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The conservation of momentum states that the amount of momentum remains constant, i.e. the momentum can neither be created nor be destroyed, however, can be changed through the action of forces as described by Newton's laws of motion.

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The initial velocity = u as it sets to motion, each particle of the fluid starts accelerating with a. It acquires a varying velocity at different times, given by,

F = ma = m Δu/Δt

∵ F = pA (p = pressure, A = area

= [(- pA)1 - (pA)2] = m (u2 - u1)/Δt

= - [(p+ ((Δp/Δx)Δx)A - pA)] = m(u + (Δu/Δx)Δx - u)/Δt

∵ m = p Δx A

= - (Δp/Δx)Δx A = m(Δu/Δx)Δx/Δt

∵ Δx/Δt = u

= - Δp/Δx = p u Δu/Δx

In Differential Form:

= - dp/dx = p u du/dx

This is an Euler’s equation, which states that the pressure drop of a fluid is proportional to the velocity and the velocity gradient.

This momentum equation gives us the form of the dynamic pressure that appears in Bernoulli's Equation.

According to this law, the resultant momentum of a system equal to the vector sum of the momentum of individual particles in the system remains conserved and is not affected by their mutual action and reaction.

We know that, p = mv

Let’s consider a system of particles, whose center of mass is given by,

The position of the center-of-mass of the system is,

rcm = m1r1 + m2r2 + …..+mnrn)/(m1 + m2 +...+mn)

On differentiating with respect to time we have velocity of the center-of-mass

vcm = (m1 v1 + m2v2 + …+mnvn)/(m1 + m2+ …+mn)

vcm (m1 + m2+ …+mn) = (m1v1 + m2v2 + …+mnvn)

Where (m1 + m2+ …+mn) = M (the sum of masses of particles of a system)

Mvcm = (m1v1 + m2v2 + …+mnvn)

Let’s say a system of particles having momentum, m1v1 , m2v2….

The resultant velocity of the system = vcm.

So, the resultant momentum of the system is,

Psystem = Mvcm

Particles in the system have infinitely small masses.

Suppose a wheel is rolling on the road while rolling, velocity will be different at different parts.

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So, to calculate the resultant momentum of the system (wheel)

Pbody(wheel) = Mvcm = constant

Hence, the resultant momentum equal to the vector sum of the momentum of individual momentum is constant.

The law of conservation of momentum states that if the net force on the system is zero, the momentum of a system remains constant.

This momentum remains constant until or unless an external force acts on the system.

According to Newton’s second law of motion:

F = dp/dt

∵ Fnet = 0

∴ dp = 0

P = constant

As we know that,

vcm = (m1 v1 + m2v2 + …+mnvn)/(m1 + m2+ …+mn)

Let’s say, the net force on the system is,

Fnet = 0

Here, Fnet doesn’t mean the force on individual particles is zero. It means the net force on the system as a whole is zero.

∴ acm = 0

Here, individual particles may have an acceleration canceling out each other to give an overall acceleration as zero.

While vcm = constant

However, the velocity of individual particles may vary.

So, we can write:

(m1 v1 + m2v2 + …+mnvn)/(m1 + m2+ …+mn) = constant

When Fnet = 0.

m1 v1 + m2v2 + …+ mn vn = constant

P1 + P2 +....+Pn = constant

P system = constant

The total momentum of a system is constant only when the total force of the system is zero.

However, the momentum of individual particles may vary.

Recoiling of a Gun

When a bullet is fired from the gun, it recoils, which means the gun moves in a direction opposite to that of the direction of motion.

The recoil velocity of the gun can be calculated by applying the principle of conservation of linear momentum.

Let,

m1 = mass of a bullet and m2 = mass of the gun

v1 = velocity of a bullet and v2 = velocity of the gun

So, the vector sum of the momentum of the gun and the bullet is,

= m1 v1 + m2v2

Or m1 v1 = - m2v2

We get,

v2 = - m1 v1 /m2

This negative sign shows the direction of v is opposite to the direction of v, i.e. the gun recoils.

Here, v2 α 1/m2

It means a gun with heavier velocity will recoil with a small velocity and vice versa.

According to Newton’s Third Law of Motion:

F12 = - F21

F12 + F21 = 0

ma + ma = 0 (Newton’s second law)

m1 dv1/dt + m2 dv2/dt = 0 (definition for acceleration)

d(m1 v1)/dt + d(m2v2)/dt = 0 (if masses are constant)

d/dt(m1 v1+ m2v2) = 0 (if the derivative wrt time is zero, then the sum is constant)

∴ p = mv (definition for the linear momentum of a particle)

When a lady jumps out a boat to the shore, the boat is pushed slightly away from the shore: It's because the momentum of the boat is equal and opposite to that of the lady following the law of conservation of linear momentum.

The explosion of a bomb: When the bomb explodes, its pieces are scattered in all directions such that the vector sum of the momentum of these pieces becomes zero.

FAQ (Frequently Asked Questions)

Q1: Does Linear Momentum have a Direction?

Ans: Linear momentum is a vector quantity that remains conserved in every direction. In the center of mass, the total momentum remains zero before and after the interaction in any direction.

Q2: Why is Linear Momentum Important?

Ans: It is important to study linear momentum because it is a conserved quantity.

In a closed system, when no external force acts on the system, its total linear momentum remains constant.

Q3: A Bullet of 20 gm is Shot from a 10 kg gun with a Velocity of 400 m/s. What is the Recoil of the Gun?

Ans: Here, m_{1} = 20 g = 20/1000 kg, m_{2} = 5 kg and v_{1}_{ }= 400 m/s

By the formula, v_{2} = - m_{1} v_{1} /m_{2}

v_{2} = 20/1000 x 400/5 = 1.6 m/s

Q4: What is the Dimension of Linear Momentum?

Ans: Since F = dp/dt or p = F dt

F = [M L T^{-2}] and T = [T]

So, dimensional formula for p is [M L T^{-1}].