Linear Momentum

Velocity tells you how fast something moves, but a heavy truck at 1 m/s is far harder to stop than a tennis ball at 1 m/s. Newton recognized that the right "quantity of motion" combines both mass and velocity. He called it momentum.

Concept

The linear momentum of a particle of mass $m$ moving with velocity $\vec{v}$ is

$\vec{p} = m\vec{v}$

It is a vector quantity, pointing in the direction of velocity. Its SI unit is kg·m/s (equivalently N·s).

Momentum is the natural variable of Newton's laws. The second law is most cleanly written

$\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt}$

Conservation of momentum: If the net external force on a system is zero,

$\frac{d\vec{p}_{\text{total}}}{dt} = 0 \quad \Longrightarrow \quad \vec{p}_{\text{total}} = \text{constant}$

This is one of the most important conservation laws in physics. It applies to collisions, explosions, recoil — any process where internal forces (which come in third-law pairs) redistribute momentum but do not change the total.

Derivation

For a system of two particles with momenta $\vec{p}_1$ and $\vec{p}_2$, the total momentum is

$\vec{P} = \vec{p}_1 + \vec{p}_2$

Differentiating with respect to time,

$\frac{d\vec{P}}{dt} = \frac{d\vec{p}_1}{dt} + \frac{d\vec{p}_2}{dt} = \vec{F}_1 + \vec{F}_2$

Each force can be split into the external force on that particle and internal forces from the other particle:

$\vec{F}_1 = \vec{F}_1^{\text{ext}} + \vec{F}_{21}, \qquad \vec{F}_2 = \vec{F}_2^{\text{ext}} + \vec{F}_{12}$

By Newton's third law, $\vec{F}_{12} = -\vec{F}_{21}$, so the internal forces cancel:

$\frac{d\vec{P}}{dt} = \vec{F}_1^{\text{ext}} + \vec{F}_2^{\text{ext}} = \vec{F}_{\text{net}}^{\text{ext}}$

If $\vec{F}_{\text{net}}^{\text{ext}} = 0$, then $\vec{P}$ is constant. This is conservation of linear momentum.

Worked Example

A gun of mass $M = 4 \, \text{kg}$ fires a bullet of mass $m = 20 \, \text{g}$ with muzzle velocity $v = 400 \, \text{m/s}$. Find the recoil velocity of the gun.

Solution:

Before firing, both gun and bullet are at rest, so total momentum is zero:

$P_i = 0$

After firing, the bullet has momentum $mv$ forward and the gun has momentum $MV$ backward (with $V$ unknown):

$P_f = mv + MV$

By conservation (no horizontal external force during firing):

$0 = mv + MV \quad \Longrightarrow \quad V = -\frac{mv}{M}$

Plugging in:

$V = -\frac{(0.020)(400)}{4} = -2 \, \text{m/s}$

The gun recoils at 2 m/s in the direction opposite to the bullet.

Common Confusions

• Momentum vs kinetic energy. Both depend on mass and velocity, but momentum is linear in $v$ (and vector), while KE goes as $v^2$ (and scalar). A 1 kg bullet at 400 m/s and a 400 kg cart at 1 m/s have the same momentum but very different kinetic energies.
• "Momentum is always conserved." Only if the net external force is zero. A ball falling freely gains momentum because gravity is an external force.
• Direction matters. When applying conservation, treat momentum as a vector. Equal-and-opposite momenta sum to zero.
• Internal forces. They never change the total momentum of a system, however violent. An explosion preserves total momentum.

Key Takeaways

• $\vec{p} = m\vec{v}$ is a vector with units kg·m/s.
• Newton's second law in its most general form is $\vec{F} = d\vec{p}/dt$.
• If the net external force vanishes, total momentum is conserved.
• Internal forces (third-law pairs) cancel and cannot change total momentum.
• Conservation of momentum is the key tool for collisions, recoil, and explosions.