If Every Push Has an Equal Pull Back, How Does Anything Move?

A horse pulls a cart forward, and by the third law the cart pulls the horse backward just as hard. So how does the pair ever move?

The answer lies in which body each force acts on. Getting this right unlocks momentum conservation, free-body diagrams and collisions.

This part covers the third law, conservation of momentum, equilibrium of concurrent forces, and collisions and connected bodies.

**Newton's third law says that when A exerts a force on B, B exerts an equal and opposite force on A, ${\vec{F}}_{AB}=-{\vec{F}}_{BA}$; the two forces act on different bodies, so they can never cancel each other.

Identifying pairs — a book on a table:

- Earth pulls the book down** (weight) $\leftrightarrow$ the book pulls Earth up
- Table pushes the book up (normal force) $\leftrightarrow$ the book pushes the table down

Weight and normal force both act on the book, so they are not a pair — they are equal only because the book is in equilibrium.

Worked example — two skaters. Skaters of $50$ kg and $75$ kg push each other with $150$ N.

$$a_1=\frac{150}{50}=3{\text{ m/s}}^2{a}_2=\frac{150}{75}=2{\text{ m/s}}^2\text{ (opposite)}$$

Equal forces, unequal accelerations.

The horse and cart. The cart's pull acts on the horse, but the ground's forward push on the horse's hooves is larger, so the horse accelerates.

An everyday example. When you walk, your foot pushes the ground backward and the ground pushes you forward.

The substance. Both forces of a pair are of the same kind — both contact, or both gravitational.

**Internal forces come in third-law pairs, so their momentum changes cancel: $\frac{d}{dt}({\vec{p}}_1+{\vec{p}}_2)={\vec{F}}_{12}+{\vec{F}}_{21}=0$, and the total momentum of an isolated system stays constant.

Derivation.** During an interaction lasting $\Delta t$:

$$\Delta {\vec{p}}_1={\vec{F}}_{12}\Delta t\Delta {\vec{p}}_2={\vec{F}}_{21}\Delta t=-{\vec{F}}_{12}\Delta t\Rightarrow \Delta {\vec{p}}_1+\Delta {\vec{p}}_2=0$$

Worked example 1 — recoil. A $4$ kg gun fires a $0.02$ kg bullet at $400$ m/s. Initially everything is at rest:

$$0=0.02\times 400+4V\Rightarrow V=-2\text{ m/s}$$

Worked example 2 — explosion at rest. A $3$ kg shell at rest bursts into $1$ kg moving at $12$ m/s and $2$ kg:

$$0=1\times 12+2v\Rightarrow v=-6\text{ m/s}$$

Worked example 3 — explosion in flight. A $3$ kg shell moving east at $10$ m/s splits; a $1$ kg piece stops dead:

$$3\times 10=1\times 0+2v\Rightarrow v=15\text{ m/s east}$$

An everyday example. Stepping off a small boat onto a jetty pushes the boat away from the shore.

The substance. Momentum is conserved in an explosion even though kinetic energy increases — the extra energy comes from the chemical energy released.

A body acted on by forces through one point is in equilibrium when their vector sum is zero, so on a free-body diagram the components along any two perpendicular axes must each add to zero.

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

Worked example 1 — a hanging lamp. A $60$ N lamp hangs from two strings making $30°$ and $60°$ with the horizontal, with tensions $T_1$ and $T_2$.

$$\text{horizontal: }{T}_1\cos 30°=T_2\cos 60°\Rightarrow T_2=\sqrt{3}{T}_1$$

$$\text{vertical: }{T}_1\sin 30°+T_2\sin 60°=60\Rightarrow 0.5T_1+1.5T_1=60$$

$$T_1=30\text{ N}{T}_2=30\sqrt{3}\approx 52\text{ N}$$

Worked example 2 — axes along an incline. A $10$ kg block rests on a smooth $30°$ incline, held by a rope parallel to the slope. With $g=10$ m/s${}^2$:

$$T=mg\sin 30°=50\text{ N}N=mg\cos 30°\approx 86.6\text{ N}$$

An everyday example. A photo frame hung from a nail by two strings stays still because the tensions and its weight add to zero.

The substance. The steeper string carries more tension — here $52$ N against $30$ N.

When no external force acts along a line, the total momentum along it before a collision or separation equals the total after, whatever internal forces act — springs, threads or impacts.

Worked example 1 — bullet into a block. A $0.01$ kg bullet at $500$ m/s embeds in a $2$ kg block at rest.

$$0.01\times 500=2.01v\Rightarrow v\approx 2.49\text{ m/s}$$

Kinetic energy falls from $1250$ J to about $6.2$ J — most becomes heat and deformation.

Worked example 2 — head-on. A $1$ kg ball at $4$ m/s meets a $2$ kg ball at $-1$ m/s. Afterwards the $1$ kg ball moves at $-2$ m/s:

$$1(4)+2(-1)=1(-2)+2v\Rightarrow v=2\text{ m/s}$$

Worked example 3 — connected by a spring. Blocks of $2$ kg and $3$ kg on a smooth floor are tied together with a compressed spring between them. The thread is burnt, and the $3$ kg block moves off at $2$ m/s:

$$0=3(2)+2v\Rightarrow v=-3\text{ m/s}$$

An everyday example. A striker hitting a coin on a carrom board passes on momentum while the total along the line stays fixed.

The substance. Momentum is conserved in every collision; kinetic energy is conserved only in elastic ones.