Car a rear ends car b, which has twice the mass of a, on an icy road at a speed low enough so that the collision is essentially elastic. car b is stopped at a light when it is struck. car a has mass m and speed v before the collision. find the speed of car b after the collision (in terms of v).

Vb = v/2

Half of the speed of Car A before the collision

Explanation:

This problem can be solved by using conservation of momentum.

Since car B was stopped before the collision and its mass is twice the mass of A:

Also, since it is an elastic collision, car A stops after hitting car B. Rewriting the equation:

Therefore, the speed of car B after the collision is half of the speed of car A before the collision (v)

Vb = v/2

to be honest as much as i would love to answer this. i feel as if something's missing. just saying.

  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