A 4000-kg car bumps into a stationary 6000kg truck. The Velocity of the car before the collision was +4m/s and -1m/s after the collision. What is the velocity of the truck?

-4.33m/s

Explanation:

Using the law of conservation of momentum

m1u1+m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and u2 are their respective velocities

v is the final velocity

From the question;

m1 = 4000kg

m2 = 6000kg

u1 = 4m/s

u2 = ?

v =-1m/s

Substitute into the formula and get u2

4000(4)+6000u2 = (4000+6000)(-1)

16000 + 6000u2 = -10000

6000u2 = -10000-16000

6000u2 = -26000

6u2 = -26

u2 = -26/6

u2 = -13/3

u2 = -4.33m/s

Hence the velocity of the truck is -4.33m/s

  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