Conservation of Momentum

Conservation of linear momentum of System of particles :

According to Newton’s Second law of motion the time rate of change of momentum is equal to the applied force.

$ \displaystyle \frac{d\vec{P}}{dt} = \vec{F} $

$ \displaystyle m \frac{d\vec{v}}{dt} = \vec{F} $

If the system consists of n bodies ,

Since , $ \displaystyle m_1 \frac{\vec{dv_1}}{dt}+ m_2 \frac{\vec{dv_2}}{dt} + ….+ m_n \frac{\vec{dv_n}}{dt} = M\vec{a_{cm}} $

$ \displaystyle \frac{d}{dt}(m_1 \vec{v_1} +m_2 \vec{v_2}+ ….+ m_n \vec{v_n} ) = M\vec{a_{cm}} $

$ \displaystyle \frac{d}{dt}(\vec{P_1}+\vec{P_2} + ….+ \vec{P_n}) = \sum \vec{F_{ext}}$ ;

Where $\vec{P_1}$ , $\vec{P_2}$  , $\vec{P_3}$ ……….$\vec{P_n}$ are the linear momenta of the particles .

” In the absence of external forces, the total momentum of the system is conserved ”.

or, For an isolated system, the initial momentum of the system is equal to the final momentum of the system.

$ \displaystyle \sum \vec{F_{ext}} = 0 $

$\large \vec{P_1} + \vec{P_2} + \vec{P_3} + …..+ \vec{P_n} = Constant$

Motion of a Shot and the Gun

Before firing, both the bullet and the gun are at rest. Hence, the momentum of the system, before firing, is zero. When the bullet is fired, it moves in the forward direction with a large velocity. To conserve momentum gun moves backward with a velocity that the final momentum of the system is also zero.
Let m = mass of bullet
M = mass of gun
v = velocity of bullet
V = velocity of recoil of gun

com

Since no external force is acting on the system hence the momentum should be same as before i.e. must be zero.

$ \displaystyle \vec{P_f} = \vec{P_i}$

$ \displaystyle m \vec{v} = M\vec{V}$

$ \displaystyle \vec{V} = -\frac{m}{M}\vec{v}$

–ve sign indicates that velocity of recoil is opposite to the bullet.

When bullet hits and enters the wooden block

com

A bullet of mass m and velocity v enters the wooden block of mass M and stops. Block is free for motion on frictionless surface. Now the velocity of the system (Bullet and block) is V then

Momentum before collision = momentum after collision

m v = (m + M) V

$ \displaystyle V = \frac{m v}{m + M} $

Initial kinetic energy $ \displaystyle E_i = \frac{1}{2}m v^2 $

Final kinetic energy $ \displaystyle E_f = \frac{1}{2}\frac{m^2 v^2}{(m+M)} $

$ \displaystyle \frac{E_f}{E_i} = \frac{m}{m+M} < 1$

Hence some part of the energy is lost.

When bullet travels across the block.

com

A bullet of mass m travels across the block of mass M. Now the block starts traveling with velocity V.

m v = M V + m v1 … (1)

Loss of Energy $ \displaystyle = \frac{1}{2}m v^2 – (\frac{1}{2}MV^2 + \frac{1}{2}m v_1 ^2)$

= (Initial energy) – (Final energy)

Explosion of a Bomb:

Suppose a bomb is at rest before it explodes. Its total momentum is zero. When the bomb explodes, it breaks up into many part carrying a particular momentum. For a part flying in one direction with a certain momentum, there is another part moving in the opposite direction with the same momentum.
The bomb explodes into two parts whose mass is m1 and m2 velocity is v1 and v2.

Since no external force is acting on the bomb hence linear momentum is conserved.

m1 v1 + m2 v2 = 0

$ \displaystyle \frac{v_1}{v_2} = – \frac{m_1}{m_2} $

Hence velocity of parts is inversely proportional to their masses and will be in opposite direction to each other.

Note :

If the bomb is in motion before explosion then initial momentum will not be zero.

Conservation of motion of centre of mass :

In absence of a net external resultant force along a certain direction there will be no change in the state of motion of the centre of mass. i.e. if it had been at rest it will remain at rest and if it had been moving with uniform velocity, it will continue moving with the same velocity.

Illustration : A body explodes in mid-air. Does its momentum remain conserved ?

Solution: During explosion, the net force acting on the system is the weight of the system, that is mg, where m is the mass of the body. In this case we see the change in momentum in time Δt is given as

$ \displaystyle \Delta P = \int_{0}^{\Delta t} F_{ext}dt $

$ \displaystyle \Delta P = F_{ext} \Delta t $

Since Δt → 0 ⇒  ΔP→ 0

The momentum of the system first before and after the impact remains practically equal. Then the external force, that is gravitational force changes the momentum of the c.m. of the system for considerable time interval not during explosion.

Exercise : A body when thrown upwards, changes its velocity, therefore it changes its momentum. Does this violate the conservation of momentum principle? Justify.

Characteristic of linear momentum

It depends on the frame of reference, e.g., the linear momentum of a body at rest in a moving train, is zero relative to a person sitting in the train while it is not zero for a person standing on the ground.

Two bodies of same mass and moving with same speed will have different momenta unless their directions of motion are same.

Relation between kinetic energy and momentum.

$ \displaystyle K = \frac{P^2}{2m} \; or , P = \sqrt{2mK} $

(a) If p = constant

$ \displaystyle K \propto \frac{1}{m}$

i.e. if different bodies have same momentum, KE will be maximum for the lightest one

(b) If K = constant

$ \displaystyle P \propto \sqrt{m} $

i.e. if different bodies have same KE, the heaviest body will have the maximum momentum

(c) If m = constant

$ \displaystyle P \propto \sqrt{K} $

i.e. if different bodies have same mass, the body having maximum KE will have maximum momentum.

Exercise : A boy of mass 30 kg is standing on a flat boat so that he is 20 meters from the shore. He walks 8 m on the boat towards the shore and then stops. The mass of the boat is 90 kg and friction between the boat and the water surface is negligible. How far is the boy from the shore now ?

Also Read :

∗ Centre of mass of discrete & continuous mass distribution
∗ Velocity & Acceleration of Centre of mass

←Back Page

Leave a Reply