2. force and motion - · pdf filejpn teacher’s guide physics module form 4

JPN Teacher’s guide Physics Module Form 4 Chapter 2 : Force and Motion

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2. FORCE AND MOTION

2.1 ANALYSING LINEAR MOTION

Distance and displacement

1. Types of physical quantity:

(i) Scalar quantity: ………………………………………………………………….

(ii) Vector quantity: …………………………………………………………………

2. The difference between distance and displacement:

(i) Distance: …………………………………………………………………………

(ii) Displacement: ……………………………………………………………………

3. Distance always longer than displacement.

4. Example: The following diagram shows the location of Johor Bahru and Desaru. You can travel by car using existing road via Kota Tinggi, or travel by a small plane along straight path. Calculate how far it is from Johor Bahru to Desaru if you traveled by: a. The car

b. The plane

Solution:

Hands-on Activity 2.2 pg 10 of the practical book. Idea of distance and displacement, speed and velocity. Speed and velocity

1. Speed is ..…………………………………………………………………………………

2. Velocity is: ..……………………………………………………………………………...

3. Average of speed: ………………………………………………………………………

4. Average of velocity: ……………………………………………………………………...

has only a magnitude

has both magnitude and direction

length of the path taken

distance of an object from a point in a certain direction

the distance traveled per unit time or rate of change of distance

the speed in a given direction or rate of change of displacement

total distance traveled, s (m) , v = s m s-1

time taken, t (s) t

displacement, s (m) , v = s ms-1

Time taken, t (s) t

Kota Tinggi

The path traveled by the plane is shorter than travelled by the car.

a. by car = 41 + 53 = 94 km b. by plane = 60 km

So, Distance = 94 km Displacement = 60 km

60 km

41 km 53 km

Desaru Johor Bahru


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5. Example:

An aeroplane flies from A to B, which is located 300 km east of A. Upon reaching B, the aeroplane then flies to C, which is located 400 km north. The total time of flight is 4 hours. Calculate

i. The speed of the aeroplane ii. The velocity of the aeroplane

Solution:

Acceleration and deceleration

1. Study the phenomenon below;

Observation: ………………………………………………………………………………

2. Acceleration is, ……………………………………………………………………….

Then, a =

3. Example of acceleration;

20 m s-1 0 m s-1 40 m s-1

The velocity of the car increases.

the rate of change of velocity

Final velocity – initial velocity Time of change

Or, a = v – u t

20 m s-1 0 m s-1 40 m s-1

A B C t = 2 s t = 2 s

C

A B 300 km

i. Speed = Distance Time

= 300 + 400 4 = 175 km h-1

400 km

ii. velocity = displacement time (Determine the displacement denoted by AC and its direction)

= 125 km h-1 (in the direction of 0530) A 300 km

= . 500 . 4

B

C

400 km


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Calculate the acceleration of car;

i) from A to B

ii) From B to C

4. Deceleration happens ...…………………………………………………………………

………………………………………………………………………………………………

5. Example of deceleration;

A lorry is moving at 30 m s-1, when suddenly the driver steps on the brakes and it stop 5 seconds later. Calculate the deceleration of lorry.

Analysing of motion

1. Linear motion can be studied in the laboratory using a ticker timer and a ticker tape.

Refer text book photo picture 2.4 page 26.

(i) Determination of time:

(ii) Determination of displacement as the length of ticker tape over a period of time. x y

(iii) Determine the type of motion;

………………………………………………………………………………………..

...……….……………………………………………………………………………..

.……………………………………………………………………………………..

aAB = 20 – 0 = 10 m s-2 2 aBC = 40 – 20 = 10 m s-2

2

when the velocity of an object decreases, In calculations, a

will be negative

Answer : v = 0 m s-1, u = 30 m s-1, t = 5 s Then , a = 0 – 30 = -6 m s-2

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. . . . . . . . the frequency of the ticker timer = 50 Hz ( 50 ticks in 1 second) so, 1 tick = 1 second = 0.02 seconds 50

xy = displacement over time t t = 7 ticks = 0.14 s

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

Uniform velocity Acceleration Acceleration, then deceleration


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(iv) Determination of velocity

displacement = ……………………… time = ………………………………..

Velocity, v =

(v) Determine the acceleration

The equation of motion

1. The important symbols : ………………………………………………………………..

………………………………………………………………………………………………

2. The list of important formula;

3. Example 1 : A car traveling with a velocity of 10 m s-1 accelerates uniformly at a rate of 3

m s-2 for 20 s. Calculate the displacement of the car while it is accelerating.

Length/cm

8 7 6 5 4 3 2 1 0

u

v a = = = =

v – u t 40.0 – 15.0 .. 5(0.2) 25.0 1.0 25.0 m s-2

ticks s : displacement, v : final velocity

u : initial velocity, t : time, a : acceleration

1. tvus )(21

2. t

uva

3. atuv 4. 2

21 atuts

5. asuv 222

given : u = 10 m s-1 , a = 3 m s-2 , t = 20 s. s = ?

s = ut + ½ at2

s = (10)(20) + ½ (3)(20)2

. . . . . . . . 12.6 cm 10 x 0.02 = 0.2 s 12.6 = 63.0 cm s-1 0.2


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Example 2 : A van that is traveling with velocity 16 m s-1 decelerates until it comes to rest. If the distance traveled is 8 m, calculate the deceleration of the van.

Execise 2.1

1. Figure 2.1 shows a tape chart consisting of 5-tick strip. Describe the motion represented by AB and BC. In each case, determine the ;

(a) displacement

(b) average velocity Figure 2.1

(c) acceleration 2. A car moving with constant velocity of 40 ms-1 . The driver saw and obtacle in front and

he immediately stepped on the brake pedal and managed to stop the car in 8 s. The distance of the obstacle from the car when the driver spotted it was 180 m. How far is the obstacles from the car has sttoped.

given : u = 16 m s-1 , v = 0(rest) , s = 8 m a = ?

v2 = u2 + 2 as

02 = 162 + 2 a(8)

a = -16 ms-2

Length / cm

16

12

8

4

0 A B C Time/s

s = 4 + 8 + 12 + 16 + 16 + 16 = 72.0 cm

vaverage = )1.0(6

0.72

= 120.0 cm s-1

Note : v = 1.00.16 = 1.6 cm s-1

a = t

uv = 5.0

4.06.1 u = 1.00.4 = 0.4 cm s-1

= 2.4 cm s-2 t = 5 (0.1) = 0.5 s

u = 40 ms-1 v = 0 t = 8 s s initial = 180 m (from car to obstacle when the driver start to step on the brake) sfinal = ? ( from car to abstacle when the stopped)

obstacle sinitial s sfinal

s = mtvu 160804021

21

sfinal = sinitial – s = 180 – 160 = 20 m


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2.2 ANALYSING MOTION GRAPHS

The data of the motion of the car can be presented………………………………….

The displacement-time Graph

0m 100m 200m 300m 400m 500m displacement 0s 10s 20s 30s 40s 50s time

a) displacement (m) Graph analysis: ……………………………………………………………… ……………………………………………………………… time (s) ……………...………………………………………………

b) displacement (m) Graph analysis:

……..………………………………………………………… ………………………………………………………………… time (s) ……….………………………………………………………… c) displacement (m) Graph analysis:

…….…………………………………………………………… ………………………………………………………………… time (s) ..…………………………………………………………………

The object moves with uniform velocity for t seconds. After t seconds, the object returns to origin (reverse) with uniform velocity Total displacement is zero Graph is quadratic form . Displacement increases with time. Graph gradient increases uniformly The object moves with increasing velocity with uniform acceleration.

d) Displacement (m) Graph analysis:

…………………………….……………………………………… ……………………………………………..……………………… time (s) ………………………………………………………………………

…………………………………………………………………

in the form of graph called a motion graphs

Uniform displacement all the time

Graph gradient = velocity = 0

The object is stationary or is not moving

Displacement increases uniformly

Graph gradient is fixed

The object move with uniform velocity


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The velocity-time Graph

c) v (m s-1) Graph analysis:

…………………………………..………………….

………………………………………………………

………………………………………………………

t1 t2 t (s)

e) displacement (m) Graph analysis:

…………………………………………………………..

…………………………………………………………..

………………………………………………………….. time (s)

………………………………………………………….. f) displacement (m) Graph analysis:

A B ………………………………………………………….. …………………………………………………………..

…………………………………………………………… O C time (s)

a) v/ m s-1 Graph analysis:

…………………………………………………………..

…………………………………………………………..

……………………………………………………………

t t / s

b) v/ m s-1 Graph analysis:

………………………………………..………………..

…………………………………………………………

…………………………………………………………

t t / s …………………………………………………………

Its velocity increases uniformly

The graph has a constant gradient

The object moves with a uniform acceleration

The area under the graph is equal to the displacement, s of the moving object : s = ½ ( v x t)

Graph is quadratic form. Displacement increases with time. Graph gradient decreases uniformly The object moves with decreasing velocity, with uniform deceleration. OA = uniform velocity (positive – move ahead) AB = velocity is zero (rest) BC = uniform velocity (negative – reverse)

No change in velocity

Zero gradient the object moves with a constant velocity or

the acceleration is zero.

The area under the graph is equal to the displacement of the moving object :

s = v x t

The object moves with a uniform acceleration for t1 s

After t1 s, the object decelerates uniformly (negative

gradient ) until it comes to rest.

The area under the graph is equal to the displacement of

the moving object : s = ½ vt2


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d) v (m s-1) Graph analysis:

...…………………………………..………………..

……………………………………………………..

………………………………………………………

t (s) ………………………………………………………

.……………………………………………………...

e) v (m s-1) Graph analysis:

………..…………………………..………………..

……….……………………………………………..

………………………………………………………

t (s) ………………………………………………………

Examples ………………………………………………………

1. s/m

2.

O

P Q

R

S

0 2 4 6 8 t/s

O

P Q

R

v/m s-1

10 5

0 2 4 6 8 10 t/s

Calculate:- (i) acceleration,a over OP, PQ and QR (ii) Displacement

Solution : Given : VO = 0 m s-1, VP = 10 m s-1 ,

VQ = 10 m s-1 VR = 0 m s-1 tOP = 4 s tPQ = 4 s tQR = 2 s

(i) aOP = 22.5ms4

010 aPQ= 2

ms 0

41010

aQR = 2 ms 5.02100

(ii) S = 70.0m10) ( 10)( 421

The shape of the graph is a curve

Its velocity increases with time.

The gradient of the graph increases.

The object moves with increasing acceleration.

The area under the graph is equal to the total displacement of the moving object. The shape of graph is a curve

Its velocity increases with time.

The gradient of the graph decreases uniformly.

The object moves with a decreasing acceleration.

The area under the graph is the total displacement of the moving object. Given : SOP = 20 m SOQ = 20 m SOR = 0 m SOS = - 10 m tOP = 2 s tPQ = 3 s tQR = 2 s tRS = 1 s

(i) VOP = 110ms220 VQR = 110ms

2200

VRS = 110ms1

010-

(ii) S = 20 – 10 = 10 m

Calculate:- (i) Velocity over OP, QR and RS (ii) Displacement Solution :


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Excercise 2.2

1. (a) s/m (b) s/m (c) s/m

10

t/s 0 2 4 t/s t/s

-5 -10

Figure 2.21

Describe and interpret the motion of a body which is represented by the displacement time graphs in Figure 2.21

2. Describe and interpret the motion of body which is represented by the velocity-time

graphs shown in figure 2.22. In each case, find the distance covered by the body and its displacement

(a) v/m s-1 (b) v/m s-1

10

t/s 0 2 4 t/s

-5 -10

Figure 2.22

a) The body remains in rest 5 m at the back of initial point b) The body start move at 10 m infront of the initial point, then back to initial

point in 2 s. The body continue it motion backward 10 m.. The body move with uniform velocity.

c) The body move with inceresing it velocity.

(a) The body move with uniform velocity , 5 m s-1 backward. (b) The body start it motion with 10 m s-1 backward and stop at initial

point in 2 s, then continue it motion forward with increasing the velocity until 10 m s-1 in 2 s.


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2.3 UNDERSTANDING INERTIA

Idea of inertia

1. ………………………………………………………………………………………………

2. ………………………………………………………………………………………………

3. ………………………………………………………………………………………………

Hand-on activity 2.5 in page 18 of the practical book to gain an idea of inertia

4. Meaning of inertia :

…………..………………………………………………………………………………….

………………………………………………………………………………………………

Mass and inertia

1. Refer to figure 2.14 of the text book, the child and an adult are given a push to swing.

(i) which one of them will be more difficult to be moved ……………………...

(ii) which one of them will be more difficult to stop? …………………………….

2. The relationship between mass and inertia :

……………………………….……………………………………………………………..

3. The larger mass ………………………………………………………………………….

………………………………………………………………………………………………

Effects of inertia

1. Positive effect : …………………………………………………………………………

(i) ………………………………………………………………………………………

(ii) ………………………………………………………………………………………

(iii) ………………………………………………………………………………………

2. Negative effect : ………………………………………………………………………….

(i) ……………………………………………………………………………………...

……………………………………………………………………………………..

(ii) ………………………………………………………………………………………

………………………………………………………………………………………

(iii) ………………………………………………………………………………………

………………………………………………………………………………………

A pillion rider is hurled backwards when the motorcycle starts to move.

Bus passengers are thrust forward when the bus stop immediately.

Large vehicle are made to move or stopped with greater difficulty.

The inertia of an object is the tendency of the object to remain at rest or, if moving, to

continue its uniform motion in a straight line

An adult

An adult

The larger the mass, the larger its inertia.

have the tendency to remain its situation either at rest or in

moving.

Application of inertia

Drying off an umbrella by moving and stopping it quickly.

Building a floating drilling rig that has a big mass in order to be stable and safe.

To tight the loose hammer

We should take a precaution to ovoid the effect.

During a road accident, passengers are thrust forward when their

car is suddenly stopped.

Passengers are hurled backwards when the vehicle starts to move and are hurled

forward when it stops immediately.

A person with a heavier/larger body will find it move difficult to stop his movement.

A heavier vehicle will take a long time to stop.


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(iv) ………………………………………………………………………………………

Execise 2.3

1. What is inertia? Does 2 kg rock have twice the inertia of 1 kg rock?

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

2.

Figure 2,3 A wooden dowel is fitted in a hole through a wooden block as shown in figure 2.31. Explain what happen when we

(a) strike the top of the dowel with a hammer,

………………………………………………………………………………………

………………………………………………………………………………………

(b) hit the end of the dowel on the floor. ………………………………………………………………………………………

……………………………………………………………………………………

2.4 ANALYSING MOMENTUM

Idea of momentum

1. When an object ic moving, …...…………………………………………………………

2. The amount of momentum ...……………………………………………………………

3. Momentum is defined…………………………………………………………………….

………………………………………………………………………………………………

it has momentum.

defends on its mass and velocity.

as the product of its mass and its velocity, that is

Momentum, p = m x v Unit= kg m s-1

Inetia is the tendency of the object to remain at rest or, if moving, to continue its

uniform motion in a straight line.

Yes, the inertia increase with the mass increased.

A wooden block move up of a wooden dowel.

A wooden block has inertia to remains at rest.

The wooden block move downward of a wooden dowel.

A wooden block has inertia to continue it motion.


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Conservation of momentum

The principle of conservation of momentum :

………………………………………………………………………………………………………

………………………………………………………………………………………………………

1. Elastic collision .…………………………………………………………………………..

Before collision after collision

(mb + mg)

mg

vg = 0

mb

vb&g

Momentum = mbvb

Momentum = (mb+mg)vb&g

Starting position before she catches the ball

vb

Receiving a massive ball

mb vb

mg

vg

Momentum = mbvb

Momentum = - mgvg

Starting position before she throws the ball

Throwing a massive ball

In the absence of an external force, the total momentum of a system remains

unchanged.

The colliding objects move separately after collision.

Momentum : m1u1 + m2u2 = m1v1 + m2v2

m1 m2 m1

m2

u2 v2

u1


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2. Inelastic collision :………………………………………………………………………...

Before collision after collision

3. explosion : …….....…………………………………………………………………...

Before explosion after explosion

Example 1 :

Car A Car B

Car A of mass 100 kg traveling at 30 m s-1 collides with Car B of mass 90 kg traveling at 20 m s-1 in front of it. Car A and B move separately after collision. If Car A is still moving at 25 m s-1 after collision, determine the velocity of Car B after collision.

Solution :

Example 2 :

Car A of mass 100 kg traveling at 30 m s-1 collides with Car B of mass 90 kg traveling at 20 m s-1 in front of it. Car A is pulled by Car B after collision. Determine the common velocity of Car A and B after collision.

Solution :

m1 m2 m1 + m2

u2 = 0

u1 v

The colliding objects move together after collision. Momentum : m1u1 + m2u2 = (m1 + m2) v

The objects involved are in contact with each other before explosion and are separated after the explosion.

Momentum : (m1 + m2)u = m1 vv - m2 v2 Given : mA = 100 kg , uA = 30 m s-1, vA = 25 m s-1, mB = 90 kg,

uB = 20 m s-1 , vB = ? mAuA + mBuB = mAvA + mBvB (100)(30) + (90)(20) = (100)(25) + (90)(vB) vB = 25.56 m s-1

(m1 + m2), u = 0 v1 m2

v2

Given : mA = 100 kg , uA = 30 m s-1, mB = 90 kg, uB = 20 m s-1 , v(A+B) = ? mAuA + mBuB = (mA + mB ) v (B+A) (100)(30) + (90)(20) = (100 + 90) v (B+A) v(A + B) = 25.26 m s-1


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Example 3 :

A bullet of mass 2 g is shot from a gun of mass 1 kg with a velocity of 150 m s-1 . Calculate the velocity of the recoil of the gun after firing. Solution :

Exercise 2.4

1. An arrow of mass 150 g is shot into a wooden block of mass 450 g lying at rest on a smooth surface. At the moment of impact, the arrow is travelling horizontally at 15 ms-1. Calculate the common velocity after the impact.

2. A riffle of mass 5.0 kg fires a bullet of mass 50 g with a velocity of 80 m s-1 .Calculate the recoil velocity. Explain why the recoil velocity of a riflle is much less than the velocity of the bullet.

2.5 UNDERSTANDING THE EFFECT OF A FORCE

Idea of force

1. What will happen when force act to an object?

………………………………………………………………………………………………

………………………………………………………………………………………………

Given ; mb = 2 g = 0.002 kg, mg = 1 kg, u(g+b) = 0 , vb = 150 m s-1 vg = ?

0 = mgvg – mb vb, 0 = (1)(vg) – (0.002)(150), vg

= 0.3 m s-1

Force can make an object;

1. Move 2. Stop the moving

3. Change the shape of the object 4. Hold the object at rest

ma = 150 g mwb = 450 g m (a+wb) = 600 g va = 15 m s-1 vwb = 0 v(a+ wb) = ? mava + mwbvwb = m(a+wb)v(a+wb) , (0.15 x 15) + (0.450 x 0) = 0.6 v(a+ wb) v(a+ wb) = 3.75 m s-1

mr = 5.0 kg mb = 50 g vr = ? vb = 80 m s-1 mr vr = mb vb , ( 5.0 ) vr = ( 0.05)(80) vr = 0.8 m s-1


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………………………………………………………………………………………………

Idea of balanced forces

1. An object is said to be in balance when it is:

………………………………………………………………………………………………

………………………………………………………………………………………………

2. Stationary object

……………………………… explanation :

………………………………………………

………………………………………………

……….……………………………………..

…………………………………………

3. An object moving with uniform velocity

…………………………….. explanation :

…..……………. …………… ……………………………………………..

……………………………………………..

……………………………………………..

……………………………… ………..…………………………………….

……………………………………………..

……………………………………………..

Idea of unbalanced forces

1. A body is said to be in unbalanced..……………………………………………………

2. ……………………….. Explanation;

………………………………………………

………………………………………………

………………………………………………

……… …….. ………………………………………………

Relationship between forces, mass and acceleration (F = ma)

Experiment 2.2 page 29. Aim : To investigate the relationship between acceleration and force applied on a constant mass.

Experiment 2.3 page 31 Aim: To investigate the relationship between mass and acceleration of an object under constant force.

Stationary object

Normal reaction, N

Frictional force Force, F Force , F = Friction

Resultant = F – Friction

= 0 (object is in equilibrium)

weight, w = mg Examples :

1.A car move at constant velocity.

2.A plane flying at constant velocity.

when it is moving in acceleration.

Resultant force

The ball move in acceleration

because the forces act are not balanced.

F > F’

F F’ So, the ball move in F direction

1. In a stationary state

2. Moving at uniform velocity

Normal reaction, N

Magnitude R = W but R acts in an

opposite direction to the weight.

( object is in equilibrium )

weight, w = mg


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1. Refer to the result of experiment 2.2 and 2.3,

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

2. 1 newton (F = 1 N) is defined as the force required to produce an acceleration of 1 m s-2 (a=1 m s-2) when its acting on an object of mass 1 kg ( m = 1 kg)

So, …………………………………………………………………………………………

3. Example 1 : Calculate F, when a = 3 m s-2 dan m = 1000 kg

Example 2 :

Calculate the acceleration, a of an object.

Exercise 2.5

1. A trolley of mass 30 kg is pulled along the ground by horizontal force of 50 N. The opposing frictional force is 20 N. Calculate the acceleration of the trolley.

2. A 1000 kg car is travelling at 72 km h-1 when the brakes are applied. It comes to a stop in

a distance of 40 m. What is the average braking force of the car?

mm == 2255 kkgg FF == 220000 NN

F = ma

F = ma

F = (1000)(3)

F = 3000 N F = ma 200 = 25 a a = 8.0 ms-2

it is found that; a F when m is constant and a 1/m when F is constant.

Therefore, a F/m

From a F/m,

F ma

Therefore, F = kma … k =constant

m = 30 kg , F = 50 N , Ff = 20 N , a = ? F – Ff = ma , 50 – 20 = 30 a a = 1.0 m s2 m = 1000 kg , u = 72 km h-1, v = 0, s = 40 m, F = ? Note : u = 72 km h-1 =20 m s-1 F = ma, v2 = u2 + 2as = 1000 x 5.0 0 = 202 + 2a(40) = 5000.0 N a = 5.0 m s2


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2.6 ANALYSING IMPULSE AND IMPULSIVE FORCE

Impulse and impulsive force

1. Impulse is ……………………………………………………………………………….

2. Impulsive force is ………………………………………………………………………

………………………………………………………………………………………………

3. Formula of impulse and impulsive force:

Refer, F = ma

Example 1; v u

wall

If ; u = 10 m s-1 , v = - 10 m s-1 , m = 5 kg and t = 1 s

Impulse, Ft = and impulsive force, F =

Example 2; v u

Wall with a soft surface

If ; u = 10 m s-1 , v = - 10 m s-1 , m = 5 kg and t = 2 s

Impulse, Ft = and impulsive force, F =

4. The relationship between time of collision and impulsive force.

………………………………………………………………………………………………

………………………………………………………………………………………………

The change of momentum

The large force that acts over a short period of time during

collision and explosion.

It is known that a = ( v – u ) / t

Therefore, F = m( v – u) t So, Ft = mv – mu , Unit = N s

Ft is defined as impulse, which is the change in momentum. F = mv – mu ,

t Ft = mv – mu Unit : newton (N) F is defined as impulsive force which is the rate of change of momentum over the short period of time

5(10) - (- 5(10)) 100 = 100 N

= 100 Ns 1

5(10) - (- 5(10)) 100 = 50 N

= 100 Ns 2

Impulsive force , F 1 / t

Therefore, F decreases when the time of collision increases ( refer to examples )


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Exercise 2.6

1. A force of 20 N is applied for 0.8 s when a football player throws a ball from the sideline. What is the impulse given to the ball?

2. A stuntman in a movie jumps from a tall building an falls toward the ground. A large

canvas bag filled with air used to break his fall. How is the impulsive force reduced? 2.7 BEING AWARE OF THE NEED FOR SAFETY FEATURES IN VEHICLES

Safety features in vehicles

Crash resistant door pillars

Anti-lock brake system (ABS)

Traction control bumpers

Windscreen

Air bags

Head rest

Crumple zones

Reinforced passenger compartment

Fimpulse = Ft = 20 x 0.8 = 16.0 Ns

1. A large canvas bag will increase the time of collision. 2. When the time of collision increase the impulsive force will decrease.


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Importance of safety features in vehicles

Safety features Importance

Padded dashboard Increases the time interval of collision so the impulsive force produced during an impact is thereby reduced

Rubber bumper Absorb impact in minor accidents, thus prevents damage to the car.

Shatter-proof windscreen Prevents the windscreen from shattering

Air bag Acts as a cushion for the head and body in an accident and thus prevents injuries to the driver and passengers.

Safety seat belt Prevents the passengers from being thrown out of the car. Slows down the forward movement of the passengers when the car stops abruptly.

Side bar in doors Prevents the collapse of the front and back of the car into the passenger compartment. Also gives good protection from a side-on collision.

Exercise 2.7

1. By using physics concepts, explain the midifications to the bus that help to improve that safety of passengers and will be more comfortable.

- The absorber made by the elastic material : To absorb the effect of impact (hentaman) during it moving

- Made by the soft material of bumper : To increase the time during collision, then the impulsive force will be decreased.

- The passenger’s space made by the strength materials. : To decrease the risk trap to the passenger during accident. - Keep an air bag at the in front of dash board and infront of passengers

: Acts as a cushion for the head and body in an accident and thus prevents injuries to the driver and passengers.

- Shatter-proof windscreen : Prevents the windscreen from shattering.


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2.8 UNDERSTANDING GRAVITY

Carry out hands-on activity 2.8 on page 35 of the practical book.

Acceleration due to gravity.

1. An object will fall to the surface of the earth because………………………………...

2. The force of gravity also known ………………………………………………………...

3. When an object falls under the force of gravity only, ………………………………...

………………………………………………………………………………………………

4. The acceleration of objects falling freely ………………………………………………

5. The magnitude of the acceleration due to gravity depends ………………………...

………………………………………………………………………………………………

Gravitational field

1. The region around the earth is ………………………………………………………….

2. The object in gravitational field …………………………………………………………

3. The gravitational field strength is defined ……………………………………………..

4. The gravitational field strength, g can be calculate as;

5. At the surface of the earth,

…………….………………………………………………………………………………..

6. This means

……………………………………………………………………………………………..

7. Example 1. Can you estimate the gravitational force act to your body? mass = 60 kg, g = 9.8 N kg-1, F = ? Example 2,

A satellite of mass 600 kg in orbit experiences a gravitational force of 4800 N. Calculate the gravitational field strength.

It pulled by the force of gravity.

as earth’s gravitational force.

the object is said to be free

falling

is known as acceleration due to gravity.

on the strength of the gravitational field

.

the gravitational field of the earth.

is on the force of gravity.

as the gravitational force acting on a 1 kg mass.

. g = F . where, F : gravitational force m m : mass of an object

g = 9.8 N kg-1

that an object of mass 1 kg will experience a gravitational force of 9.8 N.

Solution : F = mg = (60) (9.8)

= 588.0 N

Given : m = 600 kg. F = 4800 N, g = ?

g = F = 4800 . = 8 N kg-1

m 600


21

Example 3, Example 3: A stone is released from rest and falls into a well. After 1.2 s, it hits the bottom of the well.

(a) What is the velocity of the stone when it hits the bottom?

(b) Calculate the depth of the well.

Weight

1. The weight of an object is defined ……………………………………………………..

2. For an object of mass m, the weight can be calculate as :

Example : The mass of a helicopter is 600 kg. What is the weight of the helicopter when it land on the peak of a mountain where the gravitational field is 9.78 N kg-1?

Exercise 2.8

1. Sketch the following graphs for an object that falling freely.

(a) Displacement-time graph, (b) Velocity-time graph (c) Acceleration-time graph

as the gravitational force acting on the object.

weight, W = mg

where, g = acceleration due to gravity.

W = mg

= 6000 x 9.78

= 58 680 N

Given : u = 0 ms-1, t = 1.2 s, a = g = 9.8 ms-2

(a) v = ? v = u + at

= 0 + (9.8)(1.2)

= 11.76 ms-1

(b) Depth = s = ? s = ut + ½ at2

= (0)(1.2) + ½ (9.8)(1.2)2

= 7.1 m

(a) s / m (b) v / m s-1 (c) a / m s2 t / s t / s t / s


22

2. The following data was obtained from an experiment to measure the acceleration due to gravity.

Mass of steel bob = 200 g, distance covered = 3.0 m, time of fall = 0.79 s. Calculate the acceleration due to gravity of steel bob.

Give the explanation why your answer different with the constant of gravitational acceleration, g = 9.8 m s-2.

2.9 IDEA OF EQUILIBRIUM FORCES

An object is in equilibrium when :

1. ………………………………………………………………………………………………

2. ………………………………………………………………………………………………

stationary object

An object moving with uniform velocity

It is in a stationary state

It is moving with uniform velocity

Normal reaction, R Normal reaction, R

Weight, W=mg weight, W=mg

Magnitude of R = W Magnitude of R = mg cos R and W acts in opposite direction. And acts in opposite direction. So, Resultant force = W – R = 0 So,Resultant force = mg cos – R = 0

( object in equilibrium ) ( object in equilibrium ) normal reaction, R friction force force, F Weight, W

Force , F = Frictional force Resultant force = F – Frictional force

= 0 (object in equilibrium)

m = 200 g s = 3.0 m t = 0.79 s u = 0 g = ? = 0.2 kg s = ut + ½ g t2 3.0 = 0 (0.7) + ½ g (0.792) g = 9.6 m s-2

The answer less than the constant because of the air frictional force.


23

Addition of Force

1. Addition of force is defined as ...……………………………………………………..

………………………………………………………………………………………………

………………………………………………………………………………………………

Examples : the forces are acting in one direction

F1 = 10 N

F2 = 5 N

Resultant force, F

Example : the forces are acting in opposite directions

F1 = 10 N

F2 = 5 N

Resultant force, F

Example : the forces are acting in different directions

F2 = 5 N

500 F

F1 = 10 N

Parallelogram method:

1. Draw to scale.

2. Draw the line parallel with F1 to the edge of F2, and the line parallel with F2 to the

edge of F1

3. Connect the diagonal of the parallelogram starting from the initial point.

4. Measure the length of the diagonal from the initial point as the value of the

resultant force.

a resultant force is a single force the

represents in magnitude and direction two or more forces acting on an object

F resultant = the total of forces (including the directions of the forces)

= F1 + F2 = 10 + 5 = 15 N

= F1 - F2 = 10 - 5 = 5 N


24

F2

F

F1

Triangle method

1. Draw to scale.

2. Displace one of the forces to the edge of another force.

3. Complete the triangle and measure the resultant force from the initial

point.

Example 1: During Sport Day two teams in tug of war competition pull with forces of

6000 N and 5300 N respectively. What is the value of the resultant force?

Are the two team in equilibrium?

Example 2: A boat in a river is pulled horizontally by two workmen. Workmen A

pulls with a force of 200 N while workmen while workmen B pulls with a

force of 300 N. The ropes used make an angle 250 with each other. Draw a

parallelogram and label the resultant force using scale of 1 cm : 50 N.

Determine the magnitude of resultant force.

Solution : Resultant force, F = 6000 – 5300 =700 N They mere not in equilibrium Resultant force, F = 10.5 x 50 = 525 N

250

10.5 cm


25

Resolution of a force

1. Resolution of a force is …………………………………………………………………

Refer to trigonometric formula:

Example : The figure below shows Ali mopping the floor with a force 50 N at an angle of 600 to the floor.

F = 50 N

Example of resolution and combination of forces

reverse process of finding the resultant force

Fy F is the resultant force of Fx and Fy Therefore, F can be resolved

into Fx and Fy F

Vertical Component

Fx horizontal component

Cos = FFx , therefore Fx = F cos

Sin = FFy , therefore Fy = F sin

Fx = F cos = 50 cos 60 Fx = 50 (0.5)

= 25 N Fy = F Sin = 50 sin 600

Fy = 50 (0.8660) = 43.3 N

F = mg sin 400 + 200

= 800(0.6427) + 200 = 514.2 + 200 = 714.2 N

mg = 800 N

600

F = ?

200 N

400

400


26

Problem solving

1. When a system is in equilibrium, ……………………………………………………….

2. If all forces acting at one point are resolved into horizontal and vertical

components, ……………………………………………………………………………

3. Example 1; Show on a figure;

a) the direction of tension force, T of string b) the resultant force act to lamp c) calculate the magnitude of tension force, T a)

mlamp = 1.5 kg

Wlamp = 14.7 N

Exercise 2.9

1. Two force with magnitude 18 N and 6 N act along a straight line. With the aid of diagrams, determine the maximun possible value and the minimum possible value of the resultant force.

2. A football is kicked simultaneously by two players with force 220 N and 200 N respectively, as shown in Figure 2.9. Calculate the magnitude of the resultant force.

220 N

900

200 N

T b) T’ T 700 700

the resultant force is equal to zero.

the sum of each component is equal to zero.

(c ) T’ = 2T sin 700 Therefore, mlampg = 2T sin 700

T = 0lamp

2sin70

gm

= 02sin701.5(9.8) = 7.82 N

Fmaximum when both of forces act in same direction; Fmaximum = 18 + 6 18 N 24 N = 24 N 6 N Fminimum when the forces act in opposite direction ; Fminimum = 18 – 6 18 N 12 N = 12 N 6 N F = Resultant of Force F2 = 2202 + 2002 F = 297.32 N F


27

2.10 UNDERSTANDING WORK, ENERGY AND EFFICIENCY

Work

1. Work is done, ……………………………………………………………………………..

………………………………………………………………………………………………

2. WORK is the product.…………………………………………………………………….

………………………………………………………………………………………………

3. The formulae of work;

4. Example 1;

Example 2;

80 N

600

s = 5 m

When a force that acts on an object moves the object through a

distance in the direction of the force.

of a force and the distance traveled in the direction of

the force.

WORK = FORCE X DISPLACEMENT

W = F x s

W : work in Joule/J

F : force in Newton/N

s : displacement in meter/m

W = Fs

If, F = 40 N and s = 2 m

Hence, W = 40 x 2

= 80 J

Force, F

s

W = Fs

= 80 cos 600 (5)

= 80 (0.5) (5)

= 200 J


28

Example 3;

Example 4;

F = 600 N

S = 0.8 m

Energy

1. Energy is .................................................................................................................

2. Energy cannot be ....................................................................................................

3. Exist in various forms such as …………………...……………………………………

………………………………………………………………………………………………

4. Example of the energy transformation;

………………………………………………………………………………………………

………………………………………………………………………………………………

5. ………………………………………………………………………………………………

Example :

………………………………………………………………………………………………

T T

F = 30 N

h = 1.5 m W = F s = F h

= 30 (1.5)

= 45.0 J

W = F s = 600 x 0.8 = 480 J

It is the potential to do work.

created nor be destroyed.

potential energy, kinetic energy, electrical

energy, sound energy, nuclear energy and chemical energy.

When we are running up a staircase the work done consists of energy change from

Chemical Energy Kinetic Energy Potential Energy

The energy quantity consumed is equal to the work done.

If 100 J of work is done, it means 100 J of energy is consumed.


29

Work done and the change in kinetic energy

1. Kinetic energy is …………………………………………………………………………

2. Refer to the figure above,

3. Example 1; A small car of mass 100 kg is moving along a flat road. The resultant force on the car is 200 N. a) What is its kinetic energy of the car after moving through 10 m? b) What is its velocity after moving through 10 m?

Work done and gravitational potential energy

h = 1.5 m

1. Gravitational potential energy is………………………………………………………...

………………………………………………………………………………………………

2. Refer to the figure above;

3. Example; If m = 10 kg

s

Force, F

Through, v2 = u2 +2as u = 0 and, as = ½ v2

energy of an object due to its position.

(possessed by an object due to its position in a gravitational field)

W = Fs = mg h where, F = mg So, Gravitational energy, Ep = mgh

energy of an object due to its motion.

Work = Fs

= mas

= m ( ½ v2)

The formulae of Kinetic energy, Ek = ½ mv2

Solution : Given : m = 100 kg , F = 200 N

a. Kinetic energy, Ek = Fs

= 200 x 10= 2000 J

b. Velocity, v ½ mv2 = 2000

v = 6.32 m s-1

W = 10 (10) 1.5 = 1500 J

Therefore Work done = 1500J And, Ep = 1500 J


30

Principle of conservation of energy

Carry out hands-on activity 2.10 on page 38 of the practical book.

To show the principle of conservation of energy.

1. Energy cannot be ………………………………………………………………………

……………………………………………………………………………………………

2. Example : a thrown ball upwards will achieve a maximum height before changing its direction and falls

3. Example in calculation : A coconut falls from a tree from a height of 20 m. What is the velocity of coconut just before hitting the earth?

Power

1. Power is …………………………………………………………………………………

2. A weightlifter lifts 180 kg of weights from the floor to a height of 2 m above his head in a

time of 0.8 s. What is the power generated by the weightlifter during this time? g = 9.8 ms-2)

created or destroyed but can be changed from one form to

another form.

Maximum Potential energy

Kinetic energy decrease potential energy decrease and potential energy and kinetic energy Increase increase

Maximum kinetic energy

Given : h = 20 m, u = 0 , g = 9.8 ms-2 , v = ?

Ep = Ek

mgh = ½ mv2

m(9.8)(20) = ½mv2

v2 = 392, v = 19.8 m s-1

the rate of doing work.

Therefore, power, P = timetakenworkdone , so, P =

tW

Where, P : power in watt/W W : work in joule/J

t : time to do work in seconds/s

Solution : Given : m = 180 kg, h = 2 m, t = 0.8 s and g = 9.8 ms-2. P = ?

P = t

W = t

mgh

= 0.8

29.8180 = 4 410 W


31

Efficiency

1. Defined……..…………………………………………………………………………….

2. Formulae of efficiency :

3. Analogy of efficiency;

Energy transformation

4. Example; An electric motor in a toy crane can lift a 0.12 kg weight through a height of

0.4 m in 5 s. During this time, the batteries supply 0.8 J of energy to the motor. Calculate (a) The useful of output of the motor.

(b) The efficiency of the motor

Carry out hands-on activity 2.11 on page 39 of the practical book to measure the power.

Device/ mechineDevice/ mechine

as the percentage of the energy input that is transformed into useful energy.

%100inputEnergy

outputenergyUsefulEfficiency

unwanted energy

Energy input, Einput Useful energy, Eoutput

Solution : Given : m = 0.12 kg, s= 0.4 m, t = 5 s, Einput = 0.8 J

(a) Eoutput = ?

Eoutput = F x s

= (0.12 x 10) x 0.4

= 0.48 J

(b) Efficiency = ?

Efficiency %100xEE

input

output

100%x0.800.48 %60


32

Exercise 2.10

1. What is the work done by a man when he pushes a box with a force of 90 N through a distance of 10 m? State the amount of energy transferred from the man to the force.

2. A sales assistant at a shop transfers 50 tins of milk powder from the floor to the top shelf.

Each tin has a mass of 3.0 kg and the height of thee top shelf is 1.5 m.

(a) Calculate the total work done by the sales assistant.

(b) What is his power if he completes this work in 250 s? 2.11 APPRECIATING THE IMPORTANCE OF MAXIMISING THE EFFICIENCY

OF DEVICES

1. During the process of transformation the input energy to the useful output

energy,……………………………………………………………………………………..

2. .……………………………………………………………………………………………..

3. ………………………………………………………………………………………………

Example of wasting the energy;

………..…………………

Input enegy output from the petrol energy

…………………… ……………. ……………… …………………….

..………………….. …………….. ………………….. …………………….

..………………….. ……………. …………………. …………………….

4. The world we are living in face acute shortage of energy.

some of energy transformed into unwanted forms of energy.

The efficiency of energy converters is always less than 100%.

The unwanted energy produced in the device goes to waste.

Kinetic energy

Energy loss due to Energy loss Energy loss Energy loss due to friction at

friction in as heat as sound other parts in the

moving parts engine

W = F s The energy transferred to the force = 900 J = 90 x 10 = 900 J m = 3.0 x 50 = 150 kg h = 1.5 m W = mhg = 150 x 9.8 x 1.5 = 2205 J

P = t

W = 250

2205 = 8.82 w


33

5. It is very important that a device makes

…………………………………………………………………

Ways of increasing the efficiency of devices

1. Heat engines ……………………..………………………………………………………

………………………………………………………………………………………………

2. Electrical devices. ...……………………………………………………………………...

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

Refrigerators

- choose the capacity according to the size of the family.

- installed away from source of heat and direct sunlight.

- the door must always be shut tight.

- more economical use a large capacity refrigerator.

- use manual defrost consumption.

Washing machines

- use a front loading as such more economical on water and electricity

- front loading use less detergent as compared to a top loading machine.

the best possible use of the input energy.

Engine must be designed with the capability to produce greater amount of mechanical

work.

Light Fittings

- replace filament light bulb with fluorescent lamps which have higher efficiency.

- use a lamp with a reflector so that the illumination can be directed to specific areas

of the user.

Air-conditioners.

- choose a model with a high efficiency.

- accommodate the power of air-conditioner and the size of the room

- Ensure that the room totally close so that the temperature in the room can be

maintained.


34

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

Operation of electrical devices

1. The electrical devices increase the efficiency………………………………….……

2. Proper management ….....………………………………………………………………

3. …………..………………………………………………………………………………

………………………………………………………………………………………………

2.12 UNDERSTANDING ELASTICITY

Carry out Hands-on activity 2.12 page 40 of the practical book.

1. Elasticity is ……………………………………………………………………………...

………………………………………………………………………………………………

2. Forces between atoms …………………………………………………………………..

………………………………………………………………………………………………

3. Forces between atoms in equilibrium condition

Explanation :

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

4. Forces between atoms in compression

Explanation ;

………………………………………………………………………………………………

………………………………………………………………………………………………

when they are in good operating

condition.will increase the life span of device.

Example : -the filter in an air-conditioner and fins of the cooling coil of a

refrigerator must be periodically cleaned.

Force of repulsion

Force of attraction

Force of repulsion

compressive force compressive force

Force of repulsion Force of repulsion

1. Force of repulsion takes effect.

2. When the compressive force is removed, force of repulsion between the atoms

pushes

the atom back to their equilibrium positions.

the property of an object that enables it to return its original shape and

dimensions after an applied external force is removed.

The property of elasticity is caused by the existence of forces of

repulsion and attraction between molecules in the solid material.

1. The atoms are separated by a distance called the equilibrium distance and vibrate

at it position.

2. Force of repulsion = Force of attraction


35

………………………………………………………………………………………………

5. Forces between atoms in tension

force of attraction

stretching force stretching force

Explanation ;

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

Carry out Experiment 2.4 on page 41 of the practical book To investigate the relationship between force and extension of a spring Hooke’s Law

1. Hooke’s Law states ………………………………………………………………………

………………………………………………………………………………………………

2. Elastic limit of a spring is defined……………………………………………………….

………………………………………………………………………………………………

3. The spring is said to have a permanent extension,...…………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

4. The elastic limit is not exceeded,…………………………………………….…………

………………………………………………………………………………………………

………………………………………………………………………………………………

5. GGrraaff FF aaggaaiinnsstt xx

F/ N

1. Force of attraction takes effect.

2. When the compressive force is removed, force of repulsion between the

atoms pushes the atom back to their equilibrium positions.

x (cm) 0

F = kx Spring obeying Hooke’s Law

E Q P

Spring not obeying Hooke’s law (exceeded the elastic limit)

R

Force constant, k = F with unit N m-1, N cm-1 or N mm-1 x

When the spring obey Hooke’s Law.

The mathematical expression for Hooke’s Law is :

F x

F = kx, k = Force constant of the spring

Force constant, k = F with unit N m-1, N cm-1 or N mm-1 x

that the extension of a spring is directly proportional to the applied

force provided that the elastic limit is not exceeded.

as the maximum force that can be applied to

spring such that the spring will return to its original length when the force released.

when the length of the

spring longer than the original length even though the force acts was released and the

elastic limit is exceeded.


36

6. Spring Constant, k

F/N

0.8

0 8 x/cm

Example 1; A spring has an original length of 15 cm. With a load of mass 200 g

attached, the length of the spring is extend to 20 cm. a. Calculate the spring constant. b. What is the length of the spring when the load is in increased

by 150 g? [assume that g = 10 N kg-1]

Example 2;

k is the gradient of the F - x graph

k = xF

= 80.8

= 0.01 N cm-1

Given : lo = 15 cm, m = 200 g , F = 2.0 N, l = 20 cm x = 5 cm

a. k = ?, k = 10.4Ncm52.0

xF

The graph shows the relationship between the stretching force, F and the spring extension, x. (a) Calculate the spring constant of P and Q. (b) Using the graph, determine the stretching force acts to spring P and spring Q, when their extension are 0.5 cm Solution a. Spring constant, k = gradient of graph

kP = 1cm N 15.790.38

6

kQ = 1cm N 6.00.53

b. When x = 0.5, FP = 8.0 N ( extrapolation of graph P) FQ = 3.0 N

b. l = ? , when m = 150 g, F = 1.5 N From a, k = 4.0 N cm-1

x = cm75.30.41.5

kF

l = 15 + 3.75 = 18.75 cm

F (N)

x (cm)

P

Q

8

7

6

5

4

3

2

1 0 0.1 0.2 0.3 0.4 0.5

Graph F against x of spring P and spring Q


37

Elastic potential energy

1. Elastic potential energy ………………………………………………………………..

spring with the original length

F compression

x spring compressed x

F x = compression x

x F spring extended

x = extension F, extension

Other situation where the spring extended and compressed

Relationship between work and elastic potential energy

Graph F against x

Example ;

Factors that effect elasticity

Hands-on activity 2.13 on page 42 the practical book to investigate the factors that affect elasticity.

Type of material different same same same

Diameter of spring wire same different same same

Diameter of spring same same different same

Length of spring same Same same different

Summarise the four factors that affect elasticity

the energy stored in a spring when it is extended or compressed

x / cm

F/N

F x

Area under the graph = work done = ½ Fx So, Elastic potential energy = ½ Fx

15 cm

5 kg

8 cm

x = 15 – 8 = 7 cm = 0.07 m Force act to the spring, F = 5 x 10 = 50 N Elastic potential energy = ½ Fx

= ½ 50 (0.07) = 1.75 J


38

Factor Change in factor Effect on elasticity

Length Shorter spring Less elastic

Longer spring More elastic

Diameter of spring Smaller diameter Less elastic

Larger diameter More elastic

Diameter of spring wire Smaller diameter More elastic

Larger diameter Less elastic

Type of material the elasticity changes with the type of materials

Exercise 2.12

1. A 6 N force on a spring produces an extension of 2 cm. What is the extension when the force is increased to 18 N? State any assumption you made in calculating your answer.

2. If a 20 N force extends a spring from 5 cm to 9 cm,

(a) what is the force constant of the spring?

(b) Calculate the elastic potential energy stored in the spring. Reinforcement Chapter 2 Part A : Objective Questions 1. When a coconut is falling to the

ground, which of the following quantities is constant?

A. Velocity B. Momentum C. Acceleration D. Kinetic energy

To solve the problem, determine the spring constant to use the formula F = k x F = 6 N , x = 2 cm F = kx When, F = 18 N, x = ? 6 = k (2) 18 = 3 x k = 3 N cm-1 x = 6 cm F = 20 N, x = 9 – 5 = 4 cm, k = ? F = kx 20 = k (4) k = 5 N cm-1 E = ½ Fx = ½ (20)(4) = 40 J


39

2. In an inelastic collision, which of the following quantities remains constant before and after the collision?

A. Total acceleration B. Total velocity C. Total momentum D. Total kinetic energy 3. Calculate the weight of a stone with

mass 60 g on the surface of the moon. (The gravitational acceleration of the moon is 1/6 that of the Earth.)

A. 0.1 N B. 0.2 N C. 0.4 N D. 0.6 N E. 0.8 N 4. The momentum of a particle is

dependent on A. mass and acceleration B. weight and force C. mass and velocity 5. Which of the following diagrams

shows a body moving at constant velocity?

A. 2 N 2N B. 12 N 7 N C. 12 N 14 N D. 20 N 17 N

6. The graph below shows the motion of a trolley with mass 1.5 kg.

Velocity / ms-1 4 0 2 4 6 Time / s

Calculate the momentum of the trolley from t = 2s to t = 4s.

A. 1.5 kg m s-1 B. 3.0 kg m s-1 C. 4.0 kg m s-1 D. 6.0 kg m s-1 E. 7.5 kg m s-1

7. This figure shows an aircraft flying in the air.

8. m = 0.3 kg

5 m

What is the momentum of the stone just before it hits the ground?

A. 0.15 kg m s-1 B. 0.3 kg m s-1 C. 1.5 kg m s-1 D. 3.0 kg m s-1 E. 15.0 kg m s-1

Solution :

60 g = 0.06 kg W = 0.06 (1/6)(10) = 0.1 N

P = mv = 1.5 x 4 = 6.0 kg ms-1

Lift Thrust Air friction Weight

The aircraft above accelerates if A. Lift Weight B. Thrust Lift C. Lift Air friction D. Thrust Air friction

P = mv (find v first to calculate the P) Ep = Ek mgh = ½ mv2 (0.3)(10)(5) = ½ (0.3) v2 v = 10 m s-1 P = (0.3)(10) = 3.0 kg m s-1


40

9. A big ship will keep moving for some distance when its engine is turned off. This situation happens because the ship has A. great inertia B. great acceleration C. great momentum D. great kinetic energy 10. An iron ball is dropped at a height of

10 m from the surface of the moon.

Calculate the time needed for the iron ball to land. (Gravitational acceleration of the moon is 1/6 that of the Earth and g = 9.8 N kg-2)

A 0.6 s B 1.4 s C 1.7 s D 3.5 s E 12.0 s

Part B : Structure Questions 1.

(i) Car A (ii) Car B

Diagram 1.1

Diagram 1.1(i) and (ii) show two methods used by the mechanic to move a breakdown car. A constant force, F = 500 N is used to push and pull the car in method A and B. (a) (i) Which method is easier to move the car?

………………………………………………………………………………

(ii) State a reason for your answer in (a)(i).

………………………………………………………………………………

………………………………………………………………………………

(b) The frictional force acting between the car and track surface in both methods is 200 N. Calculate, the

(i) horizontal resultant force in method A.

(ii) horizontal resultant force in method B.

s = ut + ½ gt2 = (0)t + ½ (9.8/6)t2 t = 3.5 s

Method (a)

The forces given parallel with the surface of motion,

So, all the forces given used to move the car.

F = Fgiven - Ffriction

= 500 – 200

= 300 N

F = Fgiven Cos 500 – Ffriction

= 500 cos 600 – 200

= 50.0 N

F = m a

50.0 = 1000 a

a = 0.05 m s-2


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(iii) acceleration of the car in method B.

( c ) Suggest a method to move Car B so that the acceleration produced is equal to that of method A. ……………………………………………………………………………..………..

………………………………………………………………………………………

2. ceiling

Tin water M N hand P Q R (i) Diagram 2.1 (ii)

a) Diagram 2.1(i) shows tin P that is empty and tin Q that is filled with water. A student find difficult to pushed tin Q. Write the inference about the observation.

………………………………………………………………………………………

b) Diagram 2.1(ii) shows a tin being released from the different positions M and N. The hand of a student at position R needs greater force to stop the motion of the tin falling from position M. Explain this observation. ………………………………………………………………………………………

……………………………………………………………………………………… c) Based on the observation (i) and (ii), state two factors that affect the magnitude of

the momentum of the object.

……………………………………………………………………………………… d) If water flows out from a hole at the bottom of the tin Q, how would the inertia of

Tin Q depends on time ?

……………………………………………………………………………………

The difficulty to move the tin depends to its mass. From position M the velocity of tin is more than the velocity compare when it is

from N. Ek increase then the force to stop it will be increased. mass and velocity inertia of tin Q will decrease because the mass of tin decreased.


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3. 2 ms-1

P iron ball ( 2 kg ) S T 3.0 m smooth surface 1.0 m 2.0 m

Q R

Diagram 3 Rough surface

The figure shows a iron ball that is rolled through PQRST. The rough surface of QR has frictional force of 4 N. a) Calculate

(i) the kinetic energy of the iron ball at P. (ii) the potential energy of the iron ball at P.

(iii) the total of energy of the iron ball at P.

b) c) (i) Calculate the total of energy of the iron ball when it reaches at Q ?

(ii) Calculate the work done against friction along QR.

d) Calculate the total kinetic energy of the iron ball at S.

e) Calculate the speed of the ball at position T.

Ek = ½ mv2 = ½ (2)(22) = 4.0 J

Ep = mgh = (2) (10) (3.0) = 60.0 J E = Ek + Ep = 4.0 + 60.0 = 64.0 J 64.0 J ( the conservation of energy )

W = Ff x s = 4 x 1.0 = 4.0 J

Es = E – Ef Ek at S = Es - Ep at s = 64.0 – 4.0 = 60.0 – (2)(10)(2.0) = 60.0 J = 20.0 J Ek at T = 20.0 J v2 = 20 = ½ m v2 v = 4.5 m s-1 = ½ (2)(v2)


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Part C : Essay Questions 1.

(i) (ii) Diagram 1.1

Diagram 1.1(i) shows the condition of a car moving at high velocity when it suddenly crashes into a wall. Diagram 1.1(ii) shows a tennis ball hit with racquet by a player.

a) (i) What is the meaning of momentum?

(ii) Based on the observations of Diagram (i) and (ii), compare the characteristics of car when it crashes into the wall and the tennis ball when it is hit with a racquet. Hence, relate these characteristics to clarify a physics concept, and name this concept.

b) Explain why a tennis player uses a taut racquet when playing.

c) In launching a rocket, a few technical problems have to be overcome before the rocket can move upright to the sky. By using appropriate physics concepts, describe the design of a rocket and the launch techniques that can launch the rocket upright.

Answer

a) (i) momentum is product of mass and velocity (ii) - The shape of car changed but the shape of wall remained.

- The shape of ball remained but the shape of the racquet string was changed. (The racquet string is elastic but the wall is harder) - The time taken of collision between the ball and racquet string more than the time taken when the car hit the wall.

- The impulsive force will decrease when the time of collision increased. - The concept is the impulsive force.

b) - To decrease the time of collision between the ball and the racquet string. - Impulsive force will be increased. - The force act to the ball will be increased. - The velocity of ball will be increased.

c) - Make a gradually narrower at the front shape (tapering) : To decrease air friction


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- Made by the high strength and high rigidity of materials : To decrease the probability to become dented (kemik). - Made by the low density of material. : To reduce the mass/weight - The structure is fractional engine : The mass will be decreased and the velocity will increase. - Made by the high of heat capacity of materials : It will be high heat resistance. 2.

Brand Reaction time / s Mass / kg Engine thrust

force / N Resistance force

/ N A 0.3 1.5 10.0 4.0 B 0.5 1.8 12.5 2.4 C 0.2 0.9 6.5 2.2 D 0.6 2.5 16.0 6.5

In a radio-controlled car racing competition, 4 mini-cars branded A, B, C and D took part. The information of the 4 cars is given in the table above. Details of the above information are given as below;

Reaction time - Duration between the moment the radio-controlled is switched on and the moment the car starts moving. Resistance - Average value of opposing forces includes the friction between

wheels and track, and air resistance.

(a) What is the meaning of acceleration?

(b) Draw a graph of velocity against time that shows a car moving initially with constant acceleration, then moving with constant velocity and followed by constant deceleration until it stops.

(c) Explain the suitability of the properties in the above table in constructing a radio-controlled car racing purpose. Hence, determine which brand of car will win the 50-metre race.

(c) If Car B in the above table is moved up the plane at the angle of 30o to the horizon, (i) Show that the car is able to move up the plane. (ii) Determine the acceleration of the car.

Answer : (a) Increase the velocity (b) v / ms-1

displacement = area under the graph t / s

Properties


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(c) - time reaction mast be short : fast to detect the signal to start its move - has a small of mass : to decrease the inertia, then easier to start move and to

stop its moving. - thrust force is high : has more power during its moving / increase the

acceleration - friction force is low : decrease the lost of force - the best car is A : because it has short of time reaction, small of mass, high of

thrust force and low friction of force. (d) (i) EB = (12.5 – 2.4 ) (50) = 505.0 J 50 m 50Sin300 E (suitable to move up) = 1.8 (10)(50Sin300) = 450 .0 J EB> E ( car B can move up the plane) (ii) F = ma , 12.5 – 2.4 = 1.8 a, a = 5.61 ms-1

300

2. force and motion - · pdf filejpn teacher’s guide physics module form 4

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