Worked exercise — explosion into two fragments

Problem

A firecracker ($m_{\text{tot}} = 2\;\text{kg}$) explodes releasing $E_{\text{scoppio}} = 500\;\text{J}$ and splits into two fragments of mass $m_1 = 0{,}5\;\text{kg}$ and $m_2 = 1{,}5\;\text{kg}$. Calculate the velocities.

Solution

Conservation of $\vv{p}$ (the firecracker is initially at rest, so $p_\text{tot} = 0$ before and after): $0 = m_1 v_1 + m_2 v_2 \quad\Rightarrow\quad v_2 = -\frac{m_1}{m_2}\,v_1 = -\frac{0{,}5}{1{,}5}\,v_1 = -\frac{1}{3}\,v_1$ The two fragments fly off in opposite directions (the minus sign).

Energy balance: all the released energy becomes kinetic energy of the fragments: $E_{\text{scoppio}} = \tfrac{1}{2}m_1 v_1^2 + \tfrac{1}{2}m_2 v_2^2$ Substituting $v_2 = -\tfrac{1}{3}v_1$: $500 = \tfrac{1}{2}\cdot 0{,}5\cdot v_1^2 + \tfrac{1}{2}\cdot 1{,}5\cdot \frac{v_1^2}{9} = v_1^2\!\left(\frac{0{,}5}{2} + \frac{1{,}5}{18}\right) = \frac{v_1^2}{3}$ Solving: $v_1 = \sqrt{1500} \approx 38{,}7\;\text{m/s}, \qquad v_2 = -\tfrac{1}{3}v_1 \approx -12{,}9\;\text{m/s}$ $\ev{v_1 \approx 38{,}7\;\mathrm{m/s},\quad v_2 \approx -12{,}9\;\mathrm{m/s}}$ The lighter fragment goes three times faster: despite having the same momentum (in magnitude) as the other, it carries away most of the kinetic energy.

Links

Topics: Quantità di moto e urti Concepts: Esplosione · Conservazione della quantità di moto · Energia cinetica Skills: Conservazione della quantità di moto · Metodo delle due tabelle