Physics 9702 · Paper 2

Newton's Second Law

The first law says a resultant force changes motion. The second law says by how much.
For a constant mass, the resultant force and the acceleration are linked by one short equation.

The equation

F = ma

Symbols

$F$ = resultant force (N)
$m$ = mass (kg)
$a$ = acceleration (m s⁻²)

$F$ is always the resultant force: add every force on the object first, with directions.
The acceleration is in the same direction as the resultant force.
The same force gives a small mass a large acceleration and a large mass a small one: $a = F/m$ .
This is the constant-mass form of Newton's second law. The full law uses momentum, and lesson 3.11 comes back to it.

The equation also defines the unit of force. One newton is the force that gives a mass of 1 kg an acceleration of 1 m s⁻².
So $1\ \text{N} = 1\ \text{kg m s}^{-2}$ . These are the base units of force used in homogeneity checks (chapter 1).

Worked example

Smallest case: one force, one mass

A resultant force of 6.0 N acts on a 2.0 kg mass. Find the acceleration.

$a = \dfrac{F}{m} = \dfrac{6.0}{2.0} = 3.0 \text{ m s}^{-2}$ .

Answer

a = 3.0 \text{ m s}^{-2}

Find the resultant force first

Most exam questions give more than one force, so the working always has the same first move: add the forces with signs.

Worked example

One change: two opposite forces

A car of mass 500 kg has a forward driving force of 300 N and air resistance of 200 N. Find the acceleration.

Resultant force, taking forward as positive: $F = 300 - 200 = 100 \text{ N}$ .
$a = \dfrac{F}{m} = \dfrac{100}{500} = 0.20 \text{ m s}^{-2}$ .

Answer

a = 0.20 \text{ m s}^{-2}

Worked example

One change: slowing down (negative answer)

A car of mass 500 kg travels at 20 m s⁻¹. The driver brakes and the car stops in 10 s. Find the braking force.

Acceleration first: $a = \dfrac{v-u}{t} = \dfrac{0-20}{10} = -2.0 \text{ m s}^{-2}$ .
$F = ma = 500 \times (-2.0) = -1000 \text{ N}$ .
The minus sign means the force points against the motion. The brakes provide 1000 N backwards.

Answer

F = 1000 \text{ N opposite to the motion}

Common mistake

Putting a single force into

F = ma

when the object has several forces on it.

F

must be the resultant. If a question gives a driving force and a resistance, subtract before you divide by the mass.

The exam version: a force at an angle

In Paper 2 the pull often acts at an angle, along a rope or wire, so one resolving step comes first.

Only the horizontal part of the pull (T cos 28° = 212 N) drives the skier forward; the vertical part (T sin 28° = 113 N) just lifts a little. Split the pull first, then use 212 − R = ma to get the resistance.

Worked example

Kite-skier: resolve, then F = ma

A skier of mass 89 kg is pulled over level snow by a wire at 28° to the horizontal. The tension in the wire is 240 N and the skier accelerates at 0.80 m s⁻². Find the total resistive force R on the skier. (9702/23/M/J/20 Q2(b)(iii))

Horizontal part of the tension: $240\cos 28^\circ = 212 \text{ N}$ .
Resultant force needed for the acceleration: $F = ma = 89 \times 0.80 = 71.2 \text{ N}$ .
The resultant is the pull minus the resistance: $71.2 = 212 - R$ .
$R = 212 - 71.2 = 141 \approx 140 \text{ N}$ .

Answer

R = 140 \text{ N}

Ways the exam changes this question: find the acceleration instead of the force; find one missing force when $a$ is given; feed the acceleration into a SUVAT equation from chapter 2 to get a speed or a distance.
Vertical versions use weight as one of the forces. Weight is the next lesson.

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Newton's Second Law

The first law says a resultant force changes motion. The second law says by how much.
For a constant mass, the resultant force and the acceleration are linked by one short equation.

The equation

F = ma

Symbols

$F$ = resultant force (N)
$m$ = mass (kg)
$a$ = acceleration (m s⁻²)

$F$ is always the resultant force: add every force on the object first, with directions.
The acceleration is in the same direction as the resultant force.
The same force gives a small mass a large acceleration and a large mass a small one: $a = F/m$ .
This is the constant-mass form of Newton's second law. The full law uses momentum, and lesson 3.11 comes back to it.

The equation also defines the unit of force. One newton is the force that gives a mass of 1 kg an acceleration of 1 m s⁻².
So $1\ \text{N} = 1\ \text{kg m s}^{-2}$ . These are the base units of force used in homogeneity checks (chapter 1).

Worked example

Smallest case: one force, one mass

A resultant force of 6.0 N acts on a 2.0 kg mass. Find the acceleration.

$a = \dfrac{F}{m} = \dfrac{6.0}{2.0} = 3.0 \text{ m s}^{-2}$ .

Answer

a = 3.0 \text{ m s}^{-2}

Find the resultant force first

Most exam questions give more than one force, so the working always has the same first move: add the forces with signs.

Worked example

One change: two opposite forces

A car of mass 500 kg has a forward driving force of 300 N and air resistance of 200 N. Find the acceleration.

Resultant force, taking forward as positive: $F = 300 - 200 = 100 \text{ N}$ .
$a = \dfrac{F}{m} = \dfrac{100}{500} = 0.20 \text{ m s}^{-2}$ .

Answer

a = 0.20 \text{ m s}^{-2}

Worked example

One change: slowing down (negative answer)

A car of mass 500 kg travels at 20 m s⁻¹. The driver brakes and the car stops in 10 s. Find the braking force.

Acceleration first: $a = \dfrac{v-u}{t} = \dfrac{0-20}{10} = -2.0 \text{ m s}^{-2}$ .
$F = ma = 500 \times (-2.0) = -1000 \text{ N}$ .
The minus sign means the force points against the motion. The brakes provide 1000 N backwards.

Answer

F = 1000 \text{ N opposite to the motion}

Common mistake

Putting a single force into

F = ma

when the object has several forces on it.

F

must be the resultant. If a question gives a driving force and a resistance, subtract before you divide by the mass.

The exam version: a force at an angle

In Paper 2 the pull often acts at an angle, along a rope or wire, so one resolving step comes first.

Worked example

Kite-skier: resolve, then F = ma

Horizontal part of the tension: $240\cos 28^\circ = 212 \text{ N}$ .
Resultant force needed for the acceleration: $F = ma = 89 \times 0.80 = 71.2 \text{ N}$ .
The resultant is the pull minus the resistance: $71.2 = 212 - R$ .
$R = 212 - 71.2 = 141 \approx 140 \text{ N}$ .

Answer

R = 140 \text{ N}

Ways the exam changes this question: find the acceleration instead of the force; find one missing force when $a$ is given; feed the acceleration into a SUVAT equation from chapter 2 to get a speed or a distance.
Vertical versions use weight as one of the forces. Weight is the next lesson.