Boron naturally occurs in two isotopic forms. the more common isotope is 11b (atomic mass 11.01 amu), which is 80.00% abundant. the average atomic mass of boron is 10.81. what is the mass of the other isotope?

Answer is: mass of the other isotope of boron is 10.01 amu.

Ar₁(B) = 11.01 amu; the average atomic mass of isotope ¹¹B.  

Ar₂(B) =?; the average atomic mass of unknown isotope of boron.  

ω₁(B) = 80.00%; mass percentage of isotope ¹¹B.

ω₂(B) = 100.0% - 80.00%.  

ω₂(B) = 20.00%.; mass percentage of unknown isotope of boron.

ω₂(B) = 20.00% ÷ 100% = 0.20.

Ar(B) = 10.81 amu; average atomic mass of boron.

Ar(B) = Ar₁(B) · ω₁(B) + Ar₂(B) · ω₂(B).  

10.81 amu = 11.01 amu · 0.8 + Ar₂(B) · 0.2.

Ar₂(B) = 2.002 amu ÷ 0.2.

Ar₂(B) = 10.01 amu.

i just learnt this recently but i need to look through my notes again

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied