Explosions and recoil: zero before, zero after

5 minAS Level3.3.13.3.23.3.4

AI-drafted, independently checked · not yet teacher-reviewed

By the end of this lesson

By the end you can use conservation of momentum for explosions and recoil, where objects start at rest and push apart, and explain where their kinetic energy comes from.

Key to the symbols

The main symbols used in this lesson:

Symbol	Meaning
$m_1$	mass of the first object
$v_1$	final velocity of the first object
$m_2$	mass of the second object
$v_2$	final velocity of the second object
$v$	velocity

Something from nothing?

Fire a cannon and the cannon jumps backwards. Here's why:

Before firing, the cannon and the ball are at rest: total momentum zero.
With no resultant external force, the total momentum after firing must still be zero.

The only way two moving objects can have zero total momentum is if their momenta are equal in size and opposite in direction:

$m_1v_1 + m_2v_2 = 0 \quad\Rightarrow\quad m_1v_1 = -m_2v_2$

Equal and opposite momenta, but not equal speeds. Three times less mass means three times the speed.

Lighter means faster

Equal-sized momenta means $m_1v_1 = m_2v_2$ (in size), so

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

The speeds are in the inverse ratio of the masses:

the $25\ \text{kg}$ skater has a third of the mass, so she gets three times the speed;
a bullet has a tiny mass, so it flies off fast, while the much heavier rifle recoils slowly.

Worked example: Rifle recoil

A rifle of mass $4.0\ \text{kg}$ fires a bullet of mass $0.010\ \text{kg}$ (10 g) horizontally at $400\ \text{m s}^{-1}$ . Find the recoil velocity of the rifle, assuming it is free to move.

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Quick check

Multiple choiceQuestion 1 · Recall

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