An explosion blows a rock into three parts. Two pieces go off at right angles to each 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 (in kg).

Let m₁, m₂ and m₃ be the masses of the three pieces. Given m₁ =1.0 kg, m₂ = 2.0 kg

Let  v₁ =12m/s, v₂ =8m/ s, v₃ =40 m/s. Let v₁ and v₂ be directed along x- and y-axes, respectively, and v₃ be directed as shown.

By the principle of conservation of momentum, initial momentum is zero. Hence,

$\text{ along }\mathrm{x}\text{-axis: }0={\mathrm{m}}_1{\mathrm{v}}_1-{\mathrm{m}}_3{\mathrm{v}}_3\cos \mathrm{\theta }$

$\text{ along }\mathrm{y}\text{-axis: }0={\mathrm{m}}_2{\mathrm{v}}_2-{\mathrm{m}}_3{\mathrm{v}}_3\sin \mathrm{\theta }$

$\Rightarrow {\mathrm{m}}_1{\mathrm{v}}_1={\mathrm{m}}_3{\mathrm{v}}_3\cos \mathrm{\theta }\text{ and }{\mathrm{m}}_2{\mathrm{v}}_2={\mathrm{m}}_3{\mathrm{v}}_3\sin \mathrm{\theta }$

By squaring and adding, we get

$\Rightarrow {\mathrm{m}}_3^2=\frac{1^2(12{)}^2+(2{)}^2(8{)}^2}{(40{)}^2}\Rightarrow {\mathrm{m}}_3=0.5\mathrm{kg}$