But in our syllabus all question will be based upon the collide cases in which the momentum will exchange.So collision is an event in which two bodies strike each other applying large force for a small time.

So all the story of collision will be based upon the momentum, Hence any kind of collision you will apply conservation of momentum to the system, Why system ? because a single body can not collide itself at least two body required so two body system.

Hence the initial momentum of two body will be same as final momentum of two body and will also be same in between initial and final position.

So momentum remains conserved throughout the collision

We see example of collision in day to day life like when we drop a ball on the ground it reaches the ground and strike the ground then bounce back to some height. The time at which ball strike the ground that event is called collision see below picture.

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#### What happens during collision ?

In collision elasticity comes into play, see how elasticity work ? when ball strike the ground then ball compress like spring, then inside the ball elasticity forces is produce and to restore the ball shape elongate like spring and ball bounce back some height, This happens in every collision and body separate away in this way after collision.

What are changes during collision ? mass of the body does not change, Velocity of the body changes it may decrease or increase depend upon the collision condition.

Now related to velocity we study in Physics two important quantity momentum (mv) and kinetic energy (1/2mv²) so there may be changes in momentum or there may be changes in kinetic energy and on the basis of this we study and have done classification of collision.

### Why necessary to study collision ?

Collision study is necessary because in our daily life, we see many objects collide with other objects so around us many thing are colliding and collision is happening.

These collision may be for very small particles like air particle, Which we are unable to see but these air particle collide. We feel that air comes from one side and collide with wall after that we don’t see what happens ?.

So after study collision, we will see what can happens after collision. In this way big object also collide.

Hence small and big both object collide, What will be their result that is we are going to study in collision.

Now see the above picture, When a ball is drop from some height h1 to ground its velocity increase with time time due to acceleration of gravity, When the ball approaches ground this velocity is called relative velocity of approach, Then ball hit the ground and bounce back with some velocity now this ball is separating from ground this velocity is called relative velocity of separation. If relative velocity of approach and relative velocity of separation values becomes same then ball must bounce back the same height but we observe normally this not happens, Ball bounce less height so it is clear that relative velocity of separation becomes less than relative velocity of approach.

Hence in any event of collision there is a relative velocity of approach then collision occur and then relative velocity of separation .

These relative velocity of approach and relative velocity of separation may or may not be same So we made a quantity to measure this value which is called coefficient of restitution.

Which is define as ration of relative velocity of separation to relative velocity of approach,

**e = Coefficient of restitution = (Velocity of separation)/(velocity of approach)**

Now from above picture value of e may be 0, e≤1, maximum 1 and based on it we categories collision types which we will study now.

### Types of collision

There are mainly two types of collision, which we will study in this topic.

**1 Elastic collision **

**In Collision there are three quantity involve, first is momentum (mv) second is kinetic energy (1/2mv²) third is total energy.**

(a) In this collision momentum of system is conserved that is m₁u₁+m₂u₂ = m₁v₁ + m₂v₂.

(b) Kinetic energy is also conserved that 1/2m₁u₁² +1/2m₂u₂² = 1/2m₁v₁² +1/2m₂v₂².

(c) Total energy is also conserve.

Important point kinetic energy does not remain constant through the collision at the time of collision kinetic energy loss in deformation of body and again regain the original shape of body then kinetic energy regain, Hence initial and final kinetic energy will be same in elastic collision.

We assume that this is perfectly elastic collision. now we will solve momentum equation and kinetic energy equation to derive important facts about elastic collision.

m₁u₁+m₂u₂ = m₁v₁ + m₂v₂ arrange this equation m₁ one side and m₂ other side so we can write as m₁( u₁- v₁) = m₂(v₂- u₂) say this equation number 1.

Now take kinetic energy equation 1/2m₁u₁² +1/2m₂u₂² = 1/2m₁v₁² +1/2m₂v₂² cancel 1/2 and arrange same like m₁ one side and m₂ other side then we can write

m₁( u₁²- v₁²) = m₂(v₂² –u₂²) or m₁(u₁- v₁)*(u₁+ v₁) = m₂(v₂- u₂)*(v₂+ u₂) say this equation number 2,

Now divide equation 2 by equation 1 we will get like this

m₁(u₁- v₁)*(u₁+ v₁)/m₁( u₁- v₁) = m₂(v₂- u₂)*(v₂+ u₂)/m₂(v₂- u₂)

(u₁+ v₁) = (v₂+ u₂) or arrange this ( u₁- u₂) = (v₂- v₁) this is very important equation .

( u₁- u₂) is called velocity of approach

(v₂- v₁) is called velocity of separation

Hence in case of perfectly elastic collision velocity of approach = velocity of separation or

**(v₂- v₁)/****( u₁- u₂)**** = velocity of separation/velocity of approach = 1 =e**

** **

As from the first question we have to find v₁ and v₂ after collision, We assume that this is perfectly elastic collision.

Hence use the equation u₁- u₂ = v₂- v₁ now put the value of u₁ and u₂ in this equation we will get 5- 2 = v₂- v₁ or 3 = v₂- v₁ say this equation number (1).

Now here two variable v₁ and v₂ hence two equation required to solve the value of v₁, v₂.

So use one momentum equation m₁u₁+m₂u₂ = m₁v₁ + m₂v₂ because momentum is conserved so we can write .

2*5 +3*2 = 2v₁ +3v₂ or 16 = 2v₁ +3v₂ say this equation number (2) .

Now using equation (1) and equation (2) from (1) v₂ = 3+v₁ , now put the value of v₂ in equation (2) we will get 16 = 2v₁ +3(3+v₁) or 16 = 2v₁+9+3v₁

16 – 9 = 5v₁ or 7 = 5v₁ or v₁ = 7/5 = 1.4 or v₁ = 1.4m/s

Now put the value of v₁ in v₂ = 3+v₁ so v₂ = 3+1.4 = 4.4 hence v₂ = 4.4m/s.

So this is velocity of ball after collision.

Now for second question we have to find mass m which is at rest .

This is also elastic collision hence we can apply similar equation to find the value of m.

u₁- u₂ = v₂- v₁ now put the value here u₁ = u , u₂ =0 , v₂ = v , v₁ = -u/3 see from the above picture given value put in the equation u₁- u₂ = v₂- v₁ .

u-0 = v-(-u/3) or u = v+u/3 or 3u = 3v+u or 2u = 3v say this equation (1)

Now use the equation of momentum conservation

1*u +m*0 = 1*(-u/3) +m*v or u = -u/3 +mv or mv = 4u/3 say this equation (2)

Now from equation (1) v = 2u/3 put v value in equation (2) we will get

m*2u/3 = 4u/3 or m =2kg this is mass of second body at rest.

#### Elastic head on collision case

Now take the equation u₁- u₂ = v₂- v₁ (1) and m₁u₁+m₂u₂ = m₁v₁ + m₂v₂ (2)

from (1) we can write v₂ = u₁- u₂ +v₁ now put value of v₂ in equation (2) so we will get

m₁u₁+m₂u₂ = m₁v₁ +m₂(u₁- u₂ +v₁) or

m₁u₁+m₂u₂ = m₁v₁+m₂u₁ –m₂u₂+m₂v₁ now arranging we can write u₁ one side and v₁ other side

m₁u₁ – m₂u₁ + m₂u₂ + m₂u₂ = m₁v₁+ m₂v₁ or

u₁(m₁-m₂) +2m₂u₂ = v₁(m₁+m₂)

**v₁ = u₁(m₁-m₂)/(m₁+m₂) +2m₂u₂/(m₁+m₂) say equation (3)**

Now putting the value of **v₁ **then we will get the value of v₂

**v₂ = u₂(m₂-m₁)/(m₁+m₂) +2m₁u₁/(m₁+m₂) say equation (4)**

Now we will apply conditions if mass of two bodies are same m₁ = m₂ = m then see what will happens v₁ and v₂ in equation (3) and (4)

v₁ = u₂ and v₂ = u₁ this is important result for elastic collision here we see that velocity is exchange after collision if mass of two body are same. Now second condition if m₂>>> m₁ and m₂ is initially at rest then what will happen find out v₁ from equation (3) here m₂ is at rest means u₂ = 0 and m₁→0 comparison of m₂ so we will get v₁ = – u₁ this indicate will return with same speed in opposite direction

example for this case a heavy truck is at rest and one person on bicycle going to hit the track then what will happen truck will not move at all, but the bicycle person will return back with same hitting speed.

Now you find v₂ it will becomes zero v₂ = 0 its clear that truck will not move.

Now take opposite condition if m₁>>> m₂ and m₂ is at rest ( u₂ =0 ) then find v₁ and v₂ .

This case is like a person are standing or is at rest and truck is hitting the person what will happen here you calculate v₁ = u₁ and v₂ = 2u₁ this indicate that truck will move with same velocity but standing person will get twice velocity of truck.

So this is conclusion that heavier mass has no effect of change of its momentum when it collide with lighter mass in both the cases lighter mass has got effected.

So i hope that you have got the answer when air particle collide with wall what will happen ?

**2 In-elastic collision **

**(a) In this collision momentum is conserved m₁u₁+m₂u₂ = m₁v₁ + m₂v₂**

(b) Kinetic energy is __not conserved__

(c) Total energy is conserved

Now we will continue in next post. I hope that you have enjoyed learning What is meaning of collision in physics. If you like comment and share thanks for reading and sharing. Learn and grow.

dated 23rd Nov 2018

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