prove that in an elastic collision in one dimension the relative velocity of aaproach before imapct is equal to relative velocity of separation after impact
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In an elastic collision, total linear momentum is conserved.
Therefore ,m1u1 + m2u2 = m1v1 + m2v2......(1)
Similarly, kinetic energy is also conserved.
Therefore ,m1u1 + m2u2 = m1v1 + m2v2......(1)
Similarly, kinetic energy is also conserved.
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Let u1, u2 , v1 , v2 be the velocities (positive if in +ve x direction). m1 and m2 are masses. There is a collision, so v1 ≠ u1.
We apply the law of conservation of linear momentum.
m1 u1 + m2 u2 = m1 v1 + m2 v2
=> m1 (u1 - v1) = m2 (v2 - u2) ---- (1)
Law of conservation of KE (as the collision is elastic):
1/2 m1 u1² + 1/2 m2 u2² = 1/2 m1 v1² + 1/2 m2 v2²
=> m1 (u1² - v1²) = m2 (v2² - u2²) ---- (2)
Divide (2) by (1):
u1 + v1 = v2 + u2
u1 - u2 = v2 - v1
u2 < u1 , otherwise there is no collision.
So the Relative velocity of m1 approaching m2 is equal to velocity of m2 more than that of m1 (separation of m2 from m1).
We apply the law of conservation of linear momentum.
m1 u1 + m2 u2 = m1 v1 + m2 v2
=> m1 (u1 - v1) = m2 (v2 - u2) ---- (1)
Law of conservation of KE (as the collision is elastic):
1/2 m1 u1² + 1/2 m2 u2² = 1/2 m1 v1² + 1/2 m2 v2²
=> m1 (u1² - v1²) = m2 (v2² - u2²) ---- (2)
Divide (2) by (1):
u1 + v1 = v2 + u2
u1 - u2 = v2 - v1
u2 < u1 , otherwise there is no collision.
So the Relative velocity of m1 approaching m2 is equal to velocity of m2 more than that of m1 (separation of m2 from m1).
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