Why a Rifle Kicks Back When It Fires a Bullet

A bus braking suddenly throws passengers forward, a loaded cart needs a harder push than an empty one, and a small boat drifts back as someone jumps ashore. Newton's three laws, together with the conservation of momentum, explain all of these.

This lesson covers Newton's laws applied to connected bodies, conservation of linear momentum and impulse, and the equilibrium of a particle under concurrent forces.

**Newton's first law says a body keeps its state of rest or uniform motion unless a net force acts, the second says the net force equals the rate of change of momentum, $F=ma$ for constant mass, and the third says forces come in equal and opposite pairs on different bodies; for connected bodies, apply $F=ma$ to each body separately.

The three laws:

- First law — a body resists changes in its motion, and mass measures this inertia
- Second law** — $\vec{F}=\frac{d\vec{p}}{dt}=m\vec{a}$, with $1\text{ N}=1{\text{ kg m s}}^{-2}$
- Third law — if A pushes B, B pushes A equally and oppositely

Method. Draw a free-body diagram for each body, choose a positive direction, and write $F_{\text{net}}=ma$ for each body; bodies joined by a taut, inextensible string share one acceleration.

Worked example 1 — blocks in contact. A 30 N force pushes blocks of 4.0 kg and 2.0 kg along a smooth floor:

$$a=\frac{30}{4.0+2.0}=5.0{\text{ m s}}^{-2}$$

The 4.0 kg block pushes the 2.0 kg block with $2.0\times 5.0=10$ N, and by the third law receives 10 N back.

Worked example 2 — an Atwood machine. Masses of 5.0 kg and 3.0 kg hang over a light, frictionless pulley, with $g=9.8$ m s${}^{-2}$:

$$a=\frac{(m_1-m_2)g}{m_1+m_2}=\frac{2.0\times 9.8}{8.0}=2.45{\text{ m s}}^{-2}T=\frac{2m_1{m}_{2}g}{m_1+m_2}=\frac{2\times 5.0\times 3.0\times 9.8}{8.0}=36.8\text{ N}$$

An everyday example. A tractor pulling two loaded trolleys accelerates all three together, and the coupling behind the tractor carries more tension because it must pull both trolleys.

The substance. Action and reaction never cancel each other — they act on different bodies, so only forces on the same body are added to find its acceleration.

**The total linear momentum of a system stays constant when no external force acts on it, and the impulse of a force, $F\Delta t$, equals the change in momentum it produces.

Conservation of momentum.** With momentum $\vec{p}=m\vec{v}$:

$$m_1{\vec{u}}_1+m_2{\vec{u}}_2=m_1{\vec{v}}_1+m_2{\vec{v}}_2$$

It follows from the second and third laws, because internal forces are equal and opposite, and it applies to collisions, explosions and recoil whenever external forces are negligible.

Worked example 1 — rifle recoil. A 4.0 kg rifle fires a 10 g bullet at 400 m s${}^{-1}$, starting from zero total momentum:

$$0=(0.010)(400)+(4.0)v\Rightarrow v=-1.0{\text{ m s}}^{-1}$$

The rifle recoils at 1.0 m s${}^{-1}$, opposite to the bullet.

Impulse. $J=F\Delta t=\Delta p=mv-mu$, so the same change in momentum can come from a large force for a short time or a small force for a long time; impulse also equals the area under a force-time graph.

Worked example 2 — hitting a ball. A 0.15 kg ball arrives at 20 m s${}^{-1}$ and leaves the bat at 30 m s${}^{-1}$ the other way, with contact lasting 0.010 s:

$$\Delta p=0.15\times 30-0.15\times (-20)=7.5{\text{ kg m s}}^{-1}F=\frac{7.5}{0.010}=750\text{ N}$$

An everyday example. Airbags and crumple zones in cars stretch out the time over which passengers are stopped, so the same change in momentum involves a much smaller force.

The substance. Momentum is conserved even when kinetic energy is not — when two colliding vehicles lock together, momentum is unchanged but much of the kinetic energy becomes heat and sound.

A particle is in equilibrium when the vector sum of all forces on it is zero, so their components along two perpendicular directions each add to zero; for three forces, Lami's theorem relates each force to the sine of the angle between the other two.

Conditions for equilibrium:

$$\sum F_x=0\sum F_y=0$$

- Two forces balance only if they are equal, opposite and along the same line
- Three forces in equilibrium form a closed triangle when drawn head to tail

Lami's theorem. For three concurrent forces in equilibrium, where each angle is the one between the other two forces:

$$\frac{F_1}{\sin \alpha }=\frac{F_2}{\sin \beta }=\frac{F_3}{\sin \gamma }$$

Worked example — a hanging weight. A 50 N weight hangs from two strings making 30° and 60° with the vertical, so the strings are 90° apart. The angle between $T_1$ and the weight is 150°, and between $T_2$ and the weight is 120°:

$$\frac{T_1}{\sin 120^{\circ }}=\frac{T_2}{\sin 150^{\circ }}=\frac{50}{\sin 90^{\circ }}\Rightarrow T_1=43.3\text{ N},T_2=25\text{ N}$$

The horizontal components, $43.3\sin 30^{\circ }$ and $25\sin 60^{\circ }$, are both 21.7 N and cancel.

An everyday example. A clothes line pulled almost straight between two rooftop poles can snap under a few wet saris, because a nearly horizontal line needs a huge tension to support even a small weight.

The substance. Equilibrium does not require rest — a body moving at constant velocity is also in equilibrium, because the net force on it is zero.