Explosions

In an explosion a body initially at rest splits into several fragments thanks to internal energy (chemical, elastic, …) that is released. The forces pushing the pieces apart are all internal, so the total momentum does not change: if the body was at rest, the sum of the fragments’ momenta stays zero.

Explosion: the stationary body splits. The total momentum stays zero; the lighter fragment goes faster.

For just two fragments, conservation requires $m_1\vv{v}_1 + m_2\vv{v}_2 = \vv{0}$, that is, the two pieces fly off in opposite directions with equal and opposite momenta. Since $\lvert v_1\rvert/\lvert v_2\rvert = m_2/m_1$, the lighter fragment shoots off faster.

Who carries away the energy

In an explosion the lighter fragment, despite having the same momentum (in magnitude) as the other, carries away most of the kinetic energy. For equal $p = mv$, in fact $E_\text{cin} = \tfrac{p^2}{2m}$: energy inversely proportional to mass. That is why, in a collision or a recoil, it is the small piece that is “dangerous”.

The full worked numerical example is in Esercizio svolto — esplosione in due frammenti.

Links

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

Related exercises: Esercizio svolto — esplosione in due frammenti · Rinculo di una pistola · Problema — Esplosione 2D con tre frammenti