A4.0-kg mass, initially at rest on a horizontal frictionless surface, is struck by a 2.0-kg mass moving along the x axis with a speed of 8.0 m/s. after the collision, the 2.0-kg mass has a speed of 4.0 m/s at an angle of 37° from the positive x axis. what is the speed of the 4.0-kg mass after the collision?

here we will use momentum conservation along X and Y direction.

for X direction we will have

Now in Y direction we will apply momentum conservation

Now net speed of the object is given by

  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

c

explanation: