Newton’s Second Law: Relationships and Calculations

For an object of constant mass, doubling the resultant force doubles its acceleration. Tripling the force triples the acceleration. This is direct proportionality, written:

$$a \propto F \qquad \text{at constant mass}$$

The symbol $\propto$ means ‘is proportional to’. It describes a particular relationship, not just that two quantities increase together.

For a constant resultant force, doubling the mass halves the acceleration. Tripling the mass reduces the acceleration to one-third. This is inverse proportionality:

$$a \propto \frac{1}{m} \qquad \text{at constant resultant force}$$

Imagine pushing an empty shopping trolley and a loaded trolley with the same resultant force. The loaded trolley has a larger mass, so its acceleration is smaller. To give both trolleys the same acceleration, the loaded trolley needs a greater resultant force.

Calculating force, mass and acceleration

The two relationships are combined in the equation:

$$F = ma$$

Here, $F$ is the resultant force in newtons (N), $m$ is the mass in kilograms (kg), and $a$ is the acceleration in metres per second squared (m/s²).

For example, a car of mass $1200\text{ kg}$ accelerates at $2.5\text{ m/s}^2$. Its resultant force is:

$$F = 1200 \times 2.5 = 3000\text{ N}$$

This is the overall force accelerating the car. If resistive forces also act, the driving force must be greater than this resultant force.

To find acceleration, divide the resultant force by mass:

$$a = \frac{F}{m}$$

To find mass, divide the resultant force by acceleration:

$$m = \frac{F}{a}$$

For example, a resultant force of $150\text{ N}$ gives an astronaut an acceleration of $1.5\text{ m/s}^2$. The astronaut's mass is $150 \div 1.5 = 100\text{ kg}$.

Estimating large accelerations in road transport

An estimate uses sensible approximate values rather than exact measurements. The symbol $\sim$ indicates an approximate value or answer. For road transport, a car speed of about $20\text{ m/s}$ and an adult passenger mass of about $70\text{ kg}$ are useful reasonable estimates.

First estimate acceleration using:

$$a = \frac{\text{final velocity} - \text{initial velocity}}{\text{time taken}}$$

Then use $F = ma$ to estimate the resultant force on the object you are considering.

Consider a collision in which a passenger travelling at approximately $20\text{ m/s}$ comes to rest in $0.1\text{ s}$. Taking the original direction of motion as positive:

$$a \sim \frac{0 - 20}{0.1} = -200\text{ m/s}^2$$

For a passenger of mass approximately $70\text{ kg}$:

$$F \sim 70 \times (-200) = -14\,000\text{ N}$$

The estimated force has a magnitude of $14\,000\text{ N}$ and acts opposite to the passenger's motion. The very short stopping time produces a large acceleration and therefore a large force. For the same change in velocity and mass, a longer stopping time would reduce both magnitudes.

Inertial mass — Higher Tier only

Inertial mass is a measure of how difficult it is to change an object's velocity. Changing velocity includes speeding up, slowing down or changing direction.

It is defined as the ratio of resultant force to the acceleration produced:

$$m = \frac{F}{a}$$

For example, a resultant force of $18\text{ N}$ produces an acceleration of $3.0\text{ m/s}^2$, giving an inertial mass of $6.0\text{ kg}$.

A larger inertial mass means that the same resultant force produces a smaller acceleration. Equivalently, a greater resultant force is needed to produce the same acceleration. It does not mean that the object cannot accelerate.

$$F = ma \qquad a = \frac{F}{m} \qquad m = \frac{F}{a}$$

$F$: resultant force, N; $m$: mass, kg; $a$: acceleration, m/s².

Road-transport estimates

Choose sensible speeds, times and masses, then use:

$$a = \frac{v-u}{t}, \qquad F = ma$$

$\sim$ means approximately. A shorter time for the same velocity change produces a larger acceleration magnitude and force magnitude.

Inertial mass — HT only

Measures how difficult it is to change velocity; defined by $m = F/a$.

Greater inertial mass → smaller acceleration for the same resultant force.