8 Newton's laws of motion 8.1 Force and acceleration

learning objectives: -+ Describe what effect a

resultant force produces.

-+ Describe what would happen

to a body that was already

in motion ifthere was no

resultant force acting on it.

-+ Explain how weight is

different from mass.

Specification reference:3.4.1. 5

.A Figure 1 Overcomingfriction

•

Motion without force Motorists on icy roads in winter need to be ve ry careful, because the tyres of a car have little or no grip on the ice. Moving from a standstill on ice is very difficult. Stopping on ice is almost impossible, as a car moving on ice will slide when the brakes are applied. Friction is a hidden force that we don't usually think about until it is absent!

rf you have ever pushed a heavy crate across a rough concrete floor. you will know all about friction. The push force is opposed by friction, and as soon as you stop pushing, friction srops the crate moving. lf the crate had been pushed onto a patch of ice, it would have moved across the ice without any furthe r push needed.

Figure 2 shows an air track wh ich allows motion to be observed in the absence of friction. The glide r on the air track floats on a cushion of air. Provided the track is level, the glider moves at constant velocity along the track because friction is absent.

,/o o o o o o o o o o o!..[\,

.A Figure 2 The linear air track

Newton's first law of motion Objects either stay at rest or moves with constant velocity

unless acted on by a force.

Sir Isaac Newton was the first person to realise that a moving object remains in uniform motion unless acted on by a force. He recognised that when an object is acted on by a resultant force, the result is to change the object's velocity. In other words, an object moving at constant velocity is either

• acted on by no forces, or

• the forces acting on it arc ba lanced (so the resultant force is zero).

Investigating force and motion 9 How does the velocity of an object change if it is acted on by a constant force? Figure 3 shows how this can be investigated, using a dynamics trolley and a motion sensor connected to a computer. The computer is used to process a signal from the motion sensor and to display a graph showing how the velocity of the trolley changes with time.

The trolley is pulled along a sloping runway using one or more elastic bands stretched to the same length. The runway is sloped just enough to compensate for friction. To test for the correct slope, the trolley should move down the runway at constant speed after being given a brief push .

I

motion sensor

card fixed to trolley

.A Figure 3 Investigating force and motion

As a resu lt of pulling the trolley with a constant force, the vclocity­time graph should show that the velocity increases at a constant rate. The acceleration of the trolley is therefore constant and can be measured from the velocity-time graph. Table 1 below shows typical measurements using different amounts of force (one, two, or three elastic bands in parallel stretched to the same length each time) and difrerent amounts of mass (i.e., a single, double, or triple trolley).

TTable 1

2 1 2 31 3

1 21 1 2 2 12 24 36 6 12 18 12 24 36 12 24 36

The results in the table show that the force is proportional to the mass x the acceleration. In other words, if a resultant force Facts on an object of mass 111, the object undergoes acceleration a such that

F is proportional to ma

By defining the unit of force, the newton, as the amount of force that will give an object of mass I kg an acceleration of l ms-2, the above proponionality sta tement can be expressed as an equation

F=ma

where F =resultant force (in N), m = mass (in kg), a = acceleration (in ms-2).

This equa tion is known as Newton's second law for constant mass.

~ Worked example ~

A vehicle of mass 600 kg accelerates uniformly from rest to a velocity of 8.0ms­ 1 in 20s. Calculate rhe force needed to produce this acceleration.

Solution Acceleration a=~= 8·0 - O= 0.4ms-2

t 20 Force F= ma= 600 x 0.4 = 240N

Hint

The acceleration is always in the same direction as the resultant force. For example, a freely moving

projectile in motion experiences a force vertically downwards due to gravity. Its acceleration is therefore vertically downwards, no matter what its direction of motion is.

Study tip

The mass m must be in kg and a in m s· 2 when calculating a force in N.

Maths link Iii J

Why does a heavy object fall at

the same rate as a lighter object?

As outlined in Topic 7.4, Free fall, Galileo proved that objects fall at the same rate, regardless of their weight. He also reasoned that if two objects joined by a string were released and they initially fell at different rates, the faster one would be slowed down by the slower one, which would be speeded up b\J the faster one until they fell at the same rate. Many years later, Newton explained their identical fal ling motion because, for an object of mass m in free fall,

acceleration = force of gravity mg massm

=gwhich is independent of m

Synoptic link

You learnt about Galileo in Topic 7.4, Free fall.

•

8.1 Force and acceleration

spnng

weight of parcel =5.3 N

this part is thecylinder inside the balance tube

lhat slides out when weight is added

parcel

A Figure 4 Using a newtanmeter ta weigh an abject

A Figure 5 An inertia trick

•

Weight • The acceleration of a falling object acted on by gravity only is g.

Because the force of gravity on the object is the only force acting on it, its weight (in newtons), W = mg, where m = the mass of the object (in kg).

• When an object is in equilbrium, the su pport force on it is equal and opposite to its weight. Therefore, an object placed on a weighing balance (e.g .. a newtonmeter or a top pan balance) exerts a force on the balance equal to the weight of the object. Thus the balance measures the weight of the object. See Figure 4.

• g is also referred to as the gravitational field strength at a given position, as it is the force of gravity per unit mass on a small object al that position. So the gravitational field strength at the Earth's surface is 9.81 N kg- 1• Note that the weight. of a fixed mass depends on its location. For example, the weight ol' a l kg object is 9.81 Non the Earth's surface and 1.62 Non the Moon's surface.

• The mass of an object is a measure ol' its inertia. which is its resistance to change of motion. More force is needed to give an object a certain acceleration than to give an object with Jess mass the same acceleration. Figure 5 shows an entertaining demonstration of inertia. When the card is flicked, the coin drops into the glass because the force or fri ction on it due to the moving card is too small to shift it sideways.

• The scale of a top pan balance is usually calibrated for convenience in grams or kilograms.

Summary questions

g = 9.81 N kg- 1

1 G A car of mass 800 kg accelerates uniformly along a straight line

from rest to a speed of 12 m s- 1 in 50 s. Calculate:

a the acceleration of the car

b the force on the car that produced this acceleration

c the ratio of the accelerating force to the weight of the car.

2 G An aeroplane of mass 5000 kg lands on a runway at a speed of

60 m s-1 and stops 2S slater. Calculate:

a the deceleration of the aeroplane

b the braking force on the aircraft.

3G a A vehicle ofmass 1200 kg on a level road accelerates from rest to a

speed of 6.0 m s-1 in 20 s, without change of direction. Calculate the

force that accelerated the car.

b The vehicle in a is fitted with a trailer of mass 200 kg. Calculate the time

taken to reach a speed of6.0 m s-1 from rest for the same force as in a.

4 G A bullet of mass 0.002 kg travelling at a speed of 120 ms 1 hit a tree

and penetrated a distance of SS mm into the tree. Calculate:

a the deceleration of the bullet

b the impact force on the bullet.

Two forces in opposite directions When an object is acted on by two unequal forces acting in opposite directions, the object accelerates in the direction of the larger force. If the forces are F1 and F2, where F1 > F2

resultant force, F 1 - F2 =ma,

where mis the mass of the object and a is its acceleration, which is in the same direction as F1•

rf the object is on a horizontal surface and F1and F are ho1izontal and2 in opposite directions, the above equation still applies. The support force on the object is equal and opposite to its weight.

• pull of tractor

mud

A Figure 1 Unbalancedforces

Some examples are given below where two forces act in different directions on an object.

Towing a trailer Consider the example of a car of mass M fined with a trailer of mass m on a level road. When the car and the trailer accelerate, the car pulls the trailer forward and the trailer holds the car back. Assume air resistance is negligible.

~ engine thrust tension in tow bar

A Figure 2 Car ond trailer

• The car is subjected co a driving force F pushing it forwards (from its engine thrust) and the tension Tin the row bar holding it back.

Therefore the resultant force on the car= F - T =Ma

• The force on the trailer is due to the tension Tin the tow bar pulling it forwards.

Therefore T = ma.

Combining the two equations gives F= Ma+ ma= (M + m)a

Learning objectives: -+ Apply F=ma when the forces

on an object are in opposite directions.

-+ Explain why you experience less support as an ascending lift stops.

-+ Describe any situations in which F=ma cannot be applied.

Specification reference: 3.4.1.5

Study tip

Remember that it is the force of friction that pushes an object. If friction was negligible the object would not be able to move.

•

8.2 Using F =ma

engine thrust, T

acceleration, a

rocket of massm

weight, mg

.A. Figure 3 Rocket lounch

cable

tension T

r acceleration, a

total weight of lift and

occupants mg

.A. Figure 4 /no lift

Hint

Downward acceleration = upward deceleration

Study tip

Identify the separate forces acting, then work out the resultant force - show these steps by clear working [usually starting in algebra, such as T- mg =ma].

•

Rocket problems HT is the thrust of the rocket engine when its mass is rn and the rocket is moving upwards, its acceleration a is given by T- mg= ma.

Rocket thrust T =mg+ ma

The rocker thrust must therefore overco me the weight of the rocker for the rocker to take off (Figure 3).

Liftproblems Using 'upwards is positive' gives the resultant force on the lift as T - mg, where Tis the tension in the lift cable and mis the toral mass of the lift and occupants (Figure 4).

T- mg= ma, where a = acceleration.

a If the lift is moving at constant velocity, then a = 0 so T = mg (tension= weight).

b If the lift is moving up and accelerating, rhen a> 0 soT = mg + ma>mg.

c If the lift is moving up and decelerating, then a < 0 so T = mg + ma< mg.

d If the lift is moving down and accelera ting, then a< 0 (velocity and acceleration are both downwards, rheretore negative) so T = mg + ma< mg.

e If the lift is moving down and decelerating, then a > 0 (velocity downwards and acceleration upwards, therefore positive) soT=mg+ma>mg.

The t ension in the cable is less than the weight if

• the lift is moving up and decelerating (velocity> 0 and acceleration< 0 )

• the lift is moving down and accelerating (velocity < 0 and acceleration< 0).

The tension in the cable is greater than t h e weight if

• rhe lift is moving up and accelerating (velocity> 0 and acceleration > 0)

• the lift is moving down and decelerating (velocity < O and acceleration> 0).

~ Worked example ~

g =9.Sm s-2

A lifl of total mass 650kg moving downwards cabm• l direction

decelerates at 1.5ms-2 and stops. Calcu late the of motion

tension in the lift cable during the deceleration. hit

Solution .A. Figure S

The lilt is moving down so its velocity v < 0. Since it decelerates, its acceleration a is in the opposite direction 10 its velocity, so a> 0.

Therefore, inserting a=+ I. 5 ms 2 in the equation T- mg= ma gives

T = mg+ ma = (650 x 9.8) + (650 x 1.5) =7300N

Further F =ma problems Pulley problems • Consider two masses Mand m (where M > m) attached to a thread hung over a frictionless pulley, as in Figure 6. When released. mass M accelerates downwards and mass m accelerates upwards. If a is the T acceleration and Ti s the tension in the thread, then acceleration

of M • the resultant force on mass M, Mg - T =Ma f • the resultant force on mass m, T - mg = ma.

acceleralton Therefore, adding the two equations gives of m mg

Mg-mg=(M+m)a

Sliding down a slope Mg

Consider a block or mass m sliding down a slope, as in Figure 7. ..& Figure 6 Pulley problems

The component of th e block's weight down the slope = mgsin 9.

If th e fo rce of fri ction on the block = F0• then the resultan t force on friction, F0the block = mg sin 9 - F0.

Therefore mg sin 9 - = ma, where a is the acceleration of the block. F0

Note: Wi th the addition or an engine force FE' the above equation can be applied to a vehicle on a downhill slope of constant gradient. Thus the acceleration is given by FE+ mgsin9- F0 = ma, where F0 is the combined sum or the air resistance and the braking force.

mg horizontal

..& Figure 7 Sliding down a slope

Summary questions

g = 9.Bms-2 3 f)Alift and its occupants have a total mass of 1200 kg. Calculate the tension in the lift cable when1 f) A rocket of mass 550 kg blasts vertically from the the lift is:launch pad at an acceleration of 4.2 m s-2• Calculate: a stationary a the weight of the rocket b ascending at constant speedb the thrust of the rocket engines. c ascending at a constant acceleration of 0.4 m s-2

2 f) Acar of mass 1400 kg, pulling a trailer of mass 400 kg, accelerates from rest to a speed of d descending at a constant deceleration of 0.4 m s-2.

9.0 m s 1 in a time of 60 son a level road. Assuming 4 f) A brick of mass 3.2 kg on a sloping Oat roof, at air resistance is negligible, calculate: 30° to the horizontal, slides at constant acceleration a the tension in the tow bar 2.0 m down the roof in 2.0 s from rest. Calculate:

b the engine force. a the acceleration of the brick

b the frictional force on the brick due to the roof .

•

Learning objectives: ' Explain why the speed

reaches a maximum when a driving force is still acting.

' Explain what we mean by a drag force.

' Explain what determines the terminal speed of a falling object or a vehicle.

Specification reference: 3.4.1.5

ball falling dragat constant speed Tl direction

of motion

1 ~

a Falling in a jlwd

b Skydiving

• Figure 1 At terminal speed

Study tip The magnitude of the acceleration at any instant is the gradient of the speed- time curve.

•

Any object moving through a fluid experiences a force that drags on it due to the fluid. The drag force depends on

• the shape of the object

• its speed

• the viscosity of the fluid, which is a measure of how easily the fluid flows past a surface.

The faster an object travels in a fluid, the greater the drag force on it.

Motion of an object falling in a fluid 8 The speed of an object released from rest in a fluid increases as it falls, so the drag force on it due to the fluid increases. The resultant force on the object is the difference between the force of gravity on it (its weight) and the drag force. As the drag force increases, the resultant force decreases, so the acceleration becomes less as it falls. If it continues falling, it attains terminal speed, when the drag force on it is equal and opposite to its weight. Jts acceleration is then zero and its speed remains constant as it falls (Figure I).

speed- pulley

thread a Apparatus

] toac ~ l supp~nu~

tape ~ tape chart

b Making a tape chart

5 6 7

time

• Figure 2 Investigating the motion ofan objectfalling in a fluid

Figure 2a shows how to investigate the motion of an object falling in a fluid. When the object is released. the thread auached to the object pulls a tape through a tickenimer, which prints dots a1 a constant rate on the tape. The spacing between successive dots is a measure of the speed of the object as the dots are printed on the tape a t a constant ra te. A tape chart can be made from the rape to show how tl1e speed changes with time. This is done by cutting the tape at each dot and pasting successive lengths side by side on graph paper, as shown in Figure 2b. The results show that:

• the speed increases and reaches a constant value, which is the terminal speed

• the acceleration at any instant is the gradient or the speed- time curve.

At any instanc, the resultant force F = 1119 - D, where tn is the mass of the object and Dis the drag force.

Therefore, the acceleration of the object =mg - D =g - D 111 m

• The initial acceleration =g because the speed is zero, and therefore the drag force is zero, at the instant the object is released.

• At the terminal speed, the potential energy of the object is transferred, as it falls, into internal energy of the Ouid by the drag force .

Motion of a powered vehicle The top speed of a road vehicle or an aircraft depends on its engine power and its shape. A vehicle with a streamlined shape can reach a higher top speed rhan a vehicle wirh the same engine power rhar is not streamlined.

For a powered vehicle of mass m moving on a level surface, if FE represents rbe motive force (driving force) provided by the engine, the resultant force on ir = FE - FR' where FR is the resist ive force opposing rhe morion of rhe vehicle. (FR = rhe sum of the drag forces acting on rbe vehicle.)

F - FTherefore its accelera tion a= e R

111

Because the drag force increases with speed, the maximum speed (the terminal speed) of the vehicle vmax is reached when the resistive force becomes equal and opposite to the engine force, and a = 0.

{.q Worked example ~

A carol' mass 1200 kg has an engine which provides an engine force of 600 N. Ca lculate its acceleration when the resistive force is 400N.

Solution When the resistive force= 400N, the resultant force =engine force - drag force= 600 - 400 = 200N.

200Acceleration = force = = 0.17 ms-2

mass 1200

Extension

Hydrofoil physics

Some hydrofoil boats travel much faster than an ordinary boat because they have powerful jet engines that enable them to ski on its hydrofoils when the jet engines are switched on.

When the jet engine is switched on and takes over from the less powerful propeller engine, the boat speeds up and the hydrofoils are extended. The boat rides on the hydrofoils, so the drag force is reduced, as its hull is no longer in the water. At top speed, the motive force of the jet engine is equal to the drag force on the hydrofoils.

When the jet engine is switched off, the drag force on the hydrofoils reduces the speed of the boat and the hydrofoils are retracted. The speed drops further until the drag force is equal to the motive force of the propeller engine. The boat then moves at a constant speed, which is much less than its top speed with the jet engine on.

0: Describe how the force of the jet engine compares to the drag force when the boat speeds up.

friction acceleration, a and dragforces engine forces

FR +-~~

.A Figure 3 Vehicle power

Summary questions

g = 9.Sms-2

1 a G A steel ball of mass 0.15 kg, released from rest in a liquid, fa lls a distance of 0.20 m in 5.0 s. Assuming the ball reaches terminal speed within a fraction of a second, calculate I its terminal

speed, ii the drag force on it when it falls at terminal

speed.

b State and explain whether or not a smaller steel ball would fall at the same rate in the same liquid.

2 Explain why a cyclist can reach a higher top speed by crouching over the handlebars instead of sitting upright while pedalling.

3 fj A vehicle of mass 32000kghasanen~ne

which has a maximum engine force of 4.4 kN and a top speed

1of 36 ms on a level road.

Calculate a its maximum acceleration from rest, b the distance it would travel at maximum acceleration to reach a speed of 12 m s-1 from rest.

4 Explain why a vehicle has a higher top speed on a downhill stretch of road than on a level road .

•

Learning objectives: Stopping distances ' Describe what is happening Traffic accidents often happen because vehicles are being driven coo

to a vehicle's speed while a fasc and coo close. A driver needs to maintain a safe dis1ance becween driver is reacting to a hazard his or her own vehicle and the vehicle in front. If a vehicle suddenly ahead. brakes, the driver of the following vehicle needs to brake as well co

avoid a crash. ' Distinguish between braking distance and stopping Thinking distance is the distance travelled by a vehicle in che time distance. ii cakes the driver to react. For a vehicle moving at constant speed v.

' Describe how road conditions the thinking distance s1 = speed x reaction time= vt0, where t0 is che affect these distances. reaccion time of the driver. The reaction time of a driver is affected by

distraccions, drugs, and alcohol, which is why drivers in the UK areSpecification reference: 3.4.1. 5 banned from using mobile phones while driving and breathalyser tests are carried out on drivers.

Braking distance is the distance travelled by a car in the time it takes to stop safely, from when the brakes are first applied . Assuming constant deceleration, a, to zero speed from speed u,

2 2the braking distance s2 =!:!_since u = 2as2.

2a 2 Stopping distance = thinking distance+ braking distance= u t0 + !:!_

where u is the speed before the brakes were applied. ia

Figure 1 shows how thinking distance, braking distance, and st0pping distance vary with speed for a reaction time of 0.67s and a deceleration of 6.75 m s-2. Using these values for reaction time and deceleration in the above equations, Figure 1 gives the ~honest stopping distances on a dry road as recommended in the Highway Code.

"iltna'ilif?>...30 mph I aott 1 15 r1 (22JM:

50 mph l....__ SO_ ft_ __, 175 It (5:? 51 )

7o mph I 101t 315 It

(gt! 'J m)

thinking distance braking distance stopping distance

.&. Figure 1 Stopping distonces

Study tip

Acar can only move forwards by pushing backwards on the road. The same applies when you take a forward step. When you push backwards on the ground, the ground exerts an equal and opposite force on you that enables you to move forward .

•

Skidding

On a front-wheel drive vehicle, the engine turns the front wheels via the transmission system. Friction between the tyres and the road prevent wheel spin (slipping) so the driving wheels therefore roll along the road. If the driver tries to accelerate too fast, the wheels skid. This is because there is an upper limit to the amount of friction between the tyres and the road (Figure 2).

When the brakes are applied, the wheels are slowed down by the brakes. The vehicle therefore slows down, provided the wheels do not skid. If the

.6. Figure 3 Testing friction .6. Figure 4 Tyre treads

The braking distance for a vehicle depends on the speed of the vehicle at the instant the brakes are applied, on the road conditions, and on the condition of the vehicle tyres.

On a greasy or icy road, skidding is more likely because the limiting frictional force between the road and the tyres is reduced from its dry value. To stop a vehicle safely on a greasy or icy road, the brakes must be applied with less force than on a dry road otherwise skidding will occur. Therefore the braking distance is longer than on a dry road. In fast-moving traffic, a driver must ensure there is a bigger gap to the car in front so as to be able to slow down safely ifthe vehicle in front slows down.

The condition of the tyres of a vehicle also affects braking distance. The tread of a tyre must not be less

than a certain depth, otherwise any grease, oi l, or water on the road reduces the friction between the road and the tyre very considerably. In the United Kingdom, the absolute minimum legal requirement is that every tyre of a vehicle must have a tread depth of at least 1.6 mm across three-quarters of the width of the tyre all the way round. If the pressure on a tyre is too small or too great, or the wheel is unbalanced, then the tyre will wear unevenly and will quickly become unsafe. In addition, the braking force may be reduced if the tyre pressure is too high as the tyre area in contact with the road will be reduced. Clearly, a driver must check the condition of the tyres regularly. If you are a driver or hope to become one, remember that the vehicle tyres are the only contacts between the vehicle and the road and must therefore be looked after.

•

braking force is increased, the friction force between the tyres and the road increases. However, if the upper limit of friction (usually referred to as the limiting frictional force) between the tyres and the road is reached, the wheels skid. When this happens, the brakes lock and the vehicle slides uncontrollably forward. Most vehicles are now fitted with an anti-lock brake system (ABS), which consists of a speed sensor on each wheel linked to a central electronic control unit that controls hydraulic valves in the brake system. When the brakes are applied, if a wheel starts to lock, the control unit senses that the wheel is rotating much more slowly than the other wheels - as a result, the valves are activated, so the brake pressure on the wheel is reduced to stop it locking.

0: State the direction of the frictional force of the road on the driving wheels when a car accelerates.

Practical: Testing friction 8 Measure the li miting friction between the underside of a block and the surface it is on. Pull the block with an increasing force until it slides. The limiting frictional fo rce on the block is equal to the pull force on the block just be fo re sliding occurs. Find out how this force depends on the weight of the block.

pull

box lnct1on on box

I~

\~

friction due to braking

friction caused by motive force

.6. Figure 2 Stopping and starting

8.4 On the road

•

~ Worked example ~

g = 9.Sms-2

A vehicle of mass 900kg, travelling on a level road at a speed of I 5 m s- 1, ca n be brought to a standstill without skidding by a braking force equal to 0.5 x its weight. Calculate a the decele ration of the vehicle, b the braking distance.

Solution a Weight = 900 x 9.8 = 8800N

Braking force = 0.5 x 8800 = 4400 N

Deceleration = braking force = 4400 = 4 . 9 ms-2 mass 900

b u = 15ms-1, v = 0, a= -4.9 ms-2

To calculates, use v2 = u2 + 2as

- u2 -152 s=-= = 23 m

2a 2x - 4.9

Summary questions

g =9.8 ms-2

1 fiA vehicle is travelling at a speed of 18 ms 1 on a level road, when the

driver sees a pedestrian stepping off the pavement into the road 45 m

ahead. The driver reacts within 0.4 sand applies the brakes, causing 2the car to decelerate at 4.8 ms .

a Calculate i the thinking distance, ii the braking distance.

b How far does the driver stop from where the pedestrian stepped

into the road?

2 G The braking distance of a vehicle for a speed of 18 m s- 1 on a dry

level road is 24 m. Calculate:

a the deceleration of the vehicle from this speed to a standstill over

this distance

b the frictional force on a vehicle of mass 1000 kg on this road

as it stops.

3 a What is meant by the braking distance of a vehicle?

b Explain, in terms of the forces acting on the wheels of a car, why a

vehicle slows down when the brakes are applied.

4 fi The frictional force on a vehicle travelling on a certain type of level

road surface is 0.6 x the vehicle's weight. For a vehicle of mass 1200 kg, 2a show that the maximum deceleration on this road is 5.9 m s-

b calculate the braking distance on this road for a speed of 30 m s- 1 .

Impact forces

A Figure 1 Head-on collision

Measuring impacts The effect or a collision on a vehicle can be measured in terms or the acceleration or deceleration of the vehicle. By expressing an acceleration or deceleration in terms of g, the acceleration due to gravity, the force of the impact can then easily be related to the weight of the vehicle. For example, suppose a vehicle hits a wall and its acceleration is -30ms 2• In terms ofg, the acceleration= -3g. So the impact force of the wall on the vehicle must have been three times its weight (= 3 mg, where mis the mass of the vehicle). Such an impact is sometimes described as being equal to 3g. This statement, although technically wrong because the deceleration not the impact force is equal to 3g, is a convenient way of expressing the effect of an impact on a vehicle or a person.

Learning objectives: -+ Describe the force on a

moving body if it is suddenly

stopped, for example in a

road accident.

-+ Explain what should be

increased to give a smaller

deceleration from a given

speed.

-+ State which design features

attempt to achieve this in a

modern vehicle.

Specification reference: 3.4.1.5

Study tip

Remember that a negative

acceleration is equal to a positive

deceleration.

How much acceleration or deceleration can a person withstand?

The duration of the impact as well as the magnitude of the acceleration

affect the person. A person who is sitting or upright can survive a

deceleration of 20 g for a time of a few milliseconds, although not without

severe injury. A deceleration of over S g lasting for a few seconds can

cause injuries. Car designers carry out tests using dummies in remote·

control vehicles to measure the change of motion of different parts of a

vehicle or a dummy. Sensors linked to data recorders and computers are

used, as well as video cameras that record the motion to allow video clips

to be analysed.

• A Figure 2 Aside-on impact

--

8.5 Vehicle safety

Study tip When writing answers and when you are driving (especially just after passing your driving test]. don't rush - it's safer to take longer over what you are doing!

Study tip The work done by the impact force Fover an impact distances (=Fs) is equal to the change of kinetic energy of the vehicle. The impact force can also be worked out using the equation

F=change of kinetic energy impact distance

Synoptic link You will learn in Topic 10.2, Kinetic energy and potential energy, that the kinetic energy of a vehicle of

2mass m moving at speed v =i mv .

Prove for yourself that the vehicle in the example above has 200 kJ of kinetic energy at 20 m s- 1 and 72 kJ

1at 1.2 m s- . You can then show the change of kinetic energy.;. impact distance gives -32 000 Nfor the impact force .

•

Contact time and impact time When objects collide and bounce off each other, they arc in contact with each other for a certain time, which is the same for both objects. The shorter the contact time, the greater the impact force for the same initial velocities of the two objects. When two vehicles collide, they may or may not separate from each other after the collision. If they remain tangled together, they exert forces on each other until they are moving at the same velocity. The duration of the impact force is not the same as the comact time in this situation.

The impact time t (the duration of the impact force) can be worked out by applying the equations= I (u + v) t to one of the vehicles, wheres is the distance moved by that vehicle during the impact, u is its initial velocity, and vis its final velocity. If the vehicle mass is known, the impact force can also be calculated.

For a vehicle of mass m in lime t,

the impact time t =2L u+v

the acceleration a =v - u t

the impact force F =ma

~ Worked example ~

A I 000 kg vehicle moving at 20 ms 1 slows down in a distance of 4.0 m t0 a velocity of 12 m s- 1, as a result of hitting a stationary vehicle. Calculate

a the vehicle's acceleration in terms of g

b the impact force for this collision.

Solution a t = 2s = 2 x 4.0 = 0.25s

(u+v) (20+12)

The acceleration, a = ~ I

12 - 20 0.25

= -32ms- 2

= -3.39 (where g = 9 .8ms-2 ).

b The impact force F = ma = l 000 x -32 = -32 000 N.

Extension

Car safety features

front bumper rear

bumper

.A. Figure 3 Vehicle safety features

In the example on the previous page, the vehicle's deceleration and hence the impact force would be lessened if the impact time was greater. So design features that would increase the impact time reduce the impact force. With a reduced impact force, the vehicle occupants are less affected. The following vehicle safety features are designed to increase the impact time and so reduce the impact force.

• Vehicle bumpers give way a little in a low-speed impact and so increase the impact time. The impact force is therefore reduced as a result. If the initial speed of impact is too high. the bumper and/ or the vehicle chassis are likely to be damaged.

• Crumple zones The engine compartment of a car is designed to give way in a front-end impact. If the engine compartment were rigid, the impact time would be very short, so the impact force would be very large. By designing the engine compartment so it crumples in a front-end impact, the impact time is increased and the impact force is therefore reduced.

• Seat belts In a front-end impact, a correctly fitted seat belt restrains the wearer from crashing into the vehicle frame after the vehicle suddenly stops. The restraining force on the wearer is therefore much less than the impact force would be if the wearer hit the vehicle frame. With the seat belt on, the wearer is stopped more gradually than without it.

• Collapsible steering wheel In a front-end impact, the seat belt restrains the driver without holding the driver rigidly. If the driver makes contact with the steering wheel, the impact force is lessened as a result of the steering wheel collapsing in the impact.

• Airbags An airbag reduces the force on a person, because the airbag acts as a cushion and increases the impact time on the person. More significantly, the force of the impact is spread over the contact area, which is greater than the contact area with a seat belt. So the pressure on the body is less.

0: Explain why an air bag cushions an impact.

Summary questions

g =9.Sms-2

1 fi A car of mass 1200 kg travelling at a speed of 15 m s- 1 is struck from behind

by another vehicle, causing its speed to increase to 19 m s-1

in a time of0.20 s. Calculate:

a the acceleration of the car, in terms of g

b the impact force on the car.

2 fi The front end of a certain type of car of mass 1500 kg travelling at a speed of 20 ms-1

is designed to crumple in a distance of 0.8 m, if the car hits a wall. Calculate:

a the impact time if it hits a wall at 20 m s- 1

b the impact force.

3 fi The front bumper of a car of mass 900 kg is capable of withstanding an impact with a stationary object, provided the car is not moving faster than 3.0 ms-1• when the impact

occurs. The impact time at this speed is 0.40s. Calculate:

a the deceleration of the car from 3.0 m s- 1 to rest in

0.40s

b the impact force on the car.

4 fi In a crash, a vehicle travelling at a speed of 25 ms 1 stops after 4.5 m. A passenger of mass 68 kg is wearing a seat belt, which restrains her forward movement relative to the car to a distance of 0.5 m. Calculate a the deceleration of the passenger in terms of g, b the resultant force on the passenger .

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