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Explosions and recoil

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An explosion is a collision run backwards

In an explosion (or recoil, or push-apart), one object at rest separates into two that move in opposite directions.
Momentum is conserved in exactly the same way as in a collision — the internal forces are still a Newton's third-law pair.
If everything starts at rest, the total momentum before is zero, so the total momentum after must also be zero:

$0 = m_1v_1 + m_2v_2$

$m_1v_1 = -m_2v_2$

The two objects therefore have momenta that are equal in size and opposite in direction.
Their velocities are not equal — the lighter object always moves faster.

The mass–velocity trade-off

Rearranged, the result says:

$v_2 = -\frac{m_1}{m_2}v_1$

The smaller mass ends up with the larger speed, in inverse proportion to the masses.
If one object is ten times heavier, it moves at one tenth the speed of the other.
Examples to use in answers:
- a rifle recoils slowly while the light bullet leaves at high speed,
- a cannon rolls back a short distance while the shell flies far,
- a rocket pushes exhaust gas backward and moves forward,
- a person stepping off a skateboard sends the board backward,
- two skaters pushing apart on ice separate at different speeds.

Where the energy comes from

Momentum starts at zero and finishes at zero, but kinetic energy starts at zero and finishes with a large positive value.
That energy is not created — it is released from a store:
- chemical energy in the gunpowder or rocket fuel,
- elastic potential energy in a compressed spring,
- chemical energy in muscles when two skaters push apart.
This is the clearest demonstration that momentum and kinetic energy are different quantities: momentum cancels because it is a vector, while kinetic energy adds because it is a scalar.

Solving an explosion problem

Set the total momentum before equal to zero (if everything starts at rest).
Choose a positive direction.
Write the total momentum after as $m_1v_1 + m_2v_2$ .
Solve — the answer will come out negative for one object, meaning it moves the other way.

Tips and Tricks

Do not treat an explosion as a special case with its own rule. It is the same equation as a collision, $p_{\text{before}} = p_{\text{after}}$ , with the left-hand side equal to zero. Writing that zero down explicitly earns the method mark.

The minus sign in your answer is part of the physics, not an error to be dropped. It tells you the two objects move in opposite directions, which is usually what the question wants you to state.

If you are asked why the rifle's recoil is far less dangerous than the bullet, argue with kinetic energy, not momentum. The two momenta are equal in size — the difference is that $E_k = \tfrac{1}{2}mv^2$ depends on $v^2$ , so the fast, light bullet carries the overwhelming majority of the energy.

Worked ExampleRifle recoil

A $4.0$ kg rifle fires a $0.020$ kg bullet at $450$ m s⁻¹. Find the recoil velocity of the rifle.

Step 1 — Momentum before firing

Both the rifle and the bullet are at rest:

$p_{\text{before}} = 0$

Step 2 — Momentum after firing

Taking the bullet's direction as positive:

$p_{\text{after}} = m_bv_b + m_rv_r = (0.020)(450) + (4.0)v_r$

Step 3 — Apply conservation of momentum

$0 = 9.0 + 4.0v_r$ $v_r = \frac{-9.0}{4.0} = -2.25 \approx -2.3 \text{ m s}^{-1}$

$v_r = 2.3 \text{ m s}^{-1} \text{ backwards}$

Worked ExampleWhy the recoil is survivable

Using the figures above, compare the kinetic energy of the bullet with that of the rifle.

Step 1 — Kinetic energy of the bullet

$E_k = \tfrac{1}{2}mv^2 = \tfrac{1}{2}(0.020)(450)^2 = \tfrac{1}{2}(0.020)(202\,500) = 2025 \text{ J}$

Step 2 — Kinetic energy of the rifle

$E_k = \tfrac{1}{2}(4.0)(2.25)^2 = \tfrac{1}{2}(4.0)(5.06) = 10.1 \text{ J}$

Step 3 — Compare

$\frac{2025}{10.1} = 200$

The bullet carries 200 times the kinetic energy of the rifle, even though the two have identical momenta.

$E_{k,\text{bullet}} = 2.0 \times 10^3 \text{ J}, \qquad E_{k,\text{rifle}} = 10 \text{ J}$