Chapter 2 Force and Motion TEACHER's GUIDE

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Physics Module Form 4 Chapter 2 : Force and Motion

2.

FORCE AND MOTION

2.1

ANALYSING LINEAR MOTION

Distance and displacement 1.

Types of physical quantity: has only a magnitude (i) Scalar quantity: …………………………………………………………………. (ii)

2.

has both magnitude and direction Vector quantity: …………………………………………………………………

The difference between distance and displacement: length of the path taken (i) Distance: ………………………………………………………………………… (ii)

distance of an object from a point in a certain direction 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

Kota Tinggi 41 km

Solution:

a.

by car

= 41 + 53 = 94 km

b.

by plane = 60 km

53 km

Johor 60 km Desaru Bahru The path traveled by the plane is shorter than travelled by the car. So, Distance = 94 km Displacement = 60 km

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

1. 2.

the distance traveled per unit time or rate of change of distance Speed is ..………………………………………………………………………………… the speed in a given direction or rate of change of displacement Velocity is: ..……………………………………………………………………………...

3.

total distance traveled, s (m) , v = s m s-1 Average of speed: ……………………………………………………………………… time taken, t (s) t

4.

displacement, s (m) , v = s ms-1 Average of velocity: ……………………………………………………………………... Time taken, t (s) t 1

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5.

Physics Module Form 4 Chapter 2 : Force and Motion

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:

C 400 km

A

300 km

B C 400 km

A

i. Speed = Distance Time = 300 + 400 4 = 175 km h-1

ii. velocity = displacement time (Determine the displacement denoted by AC and its direction) = . 500 . 4

B

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

300 km Acceleration and deceleration 1.

Study the phenomenon below;

0 m s-1

2.

3.

20 m s-1

40 m s-1

The velocity of the car increases. Observation: ……………………………………………………………………………… the rate of change of velocity Acceleration is, ………………………………………………………………………. Final velocity – initial velocity Or, a = v – u Then, a = Time of change t Example of acceleration; t=2s t=2s A

B

0 m s-1

C

20 m s-1 2

40 m s-1

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Physics Module Form 4 Chapter 2 : Force and Motion

Calculate the acceleration of car; i)

from A to B

aAB

= 20 – 0 2

ii)

From B to C

aBC

=

= 10 m s-2 = 10 m s-2

40 – 20 2

4.

when the velocity of an object decreases, In calculations, a Deceleration happens ...………………………………………………………………… will be negative ………………………………………………………………………………………………

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. Answer : v = 0 m s-1, u = 30 m s-1, t = 5 s Then ,

a = 0 – 30 5

= -6 m s-2

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:

.

.

.

.

.

.

.

.

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

Determination of displacement as the length of ticker tape over a period of time.

.x

(iii)

.

.

.

xy = displacement over time t t = 7 ticks = 0.14 s Determine the type of motion;

.

.

.

.y

. . . . . . . . Uniform velocity ……………………………………………………………………………………….. . . . . . . . . Acceleration ...……….…………………………………………………………………………….. . . . . . . . . Acceleration, then deceleration .……………………………………………………………………………………..

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(iv)

Physics Module Form 4 Chapter 2 : Force and Motion

Determination of velocity

.

.

.

.

.

8.6 cm displacement = ……………………… Velocity, v

(v)

=

.

.

.

7 x 0.02 = 0.14 s time = ………………………………..

8.6 = 61.43 cm s-1 0.14

Determine the acceleration Length/cm v

8 7 6 5

= 25.0 m s-2

4 3

a= v–u t 40.0 – 15.0 .. = 5(0.2) = 25.0 1.0

u

2 1 0

ticks s : displacement, v : final velocity The important symbols : ……………………………………………………………….. u : initial velocity, t : time, a : acceleration ………………………………………………………………………………………………

The equation of motion 1.

2.

3.

The list of important formula; 1 1. s  (u  v)t 2. 2 3.

v  u  at

5.

v 2  u 2  2as

4.

a

vu t

1 s  ut  at 2 2

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. given : u = 10 m s-1 ,

a = 3 m s-2 , t = 20 s.

s = ut + ½ at2 s = (10)(20) + ½ (3)(20)2 = 400 m

s = 800 m 4

s=?

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Physics Module Form 4 Chapter 2 : Force and Motion

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. given : u = 16 m s-1 , 2

v = 0(rest) , s = 8 m

a=?

2

v = u + 2 as 02 = 162 + 2 a(8) a = -16 ms-2 Execise 2.1 1.

Length / cm Figure 2.1 shows a tape chart consisting of 5-tick strip. Describe 16 the motion represented by AB and BC. In each case, determine the ; 12 (a)

(b)

displacement s = 4 + 8 + 12 + 16 + 16 + 16 = 72.0 cm average velocity

8 4

72.0 6(0.1) = 120.0 cm s-1

0

vaverage =

(c)

acceleration

Note : v =

v  u 1.6  0.4 = 0.5 t = 2.4 cm s-2

u=

C Time/s

16.0 = 1.6 cm s-1 0 .1

4 .0 = 0.4 cm s-1 0 .1 t = 5 (0.1) = 0.5 s

a=

2.

Figure 2.1 A B

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. u = 40 ms-1 v=0 t=8s 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 1 1 s = u  v t  40  08  160m 2 2 sfinal = sinitial – s = 180 – 160 = 20 m

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2.2

0m 0s

Physics Module Form 4 Chapter 2 : Force and Motion

ANALYSING MOTION GRAPHS

100m 10s

200m 20s

300m 30s

400m 40s

500m 50s

displacement time

in the form of graph called a motion graphs The data of the motion of the car can be presented…………………………………. The displacement-time Graph a)

displacement,s/m

Graph analysis: Uniform displacement all the time ……………………………………………………………… Graph gradient = velocity = 0 ………………………………………………………………

The object is stationary or is not moving Time,t (s) ……………...……………………………………………… b)

displacement (m)

Graph analysis: Displacement increases uniformly ……..………………………………………………………… Graph gradient is fixed ………………………………………………………………… time (s)

c)

displacement (m)

The object move with uniform velocity ……….………………………………………………………… Graph analysis: The object moves with uniform velocity for t seconds. …….……………………………………………………………

time (s) d)

Displacement (m)

After t seconds, the object returns to origin (reverse) with ………………………………………………………………… uniform velocity Total displacement is zero ..………………………………………………………………… Graph analysis: Graph is quadratic form …………………………….……………………………………… . Displacement increases with time. ……………………………………………..………………………

time (s)

Graph gradient increases uniformly ……………………………………………………………………… The object moves with increasing velocity with uniform ………………………………………………………………… acceleration. 6

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e)

Physics Module Form 4 Chapter 2 : Force and Motion

displacement (m)

Graph analysis: Graph is quadratic form. ………………………………………………………….. Displacement increases with time. ………………………………………………………….. Graph gradient decreases uniformly …………………………………………………………..

time (s) f) displacement (m)

The object moves with decreasing velocity, with uniform ………………………………………………………….. deceleration. Graph analysis: OA = uniform velocity (positive – move ahead) …………………………………………………………..

B

A

AB = velocity is zero (rest) ………………………………………………………….. BC = uniform velocity (negative – reverse) …………………………………………………………… O

C time time(s) (s)

The velocity-time Graph a)

v/ m s-1

Graph analysis: No change in velocity ………………………………………………………….. Zero gradient the object moves with a constant velocity or ………………………………………………………….. the acceleration is zero. …………………………………………………………… t

b)

The area under the graph is equal to the displacement of the moving object : s=vxt Graph analysis: Its velocity increases uniformly ………………………………………..………………..

t /s

v/ m s-1

The graph has a constant gradient ………………………………………………………… The object moves with a uniform acceleration ………………………………………………………… t

c)

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

t /s

v (m s-1)

Graph analysis: The object moves with a uniform acceleration for t1 s …………………………………..…………………. After t1 s, the object decelerates uniformly (negative ……………………………………………………… gradient ) until it comes to rest. ………………………………………………………

t1

t2

t (s)

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

7

s = ½ vt2

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Physics Module Form 4 Chapter 2 : Force and Motion

v (m s-1)

d)

Graph analysis: 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.

t (s)

v (m s-1)

e)

Graph analysis: 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. Calculate:(i) Velocity over OP, QR and RS (ii) Displacement

t (s) Examples 1.

s/m

P

Q

Solution :

R

O

0

2

4

6

8

t/s

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 20 0  20 10ms 1 V =  10ms 1 (i) V =

2 - 10  0  10ms 1 VRS = 1 OP

S

QR

2

(ii) S = 20 – 10 = 10 m 2.

v/m s

-1

Calculate:(i) acceleration,a over OP, PQ and QR (ii) Displacement Given : VO = 0 m s-1, VP = 10 m s-1 , Q Solution : VQ = 10 m s-1 VR = 0 m s-1 tOP = 4 s tPQ = 4 s tQR = 2 s 10  0 10 10  2.5ms 2 aPQ=  0 ms 2 (i) aOP =

P

10 5 O

0

4

R

2

4

6

8

10

t/s

0  10  5.0 ms 2 aQR = 2

1 2

(ii) S = (4 10)(10) 70.0m

8

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Physics Module Form 4 Chapter 2 : Force and Motion

Excercise 2.2 1.

(a) s/m

(b) s/m

(c) s/m

10

t/s

0

-5

2

4 t/s

t/s

-10 Figure 2.21

Describe and interpret the motion of a body which is represented by the displacement time graphs in Figure 2.21 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. 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 -5

0

2

4

t/s

-10

Figure 2.22 (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

Physics Module Form 4 Chapter 2 : Force and Motion

UNDERSTANDING INERTIA

Idea of inertia A pillion rider is hurled backwards when the motorcycle starts to move. 1. ……………………………………………………………………………………………… 2. 3.

Bus passengers are thrust forward when the bus stop immediately. ……………………………………………………………………………………………… Large vehicle are made to move or stopped with greater difficulty. ……………………………………………………………………………………………… Hand-on activity 2.5 in page 18 of the practical book to gain an idea of inertia

4.

Meaning of inertia : 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 ………………………………………………………………………………………………

Mass and inertia 1.

2.

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

An adult which one of them will be more difficult to be moved ……………………...

(ii)

An adult which one of them will be more difficult to stop? …………………………….

The relationship between mass and inertia : The larger the mass, the larger its inertia. ……………………………….……………………………………………………………..

3.

have the tendency to remain its situation either at rest or in The larger mass …………………………………………………………………………. moving. ………………………………………………………………………………………………

Effects of inertia 1.

Application of inertia Positive effect : ………………………………………………………………………… off an umbrella by moving and stopping it quickly. (i) Drying ……………………………………………………………………………………… Building a floating drilling rig that has a big mass in order to be stable and safe. ……………………………………………………………………………………… tight the loose hammer (iii) To……………………………………………………………………………………… We should take a precaution to ovoid the effect. Negative effect : …………………………………………………………………………. (ii)

2.

(i)

During a road accident, passengers are thrust forward when their ……………………………………………………………………………………... car is suddenly stopped. ……………………………………………………………………………………..

(ii)

Passengers are hurled backwards when the vehicle starts to move and are hurled ……………………………………………………………………………………… forward when it stops immediately. ………………………………………………………………………………………

person with a heavier/larger body will find it move difficult to stop his movement. (iii) A ……………………………………………………………………………………… ……………………………………………………………………………………… heavier vehicle will take a long time to stop. (iv) A ………………………………………………………………………………………

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Physics Module Form 4 Chapter 2 : Force and Motion

Execise 2.3 1.

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

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, A wooden block move up of a wooden dowel. ……………………………………………………………………………………… A wooden block has inertia to remains at rest. ………………………………………………………………………………………

(b)

hit the end of the dowel on the floor. The wooden block move downward of a wooden dowel. ……………………………………………………………………………………… A wooden block has inertia to continue it motion. ……………………………………………………………………………………

2.4

ANALYSING MOMENTUM

Idea of momentum 1. 2. 3.

it has momentum. When an object ic moving, …...………………………………………………………… defends on its mass and velocity. The amount of momentum ...…………………………………………………………… as the product of its mass and its velocity, that is Momentum is defined……………………………………………………………………. Momentum, p = m x v Unit= kg m s-1 ………………………………………………………………………………………………

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Physics Module Form 4 Chapter 2 : Force and Motion

Conservation of momentum mg

mb

vg = 0

vb

Momentum = mbvb (mb + mg) vb&g

Starting position before she catches the ball

Momentum = (mb+mg)vb&g

Receiving a massive ball

vb

vg

mb Momentum = mbvb

Throwing a massive ball

mg

Starting position before she throws the ball

Momentum = mgvg

The principle of conservation of momentum : In the absence of an external force, the total momentum of a system remains ……………………………………………………………………………………………………… unchanged. ……………………………………………………………………………………………………… The colliding objects move separately after collision. 1. Elastic collision .………………………………………………………………………….. u1 m1

v2

u2 m1

m2

Before collision Momentum : m1u1 + m2u2

= m1v1

12

m2

after collision + m2v2

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2.

Physics Module Form 4 Chapter 2 : Force and Motion

The colliding objects move together after collision. Inelastic collision :………………………………………………………………………... u1 m 1

v

u2 = 0 m2

m 1 + m2

Before collision after collision Momentum : m1u1 + m2u2 = (m1 + m2) v 3.

explosion :

The objects involved are in contact with each other before explosion and …….....…………………………………………………………………... are separated after the explosion. v1 v2 (m1 + m2), u = 0 m2

Before explosion

after explosion

Momentum : (m1 + m2)u = m1 vv - m2 v2 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. , uA = 30 m s-1, vA = 25 m s-1, mB = 90 kg, Solution : Given : mA = 100 kg-1 uB = 20 m s , vB = ? mAuA + mBuB = mAvA + mBvB (100)(30) + (90)(20) = (100)(25) + (90)(vB) vB = 25.56 m s-1 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. -1 Solution : Given : mA = 100 kg , uA = 30 m s , mB = 90 kg,

mAuA + mBuB = (mA + mB ) v (B+A) (100)(30) + (90)(20) = (100 + 90) v (B+A) v(A + B) = 25.26 m s-1 13

uB = 20 m s-1 , v(A+B) = ?

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Physics Module Form 4 Chapter 2 : Force and Motion

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 :

Given ; mb = 2 g = 0.002 kg, vg = ? 0 = mgvg – mb vb, 0 = (1)(vg) – (0.002)(150),

mg = 1 kg,

u(g+b) = 0 , vb = 150 m s-1

vg = 0.3 m s-1

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

2.

(0.15 x 15) + (0.450 x 0) = 0.6 v(a+ wb) v(a+ wb) = 3.75 m s-1 -1 A riffle of mass 5.0 kg fires a bullet of mass 50 g with a velocity of 80 m s .Calculate the recoil velocity. Explain why the recoil velocity of a riflle is much less than the velocity of the bullet.

mr vr = mb vb , 2.5

mr = 5.0 kg vr = ? ( 5.0 ) vr = ( 0.05)(80) vr = 0.8 m s-1

mb = 50 g vb = 80 m s-1

UNDERSTANDING THE EFFECT OF A FORCE

Idea of force 1.

What will happen when force act to an object? Force can make an object; ……………………………………………………………………………………………… 1. Move 2. Stop the moving ……………………………………………………………………………………………… 3. Change the shape of the object 4. Hold the object at rest ………………………………………………………………………………………………

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Physics Module Form 4 Chapter 2 : Force and Motion

Idea of balanced forces 1.

An object is said to be in balance when it is: 1. In a stationary state ……………………………………………………………………………………………… 2. Moving at uniform velocity ………………………………………………………………………………………………

2.

Stationary object Normal reaction, N ………………………………

explanation : Magnitude R = W but R acts in an ………………………………………………

Stationary object

opposite direction to the weight. ……………………………………………… ( object is in equilibrium ) ……….…………………………………….. weight, w = mg ………………………………………… 3.

An object moving with uniform velocity Normal reaction, N …………………………….. Frictional force …..…………….

Force, F ……………

explanation : 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. ……………………………………………..

Idea of unbalanced forces 1. 2.

when it is moving in acceleration. A body is said to be in unbalanced..…………………………………………………… Resultant force ……………………….. Explanation; The ball move in acceleration ……………………………………………… because the forces act are not balanced. ……………………………………………… F ………

F > F’ ……………………………………………… So, the ball move in F direction ………………………………………………

F’ ……..

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.

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1.

Physics Module Form 4 Chapter 2 : Force and Motion

Refer to the result of experiment 2.2 and 2.3, 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 ………………………………………………………………………………………………

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) F = ma So, …………………………………………………………………………………………

3.

Example 1 :

Calculate F, when a = 3 m s-2 dan m = 1000 kg F = ma F = (1000)(3) F = 3000 N

Example 2 : m = 25 kg F = 200 N

Calculate the acceleration, a of an object. F = ma 200 = 25 a a = 8.0 ms-2 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. m = 30 kg , F – Ff = ma ,

2.

F = 50 N ,

Ff = 20 N ,

a =?

50 – 20 = 30 a a = 1.0 m s2

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? m = 1000 kg , u = 72 km h-1, v = 0, s = 40 m, F = ? F = ma, = 1000 x 5.0 = 5000.0 N

16

Note : u = 72 km h-1 =20 m s-1 v2 = u2 + 2as 0 = 202 + 2a(40) a = 5.0 m s2

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2.6

Physics Module Form 4 Chapter 2 : Force and Motion

ANALYSING IMPULSE AND IMPULSIVE FORCE

Impulse and impulsive force The change of momentum 1. Impulse is ………………………………………………………………………………. 2.

The rate of change of momentum during collision and explosion. Impulsive force is ……………………………………………………………………… ………………………………………………………………………………………………

3.

Formula of impulse and impulsive force: It is known that a= (v–u)/t Refer, F = ma F = m( v – u) t Ft = mv – mu , Unit = N s

Therefore, So,

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 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 = 5(-10) - ( 5(10))

and impulsive force, F =

= - 100 Ns Example 2;

v

-100 = -100 N 1

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 = 5(-10) - (5(10)) = 4.

and impulsive force, F =

100 Ns

100 = 50 N 2

The relationship between time of collision and impulsive force. Impulsive force , F  1 / t ……………………………………………………………………………………………… Therefore, F decreases when the time of collision increases ( refer to examples ) ………………………………………………………………………………………………

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Physics Module Form 4 Chapter 2 : Force and Motion

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?

Fimpulse = Ft = 20 x 0.8 = 16.0 Ns

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? 1. 2.

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

2.7 BEING AWARE OF THE NEED FOR SAFETY FEATURES IN VEHICLES Safety features in vehicles

Reinforced passenger compartment

Head rest

Crash resistant door pillars

Windscreen Crumple zones

Anti-lock brake system (ABS)

Traction control

bumpers

Air bags

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Physics Module Form 4 Chapter 2 : Force and Motion

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

Safety seat belt

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

Physics Module Form 4 Chapter 2 : Force and Motion

UNDERSTANDING GRAVITY

Carry out hands-on activity 2.8 on page 35 of the practical book. Acceleration due to gravity. 1. 2.

It pulled by the force of gravity. An object will fall to the surface of the earth because………………………………... as earth’s gravitational force.F The force of gravity also known ………………………………………………………...

3.

the object is said to be free When an object falls under the force of gravity only, ………………………………... falling ………………………………………………………………………………………………

4.

known as acceleration due to gravity. The acceleration of objects falling freely is ………………………………………………

5.

The magnitude of the acceleration due to gravity depends ………………………... on the strength of the gravitational field, g ………………………………………………………………………………………………

. field Gravitational 1. 2. 3. 4.

5.

the gravitational field of the earth. The region around the earth is …………………………………………………………. is on the force of gravity. The object in gravitational field ………………………………………………………… as the gravitational force acting on a 1 kg mass. The gravitational field strength is defined …………………………………………….. The gravitational field strength, g can be calculate as; . g = F . where, F : gravitational force m m : mass of an object At the surface of the earth, g = 9.8 N kg-1 …………….………………………………………………………………………………..

6.

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

7.

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

F = mg

= (60) (9.8) = 588.0 N

Example 2, A satellite of mass 600 kg in orbit experiences a gravitational force of 4800 N. Calculate the gravitational field strength. Given : m = 600 kg. F = 4800 N, g = ? g = F = 4800 . m

600

20

= 8 N kg-1

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Physics Module Form 4 Chapter 2 : Force and Motion

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-1of the well. Given : u = 0 ms , t = 1.2 s, (a) v = ?

a = g = 9.8 ms-2

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

Weight 1.

as the gravitational force acting on the object. The weight of an object is defined ……………………………………………………..

2.

For an object of mass m, the weight can be calculate as : weight, W = mg where, g = acceleration due to gravity. 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 = mg 9.78 N kg-1? W = 6000 x 9.78 = 58 680 N

Exercise 2.8 1.

Sketch the following graphs for an object that falling freely. (a) (b) (c)

Displacement-time graph, Velocity-time graph Acceleration-time graph (a) s / m (b) v / m s-1

t/s

(c) a / m s2

t/s

21

t/s

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2.

Physics Module Form 4 Chapter 2 : Force and Motion

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. m = 200 g = 0.2 kg

s = 3.0 m

t = 0.79 s

u=0

g=?

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. 2.9

IDEA OF EQUILIBRIUM FORCES

An object is in equilibrium when : It is in a stationary state 1. ……………………………………………………………………………………………… 2.

It is moving with uniform velocity ……………………………………………………………………………………………… Normal reaction, R Normal reaction, R

Weight, W=mg

weight, W=mg stationary object

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

An object moving with uniform velocity

Force , F = Frictional force Resultant force = F – Frictional force =0 (object in equilibrium)

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Physics Module Form 4 Chapter 2 : Force and Motion

Addition of Force 1.

a resultant force is a single force the Addition of force is defined as ...…………………………………………………….. 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) ……………………………………………………………………………………………… Examples : the forces are acting in one direction F1 = 10 N F2 = 5 N Resultant force, F

= F1 + F2

= 10 + 5 = 15 N

Example : the forces are acting in opposite directions F1 = 10 N F2 = 5 N Resultant force, F = F1 - F2

= 10 - 5 = 5 N

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.

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Physics Module Form 4 Chapter 2 : Force and Motion

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?

Solution :

Example 2:

Resultant force, F = 6000 – 5300 =700 N They mere not in equilibrium

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. Resultant force, F = 10.5 x 50 = 525 N

250 10.5 cm

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Physics Module Form 4 Chapter 2 : Force and Motion

Resolution of a force 1.

reverse process of finding the resultant force Resolution of a force is ………………………………………………………………… Fy

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

Vertical Component

 Fx

Refer to trigonometric formula:

Example :

horizontal component

Cos  =

Fx F

Sin  =

Fy

, therefore Fx = F cos 

, therefore Fy = F sin  F The figure below shows Ali mopping the floor with a force 50 N at an angle of 600 to the floor. Fx = F cos  = 50 cos 60 = 50 (0.5) = 25 N

Fx

F = 50 N

Fy

600

Fy = F Sin  = 50 sin 600 = 50 (0.8660) = 43.3 N

Example of resolution and combination of forces F=?

F = mg sin 400 + 200 = 800(0.6427) + 200 = 514.2 + 200 = 714.2 N

200 N 400 400

mg = 800 N

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Physics Module Form 4 Chapter 2 : Force and Motion

Problem solving 1.

the resultant force is equal to zero. When a system is in equilibrium, ……………………………………………………….

2.

If all forces acting at one point are resolved into horizontal and vertical the sum of each component is equal to zero. components, ……………………………………………………………………………

3.

Example 1; Show on a figure;

a) the direction of tension force, T of string b) the resultant force act to lamp 0 0 70 70 c) calculate the magnitude of tension force, T a) T b) T’ T (c ) T’ = 2T sin 700 Therefore, mlampg = 2T sin 700 mlamp g mlamp = 1.5 kg T= 2sin70 0 Wlamp = 14.7 N 1.5(9.8) = = 7.82 N 2sin700 Exercise 2.9 1.

2.

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. = mLamp g Fmaximum when both of forces act in same direction; Fmaximum = 18 + 6 18 N = 24 N 6N

24 N

Fminimum when the forces act in opposite direction ; Fminimum = 18 – 6 18 N = 12 N 6N

12 N

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. F = Resultant of Force F2 = 2202 + 2002 F = 297.32 N

220 N F

900

200 N

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2.10

Physics Module Form 4 Chapter 2 : Force and Motion

UNDERSTANDING WORK, ENERGY AND EFFICIENCY

Work 1.

When a force that acts on an object moves the object through a Work is done, …………………………………………………………………………….. distance in the direction of the force. ………………………………………………………………………………………………

2.

of a force and the distance traveled in the direction of WORK is the product.……………………………………………………………………. the force. ………………………………………………………………………………………………

3.

The formulae of work;

WORK = FORCE X DISPLACEMENT W =Fxs W : work in Joule/J F : force in Newton/N s : displacement in meter/m

4.

Example 1; Force, F s

W = Fs

If, F = 40 N and s = 2 m Hence, W = 40 x 2 = 80 J Example 2;

80 N 600 W = Fs

s= 5m

= 80 cos 600 (5) = 80 (0.5) (5) = 200 J

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Physics Module Form 4 Chapter 2 : Force and Motion

Example 3; T

T

F = 30 N h = 1.5 m W=Fs=Fh = 30 (1.5) = 45.0 J

Example 4; W=Fs = 600 x 0.8 = 480 J

F = 600 N

S = 0.8 m

Energy 1.

It is the potential to do work. Energy is .................................................................................................................

2.

created nor be destroyed. Energy cannot be ....................................................................................................

3.

potential energy, kinetic energy, electrical Exist in various forms such as …………………...…………………………………… energy, sound energy, nuclear energy and chemical energy. ………………………………………………………………………………………………

4.

5.

Example of the energy transformation; 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. ……………………………………………………………………………………………… Example : If 100 J of work is done, it means 100 J of energy is consumed. ………………………………………………………………………………………………

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Physics Module Form 4 Chapter 2 : Force and Motion

Work done and the change in kinetic energy Force, F

s 1.

energy of an object due to its motion. Kinetic energy is …………………………………………………………………………

2.

Refer to the figure above, Work = Fs = mas = m ( ½ v2)

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

The formulae of Kinetic energy, Ek = ½ mv2 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?

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 Work done and gravitational potential energy

h = 1.5 m

1.

energy of an object due to its position. Gravitational potential energy is………………………………………………………... (possessed by an object due to its position in a gravitational field) ………………………………………………………………………………………………

2.

Refer to the figure above;

3.

Example; If m = 10 kg

W = Fs = mg h where, F = mg So, Gravitational energy, Ep = mgh W = 10 (10) 1.5 = 1500 J Therefore Work done = 1500J And, Ep = 1500 J 29

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Physics Module Form 4 Chapter 2 : Force and Motion

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. created or destroyed but can be changed from one form to 1. Energy cannot be ……………………………………………………………………… another form. …………………………………………………………………………………………… 2.

Example : a thrown ball upwards will achieve a maximum height before changing its direction and falls Maximum Potential energy

Kinetic energy decrease and potential energy Increase

potential energy decrease and kinetic energy increase

Maximum kinetic energy

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? 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

Power 1.

the rate of doing work. Power is ………………………………………………………………………………… workdone W Therefore, power, P = , so, P = timetaken t Where, P : power in watt/W W : work in joule/J t : time to do work in seconds/s

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? -2 g = 9.8 ms-2) Solution : Given : m = 180 kg, h = 2 m, t = 0.8 s and g = 9.8 ms . P = ? W mgh P= = t t = 30

180  9.8  2 0.8

= 4 410 W

JPN Pahang

Physics Module Form 4 Chapter 2 : Force and Motion

Efficiency 1.

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

2.

Formulae of efficiency :

Efficiency  3.

Useful energy output  100% Energy input

Analogy of efficiency; unwanted energy Energy input, Einput

Useful energy, Eoutput

Device/ mechine

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 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 



Eoutput Einput

x 100%

0.48 x 100%  60% 0.80

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

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Physics Module Form 4 Chapter 2 : Force and Motion

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. W=Fs = 90 x 10 = 900 J

2.

The energy transferred to the force = 900 J

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.

m = 3.0 x 50 = 150 kg

h = 1.5 m

W = mhg = 150 x 9.8 x 1.5 = 2205 J (b)

What is his power if he completes this work in 250 s?

P=

W t

= 2205 = 8.82 w 250

2.11

APPRECIATING THE IMPORTANCE OF MAXIMISING THE EFFICIENCY OF DEVICES

1.

During the process of transformation the input energy to the useful output some of energy transformed into unwanted forms of energy. energy,……………………………………………………………………………………..

2.

The efficiency of energy converters is always less than 100%. .…………………………………………………………………………………………….. The unwanted energy produced in the device goes to waste. ………………………………………………………………………………………………

3.

Example of wasting the energy; Kinetic energy ………..………………… Input enegy from the petrol

output 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 ..………………….. ……………. …………………. …………………….

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Physics Module Form 4 Chapter 2 : Force and Motion

4.

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

5.

It is very important that a device makes the best possible use of the input energy. …………………………………………………………………

Ways of increasing the efficiency of devices Engine must be designed with the capability to produce greater amount 1. Heat engines ……………………..……………………………………………………… of mechanical work. ……………………………………………………………………………………………… 2.

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

Operation of electrical devices 1. 2. 3.

when they are in good operating The electrical devices increase the efficiency………………………………….…… condition.will increase the life span of device. Proper management ….....……………………………………………………………… Example : -the filter in an air-conditioner and fins of the cooling coil of a …………..……………………………………………………………………………… refrigerator must be periodically cleaned. ………………………………………………………………………………………………

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2.12

Physics Module Form 4 Chapter 2 : Force and Motion

UNDERSTANDING ELASTICITY

Carry out Hands-on activity 2.12 page 40 of the practical book. the property of an object that enables it to return its original shape and 1. Elasticity is ……………………………………………………………………………...

2.

dimensions after an applied external force is removed. ……………………………………………………………………………………………… The property of elasticity is caused by the existence of forces of Forces between atoms ………………………………………………………………….. repulsion and attraction between molecules in the solid material. ………………………………………………………………………………………………

3.

Forces between atoms in equilibrium condition Force of attraction Force of repulsion

Force of repulsion

Explanation : 1. The atoms are separated by a distance called the equilibrium distance and vibrate ……………………………………………………………………………………………… at it position. ……………………………………………………………………………………………… 2. Force of repulsion = Force of attraction ……………………………………………………………………………………………… 4.

Forces between atoms in compression compressive force

compressive force

Force of repulsion

Force of repulsion

Explanation ; 1. Force of repulsion takes effect. ……………………………………………………………………………………………… 2. When the compressive force is removed, force of repulsion between the atoms ……………………………………………………………………………………………… pushes ……………………………………………………………………………………………… 5.

thebetween atom back to their equilibrium positions. Forces atoms in tension force of attraction

stretching force

stretching force

Explanation ; 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. ………………………………………………………………………………………………

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Physics Module Form 4 Chapter 2 : Force and Motion

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 that the extension of a spring is directly proportional to the applied 1. Hooke’s Law states ………………………………………………………………………

2.

force provided that the elastic limit is not exceeded. ……………………………………………………………………………………………… as the maximum force that can be applied to Elastic limit of a spring is defined……………………………………………………….

3.

spring such that the spring will return to its original length when the force released. ……………………………………………………………………………………………… when the length of the The spring is said to have a permanent extension,...………………………………… spring longer than the original length even though the force acts was released and the ………………………………………………………………………………………………

4.

elastic limit is exceeded. ……………………………………………………………………………………………… When the spring obey Hooke’s Law. The elastic limit is not exceeded,…………………………………………….………… The mathematical expression for Hooke’s Law is : ……………………………………………………………………………………………… F x ……………………………………………………………………………………………… F = kx, k = Force constant of the spring with unit N m-1, N cm-1 or N mm-1

Force constant, k = F x 5.

Graf F against x F/ N P

Q F = kx E Spring obeying Hooke’s Law Spring not obeying Hooke’s law (exceeded the elastic limit)

Force constant, 0

6.

R

k =

F x

with unit N m-1, N cm-1 or N mm-1

x (cm)

Spring Constant, k

k is the gradient of the F - x graph

F/N

F x 0.8 = 8 = 0.01 N cm-1

k=

0.8

0

8

x/cm

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JPN Pahang

Example 1;

Physics Module Form 4 Chapter 2 : Force and Motion

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] Given : lo = 15 cm,

m = 200 g , F = 2.0 N, l = 20 cm x = 5 cm k = Fx  2.0  0.4Ncm 1

a.

k = ?,

b.

l = ? , when m = 150 g, F = 1.5 N x=

5

F 1.5   3.75cm k 0.4

l = 15 + 3.75

= 18.75 cm The graph shows the relationship between the Graph F against x of stretching force, F and the spring extension, x. spring P and spring Q (a) Calculate the spring constant of P and Q. (b) Using the graph, determine the stretching force acts to spring P and P spring Q, when their extension are 0.5 cm Solution a. Spring constant, k = gradient of graph Q 6 k = 15.79 N cm 1

Example 2; F (N) 8 7 6 5

0.38 3  6.0 N cm 1 kQ = 0.5 P

4 3 2 1 0

From a, k = 4.0 N cm-1

0.1 0.2 0.3 0.4 0.5

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

Elastic potential energy 1.

the energy stored in a spring when it is extended or compressed Elastic potential energy ……………………………………………………………….. spring with the original length F compression x

spring compressed F

x = compression

x

F

x x

spring extended x = extension

F, extension Other situation where the spring extended and compressed

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JPN Pahang

Physics Module Form 4 Chapter 2 : Force and Motion

Relationship between work and elastic potential energy F/N

Graph F against x

Area under the graph

= work done = ½ Fx So, Elastic potential energy = ½ Fx

F x

x / cm

Example ;

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

5 kg

15 cm

8 cm

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 Factor Length

Diameter of spring

Diameter of spring wire Type of material

Change in factor

Effect on elasticity

Shorter spring

Less elastic

Longer spring

More elastic

Smaller diameter

Less elastic

Larger diameter

More elastic

Smaller diameter

More elastic

Larger diameter

Less elastic

the elasticity changes with the type of materials

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Physics Module Form 4 Chapter 2 : Force and Motion

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. 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

2.

If a 20 N force extends a spring from 5 cm to 9 cm, (a) what is the force constant of the spring? F = 20 N,

(b)

x = 9 – 5 = 4 cm,

k=?

F = kx 20 = k (4) k = 5 N cm-1

Calculate the elastic potential energy stored in the spring. E = ½ Fx = ½ (20)(4) = 40 J

Reinforcement Chapter 2 Part A : Objective Questions 1.

When a coconut is falling to the ground, which of the following quantities is constant? A. B. C. D.

2.

3.

Velocity Momentum Acceleration Kinetic energy

A. B. C. D. E.

In an inelastic collision, which of the following quantities remains constant before and after the collision? A. B. C. D.

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.)

4.

Total acceleration Total velocity Total momentum Total kinetic energy

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

The momentum of a particle is dependent on A. B. C.

38

0.1 N 0.2 N 0.4 N 0.6 N 0.8 N

mass and acceleration weight and force mass and velocity

JPN Pahang

Physics Module Form 4 Chapter 2 : Force and Motion

Which of the following diagrams shows a body moving at constant velocity?

5.

A.

6.

2N

8.

m = 0.3 kg 5m

2N What is the momentum of the stone just before it hits the ground?

B.

12 N

7N

C.

12 N

14 N

D.

20 N

17 N

A. B. C. D. E.

The graph below shows the motion of a trolley with mass 1.5 kg. Velocity / ms

Solution : 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

-1

4

0 2 4 6 Time / s Calculate the momentum of the trolley from t = 2s to t = 4s. A. B. C. D. E. 7.

1.5 kg m s-1 3.0 kg m s-1 4.0 kg m s-1 6.0 kg m s-1 7.5 kg m s-1

9.

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

Thrust

A big ship will keep moving for some distance when its engine is turned off. This situation happens because the ship has A. B. C. D.

10.

This figure shows an aircraft flying in the air. Lift

Air friction

Weight The aircraft above accelerates if A. B. C. D.

0.15 kg m s-1 0.3 kg m s-1 1.5 kg m s-1 3.0 kg m s-1 15.0 kg m s-1

great inertia great acceleration great momentum great kinetic energy 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 B C D E

Lift  Weight Thrust  Lift Lift  Air friction Thrust  Air friction

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0.6 s 1.4 s 1.7 s 3.5 s 12.0 s

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

JPN Pahang

Physics Module Form 4 Chapter 2 : Force and Motion

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? Method (a) ……………………………………………………………………………… (ii)

State a reason for your answer in (a)(i). The forces given parallel with the surface of motion, ……………………………………………………………………………… So, all the forces given used to move the car. ………………………………………………………………………………

(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. F = Fgiven - Ffriction = 500 – 200 (ii)

= 300 N horizontal resultant force in method B. F = Fgiven Cos 500 – Ffriction = 500 cos 600 – 200 = 50.0 N

(iii)

acceleration of the car in method B. F=ma 50.0 = 1000 a a = 0.05 m s-2

(c)

Suggest a method to move Car B so that the acceleration produced is equal to that of method A. The acceleration of Car A = 0.3 m s-2 ……………………………………………………………………………..……….. To move Car B with the same acceleration of Car A, increase the force given ……………………………………………………………………………………… to 1000 N

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JPN Pahang

Physics Module Form 4 Chapter 2 : Force and Motion

2.

ceiling

Tin

water P



N

Q (i)

a)

M

hand

R

Diagram 2.1

(ii)

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. The difficulty to move the tin depends to its mass. ……………………………………………………………………………………… 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.

b)

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

c)

mass and velocity ……………………………………………………………………………………… If water flows out from a hole at the bottom of the tin Q, how would the inertia of Tin Q depends on time ?

d)

inertia of tin Q will decrease because the mass of tin decreased. …………………………………………………………………………………… 2 ms-1

3. P

iron ball ( 2 kg ) S 3.0 m

T

smooth surface 1.0 m Q

2.0 m

Diagram 3

R 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. Ek = ½ mv2 = ½ (2)(22) = 4.0 J

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JPN Pahang

Physics Module Form 4 Chapter 2 : Force and Motion

(ii)

(iii)

b) c)

(i) (ii)

the potential energy of the iron ball at P. Ep = mgh = (2) (10) (3.0) = 60.0 J the total of energy of the iron ball at P. E = Ek + Ep = 4.0 + 60.0 = 64.0 J Calculate the total of energy of the iron ball when it reaches at Q ? 64.0 J ( the conservation of energy ) Calculate the work done against friction along QR. W = Ff x s = 4 x 1.0 = 4.0 J

d)

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

e)

Calculate the speed of the ball at position T. v2 = 20 v = 4.5 m s-1

Ek at T = 20.0 J = ½ m v2 = ½ (2)(v2) 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. 42

JPN Pahang

Physics Module Form 4 Chapter 2 : Force and Motion

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.

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.

Answer

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 - 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.

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JPN Pahang

Physics Module Form 4 Chapter 2 : Force and Motion

2. Properties Brand

A B C D

Reaction time / s

Mass / kg

0.3 0.5 0.2 0.6

1.5 1.8 0.9 2.5

Engine thrust force / N 10.0 12.5 6.5 16.0

Resistance force /N 4.0 2.4 2.2 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 radiocontrolled 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

(c)

(d)

t/s - 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. (i) EB = (12.5 – 2.4 ) (50) = 505.0 J 0 50 m 50Sin30 E (suitable to move up) = 1.8 (10)(50Sin300) = 450 .0 J 300 EB> E ( car B can move up the plane) (ii) F = ma , 12.5 – 2.4 = 1.8 a, a = 5.61 ms-1

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