Conservation of Momentum Derivation and Principles

From Newton’s law, we know that the time rate change of the momentum of a particle is equal to the net force acting on the particle and is in the direction of that force.

Fnet=dpdtF_{net} = \frac{dp}{dt}

This clearly makes us understand that linear momentum can be changed only by a net external force. If there is no net external force, p cannot change.

Fnet=0;dpdt=0p is constant{{F}_{net}}=0;\frac{dp}{dt}=0\to p~is~constant

Implies that there is no change in the momentum or the momentum is conserved. i.e

initial  momentum=final  momentuminitial \;momentum = final\;momentum

In order to apply conservation of momentum, you have to choose the system in such a way that the net external force is zero.


In the example given below, the two cars of masses m1 and m2 are moving with velocities v1 and v2 respectively before the collision. And their velocities change to v1 and v2v_{1}^{‘}~and~v_{2}^{‘} after collision. To apply the law of conservation of linear momentum, you cannot choose any one of the cars as the system. If it so, then there is an external force on the car by another car. So we choose both the cars as our system of interest. This is why in all collisions, if both the colliding objects are considered as a system, then linear momentum is always conserved (irrespective of the type of collision).

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A bullet of mass m leaves a gun of mass M kept on a smooth horizontal surface. If the speed of the bullet relative to the gun is v, the recoil speed of the gun will be_______________________.

No external force acts on the system (gun + bullet) during their impact (till the bullet leaves the gun). Therefore the momentum of the system remains constant. Before the impact (gun + bullet) was at rest. Hence its initial momentum is zero. Therefore just after the impact, its momentum of the system (gun+bullet will be zero)

MVb+MVg=0\Rightarrow M\vec{V}_b+M\vec{V}_g = 0 m[Vbg+Vg]+MVg=0\Rightarrow m[\vec{V}_{bg}+\vec{V}_g]+M\vec{V}_g = 0 Vg=mVbgM+m,whereVbg=velocity  of  bullet  relative  to  the  gun=v\vec{V}_g = -\frac{m\vec{V}_{bg}}{M+m}, where \left | \vec{V}_{bg} \right | = velocity\;of\;bullet\;relative\;to\;the\;gun = v vg=mvM+m(opposite  to  the  direction  of  v)v_g = \frac{mv}{M+m}(opposite\;to\;the\;direction\;of\;v)

Types Of Collision

The common normal at the point of contact between the bodies is known as line of impact. If mass centers of the both the colliding bodies are located on the line of impact, the impact is called central impact and if mass centers of both or any one of the colliding bodies are not on the line of impact, the impact is called eccentric impact.

Direct impact/collision

If velocities vectors of the colliding bodies are directed along the line of impact, the impact is called a direct or head-on impact;


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Oblique impact/collision

If velocity vectors of both or of any one of the bodies are not along the line of impact, the impact is called an oblique impact.


Consider a ball of mass m1 moving with velocity v1 collides with another ball of mass m2 approaching with velocity v2. After a collision, they both move away from each other making an angle with the line of impact.

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Coefficient of restitution

When two objects collide head-on, their velocity of separation after impact is in constant ratio to their velocity of approach before impact.

e=velocity  of  separationvelocity  of  approche = \frac{velocity\;of\;separation}{velocity\;of\;approch} 0e10\le e\le 1 velocity  of  separation=e(velocity  of  approach)velocity\;of\;separation = e(velocity\;of\;approach)

The constant e is called as the coefficient of restitution.

Elastic Collision

An elastic collision between two objects is one in which total kinetic energy (as well as total momentum) is the same before and after the collision.


On a billiard board, a ball with velocity v collides with another ball at rest. Their velocities are exchanged as it is an elastic collision.

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Inelastic Collision

An inelastic collision is one in which total kinetic energy is not the same before and after the collision (even though momentum is constant).


When the colliding objects stick together after the collision, as happens when a meteorite collides with the Earth, the collision is called perfectly inelastic.

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In other inelastic collision, the velocities of the objects are reduced and they move away from each other.

Inelastic collision

Velocities of colliding bodies after collision in 1- dimension

Let there be two bodies with masses m1 and m2 moving with velocity u1 and u2. They collide at an instant and acquire velocities v1 and v2 after the collision. Let the coefficient of restitution of the colliding bodies be e. Then, applying Newton’s experimental law and the law of conservation of momentum, we can find the value of velocities v1and v2.

Conserving momentum of the colliding bodies before and the after the collision

m1u1+m2u2=m1v1+m2v2m_1u_1+m_2u_2 = m_1v_1+m_2v_2 ……(i)

Applying Newton’s experimental law

We  have  v2v1u2u1=eWe\;have\;\frac{v_2-v_1}{u_2-u_1}=-e v2=v1e(u2u1)v_2 = v_1-e(u_2 – u_1)…………………….(ii)

Putting (ii) in (i), we obtain

m1u1+m2u2=m1v1+m2v1e(u2u1)m_1u_1+m_2u_2 =m_1v_1+ m_2{v_1-e(u_2 – u_1)} v1=u1(m2em2)m1+m2+u2m2(1+e)m1+m2v_1 = u_1\frac{(m_2-em_2)}{m_1+m_2}+u_2\frac{m_2(1+e)}{m_1+m_2}……….(iii)

From  (ii)  v2=v1e(u2u1)From\;(ii)\;v_2 = v_1-e(u_2-u_1) =u1m1(1+e)m1+m2+u2(m2m1e))m1+m2=u_1\frac{m_1(1+e)}{m_1+m_2}+u_2 \frac{(m_2-m_1e))}{m_1+m_2}…………(iv)

When the collision is elastic; e = 1

Finally,  v1=(m1+m2m1+m2)u1+(2m2m1+m2)u2Finally,\;v_1 = \left ( \frac{m_1+m_2}{m_1+m_2} \right )u_1+\left ( \frac{2m_2}{m_1+m_2} \right )u_2 Finally,  v1=(2m1m1+m2)u1+(m2m1m1+m2)u2Finally,\;v_1 =\left ( \frac{2m_1}{m_1+m_2} \right )u_1+ \left ( \frac{m_2-m_1}{m_1+m_2} \right )u_2

Having derived the velocities of colliding objects in 1 dimensional collision (elastic and inelastic), let us try to set conditions and draw useful conclusions.


  1. If m1 = m2

v1 = u2 and v2 = u1

When the two bodies of equal mass collide head-on elastically, their velocities are mutually


  1. If m1 = m2 and u2 = O, then

v1 = 0, v2 = u1 

  1. If the target particle is massive;

(a) m2 >> m1  and u2 = O

v1 = -u1 and v’=O

The light particle recoils with the same speed while the heavy target remains practically at rest.

(b) m2 >> m1 and u2 ≠ O

v1 ≈ – u1+ 2u2 and v2 ≈  u2

  1. If the particle is massive: m1 >> m2

v= u1 and v,=2u1—  u2

If the target is initially at rest, u2 = O

v1 = u1 and v2 = 2 u1

The motion of the heavy particle is unaffected, while the light target moves apart at a speed twice that of the particle.

  1. When the collision is perfectly inelastic, e = O
v1=v2=u1m1m1+m2+u2m2m1+m2=m1u1+m2u2m1+m2v_{1} = v_{2} = \frac{u_{1}m_{1}}{m_{1} + m_{2}} + \frac{u_{2}m_{2}}{m_{1} + m_{2}} = \frac{m_{1}u_{1}+m_{2}u_{2}}{m_{1} + m_{2}}

Elastic Collision in Two Dimension

A particle of mass m1 moving with velocity v1 along x-direction makes an elastic collision with another stationary particle of mass m2. After the collision, the particles move with different directions with different velocities.

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Applying law of conservation of momentum,

Along x-axis:

m1V1=m1v1cosθ1+m2v2cosθ2m_1V_1 = m_1{v_1}’\cos\theta_1+m_2{v_2}’\cos\theta_2

Along y axis:

0=m1v1sinθ1+m2v2sinθ20= m_1{v_1}’\sin\theta_1+m_2{v_2}’\sin\theta_2

Conservation of kinetic energy,

12m1v12=12m1v12+12m2v22\frac{1}{2}m_1{v_1}^{2} = \frac{1}{2}m_1{{v_1}’}^{2}+\frac{1}{2}m_2{{v_2}’}^{2}

Given the values of θ1and  θ2\theta_1 and \;\theta_2 , we can calculate the values of v1and  v2{v_1}’ and \;{v_2}' by solving the above examples.

JEE Focus:

  • Understand the conservation principle
  • Make the right choice of system to apply the same

Consider a stationary system from which mass is being ejected at the rate of λ kg/second with a velocity of

vc m/s relative to the system.

The element of mass ejected in time dt = λ dt kg.

The momentum of this mass before ejection = 0

Momentum after ejection = (λ dt)ve  


Change in momentum = (λ dt)ve

Hence, Rate of change of momentum = (λdt)vedt=λve\frac{(\lambda dt)v_{e}}{dt} = \lambda v_{e} \rightarrow

From Newton’s second law, the force exerted on the ejected mass =  λ ve


The force excited by the ejected mass on the system = λve=(kgs×ms=Newton){\lambda v_{e}}\leftarrow = \left ( \frac{kg}{s}\times\frac{m}{s} = Newton\right )

If m is the instantaneous mass of the system, then dmdt=λ\frac{dm}{dt} = -\lambda

Variable Mass System

Newton’s Second Law for the Rocket:

Fmg=mdVdtF – mg = m\frac{dV}{dt}

λvemg=mdVdt\lambda v_{e} – mg = m\frac{dV}{dt}

dV=λvemdtgdtdV = \frac{\lambda v_{e}}{m} dt- gdt

Let m be the instantaneous mass and mo be the initial mass of the rocket

 ⇒ m = mo – λt

ovdV=λveotdtmoλtotgdt\int_{o}^{v} dV = \lambda v_{e}\int_{o}^{t}\frac{dt}{m_{o} – \lambda t}\int_{o}^{t}gdt

Assuming g to be uniform,

V=ve1nmomoλtgtV = v_{e}\, 1n\frac{m_{o}}{m_{o}-\lambda t} – gt

= ve1nmomgtv_{e}\, 1n\frac{m_{o}}{m} – gt

Ignoring the effect of gravity, the instantaneous velocity

and acceleration of the rocket (attributed only to the

thrust by the ejected mass) are;

V=ve1nmomV = v_{e}\, 1n\frac{m_{o}}{m} ………………………(1)

dVdt=λvem\frac{dV}{dt} = \frac{\lambda v_{e}}{m} ……………..(2)


m = mo – λt ……………..(3)

Therefore, Thrust f = λve  ……………..(4)

Rocket Equations

At any instant of time momentum of the mass (λdt) relative to the ground before ejection = 

λdtVj^\lambda dt\: V\hat{j}

After ejection = (λdt)(vej^+Vj^)(\lambda dt)(-v_{e}\hat{j} + V\hat{j})


Change in momentum = (λdt)vej^(\lambda dt)v_{e}\hat{j}

Rate of change of momentum of the ejected mass = λvej^-\lambda v_{e}\hat{j}

Force exerted on the ejected mass = λvej^-\lambda v_{e}\hat{j}

Force exerted on the system (Rocket). i.e.

Thrust F = λvej^\lambda v_{e}\hat{j}


After perfectly inelastic collision between two identical particles moving with same speed in different directions, the speed of the particles becomes (2/3)rd of the initial speed. The angle between the two particles before collision is

(a)cos1(1/3)                            (b)sin1(1/3)                            (c)cos1(1/9)                          (d)sin1(1/9)(a)\cos^{-1}(-1/3)\;\;\;\;\;\;\;\;\;\;\;\;\;\;(b)\sin^{-1}(-1/3 )\;\;\;\;\;\; \\ \;\;\;\;\;\;\;\;(c)\cos^{-1(-1/9)}\;\;\;\;\;\;\;\;\;\;\;\;\;(d)\sin^{-1}(-1/9)

Let θ\theta be the desired angle. Linear momentum of the system will remain conserved. Hence

P2=P12+P22+2P1P2cosθP^{2} = P_{1}^{2}+P_{2}^{2}+2P_1P_2\cos\theta =[2m(23)v]2=(mv)2+2(mv)2+2(mv)(mv)cosθ=\left [ 2m\left ( \frac{2}{3} \right )v \right ]^{2}=(mv)^{2}+2(mv)^{2}+2(mv)(mv)\cos\theta 169=2+2cosθ\Rightarrow \frac{16}{9} = 2+2\cos\theta 2cosθ=29\Rightarrow 2\cos\theta= \frac{-2}{9} cosθ=19\Rightarrow \cos\theta= \frac{-1}{9} θ=96.370\Rightarrow \theta = 96.37^{0}

Hence,(C) is the correct choice.

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