# Laws of Conservation of Momentum

Law of Conservation of Momentum states that the total momentum of objects before and after collision remains constant. Before stating the Laws of Conservation of Momentum, we must first learn about momentum. The momentum of an object is a physical quantity that is defined as the product of the mass and velocity of the object. Momentum a vector quantity is very useful in Newtonian physics. The laws of Conservation of Momentum explain that the momentum of any system is always constant until an external force is applied.

Let’s learn about the Laws of Conservation of Momentum, its derivation, formula, examples, and others in this article.

## Momentum Definition

Momentum can be defined as a mass in motion. In quantitative terms, momentum is defined as the product of mass and velocity. It is denoted by “**p**”. The amount of momentum possessed by an object depends upon two factors – its mass and its velocity. An object which does not have any mass will have zero momentum no matter how fast it moves. Similarly, a stationary object will always have zero momentum, whatever its mass may be.

Let’s say the mass of the object is “m” and its velocity is “v”. Then the momentum “p” is given by,

p = mv

The SI unit for momentum is **Kg-m/s.**

**Momentum: A Vector Quantity**

A vector quantity is a quantity that has both magnitude and direction. Since momentum also depends on the velocity, it is a vector quantity. For example – a ball thrown toward the north will have a velocity toward the north. In that case, momentum will also point in the direction where the ball is moving. Its direction is the same as the direction of velocity.

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## What Is Conservation Of Momentum?

Law of Conservation of Momentum is one of the basic laws of physics which is used derived from Newton’s third law of Motion. Conservation of Momentum states that the momentum of the system is always conserved, i.e. initial momentum and final momentum of the system are always conserved. We can also state that the total momentum of the system is always constant.

The law of conservation of momentum is mathematically and experimentally proven.

## Law of Conservation of Momentum

The Law of conservation of momentum says that momentum is conserved for a system but we must note that this law is applicable to isolated systems i.e. there should not be any external forces acting on the system.

The law of conservation of momentum states that,

“When no external forces are acting on a system, then the momentum of the system is conserved. Specifically, the total momentum of the system before and after any event remains the same.”

Consider a system of two point masses m_{1} and m_{2}._{ }Initially, these bodies were moving with the velocities u_{1} and u_{2}. Now they collide with each other and their final velocities become v_{1} and v_{2}. So, according to the law of conservation of momentum,

m_{1}u_{1}+ m_{2}u_{2}= m_{1}v_{1}+ m_{2}v_{2}

This is true only when there is no external force applied to the system.

For a general system with n-particles. The law is given by the equation,

m_{1}u_{1 }+ m_{2}u_{2}+ ….. + m_{2}u_{n}= m_{1}v_{1}+ m_{2}v_{2 }+ ….m_{2}v_{n}

## Derivation of Conservation of Momentum

The law of Conservation of Momentum is derived with the help of Newton’s third law of motion which states that for every action there is an equal and opposite reaction.

Consider two point masses m_{1} and m_{2}._{ }Initially, these bodies were moving with the velocities u_{1} and u_{2}. Now they collide with each other and their final velocities become v_{1} and v_{2}. Their time of collision is t.

Now the change in momentum of the mass A

**△P _{A} = m_{1}(v_{1} – u_{1})**

Now the change in momentum of the mass B

**△P _{B} = m_{1}(v_{2} – u_{2})**

From Newton’s law of motion,

F_{AB} = -F_{BA}….(1)

We also know that,

**F = △P/t**

Thus,

F_{AB} = △P_{A} / t

F_{BA} = △P_{B} / t

Now, from (1)

△P_{A} / t = -△P_{B} / t

m_{1}(v_{1} – u_{1})/t = – m_{1}(v_{2} – u_{2})/t

m_{1}(v_{1} – u_{1}) = – m_{1}(v_{2} – u_{2})

m_{1}(v_{1} – u_{1}) + m_{1}(v_{2} – u_{2}) = 0

m_{1}u_{1}+ m_{2}u_{2}= m_{1}v_{1}+ m_{2}v_{2}

**Why is Momentum Conserved? **

Momentum is conserved because of Newton’s Third Law of Motion. In a collision between two objects A and B. Object A experiences a force F_{AB} which is due to B, similarly, object B experiences force F_{BA }which is due to A. These forces must be equal according to Newton’s third law. Since the collision was for a very short time .

Now, this is a very short time. So, this is considered an impulse. An impulse is equivalent to a change in momentum.

## Conservation of Momentum Formula

The mathematical formula for the Conservation of Momentum is given as,

m_{1}u_{1}+ m_{2}u_{2}= m_{1}v_{1}+ m_{2}v_{2}where,

mis the mass of the first object_{1}mis the mass of the second object_{2}uand_{1}uare the initial velocities of m_{2}_{1}and m_{2}respectivelyvand_{1}vare the final velocities of m_{2}_{1}and m_{2}respectively

## Example of Conservation of Momentum

There are various examples that explain the law of conservation of momentum. Some of the most common examples of Conservation of Momentum are,

### Recoil of the Gun

We have noticed that whenever we fire a gun we observe a recoil which is because of the law of conservation of momentum. As the bullet gains forward momentum we observed a backward momentum(motion) which is called the recoil.

### Motion of Boat

The motion of the boat is based on the concept of the law of conservation of momentum as when we push the water backward with the help of oars the boat is pushed forward because of the momentum.

## Application of Law of Conservation of Momentum

As we have studied the Law of Conservation of Momentum is a highly useful law. It is used for various activities and has various applications some of the applications of the Law of Conservation of Momentum are,

### Rocket Propulsion

Rocket Propulsion or launching of the rocket is based on the law of conservation of momentum. As in a rocket when hot gases are expelled from the exhaust of the rocket they provide the required velocity to the rocket.

### Airbags Used in Vehicles

Airbags used in vehicles also work on the principle of the law of conservation of momentum as in case of collision they reduce the speed of the passengers diluting the energy of the impact.

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## Solved Examples on Law of Conservation of Momentum

**Example 1: Calculate the momentum of a ball thrown at a speed of 100 m/s and weighing 500g. **

**Solution: **

Given

M = 500g and V = 100 m/sMomentum is given by,

p = MVp = MV

p = (500)(100)

p = 50000 gm/sp = 5 × 10

^{4}gm/s = 50 kgm/s

**Example 2: Suppose two balls with a mass of 5 Kg and 2 Kg are moving in the same direction at 6 m/s and 2 m/s respectively collide, and after the collision, the 5 kg ball is moving at a speed of 5 m/s. What is the speed of the 2 kg ball?**

**Solution: **

Given

m_{1}= 5 kg and m_{2}= 2 kgInitial Velocities: u

_{1 }= 6 m/s and u_{2 }= 2 m/sFinal Velocities: v

_{1}= 5 m/s and v_{2}= ?According to the law of conservation of momentum

m_{1}u_{1}+ m_{2}u_{2}= m_{1}v_{1}+ m_{2}v_{2}Now,

(5)(6) + (2)(2) = (5)(5) + 2(v

_{2})30 + 4 = 25+ 2(v

_{2})2v

_{2}= 34 – 25 = 9v

_{2}= 4.5 m/s

**Example 3: Consider a cannon that weighs 500 kg. It fires a cannonball at the speed of 200 m/s. The weight of the cannonball is 2 kg. Find the speed of the recoil for the cannon. **

**Solution: **

Given

m_{1}= 500 Kg and m_{2}= 2 KgInitial Velocities,

Velocity of cannon (u

_{1})_{ }= 0 m/s

Velocity of cannon ball (u_{2})_{ }= 0 m/sFinal Velocities,

Velocity of cannon (v

_{1})_{ }= 0 m/s

Velocity of cannon ball (v_{2})_{ }= 200 m/sUsing the law of conservation of momentum

m_{1}u_{1}+ m_{2}u_{2}= m_{1}v_{1}+ m_{2}v_{2}(500)(0) + (2)(0) = (500)(v

_{1}) + 2(200)– 400 = 500(v

_{1})v

_{1}= -400/500v

_{1}= -0.8 m/s (here, – sign indicates that the recoil motion of the cannon is opposite to the motion of the cannon ball)Thus, the cannon gun will recoil at a speed of

0.8m/s after firing the cannon.

**Example 4: Find the velocity of a bullet of mass 8 grams when fired from a pistol of mass 2.4 kg. (Recoil velocity of the pistol is 1 m/s)**

**Solution:**

Mass of Bullet, m

_{1}= 8 gram = 0.008 kgMass of Pistol, m

_{2}= 2.4 kgInitial velocity of the bullet, u

_{1}= 0Initial Recoil velocity of a pistol, u

_{2}= 0Velocity of a bullet, v

_{1}=?Recoil Velocity of pistol, v

_{2}= 1 m/sUsing the law of conservation of momentum,

m_{1}u_{1}+ m_{2}u_{2}= m_{1}v_{1}+ m_{2}v_{2}(0.008)(0) + (2.4)(0) = (0.008)(v

_{1}) + (2.4)(1.5)0 = (0.008)(v

_{1}) + 3.6v

_{1}= 3.6/(0.0008) = 350 m/sHence, the recoil velocity of the pistol is 350 m/s

## FAQs on Momentum

### Q1: What is Momentum? Give Examples

**Answer:**

Momentum is a vector quantity which is defined as the product of the mass and velocity of the object. Some examples of momentum are the change in velocity of an object when two objects collide, etc.

### Q2: Is Momentum a Scalar or a Vector Quantity?

**Answer:**

Momentum of an object is a vector quantity as it has both magnitude and direction.

### Q3: What does the Law of Conservation of Momentum State?

**Answer:**

Law of conservation of momentum state that, ” The total momentum of a system is always constant.”

### Q4: What is the Formula for Law of Conservation of Momentum?

**Answer:**

The mathematical formula for the law of conservation of momentum is,

m_{1}u_{1}+ m_{2}u_{2}= m_{1}v_{1}+ m_{2}v_{2}

### Q5: What are examples of Law of Conservation of Momentum?

**Answer:**

Various examples of law of conservation of momentum are,

- Rockets Propulsion
- Firing a Bullet
- Motion of a Boat

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