Conservation of Momentum – Law, Derivation, Examples & Applications

Conservation of Momentum deals with the change in velocities of the interacting bodies whose net momentum remains the same. It gives the condition with which the change happens without altering the net momentum of the system.

What is Momentum?

Conservation of Momentum can be well understood if we first get a fair idea about the concept of momentum.

Momentum of a body is the product of the mass of the body and the velocity of the body. It is a vector quantity, so it possesses both magnitude and direction.

The \({\rm{SI}}\) unit of momentum is \({\rm{kg}}\,{\rm{m}}{{\rm{s}}^{ – 1}}\) in whereas its \({\rm{CGS}}\) unit is \({\rm{g}}{\mkern 1mu} \,{\rm{cm}}{\mkern 1mu} \,{{\rm{s}}^{ – 1}}.\)

Here, we are dealing with the translational motion in which every particle of the body has the same displacement.

So now onwards, when referring to the word momentum, we mean linear momentum that is considered while dealing with translational motion.

The Linear Momentum of a body can be explained as the product of the mass of the body and its velocity. The direction of linear momentum is along the direction of the velocity of the body.

It is very important to involve the concept of momentum while discussing the impact of force. For example, the bullet thrown by our hand may not hurt a person, but when the bullet is shot using a gun, it may kill a person. Though the mass of the bullet is the same in both cases, its velocity drastically changes while using it through the gun, and thus its impact on the person gets deadly.

What does Conservation of Momentum Mean?

Conservation of Momentum comes into the picture when bodies interact with each other without any external unbalanced force acting on them. The total momentum of the interacting bodies remains the same before and after the interaction.

For example, two balls are moving towards each other, and they collide with each other. Here, no external unbalanced force has caused the collision of the balls; rather, the balls collided with each other as they were moving towards each other.

In such a scenario, the total momentum of the balls before the collision will be equal to their total momentum after the collision. Thus, the total momentum of the balls remained conserved during their collision. This is well stated in the law formulated on the concept of conservation of momentum.

Law of Conservation of Momentum

Conservation of Momentum deals with the total momentum of the bodies interacting with each other without any external unbalanced force acting on them.

Law of conservation of momentum states that: “When two or more bodies interact with each other, the vector sum of their linear momenta remains constant and is not affected due to their mutual action and reaction. The only condition is that no external unbalanced forces should be acting on the system of bodies.”

Derivation of Conservation of Momentum

Conservation of Momentum can be verified by considering two bodies that interact with each other without any external unbalanced force acting on them.

Let us take the case of two bodies \(A\) and \(B\) having masses \({m_A}\) and \({m_B}\) as well as moving with velocities  \({u_A}\) and \({u_B}\) respectively. Here, we have taken \({u_A}\) and \({u_B}\) as they are the initial velocities of the bodies before the collision.

Let \({u_A} > \,{u_B}\) then only the body \(A\) with mass \({m_A}\) which is moving behind the body \(B\) with mass \({m_B}\)  collide without any external unbalanced force acting on them for the collision to happen, and we can study the conservation of momentum.

During the collision, the body \(A\) exerts a force \({F_{AB}}\) on the body \(B\). But the body \(B\) will also exert an equal and opposite force \({F_{BA}}\) on the body \(A\).

This is due to Newton’s third law of motion which states that, “to every action, there is always an equal and opposite reaction.”

The collision happens for a brief time \(t.\)

Due to collision, the velocities of the bodies \(A\) and \(B\) have changed to \({v_A}\) and \({v_B}\) respectively. Here, we have taken \({v_A}\) and \({v_B}\) as they are final velocities of the bodies after the collision.

The change in momentum of body \(A\) is given by \({m_A}{v_A}\, – \,{m_A}{u_A}\)

So, the rate of change of the momentum of body \(A\) is given by \(\frac{{{m_A}{v_A}\, – \,{m_A}{u_A}}}{t}\)

But this will be equal to the force \({F_{BA}}\) that body \(B\) exerts on body \(A\).

\(\therefore \) \({F_{BA}}\, = \frac{{{m_A}{v_A}\, – \,{m_A}{u_A}}}{t}\)

Similarly, the change in momentum of body \(B\) is given by \({m_B}{v_B} – {m_B}{u_B}.\)

So, the rate of change of the momentum of body \(B\) is given by \(\frac{{{m_B}{v_B}\, – \,{m_B}{u_B}}}{t}\)

But this will be equal to the force \({F_{AB}}\) that body \(A\) exerts on body \(B\)

\(\therefore \,{F_{AB}} = \frac{{{m_B}{v_B} – {m_B}{u_B}}}{t}\).

From Newton’s third law of motion, the forces \({F_{AB}}\) and \({F_{BA}}\) are equal in magnitude and opposite in direction to each other.

\(\therefore {F_{AB}} =  – \,{F_{BA}}\)

\(\therefore \frac{{{m_B}{v_B} – {m_B}{u_B}}}{t} =  – \left( {\frac{{{m_A}{v_A} – {m_A}{u_A}}}{t}} \right)\)

\(\therefore {m_B}{v_B} – {m_B}{u_B}\, = \, – {m_A}{v_A} + {m_A}{u_A}\)

\(\therefore \,{m_A}{v_A} + {m_B}{v_B} = {m_A}{u_A} + {m_B}{u_B}\)

Here, \({m_A}{u_A} + {m_B}{u_B}\) is the total momentum of the two bodies \(A\) and \(B\) before the collision and \({m_A}{v_A} + {m_B}{v_B}\) is the total momentum of the two bodies \(A\) and \(B\) after the collision.

Therefore, the total momentum of the two bodies \(A\) and \(B\) remained unchanged during the collision. So, the total momentum of the system of two bodies remained conserved and is not affected by the mutual action and reaction of the bodies.

Here, the collision has happened without any external unbalanced force acting on the bodies. Hence, the law of conservation of momentum is being verified.

Conservation of Momentum Formula

Conservation of Momentum can be represented by the formula,
\({m_A}{v_A} + {m_B}{v_B} = {m_A}{u_A} + {m_B}{u_B}\)
where,
\({m_A}\) and \({m_B}\) are the masses of the two bodies
\({u_A}\) and \({u_B}\) are the initial velocities of the two bodies
\({v_A}\) and \({v_B}\) are the final velocities of the two bodies
We can modify this relation for multiple bodies in interaction as shown below:
\({m_1}{v_1} + \,{m_2}{v_2} + {m_3}{v_3} + … = {m_1}{u_1}\, + \,{m_2}{u_2} + {m_3}{u_3} + …\,\)
where, \({m_1},\,{m_2}\) and \({m_3}\) are the masses of the bodies
\({u_1},\,{u_2}\) and \({u_3}\) are the initial velocities of the bodies
\({v_1},{v_{2\,}}\) and \({v_3}\) are the final velocities of the bodies

Solved Examples of Conservation of Momentum

Q.1. A car of mass \(1500\,{\rm{kg}}\) travelling at \({\rm{30}}\,{\rm{m}}\,{\mkern 1mu} {{\rm{s}}^{ – 1}}\) collides with another car \(B\) of mass \(1000\,{\rm{kg,}}\) travelling at \({\rm{10}}\,{\rm{m}}\,{{\rm{s}}^{ – 1}}\) in the same direction. After the collision, the velocity of car A becomes \({\rm{15}}\,{\rm{m}}\,{{\rm{s}}^{ – 1}}.\) Calculate the velocity of car B after the collision.
Sol: Given that,
The mass of car \(A\) is \({m_A}\, = \,1500\,{\rm{kg}}\)
The initial velocity of car \(A\) is \({u_A}\, = \,30\,{\rm{m}}\,{{\rm{s}}^{ – 1}}\)
The final velocity of car \(A\) is \({v_A}\, = \,15\,{\rm{m}}\,{{\rm{s}}^{ – 1}}\)
The mass of car \(B\) is \({m_B}\, = \,1000\,{\rm{kg}}\)
The initial velocity of car \(B\) is \({u_B}\, = \,10\,{\rm{m}}\,{{\rm{s}}^{ – 1}}\)
The final velocity of car \(B\) is \({v_B} = \,\frac{{{m_A}{u_A} + {m_B}{u_B} – {m_A}{v_A}}}{{{m_B}}} = \frac{{1500 \times 30 + 1000 \times 10 – 1500 \times 15}}{{1000}} = \frac{{32500}}{{1000}}\)
\(\therefore {v_B} = 32.5\,{\rm{m}}\,{{\rm{s}}^{ – 1}}\)
Thus, the final velocity of the car \(B\) after the collision \(32.5\,{\rm{m}}\,{{\rm{s}}^{ – 1}}\)

Q.2. When two bodies \(X\) and \(Y\) interact with each other, \(X\) exerts a force of \(50\;{\rm{N}}\) on \(Y\), towards the north. What is the force exerted by \(Y\) on \(X\)?
Sol: Given that,The force exerted by \(X\) on \(Y\) is \({F_{XY}} = 50\,{\rm{N}}\) towards the north.
From Newton’s third law of motion, both forces \({F_{XY}}\) and \({F_{YX}}\) are equal in magnitude but opposite in direction.
Thus, the force exerted by \(Y\) on \(X\) is \({F_{YX}} = 50\,{\rm{N}}\) towards the south.
Hence, the body \(Y\) exerts a force of \(50\,{\rm{N}}\) on body \(X\) towards the south.

Applications of Conservation of Momentum

Conservation of Momentum finds many applications in day-to-day life. Some of them are as under:

1. The recoiling of the gun. This happens when the bullet is fired from the gun, and the gun recoils in the direction opposite to the direction of the bullet. So, the shooter should hold the gun firmly to avoid hurting. This recoiling of the gun in the opposite direction is in accordance with the law of conservation of momentum.
2. Flight of rockets. The large momentum of the escaping gasses in the downward direction gives a large forward momentum to the rocket in accordance with the law of conservation of momentum.
3. The water with high momentum flowing forward from the hose of the firefighting vehicle exerts high momentum on the firefighter in the backward direction. So, to hold the hose properly, the firefighter must exert high force. The backward push on the hose is in accordance with the law of conservation of momentum.

Frequently Asked Questions (FAQs) on Conservation of Momentum

Q.1. What is the formula of conservation of momentum?
Ans: The formula of conservation of momentum is \({m_A}{v_A} + {m_B}{v_B} = {m_A}{u_A} + {m_B}{u_B}.\)

Q.2. When is the conservation of momentum applicable?
Ans: Conservation of momentum is applicable when there is no net external unbalanced force acting on a system, and the mutual interactions of the bodies in the system is due to the action and the reaction of the bodies within the system.

Q.3. What is the best example of conservation of momentum?
Ans: The recoiling of a gun is the best example of conservation of momentum. When the bullet is shot in the forward direction, the gun recoils in the backward direction due to conservation of momentum.

Q.4. What is the law of conservation of momentum?
Ans: The law of conservation of momentum states that if no net external unbalanced force acts on a system, then the total momentum of all the bodies of the system before the collision will be equal to the total momentum of all the bodies of the system after the collision.

Q.5. What is momentum?
Ans: Momentum of a body is the product of the mass of the body and the velocity of the body.

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