An explosion blows a rock into three parts. Two pieces go off at right angles to each other, `1.0 kg` piece with velocity of `12 m//s` and other , `1.0 kg` piece with a velocity of `12 m//s` and other `2.0 kg` piece with a velocity of `8 m//s`. If the third piece flies off with a velocity of `40 m//s` compute the mass of the third piece.

Explanation

Let `m_(1),m_(2)` and `m_(3)` be the masses of the three pieces.
`m_(1)=1.0 kg, m_2=2.0 kg`
Let `v_(1)=12 m//s, v_(2)=8 m//s, v_(3)=40 m//s`. Let `v_(1)` and `v_(2)` be directed along `x`- and `y`-axis respectively, and `v_(3)` be directed as shown.

By the principle of coservation of momentum, initial momentum is zero. Hence,
along `x`-axis, `0=m_(1)v_(1)-m_(3)v_(3)costheta`
along `y`-axis `0=m_(2)v_(2)-m_(3)v_(3) sin theta`
`implies m_(1)v_(1)=m_(3)v_(3) costheta` and `m_(2)v_(2)=m_(3)v_(3) sintheta`
By squaring and adding we get `m_(1)^(2)v_(1)^(2)+m_(2)^(2)v_(2)^(2)=m_(3)^(2)v_(3)^(2)`
`impliesm_(3)^(2)=(1^(2)(12)^(2)+(2)^(2)(8)^(2))/((40)^(2))impliesm_(3)=0.5kg`