9.In a perfectly elastic collision, a 0.400 kg ball moving toward the east at 3.7 m/s suddenly collides head-on with a 0.200 kg ball sitting at rest.
If the velocity of the first ball just after the collision is 4.24 m/s, what is the velocity of the second ball ?

The velocity of the second ball is 15.88 m/s East

Explanation:

The given parameters are;

The type of collision = Perfectly elastic collision

The mass of the ball moving East, m₁ = 0.400 kg

The velocity of the ball moving East, v₁ = 3.7 m/s

The mass of the ball at rest, m₂ = 0.200 kg

The initial velocity of the ball at rest, v₂ = 0 m/s

The velocity of the first ball, 'm₁', after the collision, v₃ = 4.24 m/s (The direction is assumed as being towards the West)

The velocity, 'v₄', of the second ball, 'm₂', after the collision is given by the law of conservation of linear momentum as follows;

m₁·v₁ + m₂·v₂ = m₁·v₃ + m₂·v₄

Substituting the known values gives;

0.400 kg × 3.7 m/s + 0.200 kg × 0 m/s = 0.400 kg × (-4.24 m/s) + 0.200 kg × v₄

∴ 0.200 kg × v₄ = 0.400 kg × 3.7 m/s - 0.400 kg × (-4.24 m/s)

Therefore;

0.200 kg × v₄ = 0.400 kg × 3.7 m/s + 0.400 kg × 4.24 m/s = 3.176 kg·m/s

v₄ = 3.176 kg·m/s/(0.200 kg) = 15.88 m/s

The velocity of the second ball = v₄ = 15.88 m/s East.

  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