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AS Physics 9702 unit 2: Motion, Force and Energy 1

UNIT 2 – MOTION, FORCE AND ENERGY

This unit includes topics 3, 4, 5 and 6 from the CIE syllabus for AS course.

Kinematics:

Kinematics is a branch of mechanics which describes the motion of objects without the consideration

of its masses or forces that bring that motion. Dynamics is concerned with the forces that produce

any motion to an object of mass m.

Together, kinematics and dynamics form the branch of physics known as mechanics.

Linear and non-linear motion:

Linear motion is the change in position of an object moving in a straight line and non-linear motion

is change in position of an object without following the straight path for example in circular motion.

Displacement and distance:1

The displacement is a vector quantity that points from an object’s initial position to its final position

in straight line. It has a magnitude that equals the shortest distance between the two positions. The SI

unit of displacement is meters (m). The symbols use to represent displacement are ‘s’ or ‘x’ to

represent displacement along x-axis. In case of free falling or vertical motion the displacement can be

represented by the vertical distance covered and symbol use is ‘h’ or ‘y’.

The scalar form of displacement is distance. It refers to how much ground an object has covered

during its motion. The SI unit of distance is meters (m) and the symbol use to represent distance is‘d’.

Average speed and average velocity ( ):

The average speed of an object is the total distance travel divided by the total time requires to cover

that distance.

The SI unit of average speed is ‘meters per seconds’ or ms-1

. It is a scalar quantity.

The average velocity is the total displacement of an object from point A to point B divided by the total

time taken.

The SI unit of average velocity is also

‘meters per seconds’ or ms-1

. It is a vector quantity.

The instantaneous velocity v of an object indicates how fast the object is moving and what is its direction at any particular instant of time when it is moving with non-uniform velocity. This can be found by finding the gradient of the tangent to the curved part of distance-time graph.

1 For further explanation see http://www.glenbrook.k12.il.us/GBSSCI/PHYS/CLASS/1DKin/U1L1c.html

http://www.glenbrook.k12.il.us/GBSSCI/PHYS/CLASS/1DKin/U1L1c.html

http://www.bbc.co.uk/schools/gcsebitesize/physics/forces_and_motion/representing_motionrev3.shtml

Prepared by Faisal Jaffer, revised on Nov 2011

Acceleration:

Acceleration is a vector quantity which is defined as the rate at which an object changes its velocity.

Where v is the change in velocity and t is time elapsed. Also the u and v are the initial and final

velocities of the moving object. The SI unit of acceleration is ‘meters per second square’ or ‘ms-2

’.

Note: Whenever the acceleration and velocity are in same direction then the acceleration is positive

and when acceleration and velocity are in opposite direction, it means the body is slowing down and

the acceleration is negative. It is also called retardation or deceleration.

Graphical representation of velocity, acceleration and

displacement:

Velocity – time graphs: In the following graphs the motion of an object is considered as positive

if it is moving along the x-axis towards right or y-axis in upwards direction.

(A) moving with constant acceleration in positive direction (B) moving with constant deceleration in positive

direction (C) moving with constant deceleration along negative direction (D) object is stationary (E) moving

with constant velocity in negative direction (F) moving with constant velocity in positive direction (G) object is

changing direction (H) object is momentarily stop at two occasions during its whole motion (I) object is

momentarily stop only at one occasion during its whole motion


Displacement – time graph:

(A) moving away from the mean position with constant velocity (B) moving towards the mean position with

constant velocity (C) moving towards the mean position from negative displacement with constant velocity (D)

object is stationary (E) object is stationary (F) object is stationary

Important!!!! The area under the velocity – time graph measures total displacement. The slope or gradient of a velocity – time graph represents the acceleration of the body. The slope or gradient of a displacement – time graphs represents the velocity of the body.

If the object is moving with non-uniform acceleration then the

acceleration at any particular time t can be found by finding the

gradient of the tangent to the curve at time say t=4s as shown in

the figure.

Derivation of equations of uniformly

accelerated motion:

From the definition of acceleration, it is the rate of change of velocity. Consider a body moving with

initial velocity u and after time t the velocity becomes v. the equation can be expressed as

t

uva

by arranging the above equation we can get

from the definition of average velocity that is sum of initial velocity and final velocity divided by two

and ⁄

by equating two equations we get


This equation can be derived by using two equations and

. Replace the value

of v from the first equation into the second equation.

The equation can be derived by using equation . By squaring both sides of the equation we

get

by opening the bracket

by taking common term 2a

since

, replace it in previous expression

If an object is falling freely under the influence of gravity then the only force applied on an object is

weight in downward direction and acceleration is g = 9.81ms-2

. The above equations of uniformly

accelerated motion can be rewritten as:

Application of the equations of motion:

The equations of motion can be used for any moving object as long as the acceleration of the object is

constant. However it helps to use following guidelines when solving problems:

a) Make a drawing, if require, to represent the situation being studied.

b) First decide at the start which directions are to be called positive (+) and negative (-) relative

to a conveniently choose coordinate of origin. It is a common understanding that for the

direction of motion we use Cartesian coordinates, that is moving upward or right we consider

positive direction and moving downward or left we consider negative direction.

c) In an organized way write down the values of s, v, u, a, and t with plus or minus signs and

appropriate units. This is called data.

d) Before attempting to solve a problem, verify that given information contains values for at

least three of the five variables. Once the three known and two unknown variables are

identified then select the appropriate equation.

e) If there are two segments of the motion then make sure that the final velocity of first segment

is the initial velocity of second segment.


f) Keep in mind that there may be two possible answers to visualize the different physical

situation.

Exercise: 2.1 A) Solve the following questions.

Q1) A racing car has an initial velocity of 100 m/s and it covers a displacement of 725 m in 10 s. Find its

acceleration. (Ans: -5.5 ms-2

)

Q2) An object starts from rest, moves in a straight line with a constant acceleration and covers a distance of 64

m in 4 s. Calculate a) its acceleration (ans: 8ms-2

) b) its final velocity (ans: 32ms-1

) c) at what time the object

had covered half the total distance (ans: 2.8s) d) what distance the object had covered in half the total time. (ans:

16m)

Q3) A ball is thrown vertically upwards with a velocity of 10 m/s from the balcony of a tall building. The

balcony is 15 m above the ground and gravitational acceleration is 10 m/s2. Find a) the time required for the ball

to hit the ground (ans: 3s), and b) the velocity with which it hits the ground (ans: -20ms-1

).

B) Solve assignment 5 on the website:

http://www.freewebs.com/faisalj/AS/Physics%20assignment%205%206%207%208.pdf

C) Solve following questions from past papers

May/June 2010, Paper 22, question 2


Oct/Nov 2008, Paper 2, question 2





Friction forces:

What are friction forces?

The friction force is the force exerted by a surface as an

object moves across it or makes an effort to move across it.

The friction force opposes the motion of the object. For

example, if a block of mass ‘m’ moves across the surface

of a desk, the desk exerts a friction force in the direction

opposite to the motion of the block.

As such, friction depends upon the nature of the two

surfaces and upon the degree to which they are pressed

together. There are two types of frictional forces:

a) Static or limiting frictional force fs : This is the maximum force applied by the surface

of the desk on the block until just before the block starts to move. This is the force that needs to

be overcome by force F in order for block to move. Mathematically we can write the equation as

fs = µs × N

b) Kinetic or dynamic frictional force fk: This is the frictional force which exists between

the two adjacent surfaces which are in relative motion. This is usually slightly less than the static

friction. Mathematically fk = µk × N

N is the normal reaction force of surface on mass m. μs and μk are the coefficient of static and kinetic

friction respectively. Both depend on the nature and the condition of the surfaces which are in contact

but are independent of the area of contact. For example steel on steel µs0.8 and for Teflon on Teflon

µs0.04. If two surfaces are assumed to be perfectly smooth, if there is no frictional force and µs = µk

= 0. For two surfaces which have relative motion the kinetic frictional force is directly proportional to

the motion of an object.

Exercise: 2.2

Q) A box that weighs 10.0 N is being dragged with constant velocity along a horizontal surface of the table by a

rope that is at an angle 45o with that surface. The tension in the rope is 5.0 N. What is the coefficient of friction?

(ans:

Q) Solve assignment 3b on the website:

http://www.freewebs.com/faisalj/AS/Physics%20assignment%201%202%203%203b.pdf

F fs

N

Fg= w = mg

m F fk

N

Fg = w = mg

m

http://www.freewebs.com/faisalj/AS/Physics%20assignment%201%202%203%203b.pdf


Motion of a body in uniform gravitational field without the

resistive force (air resistance or viscous drag):

When an object is falling only under the influence of gravitational force then it is said to be free

falling. Any object that is being acted upon only by the force of gravity is said to be in a state of free

fall. This force of gravity causes acceleration called acceleration due to gravity and is expressed by

letter `g`. The average numerical value for the acceleration due to force of gravity is 9.81ms-2

. There

are slight variations in this numerical value (to the second decimal place) that are dependent on the

altitude of the point under consideration.

Motion of body in uniform gravitational field with air resistance

or viscous drag:

What is air resistance?2

Air resistance is a special type of frictional force which acts upon

objects as they travel through the air in uniform gravitational field.

Like all frictional forces, the force of air resistance always opposes

the motion of the object. It is most noticeable for objects which

travel at high speeds (e.g., a skydiver or a downhill skier) or for

objects with large surface areas.

What is viscosity of fluid and viscous drag?

Any fluid is said to be viscous if it offers a resistance to the motion of any solid body passes

through it. Viscous effects are due to the frictional force which exists between two adjacent layers

of fluid which are in relative motion.

For viscous drag consider a sphere of radius r moving with velocity v through a fluid, it

experiences a drag force F which acts in opposite direction to that in which the sphere is moving.

The expression for viscous force can be stated as

(no derivation of this expression requires in this course)

This expression is called Stokes Law. The coefficient of viscosity of a fluid (liquid or

gas) is the measure of the degree to which the fluid exhibits viscous effects.

Viscosity is the property of the fluid that measures its resistance to flowing. For

example honey is more viscous then water. The higher the coefficient of viscosity,

the more viscous the fluid is. In liquids, viscosity decreases with the increase of

temperature, whereas in gases it increases with the increase of temperature. Its unit is

N.m-2

s.

Terminal velocity:

Consider a sphere falling from rest through a viscous fluid (air or liquid). The forces acting on the

sphere are its weight w acting downward, the upthrust U due to displaced fluid3, and the viscous

force R or air resistance, both acting upwards. Initially the downward force w is greater than the

upward forces U+R, and the sphere accelerates downwards. As the velocity of the sphere

increases so does the viscous force (air resistance) R, and eventually U+R becomes equal to w the

weight. Since there is now no net force acting on it, its velocity has a constant maximum value

known as terminal velocity. In the beginning w >> U+R and as the body moves down w >U+R

and eventually at terminal velocity w=U+R.

2 For further explanation http://www.physicsclassroom.com/mmedia/newtlaws/efar.html

3 It is the force that liquid exerts on the sphere. It is equal to the weight of the liquid displaced by the sphere when immersed.

w = mg

http://www.physicsclassroom.com/mmedia/newtlaws/efar.html


F vx

vx

vx

vy

vy

vy

y x

Exercise: 2.3 Solve the following questions from past papers.



Solve assignment no 6 on the website:

http://www.freewebs.com/faisalj

Projectile motion:

A projectile is any object which once projected, dropped or thrown continues its constant motion

along the x-direction (horizontal) and accelerated motion along y-direction (vertical) due to the

force of gravity.

Projectiles travel with a parabolic trajectory due to the

influence of gravity.

There are no horizontal forces acting upon projectiles

and thus no horizontal acceleration.

The horizontal velocity of a projectile is constant.

There is a vertical acceleration caused by gravity; its

value is 9.81 ms-2

, downwards.

The vertical velocity of a projectile changes by 9.81

m/s each second.

The horizontal motion of a projectile is independent of

its vertical motion.

If there was any other force acting upon an object, then

that object would not be a projectile.

Examples of projectile, in the absence of air resistance are:

1. An object dropped from rest is a projectile.

2. An object which is thrown vertically upward is also a projectile.

3. An object which is thrown upward at an angle to the horizontal is also a projectile.

A body projected with a velocity v at an angle

above the horizontal has horizontal (vx) and

vertical (vy) component of velocity

vy= v sin and vx = v cos

and horizontal (x) and vertical (y) displacements

are

and because vertical initial velocity is zero



In similar way to find the resultant velocity v from its components we use Pythagoras theorem

and direction is

If a projectile is projected at an angle θ

with the horizontal then the maximum

range of projectile can be derived from

expressions

and

Replacing y=0 in the first equation and

replacing the value of t in the second

equation we get the expression.

From the above expression we can see that the value of R will be maximum when sin 2 =1 or =45o

(because sin 90o =1). The projectile can have a maximum horizontal range if an object is thrown at

angle of 45o.

Exercise: 2.4

Q1) A soccer ball is kicked horizontally off a 22.0-metre high hill and lands a distance of 35.0 metres from the

edge of the hill. Determine the initial horizontal velocity of the soccer ball. ans: 16.5ms-1

Q2) A long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28o above the horizontal.

Determine a) the time of flight, b) the horizontal distance, and c) the peak height of the long-jumper. Ans (a)

1.1s, (b) 12.2 m, (c) 1.6 m



Solve following questions from the past papers

Oct/Nov 2009, Paper 22, question 3(a) (b)


May/June 2011, Paper 22, question 1(c)



Oct/Nov 2010, Paper 22, question 3(b)

22

yx vvv

v Vertical height, m

R, Horizontal Range, m

θ

y



Dynamics:

Dynamics is the study of motion of an object with respect to forces that produce the motion.

Force is defined as the one that changes body’s state of rest or of uniform motion in straight line.

Another definition of force is that it causes a body to accelerate. The unit of force is newton (N). 1

newton is the force required to give a mass of 1 kg an acceleration of 1 ms-2

.

Issac Newton, one of the greatest scientists (1643 – 1727) presented the most powerful three laws of

motion.

Newton’s three laws of motion:

First Law:

A body continues its state of rest or of uniform motion in a straight line unless some external force is applied to change that state.

Newton’s first law is another way of saying that all matter has a built-in opposition to being moved if

it is at rest or, if it is moving, to having its motion changed. This property of matter is called inertia

(from Latin word ‘laziness’). Therefore first law of motion is also called law of inertia.

Second law:

Whenever a net force acts on a body, it produces acceleration in the direction of the net force that is the acceleration is directly proportional to the net force and inversely proportional to the mass of the body.

Fa and m

a1

Combining both we get m

Fa

or m

Fka

where k is the constant of proportionality. By the definition of unit of force, newton, k is defined as:

1 N = k x 1 kg x 1 ms-2

thus k = 1

The newton (N) is defined as the force that produces an acceleration of 1 m/s2 when applied on a body

of mass 1 kg.

Third law:

To every action there is always an equal and opposite reaction force of two bodies acting upon each other. They are always equal and opposite in direction.

Consider a book of weight ‘w’ is placed on the table. The book

exerts a force equal to ‘w’ on the table and table exert equal and

opposite reaction force N on the book which is always perpendicular

to the surface of the table.

Other examples are:

1. Earth exerts a gravitational force of attraction on Moon; the

Moon exerts a force of the same size on the Earth.

2. A rocket moves forward as a result of the push exerted on it

by the exhaust gases which the rocket has pushed out.

3. When a man jumps off the ground it is because he a pushed down on the Earth and the Earth

has pushed up on him.

Book

w

N


Exercise: 2.5 Solve the following questions from past papers:






Mass and weight:

The following table clarifies the difference between mass and weight:

Mass Weight

The mass is the intrinsic property of matter and

does not change as an object moved from one

location to another.

The weight of the body is the force acting on its

mass due to the gravitational attraction of earth in

earth’s gravitational field.

The mass of a body is measure of its resistance of

acceleration that is it is measure of the inertia of

the body.

Weight can change if the body is moved relative

to the centre of earth. Weight of the body is

where w is the weight of the body, m is the mass

and is the acceleration due to gravity or

acceleration of free fall and its value is 9.81 ms-2

.

SI unit of mass is kilogram (kg) which is the base

unit.

SI unit of weight is newton which is derived unit

by multiplication of mass and acceleration.

Centre of gravity (or mass):

A body behaves as if its whole mass is concentrated at one point,

called its centre of mass or centre of gravity, even though the Earth

attracts every part of it.

1. The centre of mass of a body is that point at which the

mass of the body may thought to be concentrated.

2. If suspended, a body will come to rest with its centre of

mass directly below the point of suspension.

3. The centre of mass of a symmetrical body is along the axis

of symmetry.

4. If a body is not turning, the total clockwise moment must be exactly balanced by the total

anticlockwise moment about its centre of gravity.

Stability of an object: Stability is the extent to which an object resists toppling over. Stable objects

do not topple over easily. If the line of action of the weight of a body lies outside the base of the body

there will be a resultant moment and the body will tend

to topple. When designing vehicles, engineers try to

design so that the centre of mass is as low as possible.

This makes vehicles less likely to turn over when going

round corners.

For an object to start rotating it needs to have an

unbalanced moment acting on it as illustrated in the

figure.


Determining the centre of gravity of irregular shape cardboard (lamina)

The center of gravity of an irregular shaped cardboard can be found experimentally by suspending the

cardboard from different points, with a plum-line (a mass hanging on a thread) and marking the points

along the line of the thread. The point of intersection of two lines is the centre of gravity of the

cardboard. A doughnut’s CM is in the center of the hole.


May/June 2002, Paper 2, question 3(a)



Linear momentum:

Definition of linear momentum: Momentum is defined as the product of mass and velocity of a body

that is moving in straight line. It is a vector quantity and express by unit kg ms-1

or N s.

p = m v

Where p (small case) is the momentum, m is the mass of the body and v is the velocity. The direction

of the momentum is same as the direction of the velocity.

The conservation of linear momentum:

When two bodies, A and B of mass mA and mB, are involved in a collision, the body A exert force FB

on B and body B exerts force FA on body A, then the changes in the momentum, are oppositely

directed and the total change in momentum is zero. Hence the principal of conservation of linear

momentum can be stated as

The total linear momentum of a system of interacting (eg colliding) bodies remains constant if no external forces is applied.

From above diagram if the body A is moving towards right with the initial velocity uA and body B is

moving towards left with initial velocity -uB then

Initial momentum of body A before collision piA = mA uA

Initial momentum of body B before collision piB = mB(- uB)

and

Total initial momentum before collision pi = mA uA - mB uB

now after collision body A start moving with final velocity -vA towards left and B towards right with

final velocity vB

Final momentum of body A after collision pfA = mA (-vA)

Final momentum of body B after collision pfB = mB vB

Total final momentum after collision pf = -mA vA + mB vB

Principle of conservation of linear momentum is

Total initial momentum before collision = Total final momentum after collision

pi = pf

mA mB

FA FB

mA mB uA uB

mA mB vA vB


Elastic and inelastic collision:

When two bodies collide, their total momentum is conserved unless there are external forces acting on

them. The total kinetic energy Ek, however, usually decreases, since the collision converts some of its

kinetic energy to heat and/or sound. The collision in which some kinetic energy lost is known as an

inelastic collision. A collision is elastic if there is no loss of kinetic energy. Hence

Inelastic collision:

Total Ek before collision > Total Ek after collision

For elastic collision

Total Ek before collision == Total Ek after collision

Conservation of velocities in elastic momentum:

Divide equation 2 by 1

The above equation must satisfied for two bodies to collide elastically provided the object A and B are

moving towards each other and after collision they move in opposite direction. The sign convention

must be change if the bodies are moving in direction other than the one used in the derivation of the

equation.

Exercise: 2.7 Solve the following question from past papers

May/June 2010, Paper 21, question 3(b)



May/June 2009, Paper 22, question 2(c) (d)


May/June 2008, Paper 1, question 10, 11

May/June 2007, Paper 1, question 10, 11, 12





Examples of linear momentum:

A car and a truck are approaching each other and about to collide

Two trolleys of different masses travelling in the same direction one with 1ms

-1 and other with 4ms

-1 are about to collide

A ball travelling with 20 ms

-1

hit by tennis racket and bounce back with the velocity 30ms

-1

An object of mass 3m explodes

Two trolleys connected to each other by spring are about to move in opposite direction when release.

A cannon ball is fired from a cannon. Ball moves in one directon and cannon move in opposite direction.

Cases of linear momentum:

Initial conditions before collision Results after collision

Case no 1

mA = mB = m and collision is elastic

uA= initial velocity of body A

uB = initial velocity of body B

velocities of the bodies will interchange

vA=uB and vB=uA

Case no 2 mA ≠ mB and collision is elastic


uB = 0

(

)

(

)

Case no 3 mA = mB = m and collision is elastic


uB = 0

body A will be stationary and body B

will be start moving with same velocity

as initial velocity of body A

and

Case no 4 mass of body A is much smaller than B

mA<< mB and collision is elastic


uB = 0

body A will bounce back with same

velocity in opposite direction and body

B will remain stationary

and

Case no 5 mA>> mB and collision is elastic


uB = 0

body A will continue with the same

velocity and body B will start moving

with twice the velocity if body A

and


Momentum in terms of Newton’s second law:

If the steady force F acts on a body of mass m and increases its velocity from u to v in time t, the

acceleration a is given by

t

uva

)( and force applied is given by maF

Substituting the value of a

t

uvmF

)( or

t

mumvF

mv and mu are final and initial momentum then mumv is the change in momentum and t

mumv

is the rate of change of momentum. From equation it is clear that the rate of change of momentum is

equal to net force F applied which is another form of Newton’s second law of motion

Impulse:

The impulse of a constant force F acting for a time ∆t is defined as the force applied on a body for

small fraction of time ∆t. or from equation

or or

Hence the impulse is the rate of change of momentum of a body. Examples of impulse of force are the

batsman hitting the cricket ball or golfer striking the golf ball. In both situations the concept of

impulse is used where a large variable force is acting for only a short period of time.

Since the impulse of force is the product of force and time of collision, therefore if the force applied

for longer time then it increases the impulse and that will also result in gain of momentum.

DENSITY:

The density of a substance is defined as

V

m

where is the density of substance in kg/m3, m is the

mass in kg and V is the volume in m3.

The relative density of a substance is define as

C0at water ofDensity

substance ofDensity density Relative

o

Relative density has no units.

Densities of some common substances are given in the table.

Exercise: 2.8 Solve the following questions from the past papers.






Densities in (g cm-3

)

Water at 4oC 1.0

Mercury 13.6

Aluminium 2.7

Gold 19.3

Ice at 0oC 0.92

Air 0.001293

CO2 0.001977

Hydrogen 0.00009

Helium 0.000178


Pressure:

The pressure is defined as the force per unit area acting at right angles to the surface.

where p is the pressure in Pa (pascal), F is the force in N and A is the area in m2

Atmospheric pressure is equal to

1 atm = 101 x 105 Nm

-2 or Pa = 760 mm Hg = 1.01 bar

Pressure in fluids:

Consider a cylindrical object of cross-sectional area A and height h is immersed in a

fluid of density. The top of the cylinder is at the surface of the fluid and vertical

forces acting on it are its weight w and force pA. If the cylinder is in equilibrium

then

pA = w

= m g

= V g because m = Vρ

=h A g because V = h A

or p A = h A g A cancels both sides

or p = h g

a) The pressure in a fluid increases

with depth. All points at the

same depth in the fluid are at the

same pressure.

b) Any surface in a fluid

experiences a force due to the

pressure of the fluid. This force

is perpendicular to the surface no matter what the orientation of the

surface. The magnitude of the force is independent of the orientation of

the surface. Pressure acts equally in all direction.

c) The above equation is not valid for gas and when h is very high. The

density of gas decreases with height and above equation is derived on

the assumption that density of fluid (gas or liquid) is constant.

d) Mercury barometer is used to measure the atmospheric pressure and

manometer is used to measure gas pressure.







May/June 2008, Paper 2, question 4 (b) (c)

F

F F

F

pA

h

A

w


Archimedes’ principle (upthrust):

A body immersed in a fluid (totally or partially) experiences an upthrust (upward force) which is equal to the weight of the fluid displaced by the body.

Consider a cylinder of height h and cross-sectional

area A, a distance ho below the surface of the fluid of

density . According to the Archimedes Principle:

volume of fluid displaced = Volume of cylinder

V = Ah

mass of fluid = Ah [ = m/V]

weight of fluid = Ahg [W = m/g]

The fluid exerts forces pxA and pyA on the top and

bottom faces of the cylinder. The upthrust that is the

resultant force due to the fluid is therefore given by:

upthrust = pyA - pxA

= (h + ho)gA - hogA

= hgA

Upthrust = weight of fluid displaced

It is clear that upthrust on a body does not depend on

the shape of the body. If the body is more dense than the fluid in which it is immersed, then its weight

is greater than the weight of the fluid displaces and body will sink. Similarly, if the body is less dense

than the fluid in which it is immersed then it will float.

Measuring the density of liquid using Archimedes principle:

The density of liquid can be found by determining the upthrust on some suitable object when it is

immersed in the liquid and then when it is immersed in water.

in waterUpthrust

liquidin Upthrust

waterofDensity

liquid ofDensity

From the above equation the density of unknown liquid can be found.

Exercise: 2.10 Solve the following questions from the past papers




ho pxA

pyA

pressure px

pressure py

A

h


TORQUE OR MOMENT OF FORCE:

Consider a force F acting on a rod of length ‘d’ so as to

cause it to turn about a fix point called pivot or fulcrum.

This effect of force is called turning effect or moment of

force. It is depend on the size and direction of force as well

as the perpendicular distance from the fulcrum. The turning effect or moment of force is

also called torque. Mathematically it is defined as:

τ is the torque in (N.m) this is a greek letter called tau. d the perpendicular distance of the line

of action of the force from the axis (m). If F and d are not perpendicular to each other then the

expression is , where θ is the angle between F and d.

Torque due to a couple

Two forces which are equal in magnitude and which are anti-parallel are called couple.

i. There is no direction, in which couple can give rise to a resultant force, and therefore a couple

can produce only a turning effect, it cannot produce translational motion.

ii. Since a single force is bound to produce translation, it follows that a couple cannot be

represented by single force. From the figure,

Total torque about O = F x OA +F x OB= F(OA+OB) = Fd

Thus the torque about O does not depend on the position of

O and therefore it follows that:

The torque due to a couple is same about any axis and

is given by

If F and d are not perpendicular than

In the given example the couple is

Equilibrium system and moment of force:

A body is in equilibrium if it follows the following conditions:

a) The net resultant force (and acceleration) on the centre of mass of the body is zero.

The above equation means that vector sum of all the force along the x-axis is equal to zero and

vector sum of all the force along the y-axis is equal to zero.

b) Total torque about all axes is zero i.e.





Oct/Nov 2001, Paper 2, question 3(a), 3(b)


May/June 2002, Paper 2 , question 3(b)


d==

F==

F==

O

A B


WORK:

When a force ‘F’ is applied to an object than it covers a

displacement‘s’ in the direction of the force applied. It is said

that the work has been done on an object.

The unit of work done joules or J, F is the constant force applied

in newton, s is the displacement covered by the body in the

direction of the force in metre. When the force and the

displacement covered are not in the same direction then we use

the component of force along the direction of force.

where is the angle between the force applied and direction of

motion.

External work done by expanding gas:

Consider a gas enclosed in a cylinder by a frictionless piston of

cross-sectional area A. Suppose that the piston is in equilibrium

under the action of the force pA exerted by the gas and an

external force F.

Suppose the gas now expands and moves the piston outwards through a distance of x, where x is so

small that p, the pressure of the gas can be considered constant. The external work done w is given

by the equation:

but is the change in the volume of gas in the cylinder

Work done in stretching a spring:

The tension F in a spring whose extension is x and

which obeys Hooke’s law is given by

kxF

where k is the spring constant. If the extension is

increased by x where x is very small that F can

be considered constant then the work done can be

expressed as

xFw

or xkxw

The total work done in increasing the extension

from 0 to x i.e. the elastic potential energy stored in

s

N


the spring when its extension is x is given by area under

the F-x graph.

Since this work done stores energy in the spring, this

energy is called strain energy or elastic potential

energy.

Exercise: 2.12 Solve the following questions from past papers



Oct/Nov 2009, Paper 12, question 20 Oct/Nov 2009, Paper 22, question 4


Kinetic and potential energies:

Kinetic Energy:

The energy which a body of mass m possess only because it is moving is called kinetic energy Ek or

the amount of work that must have been done on an object to increase its velocity from zero to the

velocity v is

.

Derivation of Ek

If a body of mass m moves a distance s under the action of a constant force F, the work done W by

the force is given by W = Fs

and if the constant acceleration of the body is a, then according to Newton’s second law of motion

F=ma but work done is represented by

if the body has acceleration from rest to some velocity v, then initial velocity u is zero

asuv 222 or 2

2vas

Replacing this value of ‘as’ in the equation of work

2

2vmW or

2

2

1mvW

due to this work done the object gain motion and hence kinetic energy therefore this work done can be

expressed as Ek

2

2

1mvEk

m m

s

F v


Gravitational potential energy:

The energy stored in a body due to its change of position or change of

shape is called potential energy. The gravitational potential energy of a

body is the amount of work done in lifting an object of mass m to the

height h above the surface of earth.

But this work done is changing the position of mass from ground to

height h then mghEp

Relationship between gravitational potential energy and

height:

Consider a sky diver jumping from the aeroplane. He possesses a

potential energy due to his height above the ground. If no air resistance is considered then the graph

between the gravitational potential energy and height is given by straight line.

Hence this concludes that gravitational energy is directly proportional to the height.

Gravitational potential energy is directly proportional height

Law of conservation of mechanical energy:

Mechanical energy of a body is the sum of kinetic and potential energies of the

body.

Constant pk EE

pk EE ofGain of Loss

mghmv 2

2

1

or ghv 2



Oct/Nov 2009, Paper 22, question 3(c)



May/June 2008, Paper 2, question 2(b)




Electrical potential energy (Ep):

The electric potential energy due to charge Q (capital Q) at a point in an

electric field is defined as being equal to the work done in bringing a

unit positive charge q (small q) from infinity to that point. It follows

that the electrical potential energy of a charge Q at a point where the

potential is V is given by

m

m

h


Internal energy

The internal energy of a system is the sum of the kinetic and electric potential energies of the

molecules of the system. It follows from the equation wUQ

The heat energy (Q) supplied to a system is equal to the increase in the internal energy (U) of the

system plus the work done (w) by the system on its surrounding.

Temperature is a measure of the average random kinetic energy or internal energy of the molecules of

an object. When an object is moving it possess the ordered kinetic energy.

Power:

Power of a machine is the rate at which it does work or the rate at which it supplies energy. The unit

of power is joules per second or watt. t

WP

But FsW

or t

FsP

FvP

Efficiency of machines:

The ratio of useful work done, or energy output, to the work or energy input, in an energy transfer

system such as a machine or an engine.

Exercise: 2.14 Solve the following questions from past paper.





Oct/Nov 2006, Paper 1, question 17, 18


Three forces in equilibrium:


PRACTICE QUESTIONS – UNIT 2:

Laws of motion:

1) A body of mass 7.0 kg rests on the floor of lift. Calculate the force, R, exerted on the body by

the floor when the lift a) has upward acceleration of 2.0 m/s2, b) has downward acceleration

of 3.0 m/s2, c) is moving down with constant velocity. (assuming g = 10 m/s

2)

2) A train is moving along a straight horizontal track. A pendulum suspended from the roof of

one of the carriages of the train is inclined at 4o to the vertical. Calculate the acceleration of

the train (assuming g = 10 m/s2)

3) Two blocks, A of mass m and B of mass 3m, are side by side in contact with each other. They

are pushed along a smooth floor under the action of a constant force F applied to A. Find a)

the acceleration of the blocks b) the force exerted by B on A.

4) A body of mass 6.0 kg moves under the influence of two oppositely directed forces whose

magnitudes are 60N and 18N. Find the magnitude and direction of the acceleration of the

body.

5) Two forces of magnitudes 30N and 40N and which are perpendicular to each other, act on a

body of mass 25kg. Find the magnitude and direction of the acceleration of the body.

6) A body of mass 3.0 kg slides down a plane which is inclined at 30o to the horizontal. Find the

acceleration of the body. a) if the place is smooth, b) if there is frictional resistance of 9.0 N.

7) A railway truck of mass 6.0 tones moves with an acceleration of 0,050 m/s2 down a track

which is inclined to the horizontal at an angle where sin = 1/120. Find the resistance to

motion, assuming that it is constant.

8) A body hangs from a spring-balance which is suspended from the ceiling of lift. What is the

mass of the body if the balance registers a reading of 70N when the lift has an upward

acceleration of 4.0 m/s2.

9) What is the apparent weight during the take-off of an astronaut whose actual weight is 750N

if the resultant upward acceleration is 5g.

10) A body of mass 5.0 kg is pulled along smooth horizontal ground by means of 40N acting at

60o above the horizontal. Find: a) the acceleration of the body; b) the force the body exerts on

the ground.

11) A ball is thrown vertically upwards with a velocity 20m/s. Calculate a) the maximum height

reached, b) the total time for which the ball is in the air.

Projectile Motion:

12) A body is projected with a velocity of 200m/s at an angle of 30 o above the horizontal.

Calculate: a) the time taken to reach the maximum height b) its velocity after 16s.

13) A particle is projected with a speed of 25m/s at 30o above the horizontal. Find: a) the time

taken to reach the highest point of the trajectory b) the magnitude and direction of the velocity

after 2.0s.

14) A particle is projected with a velocity of 30 m/s at an angle of 40o above a horizontal plane.

Find: a) the time foe which the particle is in the air b) the horizontal distance it travels.

15) A pebble is thrown from the top of a cliff at a speed of 10m/s and at 30o above the horizontal.

It hits the sea below the cliff 6.0s later. Find: a) the height of the cliff, b) the distance from the

base of the cliff at which the pebble falls into the sea.

16) An airplane moving horizontally at 150m/s releases a bomb at a height of 500m. The bomb

hits the intended target. What was the horizontal distance of the aero plane from the target

when the bomb was released?


Moments and equilibrium

17) A uniform plank AB which is 6 cm long and has a weight of 300N is supported horizontally

by two vertical ropes at A and B. A weight of 150N rests on the plank at C where AC = 2 cm.

Find the tension in each rope. [Ta=250N, Tb=200N]

18) The system of forces is in equilibrium. Find P and Q

[P=15.6N, Q=10.2N]

19) A uniform ladder which is 5m long and has mass of 20kg

leans with its upper end against a smooth vertical wall and

its lower end on rough ground. The bottom of the ladder is

3m from the wall. Calculate the frictional force between the

ladder and the ground (g=9.8m/s2). [F=75N]

20) Two forces P and Q, acts NW and NE respectively. They are in equilibrium with a force of

50.0N acting due S and a force of 20.0 N acting due E. Find P and Q.

Energy

21) A car of mass 800kg and moving at 30m/s. along a horizontal road is brought to rest by a

constant retarding force of 5000N. Calculate the distance the car moves whilst coming to rest.

[s= 72m]

22) A small block is released from rest at A and slides

down a smooth curved track. Calculate the velocity

of the block when it reached B, a vertical distance h

below A. [v= gh2 ]

23) A car of mass 1.0 x 103 kg increases its speed from

10m/s to 20m/s whilst moving 500 m up a road

inclined at an angle to the horizontal where sin

= 1/20. There is a constant resistance to motion of

300 N. Find the driving force exerted y the engine,

assuming that it is constant. Take g = 10 m/s2

[F = 1.1 x 103 N]

A

h

40o 60o

20.0 N

Q P

B

unit 2 – motion, force and energy - webs 2 motion force and energy… · as physics 9702 unit 2:...

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