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?
What is the answer

Correct 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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  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂

uranium fissions/breaks up giving out energy and neutrons