School Physics notes: Newton's Laws of Motion - calculations

3. Newton's First Law of Motion, Newton's Second Law of Motion, Newton's Third Law of Motion, inertia and F = ma based calculations

Doc Brown's school physics revision notes: GCSE physics, IGCSE physics, O level physics,  ~US grades 8, 9 and 10 school science courses or equivalent for ~14-16 year old students of physics

This page will help you know answer questions such as ...   What is Newton's first law of motion?   What is Newton's second law of motion?    How do we do Newton's 2nd law calculations?  How do use the equation F = ma in problem solving?  What is Newton's third law of motion? and what is inertia?

(a) Newton's First Law of Motion - examples explained - resultant forces

(b) Newton's Second Law of Motion - examples of calculations using F = ma

(c) Experiments to investigate Newton's 2nd law of motion (F = ma)

(d) The concept of inertia - inertial mass and gravitational mass

(e) Newton's Third Law of Motion - examples explained

(a) Newton's Laws of Motion and resultant forces

Newton's three laws of motion are some of the earliest proposed laws of physics - indeed of science itself.

His imagination, insight and mathematical brilliance have stood the test of time since 1686 when he first proposed them - but they could not be assumed to be true i.e. valid in 1686 - repeated experimental verification is required.

BUT, countless experiments over the past few hundred years have shown them to be true and form the basis of innumerable calculations in applied physics and engineering.

Newton's First Law of Motion

Newton's First Law of Motion states that a resultant force is needed to change the motion of any object.

e.g. starting a body moving, increasing its speed (acceleration), slowing it down (deceleration), stopping it moving, changing the direction of movement (the latter is a change in velocity).

A resultant force must be an unbalanced net force of >zero on an object.

If the resultant force on a stationary body is zero, the body will remain stationary.

If the resultant force on a moving object is zero, the velocity remains unchanged, in other words the object will continue moving at the same speed and in the same direction, therefore moving with the same velocity.

If a body such as a road vehicle or an aeroplane is moving at a constant velocity (constant speed without changing direction) the driving force from the engine and the resistive forces of friction (moving parts, air resistance etc.) must be balanced i.e. the resultant force is zero.

The velocity can only be changed if a resultant non-zero force acts on the object.

A non-zero resultant force will always initially produce an acceleration or deceleration in the direction of the force (see cyclist example below).

There are five situations that you will come across when dealing with a non-zero resultant force that results in a change in velocity.

This change in velocity may involve a moving object speeding up (acceleration) or slowing down (deceleration).

The acceleration might be a stationary object being forced to move or a moving object made to stop.

These first four situations involve change in speed, and not necessarily a change in direction, but change in direction is also a 5th result of the effect on an object of a non-zero resultant force, and therefore a change in velocity.

Some of these ideas are illustrated using a cyclist On the left is a 'free body force diagram' of a cyclist showing all the forces acting on the body. If the cyclist is moving with a constant velocity then we are dealing with uniform or constant speed and no change in direction. There is no resultant force i.e. F1 = F3 and F2 = F4, so the cyclist continues in the same direction at the same speed.

Note the relative size and direction of the arrows and think of Newton's First Law of Motion.

F1 is the air resistance due to friction between the surface of the bike and cyclist and the air, also friction between the wheels and road, and, friction in moving parts of the bike. All three combined oppose the forward motion of the bike and rider.

F2 is the weight of the bike + cyclist combination due to gravity, weight of object acting on the road with the normal contact force

F3 is the thrust or push of the bike from the power generated by the cyclist.

F4 is the normal contact force of the atoms of the road surface pushing back up on the bike. If the cyclist applies more power (left free body diagram), forces F1 and F2 are unbalanced, giving a resultant force of greater than zero from right to left.

Therefore the cyclist will accelerate and increase in speed. This action does not affect forces F2 and F4 which remain balanced.  The resultant force causes the acceleration If the cyclist applies the brakes (left free body diagram), in doing so he will stop pedalling, reducing force F3 AND the increased friction from the brake pads acting on the wheel rim will increase force F1 (left diagram). Again, the arrows for forces F1 and F3 should be shown as unequal. The forces are now unbalanced and the bike and rider slow down (deceleration). Neither of these two actions affects forces F2 and F4 which remain balanced. The resultant force causes the deceleration If another force is introduced like a sudden gust of a side wind (F5), then the cyclist and bike will change direction in the direction of the wind i.e. forced to the right since the diagram indicates a gust of wind from the left. The cyclist will then apply a correcting 'balancing' force by turning the handle bars to adjust for the extra force F5 to try to maintain the same speed and direction of the cycle.

You can analyse the descent of a parachutist

This is an interesting case because it involves an acceleration, a deceleration and two terminal velocities!

1. 2. 3. The three possible 'force' situations as the parachuting person is descending after jumping out of an aeroplane.

Note the relative size and direction of the arrows and think of Newton's First Law of Motion.

1. When drag force F1 is less than the weight force F2, the parachutist is accelerating.

2. When drag force F1 equals the weight force F2, the parachutist is descending at a steady speed - a terminal velocity.

3. When drag force F1 is more than the weight force F2, the parachutist will decelerate (to another terminal velocity).

This situation is fully analysed on Acceleration, friction, drag effects and terminal velocity experiments

Any stationary object standing on a surface

The weight of an object, due to gravity, acts on the surface - normal contact force.

The atoms of the surface are compressed and push back up with an equal normal contact force.

The resultant force is zero.

Therefore from Newton's First Law of Motion, the object should stay stationary.

If there was any difference in the two forces, the object would accelerate and rise or fall.

Some examples involving circular motion

(i) Suppose you whirl an object around tied to the end of a string.

If you supply kinetic energy at a constant rate the object will whirl around with a constant speed in the same repeating circular path.

The object is held in a constant orbit by the tension in the string - the centripetal force, acting towards the centre of rotation, and balancing out the acceleration.

The forces are balanced, a zero resultant force, but what if the string breaks!?

The tension in the string has gone, so there is no balancing centripetal force, but the object is still moving and shoots off at a constant velocity in a linear direction at a tangent to the original orbit.

(ii) The Earth orbiting the Sun

A thought experiment! Suppose you could switch off gravity, what would happen to the Earth.

(OR imagine if the Sun suddenly disappeared, its the same effect, the Earth is NOT in a gravitational field.)

The centripetal gravitational force has gone, so I'm afraid the Earth would fly off at a tangent to its original planetary orbit and fly off in a straight line at a constant velocity.

Unless it hit or was hit, by another object, it would continue in a straight line at constant speed for ever!

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(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/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?

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.

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/s2

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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 left-hand 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 = v2 - v1 / 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 left-hand 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/s2)

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 wheel-road 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 non-contact force of gravity.

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 head-on collision of two objects Suppose two balls m1 and m2, of masses m1 and m2 collide head-on with a certain normal contact force.

Assume mass m1 is smaller than mass m2.

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,

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 head-on 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, mag for the 1000 kg green car = mab 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 agreen car = 500 x ablue 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

(iv) A cat leaping up from the ground to catch a butterfly!

The 'force' sequences (neglecting air resistance)

1. At the start a pair of equal non-contact gravitational forces operate - the Earth attracts the cat and the cat attracts the Earth.

2. 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.

3. 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.

4. 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.

5. 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.

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

7. The normal contact force of the cat's impact on the ground is equalled by the atoms of the ground pressing back up.

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