Explain the derivation of elastic & inelastic collision

Dear Student,

Elastic Collision in One
 Dimension

 

Consider that two perfectly elastic bodies A and B of masses M1 and M2 moving with initial velocities u1 and uundergo head on collision and continue moving along the same straight line with final velocities v1 and v2.

As in an elastic collision, momentum is conserved.

∴ M1u1 M2u2 = M1v1 + M2v2 … (i)

Since kinetic energy is also conserved in an elastic collision, we obtain

From equation (i), we obtain

M1 (u1 − v1) = M2 (v− u2) … (iii)

From equation (ii), we obtain

u1 − u2 = v2 − v1 … (v)

From equation (v), it follows that in one-dimensional elastic collision, the relative velocity of approach (u1 − u2) before collision is equal to the relative velocity of separation (v2 − v1) 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.

e = v2−v1/u1−u2

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

v2 = u1 − u​​​​​2 + v1

Substituting for v2 in equation (i), we obtain

Again from equation (v), we have

v1 v2 − u​​​​​1 + u2

Substituting for v1 in equation (i), we obtain

Special Cases

  • When the two bodies are of equal masses i.e.,

M1 = MM (say)

From equation (vi), we have

  • When the target body (B) is at rest:

In this case, u2 = 0

Substituting u2 = 0 in equations (vi) and (vii), we obtain

When M2 >> M1, in equation (viii) and (ix), M1 can be neglected in comparison to Mi.e.
M1 M2  -M2 and 
M1 + M2  M2
Therefore, we have


Kindly post the different part of the question in different thread.

Regards

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