Two friends, Al and Jo, have a combined mass of 168 kg. At an ice skating rink they stand close together on skates, at rest and facing each other, with a compressed spring between them. The spring is kept from pushing them apart because they are holding each other. When they release their arms, Al moves off in one direction at a speed of 0.91 m/s, while Jo moves off in the opposite direction at a speed of 1.1 m/s. Assuming that friction is negligible, find Al's mass.

91.94 kg

Explanation: Supoose Al's mass is m. The combined mass is 168kg, so Jo's mass is 168 - m.

The Law of Conservation of Momentum states that when two objects collide, the total momentum of both objects before the collision is the same as the total momentum of both objects after the collision.

At the beginning, Al and Jo are united without movement, so their initial momentum is zero.

After the release of their hands, Al goes in one direction and Jo moves to the opposite direction. Suppose the direction Al is moving is positive. Conservation of momentum will be

where

index i referes to initial momentum

index f to final momentum

index 1 refers to Al

index 2 to Jo

Calculating:

m = 91.94

Al has a mass of 91.94 kg.

  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