Explain the derivation of elastic & inelastic collision
Elastic Collision in One Dimension
Consider that two perfectly elastic bodies A and B of masses M_{1} and M_{2} moving with initial velocities u_{1} and u_{2 }undergo head on collision and continue moving along the same straight line with final velocities v_{1} and v_{2}.
As in an elastic collision, momentum is conserved.
∴ M_{1}u_{1}_{ }+ M_{2}u_{2} = M_{1}v_{1} + M_{2}v_{2} … (i)
Since kinetic energy is also conserved in an elastic collision, we obtain
From equation (i), we obtain
M_{1} (u_{1} − v_{1}) = M_{2} (v_{2 }− u_{2}) … (iii)
From equation (ii), we obtain
∴u_{1} − u_{2} = v_{2} − v_{1} … (v)
From equation (v), it follows that in onedimensional elastic collision, the relative velocity of approach (u_{1} − u_{2}) before collision is equal to the relative velocity of separation (v_{2} − v_{1}) after collision.
The ratio of relative velocity of separation after the collision to the relative velocity of the approach before the collision is known as coefficient of restitution or coefficient of resilience.
For perfectly elastic collision, e = 1
Calculation of velocities after collision:
Let us first find the velocity of body A after collision.
From equation (v), we have
v_{2} = u_{1} − u_{2} + v_{1}
Substituting for v_{2} in equation (i), we obtain
Again from equation (v), we have
v_{1}_{ }= v_{2} − u_{1} + u_{2}
Substituting for v_{1} in equation (i), we obtain
Special Cases

When the two bodies are of equal masses i.e.,
M_{1} = M_{2 }= M (say)
From equation (vi), we have

When the target body (B) is at rest:
In this case, u_{2} = 0
Substituting u_{2} = 0 in equations (vi) and (vii), we obtain
When M_{2} >> M_{1}, in equation (viii) and (ix), M_{1} can be neglected in comparison to M_{2 }i.e.
M_{1} −M_{2} ≈ M_{2} and
M_{1} + M_{2} ≈ M_{2}
Therefore, we have
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