case - In an isolated system, a 5 kg ball 'A' with 10 m/s velocity is made to hit a 10 kg ball 'B' at rest. After collision the balls move together in the same direction .The plane and the balls are friction less and there is vacuum in the system( zero drag).
According to conservation of energy:
Total energy after collision = Total energy of the system initially
= energy of ball A + energy of ball B
= 1/2 mv*v + 1/2 mv*v
= (1/2 x 5kg x 10m/s x 10m/s) + (1/2 x 10kg x 0m/s x 0m/s)
= 250 J --------------( 1)
According to conservation of momentum:
Total momentum of 2 balls before collision = Total momentum of balls after collision
=> mu(A) + mu(B) = v[m(A) +m(B)] { u = initial velocity ; v = final velocity of the balls}
=> 5kg x 10m/s + 10kg x 0m/s = v[5kg + 10kg]
=> 50kg m/s = v[15kg]
=> v = 10/3 m/s
Now , Total energy after collision = 1/2 * m(A+B) * v * v
= 1/2 * (5kg +10kg) *10/3 m/s *10/3 m/s
= 83.33 J --------------------(2)
(1) and (2) contradicts each other .
According to conservation of energy:
Total energy after collision = Total energy of the system initially
= energy of ball A + energy of ball B
= 1/2 mv*v + 1/2 mv*v
= (1/2 x 5kg x 10m/s x 10m/s) + (1/2 x 10kg x 0m/s x 0m/s)
= 250 J --------------( 1)
According to conservation of momentum:
Total momentum of 2 balls before collision = Total momentum of balls after collision
=> mu(A) + mu(B) = v[m(A) +m(B)] { u = initial velocity ; v = final velocity of the balls}
=> 5kg x 10m/s + 10kg x 0m/s = v[5kg + 10kg]
=> 50kg m/s = v[15kg]
=> v = 10/3 m/s
Now , Total energy after collision = 1/2 * m(A+B) * v * v
= 1/2 * (5kg +10kg) *10/3 m/s *10/3 m/s
= 83.33 J --------------------(2)
(1) and (2) contradicts each other .