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IB Physics Lesson Notes

Topic 2 - Mechanics

2.1       Kinematical Concepts

Relative Velocity

Two cars, one travelling at 60km/h and the other at 61km/h, crash. What happens?    B

 A

 B 61km/h         60km/h                                   61km/h         60km/h               +v direction

relative velocity = 1km/h                         relative velocity = 121km/h

In 1st case velocity of A relative to B is +1km/h, velocity of B relative to A is –1km/h

In 2nd case velocity of A relative to B is +121km/h, velocity of B relative to A is –121km/h

Constant Velocity   s                                                  v                                                a     t                                                   t                                 t

Constant Acceleration    s                                                  v                                                a     t                                                   t                                 t

Define Displacement, Velocity and Acceleration:

The displacement between any two points is a vector directed from one point to the other; the magnitude of this vector is the straight-line distance between the two points.

The average velocity is the displacement from start to finish of the motion divided by the time taken. (The velocity is in the direction of the displacement)

An object’s speed is not equal to the magnitude of its average velocity.

The magnitude of the instantaneous velocity of an object at a point is equal to the instantaneous speed at that point. The direction of the instantaneous velocity is the same as that of the motion of the object.

The slope of the graph of displacement verses time at any point is equal to the instantaneous velocity at that point.

The average acceleration is the change in instantaneous velocity divided by the time taken for that change to occur.

The instantaneous acceleration of an object is the rate of change of instantaneous velocity at a point.

The slope of the graph of instantaneous velocity versus time is equal to the instantaneous acceleration.     s                                    (x,y)

average velocity = (y-0) / (x-0) p              a

instantaneous velocity at point p = a / b b (0,0)                                                  t v

average acceleration = (y-0) / (x-0)  (x,y) instantaneous acceleration at point q = c / d q c

d  (0,0)                                                  t v

#  Area under graph is equal

to the displacement.

 -ve t

2.2           Linear motion with constant acceleration

(student hand-out)

Uniformly accelerated motion in one dimension

Key :     s = displacement (m)

u = initial velocity (ms-1)

v = final velocity (ms-1)

a = acceleration (ms-2)

t = time (s)

1         Average velocity = ½(u+v)

2         s = ½(u+v).t

3         v = u + a.t

4         v2 = u2 + 2.a.s

5         s = ut + ½ a.t2

1          The average velocity of an object that is accelerating uniformly in one dimension is equal to the average of the initial and final velocities … Average velocity = ½(u+v)  # Velocity

(ms-1)      v    ½(u+v) u t  t1                                 t2       time (s)

2         The distance travelled between times t1 and t2 is the same as if the object were moving at a constant speed ½(u+v) over the same time period, t. Hence the distance travelled, s = ½(u+v).t.  The areas under the two sections of the graph are identical (and equal the distance travelled).

3         The change in velocity of an object over a period of time, t, is given by  a.t  since the acceleration is the change in velocity per second, if t = 2s then the change in velocity will be 2a. If the initial velocity of the object is u then the velocity after time t will be u + a.t

4          From equation 3 we know that t = (v-u)/a. If we substitute for t in equation 2 we get;

s = (u+v).(u-v)/2a , multiplying this out we get;

s = (u2 – v2 + uv – uv)/2a , which simplifies to;

2a.s = u2 – v2 , and so either;

v2 = u2 – 2a.s  (if the displacement and acceleration are in opposite directions)

and

v2 = u2 + 2a.s  (if the displacement and acceleration are in the same direction)

the changing sign is a result of the vector nature of the problem whereas the mathematical treatment assumed scalar quantities. (the dimensions or units of a.s are ms-2.m  or m2s-2 ie. velocity2)

5          We can obtain a second derived equation by substituting for v in equation 3. From equation 2; v = u + a.t , using this in equation 2, s = ½(u+v).t  , we get;

s = ½ (u + u + a.t).t , which gives us;

s = u.t + ½ a.t2

How to approach solving problems of 1D motion (constant acceleration) – there are lots of simple examples in your text books.

s =

u =

v =

a =

t =

Write in known values and select appropriate equation of motion (ie. an equation that only contains one unknown value). Include free-fall questions and note that mass is not a factor in motion under gravity.

H/W Question Sheet

1.     Are any of the following statements true? Give an example for any true statements.

a)     An object can have a constant velocity even though its speed is changing.

b)    An object can have a constant speed even though its velocity is changing.

c)     An object can have zero velocity even though its acceleration is not zero.

d)    An object subjected to a constant acceleration can reverse its velocity.

2.     Sound travels approximately 340m/s in still air. If you shout across a canyon and hear the echo reflected from the opposite wall 3.5s after you shout, how far away is the opposite wall?

3.     Mary can run at a top speed of 4.2m/s, whereas Deema can run only at 3.4m/s. They are to race a distance of 200m, starting at the same point. If the race is to end in a tie, how much sooner should Deema start running before Mary does.

4.     A truck travelling at 22.5m/s decelerates at 2.27m/s2.

a)     How much time does it take for the truck to stop?

b)    How far does it travel whilst stopping?

c)     How far does it travel during the third second after the brakes are applied?

5.     The graph below shows how the velocity of a car varies with time. Calculate the instantaneous acceleration at points A, B and C and the average acceleration for the whole journey. Calculate the displacement of the car at points A, B and C.

 B   A  C 0           2          4           6           8        10          12        14          16        18 20          22        24

time /s

2.3       Concepts of force and mass

Forces cause velocity change or deformation, they are vectors.

Weight is the force on a mass due to gravity, W = mg and acts towards the centre of the Earth.

Consider a range of different situations involving forces, draw free-body diagrams and resolve forces into vector components (including force down a slope due to the mass of an object (mg sinq): reaction normal (perpendicular) to slope = mg cosq force down slope (parallel to slope) = mg sinq mg

Include examples of questions in which gravity is changed, for example on the moon ( g)

2.4       Newton’s 1st law of motion

A moving object continues to move with constant velocity (or if at rest, remains at rest) if the vector sum of the external forces acting on it is zero.

Examples such as a book on a table, terminal velocity in free fall and objects being pulled across horizontal surfaces (discussion of frictional forces – static and dynamic)

Types of forces:

1.     Contact Forces: tension forces, compression forces, forces of viscosity or friction

2.     Fundamental Forces: act between objects separated in space: gravitational, electrical and magnetic.

2.5       Newton’s 2nd law of motion

F = ma  if mass remains constant.

or “Force is the rate of change of linear momentum”

Practice free body diagrams and resolving force vectors to find the resultant force.

Solving problems involving lifts, friction on horizontal surfaces, pulleys with masses moving vertically and horizontally.

## H/W Questions

1.     A 900kg car accelerates from rest to 12.0m/s in 8.00s along a straight road. How large a force is required?

2.     A girl pulls on a wagon with a string at an angle of 37° to the horizontal with a force of 25.0N. As a result the wagon accelerates horizontally. The wagon has a mass of 10.4kg, and the downward force of gravity on the wagon, its weight, is 102N. Assume that there is no friction. Find the wagon’s acceleration and the upward force of compression P that the ground exerts on the wagon under these conditions.

3.     A 90kg man is standing in a lift on bathroom scales that register a weight of 900N. Calculate the force shown on the scales

(i)             When the lift is accelerating upwards at 3m/s2

(ii)           When the lift is travelling downwards at a constant speed of 2m/s

(iii)         When the lift is accelerating downwards at 2m/s2

2.6       Newton’s 3rd law of motion

Action and reaction are equal and opposite.

If object A exerts a force F on object B then object B exerts a force –F on object A

Lots of illustrative examples including that the gravitational force of attraction of a football on the Earth is the same as that of the Earth on the ball. The acceleration of the ball towards the Earth is much greater than the acceleration of the Earth towards the ball because the Earth’s mass is so much greater (F = m.a)

2.7       Projectile, simple harmonic and uniform circular motion.

A projectile follows a parabolic trajectory in two dimensions. If the motion is separated into horizontal and vertical components and air resistance is ignored then the horizontal motion is motion with constant velocity and the vertical motion can be treated as motion under constant acceleration (gravitational).  A  B

C At points A, B and C the horizontal velocity vectors are equal (zero acceleration). At A the vertical component of the velocity is zero, but with constant vertical acceleration under gravity the vertical component gets larger with time according to the equations of motion.

In problems involving horizontal launch of a projectile u = 0ms-1 and a = g (9.81ms-2).

Problem:

A toy car is pushed off a table at a horizontal velocity of 5ms-1. If the table is 1.2m high how long will the car take before hitting the ground and how far from the table will it hit the ground?

extension (link to vectors) : what is the velocity of the car when it hits the ground?

Higher Level (projectile motion)

If a projectile is launched in a direction other than horizontally then the rectangular components of its initial velocity must be established before the problem can be solved. V.sina  V a  V.cosa

The initial horizontal component of the velocity does not change throughout the motion so long as air resistance is ignored at any time during the motion it is V.cosa so at a time t, the horizontal distance covered is (V.cosa).t.

For the vertical motion the initial velocity is V.sina. The vertical motion can be treated as motion with constant acceleration in one dimension and so s, u, v, a and t at any points during the motion can be found from the equations of motion discussed earlier. As an example, the maximum height reached, s, can be found as u = Vsina, a = g and v = 0 (at the top of the parabola).

Lots of problems as practice can be found in your text books.

Simple harmonic motion

Simple harmonic motion is a periodic motion of an object about a point. The object experiences a force towards the point at all times except at the point itself.

Examples include:

1. Mass on a spring:   The mass oscillates about X, the position at which it would hang motionless with its weight exactly balanced by the restoring force of the spring (net force = 0). During its motion its displacement, velocity and acceleration change periodically:

 Displacement Velocity Acceleration A Max. positive Zero Max. downwards B Zero Max. downwards Zero C Max. negative Zero Max. upwards D Zero Max. upwards Zero

2. Circular motion: In circular motion both the force on and the acceleration of the object act towards the center of the circular path. The velocity is constantly changing although the speed is not.

If the linear component (eg. sx or sy) of the object’s displacement is plotted as a function of time then a graph identical to the one above is produced.

Both the force and acceleration are described as centripetal.

2.8       Linear Momentum

The momentum of an object is the product of its mass and velocity:

p = m.v

It is a vector quantity as it has both magnitude and direction (the same direction as the velocity).

Impulse is a change in momentum provided by the application of a force over a time interval:

Impulse = F.Dt

Impulse is also a vector quantity (its direction is the same as the applied force). Note the units of impulse are the same as those of momentum…Kg.m.s-1.

We can use the statement of Newtons’s 2nd law “force is the rate of change of linear momentum of a body” to derive the equation F = m.a:

Fnet = = = = = = mâa

providing mass remains constant.

In any collision the sum of the linear momenta of the interacting bodies (the system) is conserved. Momentum before collision = momentum after collision.

In a perfectly elastic collision KE is also conserved with none being absorbed internally (eg. by deformation) or lost as heat and sound. Snooker/pool balls collide elastically.

In an inelastic collision, momentum is conserved but KE is not, energy being used in deformation or lost through sound and heating. Crashing cars are an example of an inelastic collision.

In an explosion momentum is also conserved, the sum of the momenta of all fragments is zero.

Core students need only solve momentum (collision) problems in one dimension.

Higher students will need to use vectors to solve momentum (collision) problems in two dimensions.

2.9       Work, energy and power

The work done on an object is equal to the change in energy of the object. It is a scalar quantity whose units are Joules or Newton metres.

Work done = DEnergy

The work done on an object by a force is equal to the product of the force and the displacement of the object in the direction of the force:

Work done = Force x Displacement

Calculations may involve either one-dimensional problems or two-dimensional problems for which the component of the force parallel to the motion needs to be calculated.

The potential energy, PE, of an object is the energy it possesses due to its position in a system (eg. a ball and the Earth) or its state (eg. a stretched or relaxed rubber band)

Gravitational PE (GPE) = m.g.h

In a material that is elastic, the force applied, Fapp, to produce and extension of x is related to a constant k, which is a measure of the ‘stiffness’ of the elastic material by;

Fapp = k.x

And the energy stored in such a material that has been stretched, the elastic potential energy, EPE, is given by;

EPE = ½.k.x2

The kinetic energy, KE, of an object is the energy it possesses due to its motion.

Kinetic energy = ½.m.v2

Students need to know the different forms of energy and examples of transformations from one form to another.

The principle of conservation of energy can be used to solve problems involving changes in KE and PE in a system. For example, falling objects and roller-coaster rides where…

KE + PE = constant

This can be applied to problems of motion on an inclined plane and is sometimes a simpler method of solution than the one-dimensional equations of motion.

Power is the rate of doing work;

Power = The units of power are Watts or Joules per second.