Answer :

wiyonopaolina

The velocity of the cannon moving backward is 1.021 m/s.

If two objects collision without friction, the law of conservation of momentum will apply.

p = p'

m₁ v₁ + m₂ v₂ = m₁ v₁' + m₂ v₂'

  • m₁ = the mass of a first object (kg) = 484 kg
  • v₁ = the velocity of the first object before collision = 0 m/s
  • v₁' = the velocity of the first object after collision
  • m₂ = the mass of second object = 19 kg
  • v₂ = the velocity of the second object before collision = 0 m/s
  • v₂' = the velocity of the second object after collision = 26 m/s

The cannons and cannonballs were initially still.

(484 × 0) + (91 × 0) = (484 × v₁') + (19 × 26)

0 = 484v₁' + 494

484v₁' = - 494

v₁' = - 494 ÷ 484

v₁' = - 1.021 m/s

Learn more about law of conservation of momentum here: https://brainly.com/question/7538238

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