In a closed system with no friction, a red sphere of 2.5 kg stands stationary. A blue sphere with a mass of 5.8 kg approaches the first sphere with a speed of 4.1 m/s. The two collide. After the collision, the blue sphere begins moving forward with a speed of 1.3 m/s. What is the velocity of the red sphere after the collision? (Show all work)

m1 = the mass of the blue sphere = 5.8 kg

m2 = the mass of the red sphere = 2.5 kg

v1 = initial velocity of the blue sphere before the collision = 4.1 m/s

v2 = initial velocity of the red sphere before the collision = 0 m/s

v'1 = final velocity of the blue sphere after the collision = 1.3 m/s

v'2 = final velocity of the red sphere after the collision = ?

using conservation of momentum

m1v1 + m2v2 = m1v'1 + m2v'2

(5.8) (4.1) + (2.5) (0) = (5.8) (1.3) + (2.5) (v'2)

23.78 = 7.54 + (2.5) (v'2)

-7.54    -7.54

16.24 = (2.5) v'2

 2.5         2.5

v'2 = 6.5

Explanation:

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz