Topic 2 Mechanics

2.1 Kinematics

2.1.1 Define displacement, velocity, speed and acceleration.

Displacement- Defined as the change in position of an object. Displacement is a

quantity that has both direction and magnitude, it is a vector.

Measured in metres.

Speed- How far an object travels in a given time.

Rate of change of distance (ms-1)

Velocity- The vector version of speed. Tells us the magnitude of how fast an

object is moving and the direction in which it is moving. (ms-1)

dTimeElapse

ntDisplacemeocityAverageVel =

Acceleration- Rate of change of velocity (ms-2)

dTimeElapse

locityChangeInVeelerationAverageAcc =

2.1.2 Explain the difference between instantaneous and average

values of speed, velocity and acceleration. Instantaneous values of velocity speed and acceleration are taken at a particular moment. In

any realistic situation they will be constantly varying and is not always meaningful or useful

to know this e.g. drivers need to keep their instantaneous speed below the speed limit.

Average velocity, speed and acceleration are taken over a certain period of time or distance.

E.g. Average speed between Porto and Lisbon = distance/time = 350/km/3 hours 30 min =

100 kmh-1

2.1.3 Outline the conditions under which the equations for uniformly

accelerated motion may be applied. The following equations of motion may be applied to an object that has a constant

acceleration.

t

uva

)( =

2

21 atuts +=

asuv 222 +=

v= final velocity/ ms-1)

u= initial velocity / ms-1

a= acceleration/ ms-2

s= displacement/m

t=time/s

2.1.4 Identify the acceleration of a body falling in a vacuum near the

Earths surface with the acceleration g of free fall.

Free fall is the uniform acceleration (ignoring the effect of air resistance) in the vertical

direction of an object in a uniform gravitational field. All falling objects have the same

acceleration independent of their masses. On earth all objects accelerate towards the

ground at 9.81 ms-2.

Acceleration due to gravity can be measured in a number of

ways. In the arrangement that is shown a timer starts when

the ball is released form an electromagnet and stops when

the ball passes through a gate at the bottom. The acceleration due to gravity can then be calculated using:

2

2

t

sg =

Acceleration due to gravity can also be calculated using light gates of ticker timers.

2.1.5 Solve problems involving the equations of uniformly accelerated motion.

This question is about throwing a stone from a cliff.

Antonia stands at the edge of a vertical cliff and throws a stone vertically

upwards.

Sea

v = 8.0ms 1

The stone leaves Antonias hand with a speed v = 8.0ms1

.

The acceleration of free fall g is 10 m s2

and all distance measurements are

taken from the point where the stone leaves Antonias hand.

(a) Ignoring air resistance calculate

(i) the maximum height reached by the stone.

asuv 222 +=

s

vs

2

2

=

102

82

=s

ms 2.3=

(2)

(ii) the time taken by the stone to reach its maximum height.

t

uva

=

a

uvt

=

10

8=t

st 8.0=

(1)

The time between the stone leaving Antonias hand and hitting the sea is 3.0 s.

(b) Determine the height of the cliff.

Time to go form top of cliff to sea = 3.0 (2 x 0.8) = 1.4s

2

21 atuts +=

)4.110()4.10.8( 221

+=s

(3)

(Total 6 marks)

2.1.6 Describe the effects of air resistance on falling objects.

Air resistance acts upon all moving objects. As the velocity of an object increases the size of

the air resistance also increases.

Eventually the force of air resistance will equal the force of gravity. When this happens the

object will stop accelerating. It has reached its terminal velocity.

ms 21=

The terminal velocity of an object is dependant on its shape. A feather has a lower terminal

velocity than a hammer.

2.1.7 Draw and analyse distancetime graphs, displacementtime

graphs, velocitytime graphs and accelerationtime graphs. Displacement time graphs

0

2

4

6

8

10

12

0 2 4 6 8 10 12

time /s

dispalcement / m

The gradient of a displacement - time graph is the velocity.

Object travels at a constant speed for the first five seconds

The object then

is stationary for three seconds.

The object returns to

its original location at a faster speed.

Velocity time graphs

The gradient of a velocity time graph tells the acceleration of the object. If the line is flat

the object is travelling at a constant velocity.

The area under the graph represents the distance travelled.

Acceleration time graphs

The area under an acceleration time graph represents the change in velocity.

2.1.8 Calculate and interpret the gradients of displacementtime graphs and velocitytime graphs, and the areas under velocitytime

graphs and accelerationtime graphs.

1) The graph shows the variation with time t of the velocity v of an object.

v

t

Which one of the following graphs best represents the variation with time t of the

acceleration a of the object?

0

1

2

3

4

5

6

7

8

9

10

0 5 10 15

time /s

velocity / m/s

The objects velocity is increasing

The object is slowing down.

0 0

0 0

0 0

0 0

a a

a a

t t

t t

A. B.

C. D.

2) An athlete runs round a circular track at constant speed. Which one of the following

graphs best represents the variation with time t of the magnitude d of the

displacement of the athlete from the starting position during one lap of the track?

d

d

d

d

0

0

0

0

0

0

0

0

t

t

t

t

A.

C.

B.

D.

2.1.9 Determine relative velocity in one and in two dimensions.

Frames of Reference: If two things are moving in the same straight line but are travelling at

different speeds, then we can work out their relative velocities by simple addition or

subtraction as appropriate. For example, imagine two cars travelling along a straight road at

different speeds. If one car (travelling at 30 m/s) overtakes the other car (travelling at 25

m/s), then according to the driver of the slow car, the relative velocity of the fast car is 5

m/s. In technical terms what we are doing is moving from one frame of reference into

another. The velocities of 25 m/s and 30 m/s were measured according to a stationary

observer on the side of the road. We moved from this frame of reference into the drivers

frame of reference.

2.2 Force and Dynamics

2.2.1 Calculate the weight of a body using the expression W = mg.

The term weight can mean different things to Physics. Some define it as the gravitational

force acting on a mass, other define it as the reading on a supporting scale. Weight is a

force and as the strength of gravity varies depending on where you are the weight of an

object will also vary.

rengthnalFieldStGravitatiomassWeight =

The mass of an object is a single quantity but has tow different properties Gravitational

Mass and Inertial Mass.

The inertial mass of an object determines how that object will respond to a given force. The

gravitational mass of an object tells us how much gravitational force that an object will feel

when it is near another object.

Although the two quantities are different they turn out to be equivalent. The fact that all

objects accelerate at the same rate shows this.

2.2.2 Identify the forces acting on an object and draw free-body diagrams representing the forces acting.

Free body diagrams show the size and direction all of the forces acting on an object.

A simple situation is a bag resting on a table.

Free body diagram for the shopping Free body diagram for the table

Gravity is pulling the bag downwards and

the reaction from the table is pushing the

bag up.

2.2.3 Determine the resultant force in different situations. It is unusual for only one force to be acting on an object. Usually we have to add the forces

together to calculate the resultant force.

If the forces are acting in one dimension then they can be simply added together as shown

below.

If the forces are not in the same dimension then the resultant force can be drawn onto a

scale diagram. Force F1 has been moves so that it is nose to tail with F2. The resultant

force can then be drawn onto the diagram showing its magnitude and direction.

2.2.4 State Newtons first law of motion. Every body perseveres in its state of being at rest or of moving uniformly straight forward,

except insofar as it is compelled to change its state by force impressed.

"A particle will stay at rest or continue at a constant velocity unless acted upon by an

external unbalanced net force."

2.2.5 Describe examples of Newtons first law. When a car moves at a steady speed all forces acting on it are balanced.

2.2.6 State the condition for translational equilibrium. The change of momentum of a body is proportional to the impulse impressed on the body,

and happens along the straight line on which that impulse is impressed.

2.2.7 Solve problems involving translational equilibrium.

2.2.8 State Newtons second law of motion.

2.2.9 Solve problems involving Newtons second law.

2.2.10 Define linear momentum and impulse.

Linear momentum is defined as the product of mass and velocity.

P=mv where p= linear momentum/ kgms-1 m= mass/Kg v= velocity/ ms-1

Impulse is the product of Force and time

I= Ft I= impulse/Ns F= Force/N t=time/s

It can be shown that linear momentum and impulse have the same units.

Ns = Kgms-2s

Ns = Kgms-1

So impulse and momentum are the same quantity.

In other words:

Impulse = change in linear momentum.

This is very useful for calculations.

Force/time graphs The area under a force-time graph is the impulse.

2.2.11 Determine the impulse due to a time-varying force by

interpreting a forcetime graph.

2.2.12 State the law of conservation of linear momentum.

In a closed system the fatal linear momentum is constant

e.g. Total momentum before a collision = Total momentum after a collision

or Total momentum after an explosion = 0

2.2.13 Solve problems involving momentum and impulse.

2.2.14 State Newtons third law of motion. For a force there is always an equal and opposite reaction: or the forces of two bodies on

each other are always equal and are directed in opposite directions.

When two bodies A and B interact, the force that A exerts on B is equal and opposite to the

force that exerts on A.

2.2.15 Discuss examples of Newtons third law.

If one roller-skater pushes another, they both feel a force. The forces must be equal and

opposite, but the acceleration will be different.

A book exerting a force of 10N in a table the table will also exert 10N back.

2.3 Work, energy and power

2.3.1 Outline what is meant by work.

Work is what is accomplished when a force acts on an object as the object moves through a

distance.

Work is a form of energy and is measured in joules.

2.3.2 Determine the work done by a non-constant force by

interpreting a forcedisplacement graph.

As work is the product of force and distance, it can be represented as the area under a

graph of force as a function of distance. In the below graph the shaded region represents

the work done on an object that undergoes a constant force. Thus, the work done (area of the shaded region) can be calculated by multiplying distance (base) by force (height).

This graphical method of calculating work is also useful in estimating the work that results

from a varying force. The force applied to the object in changes over time.

The work done cannot be calculated by finding the area of a simple rectangle but it can be

estimated by dividing the area into small segments, calculating the area of each segment, and adding all of the segment areas.

Work Done = Force x Distance

W=Fs

Such areas are often divided into rectangles because the area of a rectangle is easily

calculated. However, triangles, trapezoids, or any type of segments may be used. The better the segments fit the area, the more precise the estimate.

2.3.3 Solve problems involving the work done by a force.

The diagram below shows the variation with displacement x of the force F acting on an

object in the direction of the displacement.

F

xxx

P

Q

R

S

TVW

2100

Which area represents the work done by the force when the displacement changes from x1 to x2?

A. QRS

B. WPRT

C. WPQV

D. VQRT

2.3.4 Outline what is meant by kinetic energy. Kinetic Energy: The energy of a moving body because of its movement.

The unit of Kinetic Energy is the joules /J.

K.E. = mv2

2.3.5 Outline what is meant by change in gravitational potential

energy. Gravitational potential energy is energy that is stored in an object by its height.

The unit of gravitational potential energy is the joule /J.

2.3.6 State the principle of conservation of energy.

Energy is conserved; this means that the total amount of energy in the universe is constant.

Energy cannot be made or destroyed. It can be transformed from one form to another.

2.3.7 List different forms of energy and describe examples of the

transformation of energy from one form to another. Kinetic Energy Electrostatic Potential Chemical energy Radiant energy

Gravitational Potential Thermal Energy Nuclear energy Solar energy

Elastic Potential Energy Electrical Energy Internal Energy Light energy

Situation Energy Conversion Formula

Brakes get hot K.E. - Thermal Energy mv2 = mcT

Car going up hill K.E. - GPE mv2 = mgh

Stone falling GPE K.E. mgh = mv2

2.3.8 Distinguish between elastic and inelastic collisions. Most collisions are inelastic because kinetic energy is transferred to other forms of energy

such as thermal energy, potential energy, and soundduring the collision process. If you are

asked to determine if a collision is elastic or inelastic, calculate the kinetic energy of the

bodies before and after the collision. If kinetic energy is not conserved, then the collision is

inelastic. Momentum is conserved in all inelastic collisions.

G.P.E = mgh

An elastic collision is a collision where no mechanical energy is lost. The collision of pool

balls is a good example of an elastic collision. Although some energy is lost (as sound

energy) this is a small fraction of the total energy.

For elastic collisions the relative velocity before is always equal to the relative velocity after

the collision.

Particles in an ideal gas collide elastically.

2.3.9 Define power. Power is the rate of doing work or transferring energy.

The unit of power is the Watt (W). 1 watt is equivalent to 1 Joule of energy being

transformed per second.

2.3.10 Define and apply the concept of efficiency.

In any energy transfer some of the work is transferred into a form that is not useful. This

energy is wasted. Efficiency is defined as the ratio of the useful energy to the total energy

transferred.

Efficiency does not have any units and is usually expressed as a percentage.

2.3.11 Solve problems involving momentum, work, energy and

power.

yInTotalEnerg

gyOutUsefulEnerEfficiency =

time

WorkDonePower =

1) This question is about projectile motion and the use of an energy argument to find

the speed with which a thrown stone lands in the sea.

Christina stands close to the edge of a vertical cliff and throws a stone. The diagram

below (not drawn to scale) shows part of the trajectory of the stone. Air resistance is

negligible.

Point P on the diagram is the highest point reached by the stone and point Q is at

the same height above sea level as point O.

(a) At point P on the diagram above draw arrows to represent

(i) the acceleration of the stone (label this A).

(1)

(ii) the velocity of the stone (label this V).

(1)

The stone leaves Christinas hand (point O) at a speed of 15 m s1

in the

direction shown. Her hand is at a height of 25 m above sea level. The mass of the

stone is 160 g. The acceleration due to gravity g = 10 m s2

.

(b) (i) Calculate the kinetic energy of the stone immediately after it leaves

Christinas hand.

KE = 21 mv

2

= 0.08 225

= 18 J

(1)

(ii) State the value of the kinetic energy at point Q.

18 J

(1)

(iii) Calculate the loss in potential energy of the stone in falling from point

Q to hitting the sea.

loss in PE = mgh

= 25 x 10 x 0.16

= 40J

(1)

(iv) Determine the speed with which the stone hits the sea.

Total KE = 40J + 18J = 56J

2

21 mvKE =

m

KEv

2=

16.0

562=v

127 = msv

(2)

(Total 7 marks)

2.4 Uniform circular motion

2.4.1 Draw a vector diagram to illustrate that the acceleration of a

particle moving with constant speed in a circle is directed towards the centre of the circle.

Objects perform uniform circulation motion when acted upon by a force towards the centre

of the motion ( centripetal force) e.g.

The only force on the Moon in its orbit is the pull of the Earth, which supplies the centripetal

force.

2.4.2 Apply the expression for centripetal acceleration.

2.4.3 Identify the force producing circular motion in various

situations.

2.4.4 Solve problems involving circular motion.