballistic pendulum!
Fig. 1: A ballistic pendulum uses the conservation of momentum to determine the speed of a bullet. MikeRun (CC BY-SA 4.0).
With this setup, we can calculate the system's momentum after shooting. Since momentum is conserved, the system must have had the same amount when firing the bullet, and thus, we can find the bullet's velocity. Conservation of momentum is especially helpful for understanding collisions, as sometimes they can have unexpected results.
If you have a basketball and a tennis ball, you can try this at home: hold the tennis ball on the top of the basketball and let them fall together. What do you think will happen?
Fig. 2: Letting fall a tennis ball on top of a basketball cause the tennis ball to bounce very high.
Were you surprised? Would you like to understand why this happens? If so, keep on reading. We will discuss the conservation of momentum in more detail and explore these examples and other multiple applications.
Law of conservation of momentum
Let's start by reviewing what momentum is.
Momentum is a vector quantity given as the product of the mass and velocity of a moving object.
This quantity is also known as linear momentum or translational momentum.
Remember that there are two important types of quantities in physics:
- Vector quantities: Require specifying their magnitude and direction to be well-defined.
- Scalar quantities: Only require specifying their magnitude to be well-defined.
Mathematically, we can calculate momentum with the following formula:
\[p=mv\]
where \(p\) is the momentum in kilograms metres per second \(\bigg(\dfrac{\mathrm{kg}}{\mathrm{m}\cdot \mathrm{s}}\bigg)\), \(m\) is the mass in kilograms (\(\mathrm{kg}\)) and \(v\) is the velocity in metres per second \(\bigg(\dfrac{m}{s}\bigg)\).
It is important to note that momentum is a vector quantity because it is the product of a vector quantity - velocity - and a scalar quantity - mass. The direction of the momentum vector is the same as that of the object's velocity. When calculating momentum, we choose its algebraic sign according to its direction.
Calculate the momentum of a \(15 \,\, \mathrm{kg}\) mass moving with a speed of \(8 \,\, \mathrm{m}/\mathrm{s}\) to the right.
Solution
Since the mass and the velocity are known, we can calculate the momentum directly by substituting these values in the equation for momentum and simplifying.
\[\begin{aligned} p=&mv \\ p=&(15\,\,\mathrm{kg})\bigg(8\,\,\dfrac{\mathrm{m}}{\mathrm{s}}\bigg) \\ p=& 120 \,\, \dfrac{\mathrm{kg}\cdot \mathrm{m}}{\mathrm{s}} \end{aligned}\]
The momentum of this mass turns out to be \(120\,\,\dfrac{\mathrm{kg}\cdot \mathrm{m}}{\mathrm{s}}\) to the right.
Just like the law of conservation of matter in chemistry, and the law of Conservation of Energy in physics, there is a law of conservation of momentum.
The Law of Conservation of Momentum states that the total amount of momentum in a closed system remains conserved.
As mentioned before, to keep the momentum of our system constant, we require some special conditions. Note that the Law of Conservation of Momentum clarifies that it is only valid for closed systems. But what does that mean?
Conditions for conservation of momentum
To understand the conditions for conservation of momentum, we should distinguish between internal and external forces first.
Internal forces are those exerted by objects inside the system into themselves.
Internal forces are action-reaction pairs of forces between the elements comprising the system.
External forces are forces exerted by objects from outside the system.
Having a clear distinction of the type of Force that can act on a system, we can clarify when momentum is conserved. As stated by the Law of Conservation of Momentum, this happens only for closed systems.
A closed system is one on which no external forces act.
Therefore, to observe the conservation of momentum, in our system we must only allow internal forces to interact in the system and isolate it from any external Force. Let's have a look at some examples to apply these new concepts.
Consider our system to be a billiard ball at rest. Since its velocity is zero, it has no momentum.
\[\begin{aligned} p&=mv \\ p&=m \cdot 0 \\ p&=0\end{aligned}\]
However, if a cue stick hits the ball, it applies a force making it move and changing the momentum of the ball. In this case, momentum does not remain constant. It increases because an external force applied by the cue stick was involved.
Fig. 3: The cue stick applies an external force, changing the system's momentum.
Now, for an example of a closed system, consider two billiard balls. One of them moving to the right with a certain speed and the other at rest. If the moving ball hits the one at rest, it exerts a force on this second ball. In turn, by Newton's Third Law, the ball at rest exerts a force on the first. As the balls exert forces involved in themselves that are only internal forces, so the system is closed. Therefore, the system's momentum is conserved.
Fig. 4: A billiard ball hitting another can be thought of as a closed system. Therefore, momentum gets conserved.
The system has the same total momentum before and after the impact. As the masses of both balls are the same, before and after they collide, one of them moves with the same speed to the right.
Newton's cradle is another example where we can observe the conservation of momentum. In this case, let us consider as our system the cradle and earth. The weight of the spheres and the tension of the strings are thus internal forces.
At first, the spheres are at rest, so this system has no momentum. If we interact with the system by pulling away and then releasing one of the spheres, we are applying an external force, so the system momentum changes from zero to a certain amount.
Now, leaving the system alone, the spheres start to impact each other. If we ignore air friction, only internal forces are acting on the system - those of the spheres onto themselves, the tension on the strings, and the weir weights - hence, the system can be considered to be closed.
Fig. 5: A Newton's cradle is an example of conservation of momentum. The sphere at the right hits its adjacent sphere transferring its momentum to the sphere at the left.
The first sphere collides with the second, transferring the momentum to it. Then, momentum is transferred from the second to the third sphere. It continues that way until it reaches the last sphere. As a result of the conservation of momentum, the sphere on the opposite end swings in the air with the same momentum as the ball which was pulled and released.
Conservation of momentum equation
We now know momentum is conserved when dealing with a closed system. Let's now see how we can express the conservation of momentum mathematically. Let's consider a system comprised of two masses, \(m_1\) and \(m_2\). The total momentum of the system is the sum of the momentum of each of these masses. Let's consider that they are initially moving with velocities \(u_1\) and \(u_2\), respectively.
\[\begin{aligned} \text{Total initial momentum}&= p_1+p_2 \\ \text{Total inital momentum}&=m_1\cdot u_1 + m_2 \cdot u_2 \end{aligned}\]
Then, after these masses interact with each other, their velocities change. Let's represent these new velocities as \(v_1\) and \(v_2\), respectively.
\[\begin{aligned} \text{Total initial momentum}&= p_1+p_2 \\ \text{Total inital momentum}&=m_1\cdot v_1 + m_2 \cdot v_2 \end{aligned}\]
Finally, because momentum is conserved, the final and initial momentum of the system should be the same.
\[\begin{aligned}\text{Total initial momentum}&=\text{Total final momentum} \\ m_1\cdot u_1+m_2\cdot u_2&=m_1 \cdot v_1 + m_2 \cdot v_2\end{aligned}\]
Recall that momentum is a vector quantity. Therefore, if the motion is in two dimensions, we are required to use the above equation once for the horizontal direction and another time for the vertical direction.
As part of a test, explosives are collocated in a \(50\,\,\mathrm{kg}\) mass at rest. After the explosion, the mass splits into two fragments. One of them, with a mass of \(30\,\,\mathrm{kg}\), moves to the west with a velocity of \(40\,\,\mathrm{m}/\mathrm{s}\). Calculate the velocity of the other fragment.
Solution
The mass of \(50\,\,\mathrm{kg}\) is initially at rest, so the initial momentum is zero. The final momentum is the sum of the momentum of the two fragments after the explosion. We will refer to the \(30\,\,\mathrm{kg}\) fragment as fragment \(a\) and the other fragment, of mass \(50\,\,\mathrm{kg}-30\,\,\mathrm{kg}\), will be fragment \(b\). We can use a negative sign to indicate a motion in the west direction. Thus, a positive sign means the motion is in the east direction. Let's start by identifying the quantities we know.
\[\begin{aligned} m_a &=30\,\,\mathrm{kg} \\ v_a &= -40\,\,\dfrac{m}{s}(\text{moving west})\\ m_b &=20\,\,\mathrm{kg}\\ v_b &=? \end{aligned}\]
By conservation of momentum, we know that the total momentum before and after the explosion is the same.
\[P_i=P_f\]
Moreover, we know that the initial momentum is zero as the \(50\,\,\mathrm{kg}\)mass was at rest. We can substitute this value on the left-hand side and express the final momentum as the sum of the momentum of each fragment and isolate the final velocity of the fragment \(b\).
\[\begin{aligned} P_i&=P_f \\ 0&=m_a \cdot v_a +m_a \cdot v_b \\ -m_a \cdot v_a &= m_b \cdot v_b \\ \dfrac{-m_a\cdot v_a}{m_b}&=v_b\end{aligned}\]
Now, we can substitute the values and simplify.
\[\begin{aligned} v_b &= \dfrac{-m_a\cdot v_a}{m_b} \\ v_b&= \dfrac{-30\,\,\cancel{\mathrm{kg}}\cdot -40 \,\, \dfrac{\mathrm{m}}{\mathrm{s}}}{20\,\,\cancel{\mathrm{kg}}} \\ v_b&=\dfrac{1200\,\,\dfrac{\mathrm{m}}{\mathrm{s}}}{20} \\ v_b&=60\,\,\mathrm{\dfrac{\mathrm{m}}{\mathrm{s}}}\end{aligned}\]
Therefore, the fragment \(b\), moves with a velocity of \(60\,\,\dfrac{\mathrm{m}}{\mathrm{s}}\) to the east.
Conservation of momentum during a collision
One of the most important applications of conservation of momentum happens during collisions. Collisions occur all time and allow us to model very different scenarios.
A collision refers to an object moving towards another, getting close enough to interact, and exerting a force on each other in a short amount of time.
Balls hitting each other on a pool table is an example of a collision.
Fig. 6: The concept of collision applies to balls on a pool table.
Although the concept of collision applies to a wide range of situations, what happens during or after a collision is crucial for their study. For this reason, we can categorize collisions into different types.
Elastic collisions
In an elastic collision, the objects remain separate after colliding with each other the total kinetic energy and momentum are conserved.
Two billiard balls colliding can be considered an elastic collision.
Let's go back to one of the examples we mentioned before: two billiard balls, one moving to the right and the other at rest. A billiard ball has a mass of about \(0,2\,\,\mathrm{kg}\). Consider that the ball moves to the right at \(10\,\,\dfrac{\mathrm{m}}{\mathrm{s}}\). Let's calculate the total amount of initial momentum.
\[\begin{aligned} \text{Total initial momentum}&=p_1+p_2 \\ &= m_1\cdot u_1 + m_2 \cdot u_2 \\ &=0,2\,\,\mathrm{kg} \cdot 10 \,\, \dfrac{\mathrm{m}}{\mathrm{s}}+0,2\,\,\mathrm{kg}\cdot 0 \\ &= 2\,\, \dfrac{\mathrm{kg}\cdot \mathrm{m}}{\mathrm{s}}\end{aligned} \]
We said that because of the conservation of momentum, after the collision the first ball stops, and the second one moves with the same velocity, the first one used to have, in this case, \(10\,\,\dfrac{\mathrm{m}}{\mathrm{s}}\).
Fig. 7: The white ball will stop while the blue ball should move in the right direction after collision.
This results in the same total momentum after the collision.
\[\begin{aligned} \text{Total initial momentum}&=p_1+p_2 \\ &= m_1\cdot v_1 + m_2 \cdot v_2 \\ &=0,2\,\,\mathrm{kg} \cdot 0+0,2\,\,\mathrm{kg}\cdot 10\,\,\dfrac{\mathrm{m}}{\mathrm{s}} \\ &= 2\,\, \dfrac{\mathrm{kg}\cdot \mathrm{m}}{\mathrm{s}}\end{aligned} \]
But what about this scenario: the first ball bounces back at \(10\,\,\dfrac{\mathrm{m}}{\mathrm{s}}\) while the second one starts moving at \(20\,\,\dfrac{\mathrm{m}}{\mathrm{s}}\). Let's calculate the momentum of this scenario. Since we consider the direction to the right as positive, a motion to the left is negative.
\[\begin{aligned} \text{Total initial momentum}&=p_1+p_2 \\ &= m_1\cdot v_1 + m_2 \cdot v_2 \\ &=0,2\,\,\mathrm{kg} \cdot -10\,\,\dfrac{\mathrm{m}}{\mathrm{s}}+0,2\,\,\mathrm{kg}\cdot 20\,\,\dfrac{\mathrm{m}}{\mathrm{s}} \\ &= -2\,\, \dfrac{\mathrm{kg}\cdot \mathrm{m}}{\mathrm{s}}+4\,\,\dfrac{\mathrm{kg}\cdot\mathrm{m}}{\mathrm{s}}\\ &=2\,\,\dfrac{\mathrm{kg}\cdot \mathrm{m}}{\mathrm{s}}\end{aligned} \]
Everything looks fine, right? After all, momentum conserves as well in this case. However, if you try to observe something like this by colliding two billiard balls, it won't ever happen. Can you tell why? Remember that in these collisions, not only must momentum be conserved, but energy must be conserved too! In the first scenario, the kinetic energy is the same before and after the collision because in both cases, just one ball moves at \(10\,\,\dfrac{\mathrm{m}}{\mathrm{s}}\) . But in the second scenario, both balls move after the collision, one at \(10\,\,\dfrac{\mathrm{m}}{\mathrm{s}}\) and the other at \(20\,\,\dfrac{\mathrm{m}}{\mathrm{s}}\). Therefore, the kinetic energy would be much more than at the beginning, which is not possible.
Fig. 8: This result is not possible because, although it conserves the system's momentum the kinetic energy is not conserved.
Keep in mind that no collision is truly elastic, since part of the energy is always lost. For example, if you kick a football, then your foot and the ball remain separate after colliding, but some energy is lost as heat and the sound of the impact. However, sometimes the energy loss is so small that we can model the collision as elastic without problems.
Why is Momentum Conserved?
As we mentioned before, momentum gets conserved when we have a closed system. Collisions are great examples of them! This is why momentum is essential when studying collisions. By modelling a simple collision mathematically, we can conclude that momentum must be conserved. Have a look at the figure below which shows a closed system comprised of two masses \(m_1\) and \(m_2\). The masses are heading towards each other with initial velocities \(u_1\) and \(u_2\), respectively.
Fig. 9: Two objects are about to collide.
During the collision, both objects exert forces \(F_1\) and \(F_2\) on each other as shown below.
Fig. 10: Both objects exert forces on each other.
After the collision, both objects move separately in opposite directions with final velocities \(v_1\) and \(v_2\), as depicted below.
Fig. 11: Both objects move in opposite directions with respective velocities.
As Newton's Third Law states, the forces for the interacting objects are equal and opposite. Hence, we can write:
\[F_1=-F_2\]
By Newton's Second Law, we know that these forces cause an acceleration on each object that can be described as
\[F=ma.\]
Let's use this to substitutefor each force in our previous equation.
\[\begin{aligned} F_1&=-F_2 \\ m_1 a_1 &= - m_2 a_2 \end{aligned} \]
Now, acceleration is defined as the rate of change in velocity. Therefore, acceleration can be expressed as the difference between the final velocity and the initial velocity of an object divided by the time interval of this change. Hence, by takingas the final velocity,as the initial velocity, andas the time, we get:
\[\begin{aligned} a&=\dfrac{v-u}{t} \\ m_1 a_2 &=-m_2a_2 \\ \dfrac{m_1(v_1-u_1)}{t_1}&=\dfrac{m_2(v_2-u_2)}{t_2} \end{aligned}\]
As the times t1 and t2 are the same because the time of impact between the two objects is the same. We can simplify the above equation as:
\[m_1 v_1- m_1 u_1 = m_2 u_2-m_2 v_2\]
Rearranging the above yields,
\[m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\]
Note how the left-hand side is the total momentum before the collision as it only involves the initial velocities of the masses, while the right-hand side represents the total momentum after the collision depending only on the final velocities. Therefore, the above equation states that Linear Momentum gets conserved! Keep in mind that the velocities change after impact, but the masses remain the same.
Perfectly inelastic collisions
A perfectly inelastic collision occurs when two objects collide, and instead of moving separately, they both move as a single mass.
A car crash where the cars stick together is an example of a perfectly inelastic collision.
For perfectly inelastic collisions momentum is conserved, but the total kinetic energy is not. In these collisions, the total kinetic energy changes because part of it is lost as sound, heat, changes in the Internal Energy of the new system, and bonding both objects together. This is why it is called an inelastic collision as the deformed object does not return to its original shape.
In this type of collision, we can treat the two initial objects as a single object after the collision. The mass for a single object is the sum of the individual masses before the collision. And the velocity of this single object is the vector sum of the individual velocities before the collision. We will refer to this resultant velocity as.
Initial Momentum (Before Collision) | Final momentum (After Collision) |
\(m_1 v_1 +m_2 v_2\) | \((m_1 + m_2)v_f\) where \(v_f=v_1+v_2\) |
By Conservation of Momentum |
\(m_1 v_1 +m_2 v_2=(m_1 + m_2)v_f\) |
In reality, no collision is either elastic or perfectly inelastic as these are idealized models. Instead, any collision is somewhere in between as some form of kinetic energy is always lost. However, we often approximate a collision to either of these extreme, ideal cases to make the calculations simpler.
A collision that is neither elastic nor perfectly inelastic is simply called an inelastic collision.
Conservation of momentum examples
System of gun and bullet
Initially, the gun and the bullet inside the gun are at rest, so we can deduce that the total momentum for this system before pulling the trigger is zero. After pulling the trigger, the bullet moves forward while the gun recoils in the backward direction, each of them with the same magnitude of momentum but opposite directions. As the mass of the gun is much larger than the bullet's mass, the velocity of the bullet is much larger than the recoil velocity.
Rockets and jet engines
The momentum of a rocket is initially zero. However, due to the burning of fuel, hot gases rush out at a very high speed and large momentum. Consequently, the rockets acquire the same momentum, but the rocket moves upwards as opposed to the gases as the total momentum has to remain null.
Basketball and tennis ball falling
The example presented at the beginning shows how the tennis ball is launched very high. After bouncing on the ground, the basketball transfers part of its momentum to the tennis ball. Since the basketball's mass is much bigger (around ten times the mass of the tennis ball), the tennis ball acquires a velocity much larger than the basketball would get when bouncing alone.
Conservation of Momentum - Key takeaways
- Momentum is the product of the mass and velocity of a moving object.
- Momentum is a vector quantity, so we need to specify its magnitude and direction to be able to work with it.
- Conservation of Momentum states that the total momentum in a closed system remains conserved.
- In an elastic collision, the objects remain separate after colliding.
- In an elastic collision, momentum and kinetic energy are conserved.
- In a perfectly inelastic collision, the colliding objects move as a single mass after the collision.
- In a perfectly inelastic collision, momentum is conserved but the total kinetic energy is not.
- In reality, no collision is either elastic or perfectly inelastic. These are just idealized models.
- We label the collisions that are neither elastic nor perfectly inelastic as simply inelastic.
References
- Fig. 1: Ballistic Pendulum (https://commons.wikimedia.org/wiki/File:Sketch_of_a_ballistic_pendulum.svg) by MikeRun is licensed by CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/deed.en)
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