Newton’s second law of motion focuses on rate of change of momentum. We know that momentum is mass multiplied by velocity. Law of conservation of momentum is also an important concept of physics and it is useful in understanding physical phenomena.

According to law of conservation of momentum:

**Momentum is neither created nor destroyed.**

In other words, when two or more bodies act upon one another, their total momentum remains constant or conserved provided no external forces are acting. It states that **whenever one body gains momentum, the other body loses an equal amount of momentum**.

We can understand this concept with the help of some examples.

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

Suppose the truck and car collide for a short time t and 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}. So, the momentum of the truck after collision will be m_{1}v_{1}, and the momentum of the car after the collision will be m_{2}v_{2}. So, the total momentum of truck and car after 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.

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

a_{2} = (v_{2 } – u_{2})/t

But force = mass × acceleration

F_{1} = m_{2} × (v_{2 } – u_{2})/t

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

a_{1} = (v_{1 } – u_{1})/t

But force = mass × acceleration

F_{2} = m_{1} × (v_{1 } – u_{1})/t

Now, the force F_{1} exerted by truck is ‘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}

Substituting values from above

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

Cancelling (t) from both sides we get,

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

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

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

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} is total momentum of truck and car before collision whereas m_{1}v_{1}_{ }+ m_{2}v_{2} is total momentum of car and truck after collision.

Therefore, **Total momentum before collision = Total momentum after collision**

This is law of conservation of momentum.

### Applications of law of conservation of momentum –

- The chemicals inside a rocket burn and produce high velocity blast of hot gases. These gases pass out through the tail nozzle of the rocket in the downwards direction with tremendous speed and the rocket moves up to balance the momentum of the gases.
- In jet aeroplane, a large volume of gases produced by the combustion of fuel is allowed to pass through a jet in the backward direction. Due to the very high speed, the backward rushing gases have a large momentum. To balance this momentum, jet aeroplane moves forward with great speed.
- When a bullet is fired from the gun, the bullet has a momentum. The bullet imparts an equal and opposite momentum to the gun due to which the gun recoils and jerks backwards.

### Questions related to the Law of Conservation of Momentum from NCERT textbook – Page 126 and 127

**Question 3:-**

**From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 ms ^{-1}. Calculate the initial recoil velocity of the rifle.**

**Solution –**

Momentum of bullet = mass of bullet × velocity of bullet

= (50)/(1000) Kg × 35 ms^{-1}

= 0.05 Kg x 35 ms^{-1}

= 1.75 Kg ms^{-1}

Momentum of rifle = mass of rifle × velocity of rifle

= 4 kg × v ms^{-1}

= 4 v kg ms^{-1}

According to law of conservation of momentum,

Momentum of bullet = Momentum of rifle

1.75 = 4v

v = 1.75/4

v = 0.4375 ms^{-1}

**Question 4:-**

**Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 ms ^{-1} and 1 ms^{-1}, respectively. They collide and after the collision, the first object moves at a velocity of 1.67 ms^{-1}. Determine the velocity of the second object.**

**Solution –**

Momentum of first object before collision = mass of object × velocity of object

= 100/1000 Kg × 2 ms^{-1}

= 0.1 kg × 2 ms^{-1}

= 0.2 kg ms^{-1}

Momentum of second object before collision = mass of object × velocity of object

= 200/1000 Kg × 1 ms^{-1}

= 0.2 kg × 1 ms^{-1}

= 0.2 kg ms^{-1}

Total momentum before collision = 0.2 kg ms^{-1} + 0.2 kg ms^{-1}

= 0.4 kg ms^{-1}

Momentum of first object after collision = 100/1000 Kg × 1.67 ms^{-1}

= 0.1 Kg × 1.67 ms^{-1}

= 0.167 kg ms^{-1}

Momentum of second object after collision = 200/1000 Kg × v ms^{-1}

= 0.2 v kg ms^{-1}

Total momentum after collision = 0.167 kg ms^{-1} + 0.2 v kg ms^{-1}

Total momentum after collision = total momentum before collision

0.4 = 0.167 + 0.2v

0.2v = 0.233

v = 0.233/0.2

v = 1.165 ms^{-1}

### NCERT textbook – Page 128 and 129

**Question 11:-**

**Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 ms ^{-1}**

**before the collision during which they stick together. What will be the velocity of the combined object after collision?**

**Solution –**

Mass of first object m_{1} = 1.5 kg

Velocity of first object v_{1} = 2.5 ms^{-1}

Momentum of first object = m_{1} × v_{1}

= 1.5 Kg × 2.5 ms^{-1}

= 3.75 Kg ms^{-1}

Mass of second object m_{2} = 1.5 kg

Velocity of first object v_{2} = – 2.5 ms^{-1}

Momentum of second object = m_{2} × v_{2}

= 1.5 kg × 2.5 ms^{-1}

= -3.75 Kg ms^{-1}

Total momentum before collision= 3. 75 – 3.75 = 0 kg ms^{-1}

Total mass of combined objects = m_{1}+m_{2 }= 1.5 kg+1.5 kg = 3.0 kg

Let velocity of combined objects be v

Total momentum after collision = (m_{1}+m_{2}) v

= 3v kg ms^{-1}

0 = 3v

v=0 m/s** **

**Question 13:-**

**A hockey ball of mass 200 g travelling at 10 ****ms ^{-1 }is struck bya hockey stick so as to return italong its original path with a velocity at 5 ms^{-1}. Calculate the change of momentum occurredin the motion of the hockey ball by the force applied by the hockey stick.**

**Solution –**

Mass of hockey ball m_{1 }= 200 g

= 200/1000 kg

= 0.2 kg

Initial velocity v_{1} = 10 m/s

Initial momentum = 0.2×10 kg m/s = 2 kg m/s

Mass of hockey ball = 0.2 kg

Final velocity v_{2} = -5 m/s

Final momentum = m_{2} × v_{2 }= -1 kg m/s

Change in momentum = final momentum – initial momentum

= -1 – 2 = -3 kg m/s

**Question 15:-**

**An object of mass 1 kg travelling in a straight line with a velocity of 10 ****ms ^{-1}collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.**

**Solution –**

Mass of object m_{1} = 1 kg

Velocity of object v_{1 }=10 m/s

Momentum of object = m_{1} × v_{1}

= 1×10 kg m/s

= 10 kg m/s

Mass of wooden block m_{2} = 5 kg

Velocity of wooden block v_{2} = 0

Momentum of wooden block = m_{2} × v_{2}

= 5 × 0 = 0 kg m/s

Total momentum before impact = 10+0 = 10 kg m/s

According to law of conservation of momentum, the momentum before and after the impact will be same

Total momentum = 10 kg m/s

Total mass of object and wooden block = 6 kg

Velocity of wooden block and object = v m/s

10 = 6×v

v=1.67 m/s

The law of conservation of momentum states that the momentum of a system is neither created nor destroyed. Whenever two bodies interact with each other, one body loses momentum and other body gains an equal amount of momentum. This law explains why a gun recoils when a bullet is shot from it. It also explains the movement of rocket and jet aeroplanes. This law is useful in understanding the situation when two bodies collide.