**Newton’s Third Law of the Motion and Conservation of the Momentum:Class 11CBSE Physics**

**Newton’s Third Law of the Motion and Conservation of the Momentum**: Newton’s third law of motion states that every action has an equal and opposite reaction. Action and reaction act on two different bodies. The biggest application of Newton’s law of the motion is propulsion of the rocket into space, in the rocket there is an exit at the downward or back portion of the fuel tank, and the fuel is forcibly ejected from the exit in such a way that equal reaction occurs against the action of flushed products which are produced as a result of the combustion of the fuel.The reactive force propels the rocket into the atmosphere and settles down the rocket into the space.

**Newton’s Third Law of the Motion and Conservation of the Momentum: Class 11CBSE Physics**

In Newton’s third law of motion, the action and reaction pair must have the following condition.

(i)Action and reaction in Newton’s third law are always of the same nature.

(ii)Action and reaction in Newton’s third law act on two different bodies.

(iii)Force always exist in pair.

Example: An object kept on a plane, the force exerted by an object to another object(plane) is the normal reaction which is equal to the force exerted by another object(plane) to the previous object is the normal reaction. In this case, gravitational force or weight(mg) which acts on the object in the downward direction, and normal reaction are not the action and reaction pair while both of them are equal because mg and N are not of a similar nature and also act on the same object.

In this case, the object also pulls the earth by the same force mg that makes the action-reaction pair with the weight of the object.

Here, N_{PO},N_{OP} and Mg_{EO},Mg_{OE} are the action and reaction pair in Newton’s thirld law of motion.

Where,

N_{PO }is the normal reaction from plane to object

N_{OP} is the noral reaction from object to the plane

Mg_{EO }is the gravitational force applied by the earth to object

Mg_{OE }is the gravitational force applied by the object to the earth

Conservation of the momentum:In an isolated system the momentum of the system is always constant or the change in momentum is zero. The net force on the isolated system is 0, we have

F_{net} =0

F_{net} =ΔP/ΔT

0 =ΔP/ΔT

**ΔP = 0**

Therefore P is constant

Let initial momentum of the system is P_{I } and and final momentum is P_{F}

Then

P_{I } = P_{F}

Let there are two marbles of the masses m_{1 } and m_{2 }moving with a velocities u_{1} and u_{2}colloids to each other and after collision their velocities turnes to v_{1} and v_{2}

Then

P_{I } = m_{1 } u_{1} + m_{2} u_{1}

P_{F }= m_{1 }v_{1} + m_{2} v_{2}

Therefore

**m _{1 } u_{1} + m_{2} u_{1}= m_{1 }v_{1} + m_{2} v_{2}**

Note: Isolated system means no any external force is applied on the system.

**Conservation of the momentum by the third law: ****Class 11 CBSE Physics**

Let two marbles collide to each other, then the force applied by one marble to another marble is F and according to Newton’s third law, second marble will also apply the same force to the first marble.

Let two marbles are A and B and the forces applied by them to each other are F_{AB }and F_{BA}

F_{AB }=- F_{BA}

Let the acceleration produced by the forces on the marbles are a_{1 }and a_{2}

F_{AB }=m_{1}a_{1}=m_{1}(v_{1}-u_{1})/t

F_{BA }=m_{2}a_{2}=m_{2}(v_{2}-u_{2})/t

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

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

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

**m _{1}u_{1 }+m_{2}u_{2}=m_{1}v_{1 }+ m_{2}v_{2}**

**Class 11 Physics and Chemistry Notes**

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**You can also study class 11 Physics and Chemistry Notes**

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