A water-balloon launcher with mass 2 kg fires a 0.75 kg balloon with a

velocity of 14 m/s to the west. What is the recoil velocity of the launcher?

Correct answer: V₂ = 5.25 m/s

Explanation:

Given:

m₁ = 0.75 kg mass of balloon, V₁ = 14 m/s the velocity of the balloon

m₂ = 2 kg mass of water balloon launcher

The law of impulse sustainability applies for system water balloon launcher - balloon which means:

J₁ = m₁ · V₁ Impulse of the balloon after launching

J₂ = m₂ · V₂ Impulse of the water balloon launcher

these impulses are equal

J₁ = J₂ ⇒ m₁ · V₁ = m₂ · V₂ ⇒ V₂ = (m₁ · V₁) / m₂

V₂ = (0.75 · 14) / 2 = 5.25 m/s

V₂ = 5.25 m/s

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