Conservation of Momentum

Law of conservation of momentum

Let's start by reviewing what momentum is.

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.

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.

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 action-reaction pairs of forces between the elements comprising 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.

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.

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.

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.

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

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.

Perfectly inelastic collisions

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$v_f$.

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.

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.