(b)
Newton's Second Law of Motion F = ma
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/s^{2})
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/s^{2}?
F (N)
= m (kg) x
a (m/s^{2})
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/s^{2})
The resultant forward force = 4200  3800
= 400 N
F = ma, so a = F / m = 400 / 1000
= 0.40 m/s^{2}
Q2.3 If a
body experiences a resultant force of 500 N and accelerates away at 2.0 m/s^{2},
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/s^{2}) = change in speed (m/s) / time taken
(s)
a = ∆v / ∆t = 15 /
20 =
0.75
m/s^{2}
(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/s^{2}
Q2.6
What force is required to vertically accelerate a 8000 kg rocket at 3 m/s^{2}?
(g = 9.8 m/s^{2})
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/s^{2}
F = ma, F = the net vertical force =
F2  F1, m = mass of rocket, a = acceleration of 3 m/s^{2}
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/s^{2}), 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/s^{2}
(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/s^{2}
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.
https://www.physics.usyd.edu.au/~cross/cricket.html
is an interesting page on the physics of cricket and this question is taking
a simplified view of the 'kinetic' situation!
Q2.9
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(c) Experiments to
investigate Newton's 2nd law of motion (F = ma)
The experiments described below can be used to investigate how
force and mass affect acceleration.
These experiments use light gates to
measure speed  so they need to be explained!
A light gate directs a beam of light from
one side to a detector on the other side (no details shown, just the
labels!).
When the light beam is cut by
something passing through the light gate (e.g. the a card on a trolley) the
light gate measures the time the light beam is not detected.
To make an acceleration measurement the
moving object must break the light beam twice and there are two ways of
doing this  versions 1 and 2 experiments are both described below. Yu need
two speeds and the difference between them to calculate an acceleration.
The light gate is connected to a
computer and the rest is explained via the diagram below.
In version 1, with one light gate, you
use an inverted Π shaped card (rectangle with a middle section cut out).
The lengths of the upright sections that cut
the light beam are d1 and d2 (which you can make the same).
As the trolley and card pass through the
light gate, the first section of card of length d1, cuts light beam for time t1
(for speed 1).
When the second section of the card of length
d2 passes through the light gate, the beam is cut for time t2 (for speed 2).
At this point the combined light gate also
records the time difference t3 between the two speed measurements (for
acceleration a). So for speed s ...
s1 = d1/t1, s2 = d2/t2, acceleration = change
in speed / time taken = a = (s2  s1)
/ t3
However, all you do is input d1 (= d2) into
the computer and the software does the acceleration calculation for you!
Obviously, the computer software system must
be set up to suit the way you have set up the experiment.
In version 2, with two light gates,
you use a rectangular shaped card
[ ]
of length
d.
You set up two light gates a suitable
distance apart.
When the trolley and card pass through the
first light gate, the beam is cut for t1.
After passing through the 2nd light gate,
time t2 is recorded.
At this point the combined light gates also
record the time difference t3 between the two speed measurements.
Its now a similar data situation to version 1
of the experiment, so for speed s ....
s1 = d/t1, s2 = d/t2, acceleration = change
in speed / time taken = a = (s2  s1)
/ t3
Again, the computer software system must be
set up to suit the way you have set up the experiment and will do the
acceleration calculation for you.
Example of calculation for experiment version 2
Suppose the length of the trolley is 10
cm (0.10 m) and interrupts the 1st light gate for 0.25 s.
The velocity at the 1st light gate is
0.10 / 0.25 = 0.40 m/s.
The trolley interrupts the 2nd light gate
for 0.10 s.
The velocity at the 2nd light gate is
0.10 / 0.10 = 1.0 m/s.
The time taken for the trolley to travel
between the light gates was 0.80 s.
Therefore the
acceleration =
(v  u ) / ∆t = (1.0  0.40) / 0.80 = 0.75
m/s^{2}
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Investigation version 1 using one light
gate
Error correction
NOT shown in the diagram above
Strictly speaking you should do the
experiment on a slightly tilted running board rather than the horizontal
laboratory bench shown in the diagrams above.
You tilt the running board ramp down to
the right until the trolley (without hook attached) just begins to move on
its own.
You then prop up the lefthand end at the
same height and then perform the experiments as described.
After doing this, the weight of the
trolley down the slope will compensate for the friction acting between the
wheels of the trolley and the running board ramp.
It is also possible to do these
investigations with a horizontal air track on which a 'trolley' hovers on
jets of air  this is an alternative investigation design to minimise the
effects of friction in the experiments.
The experiment is conducted on a smooth bench on which a trolley
can freely run.
You need to mark on a start line to keep
things consistent.
The total mass of the trolley and extra
weights (m) is measured and can be altered by adding
extra weights.
On the trolley is fixed a 'rectangular' Џ
shaped card, whose two arms will block the light beam twice when the trolley
passes through the light gate when it measures the time taken twice.
The trolley is connected by wire or string to
the weights via the pulley wheel, that on falling, provide the downward force to
accelerate the trolley along the 'runway'
The time measured by the light gate system is
recorded electronically and passed to the connected data logger or computer.
As the Џ card passes through the light
gate, the light beam is cut off twice, so two velocities are measured and the
time interval between them.
The principles of the F = ma
calculations for version 1.
The accelerating force F is given by
the falling weight: mass in kg x 9.8 N/kg
The mass being accelerated = total mass of
trolley plus weights (m)
The acceleration is given by the light gate
velocities divided by the time interval between them:
a = v_{2} 
v_{1} / Δt
You can do a series of experiments with
constant trolley mass m, and varying the accelerating force F by varying the falling weights
on the end of the hook.
Theoretically, a graph of force F versus
acceleration a should be linear  or look something like it!
The gradient is given by the
trolley mass
m (the constant).
You should definitely be able to show
that increasing the applied force a greater acceleration is produced,
ideally your results are a linear graph.
Newton's 2nd law equation: F = ma,
F
a, and the constant gradient should be the total mass of the trolley.
Extending the investigation
You can then investigate the effect
of varying the weights on the trolley, to vary its mass and applying a
constant force of acceleration using the same falling weight.
You measure the acceleration in
the same way as described.
You should be able to show the
greater the total mass of the trolley m (the greater its inertia), the
smaller the acceleration
produced by a constant accelerating force (F).
A graph of mass versus acceleration
should be a downward curve.
F = ma, as m gets
bigger, a gets smaller for constant F.
m = F/a and
a = F / m
You can do a linear graph analysis of
your results e.g.
A graph of m versus 1/a should be
linear with a gradient F.
A graph of acceleration a
versus 1/m should be linear with a gradient F.
Sources of error
Friction in the wheel axles and friction
between the wheels and the running board will have a small retarding effect.
This will reduce the acceleration a little bit.
One way to compensate for this is to
tilt the running board slightly until the trolley just moves, but don't
overdo it or you will record accelerations greater than you should.
You can do more accurate experiments
using an air track on which the 'trolley' is made to hover on
jets of air.
Using the light gates and a computer is
more accurate than trying to make direct observations with a stopwatch.

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Investigation version 2 using two light
gates
Error correction
NOT shown in the diagram above
Strictly speaking you should do the
experiment on a slightly tilted running board rather than the horizontal
laboratory bench shown in the diagrams above.
You tilt the running board ramp down to
the right until the trolley (without hook attached) just begins to move on
its own.
You then prop up the lefthand end at the
same height and then perform the experiments as described.
After doing this, the weight of the
trolley down the slope will compensate for the friction acting between the
wheels of the trolley and the running board ramp.
It is also possible to do these
investigations with a horizontal air track on which a 'trolley' hovers on
jets of air  this is an alternative investigation design to minimise the
effects of friction in the experiments.
As the
[ ] card goes past the light gates, the
two
time intervals for blocking the light give you an initial and final velocity,
and the time interval between these two events can be used to calculate the
acceleration of the trolley.
The investigations, calculations and graphs
you can do are described above in method 1.
Version 3 of
acceleration experiments.
You can run the trolley down an inclined ramp onto the level
running board and then past two light gates (2 and 3) to measure the final
velocity.
You can set the first light gate (1) near the start line of
the inclined ramp  the initial velocity should be close to zero.
You can vary the angle of the ramp  the steeper angle
should give a greater acceleration.
You can vary the length of the inclined ramp to see if a
longer running downhill distance increases acceleration.
You can vary the total mass of the trolley to if this has
any effect on the acceleration.
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(d) The
concept of inertia  comparing
inertial
mass and gravitational mass
Inertia can be defined as the tendency of
an object's motion to remain unchanged.
If we start by thinking about the implication
of Newtons First Law of motion ...
... a
resultant force is needed to change the motion of any object.
In other words, unless acted on by a
resultant force, anything stationary remains at rest (zero velocity), anything
moving keeps moving with the same velocity (same speed and direction).
So we can say
the tendency of an
object to keep moving with the same velocity is called inertia.
Inertial
mass measures an object's resistance to being accelerated by a force (F =
ma)
We can measure how difficult it is to change
an object's velocity by calculating its inertial mass.
An object's inertial mass is a measure of how difficult it is to change its velocity.
We can do this using the equation from
Newton's Second Law of motion (force = mass x acceleration).
F = ma, on rearrangement this
gives: m = F/a, and this expression defines inertial mass
so
inertial mass (kg) = applied force (N) /
acceleration (m/s^{2})
Another way of looking at the equation is to
consider the effect of force on acceleration.
a = F/m, some consequences are ...
as already stated, for a given mass on
object's acceleration is proportional to the force applied, but ...
if the same force F is applied to
two different masses, the smaller mass, with the smaller inertia, will experience the greater
acceleration,
and, if two objects have the same mass, then
applying the same force to each object will produce the same acceleration.
Inertia and moving objects
As well as looking at inertia from the point
of view of acceleration, think about slowing moving objects down.
If two objects of different masses are
moving at the same speed, the object of greater mass will need a bigger
force to slow it down (decelerate) due to Newton's second law.
Imagine two cars are moving at the same
speed and both drivers take the foot off the accelerator. If the two cars
experience the same air resistance and wheelroad friction forces, the car
of bigger mass would travel on further before coming to a halt. The bigger
the inertial mass, the bigger the force would be needed to bring it to a
halt in the same stopping distance as the car with the smaller mass. F = ma,
force proportional to mass.
Large objects like cargo ships or
high speed trains can take several km to come to a halt.
What is the difference between inertial
mass and gravitational mass
Inertial mass measures an object's
resistance to acceleration.
Inertial mass = force /
acceleration (m = F / a, from F = ma)
Gravitational mass determines the
gravitational attractive force it exerts on another object
mass = weight / gravitational
field constant (m = W / g, from W = mg).
Inertial mass and gravitational
mass are numerical identical.
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(e)
Newton's Third law of Motion
Newton's Third Law of motion states that
when two objects interact, the forces they exert on each other are equal in
numerical value, act in opposite directions and are of the same type.
This at first suggests that neither object
can move anywhere, but you have two objects and two outcomes!
In general, what can you can say for the
forces in a pair of objects interacting?
They are the same size.
They act in opposite directions.
They act on different objects.
They are the same type of force.
These four points will be illustrated
in the following examples.
(i)
A stationary example of Newton's 3rd
Law
A bit
messy to analyse!
Consider the flask of liquid standing
motionless on a laboratory bench.
There are two sets of forces operating shown
by the arrows of opposing direction, but the same length  same magnitude of
force.
Both sets of forces are examples of Newton's 3rd Law, but don't mix the
two up!
(i) The normal contact force due to the
weight of the object acting (pushing) down on the surface of the bench (F1) is
balanced by the bench under minute compression of the atoms pushing back up to an equal and
opposite extent onto the flask (F2).
A similar argument applies to when you
push against an object that isn't moving.
The force of your push is balanced by the compressed atoms of the object
pushing back!
(ii) At the same time both the flask and the Earth
(including the bench) are mutually attracting each other (F3 and F4) to an equal
and opposite extent due to the noncontact force of gravity.
The Earth attracts the flask and the
flask attracts the Earth.
It makes no
difference whether the objects are in contact or not, here gravity acts
throughout everything!).
In the cases described so far there is no
resultant force, everything is balanced.
If the forces were not balanced and
there was some net resultant force, the object would move or be reshaped
 something would change!
For stationary objects, if the resultant
force acting on the object is zero the object is said to be in
equilibrium (effectively means a state of balance).
Take care with equilibrium situations
which might look like examples of Newton's Third Law of motion
e.g. suppose a watch is hanging on the
end of gold chain in a stationary equilibrium situation.
Does this conform with Newton's Third
Law.
The answer is no because the
forces are not of the same type and they both acting on the watch.
The weight of the watch is acting
downwards due to gravity.
The tension in the chain is acting
upwards from the watch  the tension is due to atoms pulling on
each other.
(ii)
'Movement' examples of Newton's 3rd Law
and motion consequences
illustrating all four points
listed above
Situation 1. Ice skaters on
an ice rink.
Suppose two ice skaters, of similar
mass, are standing next to each other, and then one of them pushes the
other with a force of 80 N, what happens next?
Both skaters experience the same
normal contact force of 80 N, but in opposite directions.
Both skaters will accelerate away
from each other in opposite directions.
However, they will only
accelerate away from each other with the same acceleration if they
of the same mass.
Since F = ma and a = F/m,
if one of the skaters has a
smaller mass than the other, the skater of smaller mass will move in
the opposite direction with a greater acceleration (F is a constant
and a
1/m).
Situation 2. Astronaut with a rocket
back pack
In 'outer space' in a zero gravity
situation, an astronaut could manoeuvre around using a small jet pack
attached to the back of a spacesuit.
When fired, the gases are accelerated
one way and an equal and opposite force acts on the astronaut who is
accelerated in the opposite direction  hopefully back to the space
station or spaceship!
Situation 3. A headon
collision of two objects
Suppose two balls m1 and m2, of masses m_{1}
and m_{2} collide headon with a certain normal contact force.
Assume mass m_{1} is smaller than mass m_{2}.
Let us assume, before impact, they are
moving at the same velocity towards each other.
On impact they both fly backwards in the
opposite direction than before impact.
According to Newton's 3rd Law, both
balls experience exactly the same force, but acting in opposite directions.
However
because they have different
inertial masses, so, as a consequence of a = F/m (Newton's 2nd Law)
..
the smaller inertial mass, ball m1, will
be accelerated more and fly off with a greater velocity than ball m2,
the larger inertial mass, ball m2, will
be accelerated less and fly off with a smaller velocity than ball m1,
(iii)
A headon collision between two road
vehicles
In example (ii) the balls rebounded of
each other, but what are the consequences of two objects colliding head on
and 'crunching' together.
So, imagine a 1000 kg car travelling
crashing headon into a 500 kg car travelling in the opposite direction.
(a) What can you say about the impact
force experienced by each car?
According to Newton's 3rd law of
motion, both cars experience the same impact force, but in opposite
directions.
(b) What can you say about the relative
deceleration of each car and their relative inertial masses?
Newton's 2nd law of motion can be
stated as F = ma,
F = impact force, m =
inertial mass and a = deceleration (negative acceleration).
Since the impact force is the same,
ma_{g} for the 1000 kg green car = ma_{b}
for the 500 kg blue car.
This means that the deceleration
(negative a) for the green car of larger mass must be less than that for
the blue car of smaller mass.
In fact, we can say mathematically,
for the same impact force F
1000 x a_{green car} =
500 x a_{blue car}
Which means that the relative
deceleration of the blue/green cars is 1000/500 = 2.
So, the 'lighter' blue car
decelerates at twice the rate of the 'heavier' green car.
The green car has a much greater
inertial mass, and is more difficult to slow down than the blue car
of much smaller inertial mass, for the same impact force.
(c) What are the consequences for each
driver from your answers to (b)?
The consequences for the driver of
the blue car are potentially much more serious than the consequences for
the driver of the green car.
The blue car driver will be
accelerated forwards at twice as much as the green car driver, with
potentially greater impact injuries  although all should be wearing
safety belts, there is still the possibility of 'whiplash' injuries.
You could also presume, in the case
of either driver, the smaller the mass of the driver, the more the would
be accelerated forwards.
(d) If the cards were travelling at the
same speed prior to impact, and the crunch combination continued moving as a
single object, which direction would the 'wreck' move and why?
(i) The 'combination wreck' would
continue moving to the right.
Eventually it would come to a
halt due to friction with the road.
(ii) The 'combination wreck' would
continue moving to the right because of he heavier mass of the green
car.
The green car has twice the
momentum (mass x velocity) of the blue car.
For more on (d) see Q2.7 on
momentum
calculations and Newton's 2nd law of motion
(iv) A cat leaping up from the ground to catch a
butterfly!
The 'force' sequences (neglecting air
resistance)

At the start a pair of equal
noncontact gravitational forces operate  the Earth attracts the cat
and the cat attracts the Earth.

Coincident are the normal contact
forces operating  the weight of the cat pushes down on the ground via
the cat's paws and the compressed ground pushes back up with an equal
opposite action force.

On leaping up, the cat's muscles
exert an increased normal contact force on the ground through its back
legs and, again, the ground pushes up with an increased equal an
opposite normal contact force.

The normal contact forces are no
longer operating and the upward acting force generated by the cat's
muscles must initially exceed the weight of the cat due to gravity  the
cat's acceleration must exceed the Earth's gravitational pull on it 
otherwise it can't leave the ground.

The cat's kinetic energy store (KE)
decreases and its gravitational potential energy store (GPE) increases
to a maximum at the maximum height the cat reaches into the air  the KE
is zero just for a moment, then as it falls the cat's GPE is converted
back to KE.

The cat is attracting the Earth and
the Earth is equally attracting the cat, which now falls to the ground.

The normal contact force of the cat's
impact on the ground is equalled by the atoms of the ground pressing
back up.
See
also
Elastic and
nonelastic collisions, momentum calculations and Newton's 2nd law of motion
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Motion and associated forces notes index (including Newton's Laws of
Motion)
1. Speed and velocity  the relationship between
distance and time, distancetime graphs gcse physics
2. Acceleration, velocitytime graph interpretation and calculations,
problem solving
gcse physics revision notes
3. Acceleration,
friction, drag effects and terminal velocity experiments
gcse physics revision notes
4. Newton's First, Second and Third Laws of
Motion, inertia and F = ma calculations
gcse physics revision notes
5. Reaction times stopping distances and example
calculations
gcse physics revision notes
6. Elastic and nonelastic collisions, momentum
calculations and Newton's 2nd law of
motion
gcse physics notes