Q.

Show that in the case of one dimensional elastic collision, the relative velocity of approach of two colliding bodies before collision is equal to the relative velocity of separation after collision.

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Detailed Solution

Given:

Mass of body 1 = m1
Mass of body 2 = m2
Initial velocity of body 1 = u1
Initial velocity of body 2 = u2
Final velocity of body 1 = v1
Final velocity of body 2 = v2

Step 1: Conservation of Momentum

From the law of conservation of momentum:

m1u1 + m2u2 = m1v1 + m2v2

Step 2: Conservation of Kinetic Energy

From the law of conservation of kinetic energy:

½ m1u12 + ½ m2u22 = ½ m1v12 + ½ m2v22

Step 3: Derivation

From the momentum equation:

m1(u1 - v1) = m2(v2 - u2)

From the kinetic energy equation:

(u1 - u2)2 = (v2 - v1)2

Taking the square root of both sides:

|u1 - u2| = |v2 - v1|

Step 4: Relative Velocities

Relative velocity of approach before collision = u1 - u2

Relative velocity of separation after collision = v2 - v1

Since their magnitudes are the same:

u1 - u2 = -(v2 - v1)

Final Answer

Relative velocity of approach = Relative velocity of separation

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Show that in the case of one dimensional elastic collision, the relative velocity of approach of two colliding bodies before collision is equal to the relative velocity of separation after collision.