An explosion is a collision run backward. Pieces that start together at rest fly apart, and conservation of momentum says how their speeds compare.
Key idea
If a system starts at rest and only internal forces push the pieces apart, the total momentum stays zero: $$m_1 v_1 + m_2 v_2 = 0 \quad\Rightarrow\quad v_2 = -\frac{m_1}{m_2}v_1.$$ The pieces get equal and opposite momenta. The lighter piece moves faster, in proportion to the mass ratio.
Why it works
The push between the pieces is one third-law pair: equal forces, opposite directions, same time. So the impulses are equal and opposite, and each piece leaves with momentum of the same size. Kinetic energy is not conserved here; it starts at zero. It comes from stored energy: a compressed spring, muscles, or the chemical energy of gunpowder or rocket fuel. Because $K = \dfrac{p^2}{2m}$ and the momenta are equal in size, the lighter piece gets more of that energy.
Worked examples
Worked example: Skaters push apart
A $50\ \text{kg}$ skater and a $75\ \text{kg}$ skater stand at rest and push off. The lighter one moves west at $3.0\ \text{m/s}$. Find the other's velocity (east positive).
- $0 = (50)(-3.0) + (75)v_2$.
- $v_2 = \dfrac{150}{75} = +2.0\ \text{m/s}$: east at $2.0\ \text{m/s}$.
- Momenta: $-150$ and $+150\ \text{kg·m/s}$. Equal in size, opposite in direction.
Answer: $2.0\ \text{m/s}$ east
Worked example: Recoil of a cannon
A $1000\ \text{kg}$ cannon on frictionless wheels fires a $10\ \text{kg}$ ball forward at $150\ \text{m/s}$. Find the cannon's recoil speed and the kinetic energy of each.
- $|v_c| = \dfrac{(10)(150)}{1000} = 1.5\ \text{m/s}$ backward.
- $K_{\text{ball}} = \tfrac{1}{2}(10)(150)^2 = 112{,}500\ \text{J}$; $K_{\text{cannon}} = \tfrac{1}{2}(1000)(1.5)^2 = 1125\ \text{J}$.
- Same momentum, but the ball gets 100 times the energy, the inverse of the mass ratio.
Answer: $1.5\ \text{m/s}$ backward
Common mistake
Thinking the lighter piece gets more momentum because it moves faster. Its speed is larger exactly in proportion to its smaller mass, so $mv$ comes out the same for both. It is the kinetic energy that favors the lighter piece. Check: after any push-apart from rest, $m_1|v_1|$ should equal $m_2|v_2|$.
A rocket is a nonstop explosion: it throws exhaust backward, and the exhaust's backward momentum is matched by the rocket's forward momentum. No air to push against is needed. See it in the Rocket Builder of the Science Lab.