Why a Fielder Pulls the Hands Back While Catching a Fast Ball

Stop pushing a vegetable cart and it soon halts. It is tempting to conclude that motion needs a constant force — but that idea is wrong, and correcting it is the starting point of mechanics.

Forces do not cause motion; they cause changes in motion. Newton's laws make that precise and let us calculate forces from how velocity and momentum change.

This part covers the law of inertia, the first law, the second law, and impulse.

Aristotle's fallacy is the belief that an external force is needed to keep a body moving; the law of inertia corrects it by showing that a moving body slows only because of opposing forces such as friction, and would otherwise keep moving.

A thought experiment. A ball rolls down one smooth incline and up a facing one, rising to almost its starting height. Make the second incline flatter and the ball rolls farther to reach that height. Make it perfectly horizontal and smooth, and the ball has no reason ever to stop.

Worked example — finding the hidden force. A $0.16$ kg puck sliding on a rink slows from $10$ m/s to rest over $50$ m.

$$a=\frac{v^2-u^2}{2s}=\frac{0-100}{100}=-1{\text{ m/s}}^{2}f=ma=0.16\times 1=0.16\text{ N}$$

An everyday example. A vegetable cart stops once you let go because the ground and axles supply friction; on a smoother road it rolls farther.

The substance. Inertia is the resistance of a body to any change in its state of rest or uniform motion, and mass measures it.

Newton's first law states that a body stays at rest or in uniform straight-line motion unless a net external force acts on it; so whenever acceleration is zero, the net force is zero, even if several forces act.

Take $g=10$ m/s${}^2$.

Worked example 1 — at rest. A $1.5$ kg book lies on a table. Weight $mg=15$ N acts down, so the table's normal force must be **$15$ N up.

Worked example 2 — moving uniformly.** A $500$ kg lift moves up at a steady $2$ m/s.

$$a=0\Rightarrow T-mg=0\Rightarrow T=500\times 10=5000\text{ N}$$

The cable tension is the same as when the lift hangs at rest.

An everyday example. Passengers in a bus lurch forward when it brakes suddenly — their bodies tend to keep moving at the bus's earlier speed.

The substance. Moving does not mean a net force is acting — a car cruising at constant velocity has its driving force exactly balanced by friction and air drag.

**The net external force equals the rate of change of momentum, $\vec{F}=\frac{d\vec{p}}{dt}$ with $\vec{p}=m\vec{v}$; for constant mass this becomes $\vec{F}=m\vec{a}$.**

$$\vec{F}=\frac{d(m\vec{v})}{dt}=m\frac{d\vec{v}}{dt}=m\vec{a}1\text{ N}=1{\text{ kg m s}}^{-2}$$

Worked example 1 — F = ma. A $1000$ kg car goes from rest to $20$ m/s in $10$ s.

$$a=\frac{20}{10}=2{\text{ m/s}}^{2}F=1000\times 2=2000\text{ N}$$

Worked example 2 — vector form. $\vec{F}=3\hat{i}+4\hat{j}$ N acts on a $2$ kg body.

$$\vec{a}=1.5\hat{i}+2\hat{j}{\text{ m/s}}^2|\vec{a}|=2.5{\text{ m/s}}^2$$

Worked example 3 — changing mass. Sand drops at $2$ kg/s onto a conveyor belt moving at a steady $3$ m/s. The belt must give the new sand momentum at the rate

$$F=v\frac{dm}{dt}=3\times 2=6\text{ N}$$

Here $F=ma$ would wrongly give zero, because $a=0$.

An everyday example. A loaded auto-rickshaw picks up speed more slowly than an empty one for the same engine force, because its mass is larger.

The substance. The second law contains the first — putting $F=0$ gives $a=0$.

**Impulse is force multiplied by the time it acts, $J=F\Delta t$, and it equals the change in momentum, $J=\Delta p$; for a varying force, impulse is the area under the force-time graph.

Worked example 1 — a rebound.** A $0.15$ kg ball arrives at $20$ m/s and is hit straight back at $30$ m/s. Taking the return direction as positive:

$$\Delta p=0.15\times 30-0.15\times (-20)=7.5\text{ kg m/s}$$

If the bat touches the ball for $0.01$ s, the average force is $\frac{7.5}{0.01}=750$ N.

Worked example 2 — catching. The same ball at $20$ m/s is stopped, so $\Delta p=3$ kg m/s.

- **Stopped in $0.01$ s**: $F=300$ N
- **Stopped in $0.1$ s** by pulling the hands back: $F=30$ N

Worked example 3 — a force-time graph. A kick produces a triangular force peaking at $400$ N over $0.02$ s:

$$J=\frac{1}{2}\times 0.02\times 400=4\text{ N s}$$

An everyday example. A fielder draws the hands back while catching, stretching the stopping time so the same momentum change needs a much smaller force on the palms.

The substance. Change in momentum is a vector — a rebound changes momentum by more than a stop does.