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Two balls (X(2 ~kg ) ) and (Y(4 ~kg ) ) approach each other with equal speeds of (10 ~ms ⁻¹ ). If the collision is perfectly elastic, then the new velocities of balls (X ) and (Y ) are respectively

Options

  1. A( 50 3 ~ms ⁻¹,- 10 3 ~ms ⁻¹ )
  2. B(- 50 3 ~ms ⁻¹,- 10 3 ~ms ⁻¹ )
  3. C(- 50 3 ~ms ⁻¹, 10 3 ~ms ⁻¹ )
  4. D( 50 3 ~ms ⁻¹, 10 3 ~ms ⁻¹ )

Correct answer

C. (- 50 3 ~ms ⁻¹, 10 3 ~ms ⁻¹ )

Step-by-step solution

Given, (m₁=2 ~kg , m₂=4 ~kg ) (v₁=v₂=10 ~ms ⁻¹ ) In perfectly elastic collision, momentum and kinetic energy is conserved. ( ) By conservation of momentum, ( aligned & m₁ 10+m₂ (-10)=m₁ v_ 1 f +m₂ v_ 2 f & 2 10+4 (-10)=2 v_ 1 f +4 v_ 2 f & -20=2 (v_ 1 f +2 v_ 2 f ) & v_ 1 f +2 v_ 2 f =-10 (i) aligned ) In perfectly elastic collision, Velocity of separation (= ) Velocity of approach ( array ll & v_ 2 f -v_ 1 f =10-(-10) & v_ 2 f -v_ 1 f =20 (ii) array ) Adding Eqs. (i) and (ii), we get ( aligned & 3 v_ 2 f =10 & v_

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