Document From Marzia

8/16/2019 Document From Marzia

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A LEVEL MECHANICSA QUICK REFERENCE GUIDE FOR FINAL PREPARATION

Prepared By:

Ahmed Saya Answers available at

www.eastpk.org/cars


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KINEMATICS

Kinematics means motion of particles with respect to time.

Scalar quantities: {Distance, Speed and time} only magnitude is important irrespective of the

direction. 10

7 Total = 10 + 7 = 17 units

Vectors quantities: {Displacement, velocity, acceleration} magnitude and direction both are

important. 10

3 7 Vector = 3 units

Distance: It is the length of the movement of particles since its start irrespective of the direction.

Speed: It is the rate of change of distance i.e. the rate at which distance is covered.

Displacement: It is the shortest distance from the starting point.

Velocity: It is the rate of change of displacement but it can be both positive and negative.

Acceleration: It is the rate of change of both, velocity as well as speed.

Types of questions of Kinematics:

1. Motion of the particle travelling at a constant speed.

D= S x T

T= D

S

S = D

T

2. Motion of the particles travelling at varying speed but constant acceleration.

(Multiple phased journey)

In order to solve such questions we will make:

I. Distance-time graphs.

II. Speed-time graphs.

III. Displacement-time graphs.

IV. Velocity-time graphs.


(Single phased journey)

To solve such questions we apply the following formulas:

I. V2 – u2 = 2as

II. S = ut + 1 at2

2

III. S = 1 (u + v)t

2IV. V = u + at

4. Motion of the particle, falling freely under gravity.

Gravity is equal to constant acceleration of 9.8ms-2 or 10ms-2. For the purpose of M1 if

question is silent then g = 10ms-2

.

5. Motion of the particle travelling at varying acceleration with respect to time.

To solve such questions we apply “The calculus" that is, differentiation and integration.


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1. Motion of the particle travelling at a constant speed.

D= S x T (km or m)

T= D (hrs. or sec.)

S

S = D (Kmh-1 or ms-1)

T

To convert from:

Kmh-1 to ms-1

X 1000

3600

To convert from:

Ms-1

to kmh-1

X 36001000

Average speed = Total distance

Total time

Example 1:

A cyclist is travelling at a speed of 20 ms-1

for 50 seconds on a dirt track. He then travels for 1500 meters on

a proper track at 50 ms-1

. Finally he travels for 900 meters for 30 seconds on a wet track. Find:

i) the distance travelled on the dirt track

ii) the time taken on the proper track

iii) the speed on the wet track

iv) average speed of the whole journey

Solution:

i) distance = 20 X 50 = 1000 meters

ii) time = 1500 / 50 = 30 seconds

iii) speed = 900 / 30 = 30 ms-1

iv) Average Speed = 1000 + 1500 + 900

50 + 30 + 30

3400

110

= 30.91 ms-1


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(Multiple phased journey)

Distance – time

A – B: m of d – t graph is constant speed.

(0, 0) and (10, 25)

A = 25 – 0 = 2.5 m/s.

10 - 0

B – C: m = 0. Particle is at rest. It is not moving. S = 0.

At C: Particle has resumed the journey.

C – D: m of d – f graph is constant speed.(25, 25) (30, 75)

S = 75 – 25 = 10 ms-1

30 – 25

Average speed = Total Distance

Total Time

= 75 = 2.5ms-1

30

Speed – Time graph:

75

25

3010 25

i t

(

)

A

B C

D

9010 50

C

A D

B

50

S p e e d (

s - 1 )

Time (seconds)


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A – B: m of S – T graph is constant acceleration. It indicates increasing speed.

B – C: m = 0; constant speed. Particle is not at rest.

At C: Particle has applied brakes.

C – D: m of S – T is negative constant acceleration (Deceleration (Retardation)). Particle is slowing

down and coming to rest.

Area under the S – T graph: Distance

Displacement – Time graph:

A – B: m of Displacement – Time graph is constant velocity.

B – C: m = 0; speed = 0; particle is at rest. Particle has turned.

At B: Particle has reached the maximum displacement.

At c: Particle has resumed its motion.

C – D: Particle is travelling in the opposite direction. ‘m’ of Displacement – Time graph = V.

At D: Particle has reached the starting point.

D – E: Particle has gone beyond the starting point.

Velocity – Time graph :

A – B: ‘m’ of V – T graph is acceleration (increasing velocity).

B – C: Particle is travelling at a constant velocity. Acc = 0.

At point C: Particle has applied brakes and is slowing down.

C – D: m of V – T graph is deceleration/ retardation / negative acceleration, decreasing velocity.

At D: Particle has turned.

D – E: Particle is now moving in opposite direction.

50CB

A

10 80 120100

E

DTime (seconds)

i s l c e

e

t (

)

‘x’

A

B C

D

E

‘y’

Time (s) l

i t (

- 1 )


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25

Police car

car20

A B

Q

10 t

Ct sec

EXAMPLE 2

A car is travelling at a constant speed of 72 km h-1 and passes a stationary police car. The police car

immediately gives chase, accelerating uniformly to reach a speed of 90 km h-1 in 10s and continues at this

speed until he overtakes the other car. Find

i) The time taken by the police to catch up with the car

ii) The distance travelled by the police car when this happens.

The v – t graphs of the car and the police are shown in the following figure

We work in meters and sec so 72 km h-1

= 20 m s-1

and 90 km h-1

= 25 m s-1

.

i) Time taken by the police car to overtake the other car is t s.

when this happens, both cars have travelled the same distance, therefore area of OABC = area

of OPQC

1

2 [t + (t – 10)] x 25 = 20 x t

t = 25 s

ii) The distance travelled by both the cars is the same:

20 x 25 = 500 m

V m s-1


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Starting point

0

Turning point, v = 0. Rest

Gone movement in

Beyond opposite direction

The Starting

Point

If X > Y, the particle has not yet reached the starting point, displacement is positive.

X

Y

If X = Y, the particle is at the starting point, displacement is zero.

X

Y

If X < Y, the particle has gone beyond the starting point, displacement is negative.

X

Y


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3. Motion of the particle travelling at varying speed but constant acceleration.

(Single phased journey)

(o, u) (t, v)

v – u = acc

t – o

v – u = at

v – u = at

v = u + at

½ x [ u + v ] x t = S

S = ½ [ u + v ] t

S = ½ ( u + v ) t

S = ½ (u + u + at) t

S = ½ x ( 2ut ) + ½ at2

S = ut + ½ at

2

v = u + at

v – u = at

T = v –u

a

S = ½ ( u + v ) t

S = ½ ( v + u ) ( v – u )

a

S = v2 – u2

2a

v2 – u

2 = 2as

u

v

tTime (s)

e l o c i t y (

s - 1 )


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EXAMPLE 3

A particle is travelling at a constant acceleration of 0.06 ms-2. His initial velocity is 3 ms-1 and after t seconds

has a velocity of 6 ms-1. Find

(a) the total time taken for the journey.

(b) the total distance travelled

A to B: constant a = 0.06 ms2

a) a = v2 – v1

T . T

T . T = 6 – 3 = 50s

0.06

b) Distance = ½ x 50 x (3 + 6) = 225 m.

EXAMPLE 4

Two particles P and Q are moving on the same horizontal line towards each other. P passes a point on theline with speed 8 m s-1 and constant acceleration 0.75 m s -2. Simultaneously, Q passes a point B on the line

with speed 4 m s-1 and constant acceleration 1.25 m s -2. Given that distance AB is 64 m ,calculate

i) Time taken for the particles to collide

ii) The distance from A of the point of collision

Solution:

i) Let the time be t s

P will travel : 8t + ½ (0.75)t2 ; and Q will travel : 4t + ½ (1.25)t

2

These distances total 64 m

then 8t + ½ (0.75)t2 + 4t + ½ (1.25)t

2 = 64 ; t

2 + 12t – 64 = 0

giving t = -16 (not admissible) or t = 4.

So they collide after 4 seconds.

ii) Distance from A of the point of collision is then

8 x 4 + ½ (0.75) x 16 = 38 m


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4. Free fall under gravity:-

Gravity:

Gravity is the acceleration with magnitude 10ms-2

Dropped:

Whenever the particle is dropped, it will always have an initial speed of 0. When the particle is

dropped it will fall vertically downwards.

Thrown downwards:

When the particle is thrown downwards, it will definitely have some initial speed.

Thrown upwards:

When the particle is thrown upwards, it will always move vertically upwards. It must have some

initial speed.

When particle is moving downwards, its gradient is positive i-e. 10 m/s.

When particle is moving upwards, its gradient is negative i-e. -10 m/s.

When a particle falls freely under gravity. Its motion is symmetrical. If the motion is not obstructed

by any hurdle.

Max pointv = 0

S, t, g = -10 s, t, g = 10, v = v

u = u

When the motion of the particle is obstructed then the particle will never reach the maximum

height, instead it will reach the greatest height.

Max height

Greatest height

When the motion is obstructed, then it is not symmetrical.

Note: - Velocity at max height is zero; velocity at greatest height is not zero.

When a particle moves freely on a smooth slope then acceleration of the particle is g sin ө.


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EXAMPLE 5

A particle is projected vertically upwards with a velocity of 30 m s-1 from a point O. Find

a) The maximum height reached

b) The time taken for it to return to O

c) The time taken for it to be 35 m below O

Solution:

Take positive direction as vertically upwards; the initially velocity u is 30 m s-1

a) At max height, v = 0; acceleration is -10 m s -2.

Using equation v2 = u

2 + 2as

We get h = 45 m

hence max height is 45 m

b) When particle returns to O, displacement is 0

Using equation s = ut + ½ at2

We get 5t2 + 30t = 0; t = 0 (at start) and t = 6 (when particle returns to O)

c) When particle is 35 m below, displacement is -35 m

Using equation s = ut + ½ at2

We get t = -1 (not admissible) and t = 7

Hence particle will take 7 seconds


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5 Motion of the particle with varying acceleration.

If acceleration is varying with respect to time then velocity and displacement will also vary with

respect to time and hence in order to solve such questions, we have to apply the“Calculus” i.e.

differentiation and integration.

Displacement

∫ v = s ds = vdt

Velocity

∫ a = v dv = adt

Acceleration

Key Points to remember:

1. At initial time, t = 0.

2. At turning point, v = 0.

3. At max velocity, acc = 0.

4. When the particle is instantaneously at rest, velocity is zero.

5. Therefore, at turning points, velocity is zero.

6. At max displacement, velocity is zero.

7. At starting point, displacement is zero.

8. When particle is moving away from the starting point, displacement is increasing.

9. When particle is moving towards the starting point, displacement starts decreasing.

10. Displacement is the shortest distance from the starting point.

11. When the particle goes beyond the starting point, displacement is negative.

12. Terminal speed is the speed at which acceleration is zero.

13. When you are told to find the displacement, insert the time in the equations of

displacement.


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14. When you are told to find the distance, follow the following steps:-

Step 1: Check whether the particle is turning within the required time.

Step 2: If the particle does not possess the turning or does not turn within the

required time, the distance is equal to displacement.

Step 3: If the particle turns within the required time; the distance is not equal to

displacement and hence to find distance, the following steps are to be followed.

(a) Insert the time of turning in the equation of displacement.

(b) Insert the time at which distance is required in the equation of

displacement.

(c) If b > 0

a

b a - b

Distance = a + a – b = 2a – b

(d) If b = 0

a

a

Distance = 2a

(e) If b < 0

a

b a

Distance = 2a + b

15. When the distance from 0 is required, it means that the question is asking for displacement.


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Example 6

A particle travels in a straight line from a point A to a point B. its velocity t seconds after leaving A is v m s-1,

where

V = 4t – 0.04t3

Given that AB is 100 m, find

a) the value of t when the p

article reaches B

b) whether the particle is speeding up or slowing down at the instant that it reaches B

N = 4t – 0.04t3

S = ∫ 4t – 0.04t3 S = 4t

2 – 0.04t

4

2 4

s = 2t2 – 0.01t

4

When s = 1002t

2 – 0.01t

4 = 100

0.01t4 – 2t

2 + 100 = 0

0.01t2 – 2t + 100 = 0

T = 100

t2 = T ii) In order to determine whether the particle is speeding

t2 = 100 up or slowing down check the sign of acceleration, if acc. is positive,

t = 10sec the particle is speeding up and if acc. is negative particle is slowing

ii) v = 4t – 0.04t3 down.

a = 4 – 0.12t2

when t = 10

a = 4 – 0.12(10)

2

= - 8

As “a” is negative, the particle is slowing down.

Note: In order to determine whether the particle is speeding up or slowing down check the

sign of acceleration, if acc. is positive, the particle is speeding up and if acc. is negative particle is

slowing down.


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Exercise 1

1. A train moves with a constant speed of 20 m s-1 for 10s. it then accelerate uniformly for next 5s to

reach a speed of V m s-1 when it decelerates uniformly to rest in the next 4s.

a) Sketch a v – t graph for the journey.

b) If the total distance covered is 385m, find the value of V.

2. Two particles A and B are moving in the same direction on parallel horizontal lines. They pass a

certain point O at the same time, A moving with the speed of 3 m s-1

and constant acceleration 0.5

m s-2 and B moving with the speed of 4 m s-1 and constant acceleration 0.2 m s-2+.

If the speed of A and B are equal after t seconds, calculate

a) The value of t

b) The distance between the particles at that instant.

3. A stone is dropped from the top of a building of height 20 m. find the time it takes to reach the

ground and the velocity with which it hits the ground.

4. From the top of a vertical tower 246m high, a stone is projected vertically downwards with a speed

of 8 m s-1. After t seconds another stone is projected vertically upwards from the level of the base of

the tower with a speed of 25 m s-1. Given that two stones are first at the same height 6 seconds

after the projection of the first stone, calculate

a) The value of t

b) The velocity of the second stone at this instant

5. A particle is projected vertically upwards from a ground with a speed of 36 m s-1. Calculate

a) The time for which it is 63 m above the groundb) Speed which it has at this height on its way down

c) Total time of flight

6. A particle is travelling in a straight line with constant acceleration a ms-2. It passes a point A with

speed u ms-1

. between 3 and 4 seconds after passing A it travels 11.4m and between 4 and 5

seconds after passing A it travels 11.8 m. find the value of u and of a.


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VECTORSHow to solve vectors

Head to tail rule. Parallelogram Rule.

Example1 Example 2

One vector will be followed Both vectors will start from the

by the other vector. same point at the same instant.

2km

5km 5.44 km

5km 5.44m

21.8

0

21.8

0

2 km

Example 3

Find the resultant of the following vectors?

a) 10m 0600. Followed by 4m east.

b) 40km on the bearing of 600 and 40 km on the bearing of 110

0.

a)

120 90 4m

10m 150 60 8.5

13.6m


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Exercise 2

1. Two forces, each of magnitude 10 N , act at a point O in the directions of OA and OB, as shown in

the diagram. The angle between the forces is θ. The resultant of thes two forces has magnitude 12

N.

a) Find θ

b) Find the component of the resultant force in the direction of OA

2. Two horizontal forces, of magnitudes 10 N and P N, act on a particle. The forces of magnitude 10 N

acts due west and the forces of magnitude of P N acts on a bearing 030 (see diagram). The resultant

of these two forces acts due north. Find the magnitude of this resultant.

10 N

10 N

θ

30

P N

10 N


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FORCES IN EQUILIBRIUM What is force?

Force is a push or pull exerted on a particle in order to move the particle or stop the particle.

Types of Forces:

1. Weight:

It is the gravitational attraction of the earth on the particle. Weight always act

vertically downwards ( ) and the formula to obtain weight is:

w = m x g

N = kg x ms-2

e. g. 2kg w = 2 x 10 = 20

2. Reaction / Normal Reaction:

It is always perpendicular to the slope on which particle is lying. It

will only exist if the particle is on the surface.

R

3. Air Resistance:

If the particle is not on the surface then instead of normal reaction air resistance

exists. If the particle is falling down, air resistance will act vertically upwards and if the

particle is going up, air resistance will act vertically downwards.

Air resistance

Air Resistance

4. Tension:

When the particle is being pushed or pulled through a string, then the force exerted

on the string is called tension.

Tension


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5. Friction / Frictional force:

It is always opposite to the motion of the particle, when the particle

is moving on a rough surface. On smooth surface, friction do not exists. The formula to

calculate friction is:

ff = µ R

Where R is the Normal Reaction.

µ is the coefficient of friction.

On smooth surface µ = 0, therefore on smooth surface friction does not exist.

f f. f f.

F

What is Equilibrium?

Condition of equilibrium occurs when the particle is either travelling at a constant speed or

is at rest. In both of these conditions, acceleration is equal to zero. Hence equilibrium occurs when

acceleration = 0.

When acceleration is zero and equilibrium prevails then

1- All upward forces are equal to downward forces or sum of all vertical forces is equal

to zero.2- All right hand forces are equal to left hand forces or sum of all horizontal forces is

equal to zero.

s + x = 8

x = 3N

Y

2Y 12

y + 12 = 2y

y = 12N

5 N X N

8 N


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2x

8 + 3y 5y

X 2 8

5y = 8 + 3y

2x = x + 2 + 8

3- If the force applied to the particle is an inclined force then first we will resolve the

force and then equate the upward forces with the downward forces.

F

ө F sin ө

F cos ө

How to resolve forces in equilibrium?

Step 1: Determine your own axis (preferably consider the axis parallel to the slope as x – axis and

consider the axis perpendicular to the slope as y – axis).

Step 2: Resolve all forces applied to the particle in accordance with the axis determined.

Step 3: Equate all vertical forces to zero.

Step 4: Equate all horizontal forces to zero.

Step 5: Hence find the unknowns.

EXAMPLE 1

NR

F

5kg

F sin 30 300

Fcos30 300 50 cos 30

ff = 0 300

50 sin 3050N

Smooth Surface R = Fsin30 + 50cos30

µ = 0 = 28.9sin30 + 50cos30

ff = 0 = 57.8N

Fcos30 = 50sin30

F = 50sin30

Cos30

= 28.9N


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Note:

Angle between the weight component and resolving component will always be equal to the angle of

slope.

EXAMPLE 2:

A particle of mass 1kg rests on a horizontal floor. The coefficient of friction between the particle and the

floor is 0.5. What force is required just to make the particle move when:

a) Pulling horizontally.

b) Pulling at an angle of 300 to the horizontal.

Note:

When you are encountered with the following terms, it is an indication that the particle is in the

condition of equilibrium.

i) Particle is at rest.

ii) Particle is travelling at a constant speed.

iii) Particle is at maximum speed.

iv) Particle is at momentary rest.v) Particle is about to move.

vi) Particle is about to slip.

vii) Particle is travelling at a terminal velocity.

R

solution a)

ff = 0.5R 1kg x

10N

R = 10N 0.5R = x

0.5(10) = x

x = 5N

solution b)

F

Fsin30

300

ff = 0.5R Fcos30

10N

R + Fsin30 = 10 ff = 0.5(10 – 1 F) F = 4.48N

R = 10 – 1 F 2

2

Fcos30 = 5 – 1 F R = 10 – 0.5(4.48) = 7.76N

4


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P NT N

30 ͦ

10 N

40 ͦ

LAMI’S THEOREM: o Conditions to apply LAMI’s theorem:

1. Particle must be in equilibrium.

2. Only three forces must act on a particle.

3.

All three forces must either diverge from the particle or converge to the particle.

4. It is a modified form of sine rule.

Note:

LAMI’s theorem is an alternative method of resolving forces when the particle is in

equilibrium. The formula is:

Sin A = Sin B = Sin C

a b c

EXAMPLE 3

A particle of weight 10 is suspended by 2 strings. If these strings mae angles of 30 ͦand 40 ͦto thehorizontal. Find the tensions in the string.

Solution:

Let the tension in the string be T N and P N respectively.

Using Lami’s theorem,

T

sin0 ͦ 40 )ͦ =

sin 0 ͦ 30 ͦ) =

10

sin 110 ͦ

Giving T = 8.2 N and P =9.2 N

Where a, b and c are three forces and A, B and C are angles between the forces.

Diverge Converge


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If the F.F = D.F; the particle is at rest.

If the F.F < D.F; the particle will not be at rest.

If the F.F > D.F; the particle will be at rest.

o The resistive force is greater than the driving force; the value of resistive force will be equals

to the driving force.

F.F = 10 D.F = 10

F.F = 10 D.F = 15

7

D.F = 7f.f = 10


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Exercise 3

1. particle of mass 1kg is placed on a rough plane inclined at an angle of 300 to the horizontal. The

coefficient of friction is 2/5. Find the least force parallel to the plane that is required

a) To hold the particle at rest.

b) To make the particle slide up the plane.

2. Find sum of the components of the forces acting as shown in the figure in the direction of the dotted

lines and hence find their resultant

3. Two forces P N and Q N include an angle of 120 ͦand their resultant is 1 . if the included angle

between the forces is 0 ͦ , their resultant would be . find and

4. A box of mass 1 g is placed on an inclined plane of angle 0 ͦ and is ust held there at rest by a

horizontal force P. if the coefficient of friction is 0.4, find P.

5. A horiontal force of 10 ust prevents a mass of 2 g from sliding down a rough plane inclined at

4 ͦ to the horiontal. Find the coefficient of friction.

6. Figure shows a small block of mass 5 kg held against a rough vertical wall by a force inclined at

and angle of 0 ͦ to the wall. If the coefficient of friction is 0.4 , calculate the value of when the

block is

a) Just prevented from slipping down

b) Just about to move up the wall

10 N

30 ͦ

30 ͦ

0 ͦ

30 N

20 N

0 ͦ P N


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7. The three coplanar forces shown in the following figure are in equilibrium. Calculate the value of Ѳ

and P

8. Two rectangular boxes A and B are of identical size. The boxes are at rest on a rough horizontal floor

with A on top of B. Box A has mass 200 kgand box B has mass 250 kg. A horizontal force of

magnitude P N is applied to B (see diagram). The boxes remain at rest if ≤ 310 and start to move

if P > 3150.

a) Find the coefficient of friction between B and the floor.

The coefficient of friction between two forces is 0.2. Given that P > 3150 and that no sliding takes

place between the boxes .

b) Show that the acceleration of the boxes is not greater than 2 m s-2

c) Find the maximum possible value of P.

3P N

30 N

120°

2P N

Ѳ


26/39


27/39

If multiple forces are applied on a particle which is not in equilibrium then we will resolve the

vertical forces and the horizontal forces and then find the resultant of the vertical and horizontal forces.

Newton’s laws of motion applied to conditions of “Pulley”:

Assumption when applying conditions of Pulley:

1. Both particles must be attached through a single string.

2. The string must be weightless i.e. it should be light.

3. The string must be inextensible.

4. Pulley must be frictionless i.e. smooth.5. Pulley must be weightless.

6. At the start of the motion, under the condition of pulley, the particle must be at rest.

7. The particle will remain under the condition of pulley until and unless.

a) The string breaks.

b) One of the particles hits the ground.

c) One of the particles reaches the pulley.

8. The tension throughout the string will be same.

9. As long as particles will be under the condition of pulley, the acceleration will be the same.

10. As soon as the particles will go out of the conditions of pulley, they will come under the

condition of free fall under gravity.

11. If the question is silent, the motion of the pulley will be in the direction of the heavier particle.12. When the string breaks or one of the particles hits the ground or reaches the pulley, the string

become slack.

13. Condition of pulley is restored when the string becomes taut again.

(Hook) x

Slack tout

Example 4

70 – T = 7a

T T T – 30 = 3a

3kg 7kg 70 – T = 7a

-30 + T = 3a

40 = 10a

30 70 a = 4ms-2

T = 3(4) + 30 = 42N.

Tension exerted on the pulley will always be 2 times the tension:

x= 2(T) = 2 x 42 = 84N


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EXAMPLE 5

The following figure shows two particles: A of mass 1 kg and B of mass 2 kg connected by a light inextensible

string which passes over a smooth pulley at C. the system is held at rest with B hanging freely while A is on a

rough plane inclined at Ѳ to the horiontal where tanѲ = ¾ . The coefficient fo friction between A and the

plane is 0.2. find the magnitude of the acceleration of each particle and the tension in the string when the

system is released.

Solution:

Let the acceleration of B be a ms-2 downwards and hence A will move towards C with the same acceleration.

The string transmits a tension T. The figure below shows the forces acting on A and B.

Now consider the motion of each particle separately.

For B, since acceleration is downward,

2g – T = 2a

For A, since acceleration is along AC, and the tension is opposed by the frictional force , F and the component of

weight downslope, g sin Ѳ, we have

T – (F + g sin Ѳ) = 1a

resolving forces perpendicular to the plane , the normal reaction is

R = g cos Ѳ

the frictional force F is given by

F = µ R

= µ g cos Ѳ

Substituting F into 2nd

equation: T – (µ g cos Ѳ g sin Ѳ) = a

Now solve to obtain a and T.

Adding both the equations

2g – (µ g cos Ѳ g sin Ѳ) = 3a

2 kg

1 kg

Ѳ

C

AB

Ѳ

C

g

T

F

R

aT


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2(10) – [(0.2)(10)4

5 + (10)

3

5 ] = 3a

a = 4.13 ; and T = 11.74

Hence the acceleration of each particle is 4.13 m s-2 and the tension in the string is 11.74 N.


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B

5 m

P R

A

1.5kg

3 m2m

Exercise 4

1. A particle of mass 2.5 kg is moving at a steady speed of 12 m s-1 when it meets with a fixed

resistance of 10 N. How long does it take to come to rest?

2. A body of mass 4 kg is pulled from rest to a speed of 4.5 m s-1 in a time of 3 seconds on a rough

horiontal surface by a force of 20 which maes an angle of 10 ͦ with the horizontal. Find thecoefficient of friction.

3. PQR is a fixed wedge on level ground (as shown in the figure) where PQ = 5 m, QR = 3 m and PRQ is

a right angle. Particle A of mass 1.5 kg, lies at the foot of the smooth slope PQ, attached by a light

inextensible string passing over the smooth pulley at Q, to the particle B of mass 1 kg . B is released

from rest when it is 2 m above ground level. Find

a) The acceleration of the particles,

b) How far A will travel up the slope before coming to momentary rest.

4. A wedge has two equally rough faces each inclined at 30 ͦ to the horiontal. Masses of g and 2 g,

one on each face, are connected by a light inextensible string passing over a smooth pulley at the

top of the wedge. The coefficient of friction between each mass and the surface of the wedge is 0.2.

Find the acceleration of the masses when they are released.


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5. The following figure shows two masses of 5 kg and 7 kg respectively connected by a light

inextensible string which passes over a smooth fixed pulley. The system is released from rest and

the 7 kg mass reaches the ground after 3 second s. Calculate

a) The acceleration of masses while the string remains taut

b) The total distance moved by the 5 kg mass before it comes instantaneously to rest,

assuming that it does not reach the pulley.

6. A block A of mass 3 kg is attached to a string S1. Another block B of mass 2 kg is attached to the othe

other end of the string S1, and is also attached to another light inextensible string S2. The other end

of S2 is attached to a fixed point O and the blocks hang in equilibrium below O as shown in the figure

a) Find the tension in S1 and tension in S2.

The string S2 breaks and the particles fall. The air resistance on A is 1.6 N and on B is 4 N

b) Find the acceleration of the particles and the tension in S1.

7 kg

5 kg


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WORK POWER AND ENERGY

WORK DONE

When a force acts on a particle and causes the particle to move the work is said to be done on the

particle. The amount of work done is defined as the product of the force and the distance moved by

the particle in the direction of the force and hence:

W.D = F x S

F = 15N

10m

W.D = 15 x 10 = 150J.

Work done = Joules.

= N x m = Joules

If an inclined force is applied to the particle causing the particle to move horizontally then the work

done will be:

W.D = Fcosө x S.F

Fsinө Fcosө

S

Example1:

A force of 60N inclined at 600 to the horizontal caused the particle to move 7m horizontally.Find the work done by the particle.

60N

60

60cos60

W.D = 60cos60 x 7

= 210 J.

If an inclined force is applied to the particle causing it to move vertically upwards then the

word done will be:

W.D = Fsinө x S

F

Fsinө S


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EXAMPLE 3

A force of 50N is being applied on a particle of 10kg giving it a speed of 15 ms-1

from rest. Find the work

done by the kinetic energy.

K.E = ½ x 10 x 152 – ½ x 10 x 0

2

= 1125 J.

EXAMPLE 4

A force acting on a particle of mass 2 kg moves it from rest to a velocity of 3ms-1 over a distance of 5m. What

is the magnitude of the force applied?

M = 2kg v = 3 u = 0.

` W.D = ½ x 2 x 32 – ½ x 2 x 02

= 9J.

W.D = F x S

9 = F x 5

F = 1.8N

NOTE:When a particle travels through steady speed it does not have a kinetic energy because there is no

change in kinetic energy.

EXAMPLE 3

A car of mass 400kg travelling at 9ms-1 comes to rest in 200m. What is the resistance?

W.D = ½ x 400 x (02 – 92)

= -16200 J

NOTE:

When there is loss in kinetic energy it is an indication that some resistive force is applied to the

particle in order to stop it.

F x S = 16200

F x 200 = 16200

F = 81N

2. Potential Energy:

Potential energy is the ability to do the work because of the position of the particle

when the particle is on the ground, it has no potential energy. As particle moves up,

there is a gain in potential energy. Potential energy is also referred to as work done

against gravity and the formula to calculate P.E. is:

P.E = mgh

Where m is mass, g is gravity and h is height.

In order to find the work done by the potential energy we have to find the difference between final

potential energy and initial potential energy.

W.D by P.E = Final P.E - Initial P.E = mgH – mgh

mg (H-h)


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EXAMPLE 5

Find the work done when a load of mass 50kg is lifted vertically through 10m.

W.D by P.E = 50 x 10

= 5000kJ

EXAMPLE 6

A block is pulled through 4m at a steady speed by a force of 20N. The plane is inclined at an angle of 600.

Find

i) W.D by K.E

ii) W.D by P.E 4m 20N

h

600

i) No K.E

ii) H = 4sin60 = 23m

W.D by the particle = 4 x 20 = 80J

As there is zero K.E, all W.D will be by the P.EP.E = 80J

Mgh = 80

M 80 ÷ 10 x 23) = 2.3g.

NOTE:

When a particle travels horizontally there is no change in potential energy and hence work done

by potential energy is zero.

3. Mechanical Energy:

Mechanical energy is the sum of kinetic energy and potential energy.M.E = P.E. + K.E.

Mechanical energy throughout the motion of the particle is constant IF no external force is applied to it.

External force = Driving Force or Resistive force or Both driving and resistive force

When NO external force is applied, Law of Conservation of Energy is being applied to the particle. Law of

Conservation of Energy states that since mechanical energy does not change therefore change in kinetic

energy must correspond with change in potential energy, thus;

Gain in K.E = Loss in P.E

Loss in K.E = Gain in P.E

When an external force is applied to the particle law of conservation of energy does not prevail and hence

we will apply the following formula to find any possible unknown.

Work done by = Change in Kinetic Energy + change in potential energy + work done by resistive force


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Example 7

The diagram shows a vertical cross section of a surface. A and B are two points on the cross section. A mass

of 0.15 kg is released from rest at A.

a) Assuming that the particle reaches B with a speed of 8 m s-1 and that there are no resistances to

motion. Find the height of A above B

b) Assuming instead that the particle reaches B with a speed of 6 m s-1 and that the height of A above B

is 4 m. find the work done against the resistance to motion.

Solution

a) ½ x 0.15 x 82 = 0.15 x 10 H.H = 3.2m.

b) W.D by D.F = ∆ in K.E ∆ in P.E + W.D by R.F

0 = (1/2 x 0.15 x 82) + (-0.15 x 10 x 4)

m = 3.3J

Example 8

The top of an inclined plane is at a height of 0.7m above the bottom. A block of mass 0.2kg is released from

rest at the top of the plane and slides a distance of 2.5 m to the bottom. Find the kinetic energy of the block

when it reaches the bottom of the plane in each of the following cases:

a) The plane is smoothb) The coefficient of friction between the plane and the block is 0.15

R

0.7 ө 2 2.5m

Ө

2.4

i) K.E = mgh

= 0.2 x 10 x 0.7 = 1.4J.

ii) R = 2cosө = 2(2.4/2.5) = 1.92

ff = µR

= 0.15 x 1.92 = 0.288.

W.D b D.F = ∆ in K.E ∆ in P.E + W.D by Rf.

0 = x + (0.2 x 10 x -0.7) + (0.288 x 2.5)

0 = x – 1.4 + 0.72

X = 0.68J

B

A


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Exercise 5

1. A force acting on a body of mass 2 kg moves it from rest to velocity of 3 m s -1 over a distnace of 5 m.

what is the magnitude of the force?

2. A car of mass 1500 kg arrives at the foot of a straight hill travelling at 30 m s-1

. it reaches

the top of the hill 40 sec later at 10 3 m s-1

. The length of the hill is 1000 m, and gain in the

height is 120 m. the average resistance to the motion is 500 N. Find the work done by the

engine and the average power developed.

3. A particle of mass 0. g is proected up an incline of angle Ѳ where sin Ѳ = 0. with speed 4 ms -1 .

how far will it travel up the incline if

a) The surface is smooth

b) The coefficient of friction is ¼ ?

4. The power developed by a motorcycle as it travels on a horizontal straight road at s constant speed

of 50 m s-1

, is 25 kW. Calculate the resistance of the motion of the motorcycle.

5. A small block is pulled along a rough horizontal floor at a constant speed of 1.5 m s -1 by a constant

force of magnitude 30 acting at an angle of Ѳ upwards from the horiontal. Given that the work

done by the force in 20s is 20J. calculate the value of Ѳ.

6.

The diagram shows a vertical cross section of LMN of a fixed smooth surface. M is the lowest point of the

cross section. L is 2.45 m above the level of M and N is 1.2 m above the level of M. a particle of mass 0.5 kg

is released from rest at L and moves on the surface until it leaves it at N. Find

a) The greatest speed of the particle

b) The kinetic energy of the particle at N.

The particle is now projected from N with a speed v ms -1 , along the surface towards M.

c) Find the least value of v for which the particle will reach L.

L

2.45m

1.2m

M

N


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7. A lorry of mass 16 000 kg climbs from the bottom to the top of a straight hill of length 1000 m at a

constant speed of 10 m s-1

. the top of the hill is 20 m above the level of the bottom of the hill. The

driving force of the lorry is constant and is 5000 N. Find

a) The gain in gravitational potential energy of the lorry

b) Work done by the driving force

c) Work done against the force resisting the motion of lorry

On reaching the top of the hill, the lorry continues along a straight horizontal road against a

constant resistance of 1500 N . the driving force of the lorry is not now constant, and the speed of

the lorry increases from 10 m s-1 at the top of the hill to 25 m s -1 at the point P. the distance from P

to the top is 2000 m .

d) Find the driving force of the lorry while it travels to P from the top.

8. i) A car C of mass 1200 kg climbs a hill of length 500m at a constant speed. The hill is inlcined

at an angle of ˚ to the horiontal. The driving force exerted by C’s engine has magnitude 1800.find the work done against the resistance of the motionof C, as it climbs from the bottomof the hill

to the top.

ii) Another car D, also of mass 1200 kg, climbs the same hill with increasing speed. The speed

at the bottom is 8 ms-1

and the speed at the top is 20 ms-1

. Assuming the resistance to the motion of

D is constant and has magnitude 00, find the wor done by D’s engine as D climbs from the

bottom of the hill to the top.

iii) The driving force exerted by D’s engine is 4 times as great when D is at the top f the hill as it

is when D is at the bottom. Find the ratio of the power developed by D’s engine at the top of the hill

to the power developed at the bottom.

9. i) A lorry P of mass 15000 kg climbs a straight hill of length 800m at a steady speed. The hill is

inclined at 2˚ to the horiontal. For ’s ourney from bottom of the hill to the top, find

a) the gain in gravitational potential energy

b) the work done by the driving force

c) the work done against the force resisting the motion

ii) A second lorry, Q also has mass 15000 kg and climbs the hill as P. the motion of Q is subject

to a constant resisting force of magnitude, 00, and ’s speed f alls from 20 ms-1

at the bottom of

the hill to 10 ms-1 at the top. Find the work done by the driving force as Q climbs from the bottom of

the hill to the top.


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10. The diagram shows a vertical cross section of ABCD of a surface. The parts AB and CD are straight

and have lengths 2.5 m and 5.2m respectively. AD is horiontal and AB is inclined at 0˚ to the

horizontal. The points B and C are at the same heights abpve AD. The parts of the surface containing

AB and BC are smooth. A particle P is given a velocity of 8 ms-1 at A, in the direction AB and it

subsequently reaches D. the particle does not lose contact with the surface during the motion.

i) Find the speed of P at B

ii) Show that the maximum height of the cross section above AD is less than 3.2 miii) State briefly why ’s speed is the same as that at B

iv) The frictional force acting on the particle as it travels from C to D is 1.4N. given that the

mass of P is 0.4kg, find the speed with which P reaches D

D

5.2 m

2.5 m

A B

A

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