Solved problem: Momentum conservation in explosions

Development
Formulas

Statement

difficulty

A 10 Kg body is moving at ${\vec{v}}_{i} = \vec{i} + \vec{j} m / s$ and after its explosion it breaks into 3 pieces. The first piece is 3 Kg and goes through to ${\vec{v}}_{1} = \frac{1}{2} \cdot \vec{i} + 2 \cdot \vec{j} m / s$ , the second body weights 5 Kg and goes through to ${\vec{v}}_{2} = 2 \cdot \vec{i} + \frac{1}{2} \cdot \vec{j} m / s$ . What velocity does the third piece reach?

Solution

Data

Total mass of th body and velocity before the explosion: $m_{i} = 10 k g; {\vec{v}}_{i} = \vec{i} + \vec{j} m / s$
Mass and velocity of the first piece: $m_{1} = 3 k g; {\vec{v}}_{1} = 0.5 \cdot \vec{i} + 2 \cdot \vec{j} m / s$
Mass and velocity of the second piece: $m_{2} = 5 k g; {\vec{v}}_{2} = 2 \cdot \vec{i} + 0.5 \cdot \vec{j} m / s$

Resolution

Firstly, we can know the mass of the third piece, since the body had 10 kg before the explosion, the mass of the third piece will be m₃ = m_T - m₁ - m₂ = 2 kg. Because the linear momentum conservation theorema, the quantity of motion before and after the explosion mus be preserved:

$\sum \vec{F} = 0 \Leftrightarrow \vec{p} = constant \Rightarrow {\vec{p}}_{i} = {\vec{p}}_{f} \Rightarrow m_{T} \cdot {\vec{v}}_{i} = m_{1} \cdot {\vec{v}}_{1} + m_{2} \cdot {\vec{v}}_{2} + m_{3} \cdot {\vec{v}}_{3} \Rightarrow \Rightarrow {\vec{v}}_{3} = \frac{m_{T} \cdot {\vec{v}}_{i} - m_{1} \cdot {\vec{v}}_{1} - m_{2} \cdot {\vec{v}}_{2}}{m_{3}} = \frac{(10 - 1.5 - 10) \cdot \vec{i} + (10 - 6 - 2.5) \cdot j}{2} = - 0.75 \cdot \vec{i} + 0.75 \cdot \vec{j}$

Formulas worksheet

These are the main formulas that you must know to solve this exercise. If you are not clear about their meaning, we recommend you to check the theory in the corresponding sections. Furthermore, you will find in them, under the Formulas tab, the codes that will allow you to integrate these equations in external programs like Word or Mathematica.

Formulas

Related sections