On a frictionless air track, a 0.30 kg glider moving at 0.40m/s to the right collides with a stationary 0.80kg glider moving at 0.15 m/s to the left. The collision is cushioned by a bumper made of perfectly elastic spring steel. a. What is the velocity of each glider after the collision? b. What is the minimum amount of total kinetic energy during the collision? c. Where is the missing energy?

Explanation:

It is the case of perfectly elastic collision . So we shall apply formula of velocity after collision as follows .

Let  m₁ and m₂ be the mass colliding with velocity u₁ and u₂ and their velocity become v₁ and v₂ after collision .

Putting the values

= - 0.4 m /s

So direction of .30 kg mass will be reversed .

putting the values

= .15 m /s

The direction of .80 kg will become from left to right ie its direction will be reversed .

b ) Minimum amount of kinetic energy will be at the position when they move with common velocity

common velocity

v = .3 x .4 - .8 x .15 / (.3 + .8)

= 0

c )

Missing energy is stored as elastic potential energy in the spring .

  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

like poles repel so keep pushing north to north and repels funny to see this