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GCSE level Physics exam revision notes on
Forces & Motion Part 4
Forces and Newton's Laws of Motion 4.2
Newton's Second Law of Motion, formula F = ma and calculations
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INDEX for physics notes on
Newton's Laws of Motion: concepts, formulae, calculations and problem
solving
This page contains online questions only. Jot down
your answers and check them against the worked out answers at the end of
the page
4.2
Newton's Second Law of Motion F = ma and calculations
Examples of worked out calculation practice exam questions and
problem solving based on Newton's 2nd Law of Motion
What is Newton's second law of motion?
How do we do Newton's 2nd law
calculations?
How do you use the equation F = ma in
problem solving?
Newton's Second Law of Motion can be expressed as
the
acceleration of a body is directly proportional to the force applied to that
body. Remember that acceleration is the change in velocity over a specified
time period.
(i) The greater the resultant force acting on a body, the
greater the acceleration of that body.
In fact, you can
see from the equation for Newton's 2nd law that the acceleration
experienced by an object is directly proportional to the force acting on
it.
F
A
(ii) The acceleration of a body is also inversely
proportional to its mass:
a
1/m
Those two statements can be expressed in the general
formula:
resultant force (newtons) = mass (kilograms)
x acceleration (metres per seconds squared),
this equation is a mathematical
expression of Newton's 2nd law
F (N)
= m (kg) x
a (m/s2)
rearranging (i)
F = ma
gives (ii)
a = F ÷ m
and (iii)
m = F ÷ a
The
greater the force acting on an object, the greater the object is
accelerated.
Acceleration is inversely proportional to the mass.
If the
same force is applied to objects of different mass, the smallest mass
will be accelerated the most.
The consequence of statement (i) is that
the faster you want to speed up (accelerate) an object the greater the force
required (putting your foot down further on a car accelerator to increase
the power).
Conversely, the faster you want an object to slow down
(decelerate) the greater the resistive force you must apply (pressing the
brake pedal of a car more forcefully).
One consequence of statement (ii) is that
a specific applied resultant force accelerates a body of smaller mass more
than a body of larger mass.
Or to express it another way, you need to apply
a greater force to accelerate a larger body to the same extent as a smaller
body.
Examples of
calculations based on Newton's 2nd Law
(some
questions also need the formula for acceleration a =
∆v / ∆t)
Q2.1 What
resultant force must be applied by a cyclist (mass 70.0 kg) to the pedals of a bike (8.50
kg) to give an acceleration of 0.15 m/s2?
Worked out ANSWERS
Q2.2 The
engine of 1000 kg car generates a driving force of 4200 N.
If the car is
travelling at 60 mph and the total resistive force (friction in engine, road
contact and air resistance) is 3800 N what is the car's acceleration at this
point?
Worked out ANSWERS
Q2.3 If a
body experiences a resultant force of 500 N and accelerates away at 2.0 m/s2,
what is the mass of the body?
Worked out ANSWERS
Q2.4 A bus
of mass 2800 kg uniformly accelerates from 0 to 15 m/s in 20 seconds.
(a) Calculate the acceleration of the bus.
(b) Calculate the resultant force acting on the bus.
Worked out ANSWERS
Q2.5 The engine of a van of mass 2000 kg
generates a driving force of 6000 N.
If it experiences a drag force due to air
resistance of 500 N, calculate its acceleration.
Worked out ANSWERS
Q2.6
What force is required to vertically accelerate a 8000 kg rocket at 3 m/s2?
(g = 9.8 m/s2)
Worked out ANSWERS
Q2.7 If the force exerted on an object is quadrupled and the mass
doubled, what happens to the acceleration?
Worked out ANSWERS
Q2.8
High speed photography can be used to analyses the motion of sports equipment.
In cricket, a fast bowler sends the
cricket ball down at 36 m/s (~80 mph, ~130 km/h) towards the batsman.
The cricket ball has a mass of 160 g and
is a bit worn and slightly soft.
The batsman holds the bat still and
blocks the ball which rebounds directly back towards the bowler at 30 m/s.
(a) If the contact time was found to be
0.002 seconds, what force did both the cricket ball experience and in what
direction?
(i) You first need to calculate the
acceleration
(ii) You can now
calculate the force involved from Newton's 2nd Law equation.
(b) A new ball is
harder than a used ball and the contact time may only be 0.001 seconds.
What effect does
this have on the force experienced by a new cricket ball compared to the
worn ball in (a)?
(c) Why might the
impact time be longer for the worn ball compared to a new ball?
Worked out ANSWERS
INDEX for physics notes on
Newton's Laws of Motion
Key points
on
Newton's laws of motion:
Newton's 2nd Law of Motion F = ma
Information
sources for Doc Brown's key points: IGCSE-GCSE physics are based on
textbooks & syllabus-specifications for students taking the UK AQA, Edexcel,
OCR 21st Century Science, OCR Gateway science suite, WJEC, CCEA and CIE GCSE
physics 9-1 level science examinations
A comprehensive set of summary revision notes on
Newton’s Second
Law of Motion (F = ma), tailored to the GCSE/IGCSE
Physics specifications across major UK exam boards: WJEC,
CCEA, CIE, AQA, Edexcel, and OCR.
Newton’s Second Law of Motion –
Summary Revision Notes
Definition (Universal Across All
Boards)
The acceleration of an
object is directly proportional to the resultant force acting on it
and inversely proportional to its mass.
This is mathematically expressed
as:
F = m × a
Where:
- F
= Force (Newtons, N)
- m
= Mass (kilograms, kg)
- a
= Acceleration (metres per second squared, m/s²)
Key Concepts relating to Newton's 2nd
Law of Motion
- Resultant Force:
The net force acting on an object after all forces are considered.
- Acceleration:
Change in velocity per unit time.
- Mass:
A measure of an object’s inertia (resistance to acceleration).
- Direct Proportionality:
More force → more acceleration.
- Inverse Proportionality:
More mass → less acceleration for the same force.
Worked Example of Newton's 2nd Law of
Motion F = ma
Q:
A car of mass 900 kg accelerates from 0 to 27 m/s in 3 seconds.
Step 1: Acceleration = Δv / t = (27 - 0) / 3 =
9 m/s²
Step 2: Force = m × a = 900 × 9 =
8100 N
Typical Exam Board Syllabus content
relating to Newton's 2nd Law of Motion
|
Specific
Requirements |
| Must recall and apply F = ma;
use in context of unbalanced forces and acceleration |
| Emphasis on calculating force,
mass, or acceleration; apply to motion scenarios and inertial
mass |
| Required to define and apply F
= ma; extended tier students must rearrange and use formula
triangle |
| Apply F = ma to estimate
forces in realistic contexts; understand inertial mass and
vector quantities |
| Use F = ma in calculations and
practical applications; explain proportional relationships |
| Apply Newton’s Second Law to
free-body diagrams and vector resolution; understand inertia and
momentum |
Inertial Mass
- Definition:
A measure of how difficult it is to change an object’s velocity.
- Formula:
Inertial Mass = F / a
- Implication:
Higher inertial mass → more force needed for same acceleration.
Real-Life Applications of Newton's 2nd
Law of Motion
| Scenario |
Newton’s
Second Law in Action |
| Cyclist accelerating |
More pedalling force → faster
acceleration |
| Car crash |
Sudden deceleration → large
force on passengers |
| Spacecraft propulsion |
Small mass → large
acceleration with minimal force |
Student Tips on Newton's 2nd Law of
Motion
- Memorise the formula:
F = ma and be able to rearrange it.
- Use formula triangles
to solve for unknowns, BUT better to be able to rearrange F = ma.
- Practice with units:
N, kg, m/s².
- Estimate values:
Use approximations in context (e.g. mass of person ≈ 70 kg).
- Interpret graphs:
Velocity-time and force-acceleration relationships.
- Required Practicals:
Investigate how changing force or mass affects acceleration (e.g.
trolley and motion sensor setups).
Everyday Applications of Newton's 2nd
Law of Motion
Newton’s Second Law - F =
ma - is everywhere once you start noticing how force, mass,
and acceleration work together. Here are some clear and relatable
examples:
Everyday Applications of F = ma
Car Acceleration
- When you press the accelerator,
the engine produces more force.
- A lighter car accelerates faster
with the same force than a heavier SUV.
- Formula in action:
If a 1000 kg car accelerates at 2 m/s², the force is 2000 N.
Shopping Trolley
- Push an empty trolley →
accelerates quickly.
- Load it with groceries → needs
more force for same acceleration.
- Same force, more mass → lower
acceleration.
Throwing a Ball
- A basketball needs more force than
a tennis ball to accelerate at the same rate.
- The ball’s mass affects how much
force you must apply to change its speed.
Cyclist Climbing a Hill
- More force needed to accelerate
upward due to gravity and the bike’s mass.
- Lighter bikes require less effort
→ better acceleration with same pedalling force.
Sled on Snow
- A small sled with a child requires
less force to pull than one with an adult.
- Greater mass → greater inertia
→ more force needed for same acceleration.
Rocket Launch
- Engines must produce enormous
force to overcome the rocket’s mass and Earth’s gravity.
- As fuel burns off, mass decreases
→ acceleration increases.
Keywords, phrases and learning objectives for Newton's 2nd law of motion
Be able to know and use
Newton's Second Law of Motion.
Know that acceleration is directly proportional to force applied to an
object of a given mass and be able to solve problems using the formula
F = ma with the appropriate units.
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INDEX for physics notes on
Newton's Laws of Motion
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Worked out
ANSWERS to calculations using Newton's 2nd Law
Q2.1 What
resultant force must be applied by a cyclist (mass 70.0 kg) to the pedals of a bike (8.50
kg) to give an acceleration of 0.15 m/s2?
F (N)
= m (kg) x
a (m/s2)
F = (70 + 8.5) x 0.15
force required =
11.8 N
Q2.2 The
engine of 1000 kg car generates a driving force of 4200 N.
If the car is
travelling at 60 mph and the total resistive force (friction in engine, road
contact and air resistance) is 3800 N what is the car's acceleration at this
point?
F (N)
= m (kg) x
a (m/s2)
The resultant forward force = 4200 - 3800
= 400 N
F = ma, so a = F / m = 400 / 1000
= 0.40 m/s2
Q2.3 If a
body experiences a resultant force of 500 N and accelerates away at 2.0 m/s2,
what is the mass of the body?
F = ma, so m = F / a = 500 / 2 = 250 kg
Q2.4 A bus
of mass 2800 kg uniformly accelerates from 0 to 15 m/s in 20 seconds.
(a) Calculate the acceleration of the bus.
acceleration (m/s2) = change in speed (m/s) / time taken
(s)
a = ∆v / ∆t = 15 /
20 =
0.75
m/s2
(b) Calculate the resultant force acting on the bus.
F = ma, substituting gives resultant force = 2800 x 0.75
= 2100 N
Q2.5 The engine of a van of mass 2000 kg
generates a driving force of 6000 N.
If it experiences a drag force due to air
resistance of 500 N, calculate its acceleration.
F = ma, a = F / m = 6000 / 2000 =
3 m/s2
Q2.6
What force is required to vertically accelerate a 8000 kg rocket at 3 m/s2?
(g = 9.8 m/s2)
F1 = weight of rocket = downward
force = 8000 x 9.8 = 78 400 N
F2 = upward force of the rocket to
overcome gravity and accelerate vertically at 3 m/s2
F = ma, F = the net vertical force =
F2 - F1, m = mass of rocket, a = acceleration of 3 m/s2
F2 - F1 = ma, F2 = F1 + ma
F2 = F1 + ma = 78 400 + (8000 x 3) =
102 400 N
Q2.7 If the force exerted on an object is quadrupled and the mass
doubled, what happens to the acceleration?
F = ma, a = F/m (call it 1)
a = 4f / 2m = 2 F/m, so the
acceleration is doubled.
Q2.8
High speed photography can be used to analyses the motion of sports equipment.
In cricket, a fast bowler sends the
cricket ball down at 36 m/s (~80 mph, ~130 km/h) towards the batsman.
The cricket ball has a mass of 160 g and
is a bit worn and slightly soft.
The batsman holds the bat still and
blocks the ball which rebounds directly back towards the bowler at 30 m/s.
(a) If the contact time was found to be
0.002 seconds, what force did both the cricket ball experience and in what
direction?
(i) You first need to calculate the
acceleration - rate of change of velocity, and watch the signs, you are
dealing with vector quantities!
a =
∆v / ∆t =
(v - u) / ∆t, a = acceleration (m/s2), v
and u = final and initial velocities, ∆t = time taken.
u = +36 m/s, v
= -30 m/s, and
∆t = 0.002 s, a plus sign means the direction towards the batsman.
a = (-30) -
(+36) / 0.002 = -33 000 m/s2
(ii) You can now
calculate the force involved from Newton's 2nd Law equation.
F = ma,
F force in N, m = mass in kg, a = acceleration in m/s2
F = (160/1000)
x 33 000 = - 5280 N
The cricket ball experiences a force of 5280 N back towards the
bowler.
(b) A new ball is
harder than a used ball and the contact time may only be 0.001 seconds.
What effect does
this have on the force experienced by a new cricket ball compared to the
worn ball in (a)?
This means ∆t is
halved, the acceleration is doubled, so is the impact
force is doubled, so the cricket ball experiences a force of 2 x 5280 =
10 560 N in
the direction back towards the bowler.
(c) Why might the
impact time be longer for the worn ball compared to a new ball?
The worn ball is a
little softer and the change in velocity takes a bit longer - the
softness has a cushioning effect that reduces the impact force.
importance of Newton's
2nd Law of Motion, inertia, resultant forces, F=ma calculations
in GCSE level physics, What you need to know about Newton's 2nd Law of
Motion, inertia, resultant forces, F=ma calculations for
GCSE level
physics,
Explaining the use of Newton's 2nd Law of Motion, inertia, resultant
forces, F=ma calculations knowledge in GCSE level physics, Examples of
Newton's 2nd Law of Motion, inertia, resultant forces, F=ma calculations explained
when studying GCSE level physics, What is
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INDEX for physics notes on
Newton's Laws of Motion
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