Physics Topics can also be used to explain the behavior of complex systems, such as the stock market or the dynamics of traffic flow.

## Law of Conservation of Momentum Formula Derivation

In the second law of motion we have studied that the moving bodies possess momentum which is equal to the product of mass and velocity. That is,

Momentum = mass × velocity

We will now take one example to understand the meaning of the term ‘conservation of momentum’.

Suppose a speeding truck (fast moving truck) hits a stationary car due to which the car also starts moving. Now, in this collision, the velocity of truck decreases but the velocity of car increases. Due to this the momentum of the truck decreases, and the momentum of car increases. It has been found that the increase in the momentum of car is equal to the decrease in the momentum of truck, so that there is no loss of momentum in the collision. The momentum lost by the truck has been gained by the car. This is an illustration of the law of conservation of momentum.

According to the law of conservation of momentum : When two (or more) bodies act upon one another, their total momentum remains constant (or conserved) provided no external forces are acting. The law of conservation of momentum means that whenever one body gains momentum, then some other body must lose an equal amount of momentum. This law can also be stated as : Momentum is never created or destroyed. The law of conservation of momentum is also known as the principle of conservation of momentum. The principle of conservation of momentum is in accord with Newton’s third law of motion which says that action and reaction (forces) are equal and opposite. We will now take an example to prove the law of conservation of momentum.

Suppose two bodies, a truck and a car, are moving in the same direction (towards east) but with different speeds or velocities [see Figure]. Let the mass of the truck be m_{1} and its velocity be u_{1} so that its initial momentum is m_{1}u_{1}. Let the mass of the car be m_{2} and its velocity be u_{2} so that the initial momentum of the car is m_{2}u_{2}. Thus, the total momentum of the truck and the car before collision is m_{1}u_{1} + m_{2}u_{2}.

Suppose the truck and the car collide for a short time t [see Figure], Due to collision, the velocities of the truck and the car will change. Let the velocity of the truck after the collision be v_{1}, and the velocity of the car after the collision be v_{2} [see Figure], So the momentum of the truck after the collision will be m_{1}v_{1}, and the momentum of the car after the collision will be m_{2}v_{2}. In this way, the total momentum of the truck and the car after the collision will be m_{1}v_{1} + m_{2}v_{2}.

Suppose that during collision, the truck exerts a force F_{1} on the car and, in turn, the car exerts a force F_{2} on the truck. We will now find out the values of the forces F_{1} and F_{2}. This can be done as follows :

(i) When the force F_{1} of the truck acts on the car for a time t, then the velocity of car changes from u_{2} to v_{2}. So,

Acceleration of car, a_{2} = \(\frac{\left(v_2-u_2\right)}{t}\)

But force = mass × acceleration, so the force F_{1} exerted by the truck on the car is given by :

F_{1} = m_{2} × \(\frac{\left(v_2-u_2\right)}{t}\) ………. (1)

(ii) When the force F_{2} of the car reacts on the truck for a time t, then the velocity of the truck changes from u_{1} to v_{1}. So,

Acceleration of truck, a_{1} = \(\frac{\left(v_1-u_1\right)}{t}\)

But, force = mass × acceleration, so the force F_{2} exerted by car on the truck is given by :

F_{2} = m_{1} × \(\frac{\left(v_1-u_1\right)}{t}\) ……. (2)

Now, the force F_{1} exerted by the truck is the ‘action’ and the force F_{2} exerted by the car is the ‘reaction’. But according to the third law of motion, the action and reaction are equal and opposite. That is,

F_{1} = F_{2}

Now, putting the values of F_{1} and F_{2} from equations (1) and (2), we get \(\frac{m_2\left(v_2-u_2\right)}{t}\) = \(-\frac{m_1\left(v_1-u_1\right)}{t}\)

Cancelling t from both sides, we get :

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

or m_{2}v_{2} — m_{2}u_{2} = -m_{1}v_{1} + m_{2}u_{2}

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

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

Now, m_{1}u_{1} + m_{2}u_{2} represents total momentum of the truck and car before collision whereas m_{1}v_{1} + m_{2}v_{2} represents the total momentum of the truck and car after collision. This means that:

Total momentum before collision = Total momentum after collision

It is obvious that the total momentum of the two bodies before and after the collision is the same. This means that the momentum of the two bodies remains constant (or conserved). And this result proves the law of conservation of momentum. We will now describe some of the important applications of the law of conservation of momentum.

### Applications of the Law of Conservation of Momentum

We have already studied that the rockets and jet aeroplanes work on the principle of action and reaction. We will now describe the working of rockets and jet aeroplanes according to the law of conservation of momentum.

The chemicals inside the rocket burn and produce high velocity blast of hot gases. These gases pass out through the tail nozzle of the rocket in the downward direction with tremendous speed or velocity, and the rocket moves up to balance the momentum of the gases. Although the mass of gases emitted is comparatively small, but they have a very high velocity and hence a very large momentum. An equal momentum is imparted to the rocket in the opposite direction, so that, inspité of its large mass, the rocket goes up with a high velocity.

In jet aeroplanes, a large volume of gases produced by the combustion of fuel is allowed to escape

through a jet in the backward direction. Due to the very high speed or velocity, the backward rushing

gases have a large momentum. They impart an equal and opposite momentum to the jet aeroplane due to which the jet aeroplane moves forward with a great speed. Thus, we can also say that the rockets and jet aeroplanes work on the principle of conservation of momentum.

We will now describe how the momentum is conserved when a bullet is fired from a gun. Initially,

before a bullet is fired from a gun, both, the bullet and the gun, are at rest. So, before a bullet is fired, the initial momentum of the bullet and the gun is zero (because their velocities are zero).

Now, when a bullet is fired from a gun, then the bullet has the momentum given by: mass of bullet ×

velocity of bullet. The bullet imparts an equal and opposite momentum to the gun due to which the gun jerks backwards. The gun is said to recoil. The backward velocity of the gun is called recoil velocity. The momentum acquired by the gun is : mass of gun × recoil velocity of gun. Now, according to the law of conservation of momentum:

Momentum of bullet = Momentum of gun

or Mass of bullet × Velocity of bullet = Mass of gun × Reconcil velocity of gun

We should remember this relation because it will be used to solve the numerical problems. Let us solve some problems now.

**Example Problem 1.**

A bullet of mass 10 g is fired from a gun of mass 6 kg with a velocity of 300 m/s. Calculate the recoil velocity of the gun.

**Solution:**

Here, Mass of bullet = 10 g

= \(\frac{10}{1000}\)

= 0.01 kg

Velocity of bullet = 300 m/s

Mass of gun = 6 kg

And, Recoil velocity of gun = ? (To be calculated)

Now, putting these values in the relation :

Mass of bullet × Velocity of bullet = Mass of gun × Recoil velocity of gun

We get: 0.01 × 300 = 6 × Recoil velocity of gun

So, Recoil velocity of gun = \(\frac{0.01 \times 300}{6}\)

= 0.5 m/s

Note. The above problem can also be solved by calculating the momentum of bullet and the gun separately as follows :

Momentum of bullet = Mass of bullet × Velocity of bullet

= 0.01 × 300 = 3 kg.m/s

Now, suppose the recoil velocity of gun is v m/s.

So, Momentum of gun = Mass of gun × Recoil velocity of gun

= 6 × v kg.m/s

According to the law of conservation of momentum :

Momentum of bullet = Momentum of gun

So, 3 = 6 × v

And, v = \(\frac{3}{6}\) m/s

Recoil velocity of gun, v = 0.5 m/s

**Example Problem 2.**

The car A of mass 1500 kg, travelling at 25 m/s collides with another car B of mass 1000 kg travelling at 15 m/s in the same direction. After collision the velocity of car A becomes 20 m/s. Calculate the velocity of car B after the collision.

**Solution.**

In order to solve this problem, we will calculate the total momentum of both the cars, before and after the collision.

(a) Momentum of car A(before collision) = Mass of Car A × Velocity of car A

= 1500 × 25

= 37500 kg.m/s

Momentum of car B = Mass of car B × Velocity of car B

= 1000 × 15

= 15000 kg.m/s

Total momentum of car A and car B(before collision) = 37500 + 15000

= 52500 kg.m/s

(b) After collision, the velocity of car A of mass 1500 kg becomes 20 m/s.

So, Momentum of car A(after collision) = 1500 × 20

= 30000 kg.m/s

After collision, suppose the velocity of car B of mass 1000 kg becomes v m/s.

So, Momentum of car B(after collision) = 1000 × v

= 1000 v kg.m/s

Total momentum of car A and car B(after collision) = 30000 + 1000 v

Now, according to the law of conservation of momentum :

Total momentum before collision = Total momentum after collision

That is, 52500 = 30000 + 1000 v

1000 v = 52500 – 30000

1000v = 22500

v = \(\frac{22500}{1000}\)

v = 22.5 m/s

Thus, the velocity of car B after the collision will be 22.5 m/s.

**Example Problem 3.**

A bullet of mass 10 g moving with a velocity of 400 m/s gets embedded in a freely suspended wooden block of mass 900 g. What is the velocity acquired by the block ?

**Solution.**

Here, Mass of the bullet, m_{1} = 10 g

= \(\frac{10}{1000}\) kg

= 0.01 kg

And, Velocity of the bullet, v_{1} = 400 m/s

So, Momentum of the bullet = m_{1} × v_{1}

= 0.01 × 400 kg.m/s … (1)

Now, this bullet of mass 10 g gets embedded into a wooden block of mass 900 g. So, the mass of wooden block alongwith the embedded bullet will become 900 + 10 = 910 g. Thus,

Mass of wooden block + Bullet, m_{2} = 900 + 10

= 910g

= \(\frac{910}{1000}\) kg

= 0.91 kg

And, Velocity of wooden block + bullet, v_{2} = ? (To be calculated)

So, Momentum of wooden block + bullet = m_{2} × v_{2}

= 0.91 × v_{2} kg.m/s …… (2)

Now, according to the law of conservation of momentum, the two momenta as given by equations (1) and (2) should be equal.

So, m_{1} × v_{1} = m_{2} × v_{2}

or 0.01 × 400 = 0.91 × v_{2}

And, v_{2} = \(\frac{0.01 \times 400}{0.91}\)

= 4.4 m/s

Thus, the velocity acquired by the wooden block (having the bullet embedded in it) is 4.4 metres per second.