Two spacemen are floating together with zero speed in a gravity-free region of space. The mass of spaceman A is 140 kg and that of spaceman B is 100 kg. Spaceman A pushes B away from him with B attaining a final speed of 0.5 m/s. The final recoil speed of A is:
a -zero
b -0.38 m/s
c -0.5 m/s
d -0.67 m/s
e -1 m/s

The final recoil speed of A is -0.36 m/s

Explanation:

Given;

mass of spaceman A, m₁ = 140 kg

mass of spaceman B, m₂ = 100 kg

initial speed of spaceman A, u₁ = 0

initial speed of spaceman B, u₂ = 0

final speed of spaceman B, v₂ = 0.5 m/s

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

140 x 0 + 100 x 0 = 140v₁ + 100 x 0.5

0 = 140v₁ + 50

-50 = 140v₁

v₁ = -50/140

v₁ = -0.36 m/s

Thus, the final recoil speed of A is -0.36 m/s

The negative sign indicates that it moved in opposite direction of spaceman B

  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