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Derive the equation of Perfectly Elastic Head on Collision.

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Let two bodies of masses m1 and m2 moving initial velocities u1 and u2 in the same direction they collide such that after collision their final velocities are v1 and v2 respectively.

According to law of conservation of momentum and conservation of kinetic energy.

The ratio of relative velocity of separation and relative velocity of approach is defined as coefficient of restitution.

 For perfectly elastic collision e = 1 

 v2 – v1 = u1 – u2    [As shown in eq. (vi)] 

For perfectly inelastic collision e = 0 

∴ v2 – v1 = 0 or v2 = v

It means that two body stick together and move with same velocity.

For inelastic collision 0 < e < 1 

∴ v2 – v1 = (u1 – u2)

In short we can say that e is the degree of elasticity of collision and it is dimension less quantity.

When two bodies of equal masses undergo head on elastic collision, their velocities get interchanged.

Kinetic energy transfer during head on elastic collision : Fractional decrease in kinetic energy

Note : 

• Greater the difference in masses less will be transfer of kinetic energy and vice versa. 

• Transfer of kinetic energy in head on elastic collision (when target is at rest) is maximum when the masses of particles are equal.

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