When two bodies moving along a straight line collide with each other the collision is called the head-on collision. In a head-on collision, the initial and the final velocities are along the same straight line.

Elastic Collision:

The collision in which the total kinetic energy, as well as total momentum, is conserved is called an elastic collision.

In a perfectly elastic collision, the relative velocity of the approach before the collision is equal to the relative velocity of separation after the collision.

Proof:

Let us consider two bodies having masses m₁ and m₂ moving in the same direction, along the same straight line with velocities u₁ and u₂ respectively ( u1  >  u2). Let v₁ and v₂ be their velocities after the collision. Since all the velocities are in the same direction, we can write the equation of conservation of momentum in scalar form.

m₁u₁+ m₂u₂ = m₁v₁+ m₂v₂

∴  m₁u₁ – m₁v₁   = m₂v₂–  m₂u₂

∴  m₁(u₁ – v₁)  = m₂₍v₂–  u₂)  ……….. (1)

For elastic collision, total kinetic energy before the collision is equal to total kinetic energy after collision.

The quantity u₁ – u₂ is called the relative velocity of approach and the quantity v2 – v1 is called the velocity of separation. Thus for elastic collision, The relative velocity of the approach before the collision is equal to the relative velocity of separation after the collision.

Calculation of Final Velocities of Bodies:

Values of v1 and v2 can be found using values of u1 and u2 as follows.

We have,

u1  –  u2     = v2  – v1

v2   =   u1  –  u2   +  v1

Putting this value in equation (1) we have

    m1(u1 – v1)   =  m2( u1 + v1 – u2 – u2)

m1 u1 – m1v1   =  m2( u1 + v1 –  2 u2 )

m1 u1 – m1v1   = m2 u1 + m2 v1 –  2 m2 u2

m1 u1 – m2 u1 +  2 m2 u2  =   m1v1  + m2 v1

 m1v1  + m2 v1  =   m1 u1 – m2 u1 +  2 m2 u2

v1 ( m1  + m2 )  =   (  m1 – m2  ) u1 +  2 m2 u2

We have,

u1  –  u2     = v2  – v1

v1   =   u1  –  u2   – v2 

Putting this value in equation (1)

m1( u1 – v2 – u2 + u1)  =  m2(v2 – u2)

m1 ( 2 u1 – v2 – u2)  =  m2(v2 – u2)

2 m1u1 – m1v2 – m1u2  = m2 v2 – m2 u2

2 m1u1 – m1u2  +  m2 u2 = m2 v2 + m1v2

m2 v2 + m1v2   =  2 m1u1 – m1u2  +  m2 u2

( m2 + m1) v2   =  2 m1u1 + ( m2  –  m1 )u2

We should study following special cases.

Case – I

When two spheres have equal masses i.e. m₁ = m₂ =  m

Similarly

Thus, in this case, the two bodies exchange their velocities during a collision.

Case – II

The sphere B is at rest i.e. u₂ = 0

Similarly,

Further if m1  = m2  = m,

∴ v₁  =  0    and v₂  = u₁

Thus the body of mass m₁ is stopped cold and the body of mass m2 takes off with velocity which is equal to the initial velocity of the body of mass m1 before the collision.

Inelastic Collision:

The collision in which the total momentum is conserved but the total kinetic energy is not conserved is called the inelastic collision.

A collision between two bodies is said to be a perfectly inelastic collision if they stick to each other and moves together with common velocity after collision.

Let us consider two bodies having masses m₁ and m₂ moving in the same direction along the same straight line with velocities u₁ and u₂ respectively. ( u₁ >  u₂). Let v be the common velocity of the two bodies after the collision. Since all the velocities are in the same direction, we can write the equation of conservation of momentum in scalar form.

m₁u₁ +   m₂u₂ =   m₁v  +  m₂v

m₁u₁  +   m₂u₂ =   ( m₁ +  m₂)v

Since the quantities on the right sides are positive. The initial kinetic energy is always greater than the final kinetic energy in a perfectly inelastic collision.

When two bodies collide with each other part of its kinetic energy gets converted into sound energy, heat energy, etc. Therefore, total kinetic energy after collision decreases due to the loss of K.E.

The Coefficient of Restitution:

The ratio of the relative velocity of separation to the relative velocity of approach, in a collision between two bodies, is called the coefficient of restitution. It is denoted by e.

For perfectly elastic collision e = 1 , For perfectly inelastic collision e = 0, for all other collisions 0  <  e   <  1

Action and reaction are two forces equal in magnitude and opposite in direction. Why they do not cancel each other?

When two forces having the same magnitude and opposite direction act on the same body their effect is canceled. Action and reaction forces are equal in magnitude and opposite in direction but they act on two different bodies. Therefore they do not cancel each other.

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In this article, we shall study the law of conservation of momentum and its proof.

Statement:

When the resultant external force acting on a system is zero, the total linear momentum of the system is constant. OR When no external force acts on a system of bodies and there is collision among them then the total linear momentum before the collision equals to the total momentum after the collision.

Explanation :

Let us consider two bodies having masses m₁ and m₂ moving towards each other along the same straight line with velocities u₁ and u₂ respectively. Let v₁ and v₂ be their velocities after the collision.

By the law of conservation of momentum,The total momentum of the system before collision   = The total momentum of the system after collision

m₁u₁+ m₂u₂ = m₁v₁+ m₂v₂.

This law is universal because It is not only true for the collision of astronomical bodies but also for collision of subatomic particles.

Proof:

Let us consider two bodies having masses m₁ and m₂ moving towards each other along the same straight line with velocities u₁ and u₂ respectively.

The total momentum of the system before collision   = m₁u₁ – m₂u₂    …………….(1)

When two bodies collide body of mass m₁ exerts pressure F on the body of mass m2. This is action. By Newton’s third law of motion body of mass m2 exerts force   – F on the body of mass m₁. This is a reaction. Thus resultant force acting on the system is zero.

Let the bodies of mass m₁ and m₂ move with velocities  v₁ and v₂ respectively as shown in the diagram.

The total momentum of the system after  collision = – m₁v₂ + m₂v₂   ……….. (2)

From equations (3) and (4) Thus, the total momentum of the system before collision   = The total momentum of the system after the collision. Thus the law of conservation of momentum is proved.

Numerical Problems:

Example – 01:

A ball of mass 100 g moving at a speed of 12 m/s strikes another ball of mass 200 g at rest. After the collision both the balls stick to each other and move with a common velocity. Find the common velocity.

Given : mass of first ball = m₁ = 100 g = 0.1 kg , initial velocity of first ball = u₁= 12 m/s, mass of the second ball = m₂ = 200 g = 0.2 kg , initial velocity of second ball = u₂ = 0 m/s,

To Find: common velocity = v = ?

Solution:

Let v m/s be their common velocity after collision

Final velocitie v₁ = v₂ = v ms-1.

By Law of Conservation of momentum

m₁u₁+ m₂u₂ = m₁v₁+ m₂v₂

∴  (0.1)(12) + (0.2)(0) = (0.1)(v) + (0.2)(v)

∴  1.2 + 0 = 0.3 v

∴  v = 1.2/0.3 = 4 m/s

Ans: Their common velocity is 4 m/s

Example – 02:

A gun fires a bullet of mass 50 g with a velocity of 30 m/s because of which the gun recoils with a velocity 1 m/s. Find the mass of the gun.

Given : mass of bullet = m₁= 50 g = 0.050 kg , velocity of bullet = v₁ = 30 m/s, recoil velocity of gun = v₂ = 1 m/s.

To Find:  mass of gun = m₂ = ?

Solution:

By law of conservation of momentum

Mass of gun x Recoil velocity of gun = Mass of bullet x Velocity of bullet

m₁v₁ = m₂v₂

∴  0.050 x 30 = m₂ x 1

∴  m₂ = 1.5 kg

Ans: The mass of the gun is 1.5 kg.

Example – 03:

A railway engine of mass 2 tons moving at a speed of 72 km/hr, collides with a wagon at rest. After the collision, both have a common velocity of 36 km/hr. Find the mass of the wagon.

Given : mass of engine = m₁ = 2 T = 2000 kg, initial velocity of engine = v₁ = 72 km/hr = 72 x 5/18 = 20 m/s, Mass of wagon = m₂ = Mass of wgon

To Find: Mass of wagon = m₂ = ?

Solution:

Let ‘v m/s be their common velocity after collision

v₁ = v₂ = v  = 36 km/hr = 36 x 5/18 = 10 m/s

By Law of Conservation of momentum

m₁u₁+ m₂u₂ = m₁v₁+ m₂v₂

∴  (2000)(20) + (m2)(0) = (2000)(10) + (m₂)(10)

∴  40000 + 0  = 20000 + 10(m₂)

∴  40000 – 20000 = 10(m₂)

∴  20000 = 10(m₂)

∴  m₂ = 20000/10 = 2000 kg

Ans: The mass of the wagon 2000 kg.

Example – 04:

A bullet of mass 50 g is fired from a gun. If the bullet acquires a velocity of 100 m/s in 0.1secondsd. What is the recoil force on the gun?

Given : mass of bullet = m = 50 g = 0.05 kg, initial velocity of bullet = u = 0, final velocity of bullet = v = 100 m/s, time taken = t = o.1 s

To Find:  recoil force = F = ?,

Solution:

a = (v – u)/t = (100 – 0)/0.1 = 100/0.1 = 1000 m/ s²

Force on bullet = m.a = 0.050 X 1000 = 50 N

By Newton’s third law of motion, the recoiling force on the gun is 50N.

Ans: Recoil force on the gun is 50 N

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