Let’s suppose a speeding truck hits a stationary car due to which the car starts moving. What is actually happening behind the scene? Well here, as the velocity of the truck decreases, the velocity of the car increases and hence the momentum lost by truck is gained by car. Interesting? Let’s find out more about the momentum and its conversation below:

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## What is Momentum?

A vector quantity that is the product of the mass and velocity of an object or particle is **‘momentum’**. Momentum is measured in the standard unit of kilogram-meter per second (kg · m/s or kg · m · s ^{-1} ). The direction of momentum can be expressed in various ways, depending on the number of dimensions involved. The direction of momentum is same as the direction of velocity.

Momentum, like velocity, is relative. Let us take a 1,000-kg car moving at 20 m/s with respect to the surface of a highway, travelling northward. If the car is driven, the momentum of the car is relative to the body of the person driving the car which is zero. And if a person stands by the side of the road, the momentum of the car relative to that person is 20,000 kg · m/s northward.

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### Linear Momentum

Linear momentum is defined as a vector quantity that is the product of the mass of an object and its velocity. Any change in the mass or the velocity of the system causes a change in linear momentum.

## Conservation of Momentum

The momentum of a system is constant if there is no external force acting on the system. For a collision occurring between two objects in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

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## Derivation of Conservation of Momentum

Let us consider a situation wherein: a truck of mass m_{1}, velocity u_{1} and its momentum = m_{1}u_{1} and a car of mass m_{2}, velocity u_{2} and its momentum m_{2}u_{2}; are moving in the same direction but with different speeds. Therefore, total momentum=m_{1}u_{1} + m_{2}u_{2.}

Now suppose the car and truck collide for a short time t, their velocities will change. So now the velocity of the truck and car become v1 and v2 respectively. However, their mass remains the same. Hence, now the total momentum = m_{1}v_{1 }+ m_{2}v_{2}

Acceleration of car (a) = (v_{2}–u_{2})/t

Also, F = ma

F_{1 }= Force exerted by truck on the car

F_{1} = m_{2}(v_{2}–u_{2})/t

Acceleration of truck =(v_{1}–u_{1})/t

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

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

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

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

## Solved Example For You

Q. A shell is fired from a gun with a velocity of 300 m/s making an angle 60^{o} with the horizontal. It explodes into two fragments when it reaches the highest position. The ratio of the masses of the two pieces is 1 : 3. If the smaller piece stops immediately after the collision. Find the velocity of the other.

Sol: Velocity at the highest point = \( 300 \times \cos{60^o} \)

=150 m/s

Using momentum conservation,

\( 150 \times m = 3m/4 \times v \)

\( \Rightarrow v=200 m/s \)

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