## COLLISION

**When one body struck another body, the collision is said to occurs in two**.

### TYPE OF COLLISION

### # ELASTIC COLLISION

A
collision in which there is absolute no loss of kinetic energy is called
Elastic Collision.

Example-
The collision between two Ivary balls is an example of elastic collision.

### CHARACTERISTIC OF ELASTIC COLLISION

- Linear momentum is conserved
- Kinetic energy is conserved
- Total energy is conserved

### # INELASTIC COLLISION

**A collision in which there occurs in some loss of kinetic energy is called Inelastic Collision.**

Example
– A ball is dropped from a certain height and after collision with the ground
it could not reach at the same position then the collision of the ball with
ground is an example of Inelastic Collision.

### CHARACTERISTIC OF INELASTIC COLLISION

- Linear momentum is conserved (Because no external force is acting on the body)
- Kinetic energy is not conserved
- Total energy is not conserved

### # PERFECT INELASTIC COLLISION

A
collision in which there occurs maximum
loss of energy is called Perfect Inelastic Collision.

Example-
mud is thrown on the wall it will stick to wall is an example of Perfect
Inelastic Collision.

### CHARACTERISTIC OF PERFECT INELASTIC COLLISION

- Linear momentum is conserved
- Kinetic energy is not conserved
- Total energy is not conserved

### # COFFICENT OF RESTITUTION (e)

Before
relative velocity = U

_{1}– U_{2}
After
relative velocity = V

_{2}– V_{1}_{}

**Coefficient of restitution is defined as the ratio of relative velocity of separation after collision to the relative velocity of approach before collision.**

e
= V

_{2}– V_{1}/ U_{1}– U_{2}_{}

If
e = 1 (Then it is called Elastic Collision)

If
e = 0 (Then it is called Perfectly Inelastic Collision)

If 0 < e
< 1 (Then it is called Inelastic Collision)

### # DERIVITION OF COLLISION OF VELOCITY OF ELASTIC COLLISION

When
two body of different mass when two body as moving with different velocity in
same straight line and after collision moves in same straight line collision is
said to be called Elastic Collision.

Consider
two body mass M

_{1}and M_{2}moving with the velocity U_{1}and U_{2 }in same straight line if U_{1 }> U_{2 }they collide and after collision they move with velocity with V_{1 }and V_{2 }in a straight same line.
As
the collision is elastic , the linear momentum is conserved therefore the

Linear
momentum before collision =

_{ }Linear momentum after collision
M

_{1}U_{1 }+ M_{1}U_{2}= M_{1}V_{1 }+ M_{1}V_{2 }----------------------(1)
M

_{1}U_{1 }- M_{1}V_{1}= M_{2}V_{2 }- M_{2}U_{2 }
M

_{1}( U_{1 }- U_{2}) = M_{2}(V_{1 }- V_{2 })_{ }---------------------------(2)
As
the collision is elastic the kinetic energy is conserved

Kinetic
energy before collision = Kinetic energy after collision

½
M

_{1}U_{1}^{2}_{ }+ ½ M_{1}U_{2}^{2}= ½ M_{1}V_{1 }^{2}+ ½ M_{1}V_{2}^{2}
M

_{1}U_{1}^{2}_{ }+ M_{1}V_{1}^{2}= M_{2}V_{2 }^{2}+ M_{2}U_{2}^{2}
M

_{1 }( U_{1}^{2}_{ }+ V_{1}^{2}) = M_{2}(V_{2 }^{2}+ U_{2}^{2})
M

_{1 }( U_{1 }+ V_{1}) ( U_{1 }- V_{1}) = M_{2}(V_{2 }+ U_{2}) (V_{2 }- U_{2}) -----------(3)
Dividing equation ‘3’ by equation ‘2’

M

_{1 }( U_{1 }+ V_{1}) ( U_{1 }- V_{1}) / M_{1}( U_{1 }- U_{2}) = M_{2}(V_{2 }+ U_{2}) (V_{2 }- U_{2}) / M_{2}(V_{1 }-V_{2 })_{ }
U

_{1 }+ V_{1}= V_{2 }+ U_{2}
V

_{2 }= U_{1 }+ V_{1}- U_{2}----------------------------------(4)
Putting
the value of V

_{2}in equation ‘1’
M

_{1 }V_{1 }+ M_{2}U_{1 }+ M_{2}V_{1}– M_{2}U_{2}= M_{2}U_{1}+ M_{2}U_{2 }
M

_{1 }V_{1 }+ M_{2}V_{1 }= M_{1}U_{1}- M_{2}U_{1 }+ M_{2}U_{2 }+ M_{2}U_{2 }
V

_{1 }(M_{1 }+ M_{2}) = U_{1}( M_{1}- M_{2})_{ }+ 2 M_{2}U_{2}
V

_{1 }= U_{1}( M_{1}- M_{2})/ (M_{1 }+ M_{2})_{ }+ 2 M_{2}U_{2}/(M_{1 }+ M_{2}) -----------(5)
Putting
the vale of V

_{1 }in equation ‘4’**– In place of 1 write 2 and in place of 2 write 1.**

__TRICK__
V

_{2 }= U_{2}( M_{2}- M_{1})/ (M_{2 }+ M_{1})_{ }+ 2 M_{1}U_{1}/(M_{2 }+ M_{1}) --------(6)_{ }

### # What happen wen equal mass collide in elastic collision.

M

_{1}=M_{2}
V

_{1 }= U_{1}( M_{1}- M_{2})/ (M_{1 }+ M_{2})_{ }+ 2 M_{2}U_{2}/(M_{1 }+ M_{2})
V

_{1 }= U_{1}( M - M)/ (M_{ }+ M)_{ }+ 2 M U_{2}/(M_{ }+ M)
V

_{1 }= 0_{ }+ 2 M U_{2}/(M_{ }+ M)
V

_{1 }= U_{1}_{}

V

_{2 }= U_{2}( M_{2}- M_{1})/ (M_{2 }+ M_{1})_{ }+ 2 M_{1}U_{1}/(M_{2 }+ M_{1})_{}
V

_{2 }= U_{2}( M - M)/ (M_{ }+ M_{1})_{ }+ 2 M U_{1}/(M_{ }+ M)
V

_{2 }= 0 + 2 M U_{1}/(M_{ }+ M)_{}
V

_{2 }= U_{2}
It show before collision and after collision the velocity remain same if
the mass is equal.

### # DERIVATION OF INELASTIC COLLISION OF VELOCITY

As
the collision is inelastic the linear momentum is conserved

M

_{1}U_{1 }+ M_{2}U_{2}= M_{1}V_{1 }+ M_{2}V_{2 }----------(1)
Now
with the help of coefficient of restoration
e = V

_{2}-V_{1}/U_{1}-U_{2}
e(
U

_{1}- U_{2}) = V_{2 }- V_{1}_{ }eU

_{1}- eU

_{2}= V

_{2 }- V

_{1}

V

_{2}=_{ }eU_{1}- eU_{2}+ V_{1 }--------------(2)
Putting
the value of V

_{2}in equation ‘1’
M

_{1}U_{1 }+ M_{2}U_{2}= M_{1}V_{1 }+ M_{2}(eU_{1}- eU_{2}+ V_{1 })_{ }
M

_{1}U_{1 }+ M_{2}U_{2}- M_{2}eU_{1}+ M_{2}eU_{2}= M_{1}V_{1 }+ M_{2}V_{1 }
V

_{1 }(M_{1}+ M_{2})_{ }= M_{1}U_{1 }- M_{2}eU_{1}+ M_{2}U_{2}+ M_{2}eU_{2}
V

_{1 }= U_{1 }(M_{1 }- M_{2}e) /(M_{1}+ M_{2})_{ }+ M_{2}U_{2}(1+ e )/(M_{1}+ M_{2})------(3)
Putting
the value of V

_{1}in equation ‘2’

**– In place of 1 write 2 and in place of 2 write 1.**

__TRICK__
V

_{2 }= U_{2 }(M_{2 }- M_{1}e) /(M_{1}+ M_{2})_{ }+ M_{1}U_{1}(1+ e )/(M_{1}+ M_{2})------(4)### # If e=0 for perfectly inelastic collision

V

_{1 }= U_{1 }(M_{1 }- 0) /(M_{1}+ M_{2})_{ }+ M_{2}U_{2}(1 )/(M_{1}+ M_{2})------(5)
V

_{2 }= U_{2 }(M_{2 }- 0) /(M_{1}+ M_{2})_{ }+ M_{1}U_{1}(1 )/(M_{1}+ M_{2})------(6)
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