Form 4 Physics Chapter 2.1-2.4 - Teacher's Copy

CHAPTER 2 : FORCE AND MOTION

2.1 Linear Motion

A student is able to :2.1.1 Define distance and displacement 2.1.2 Define speed and velocity and state that v = s/t2.1.3 Define acceleration and deceleration and state that a = v - u t2.1.4 Calculate speed and velocity2.1.5 Calculate acceleration / deceleration2.1.6 Solve problems on linear motion with uniform acceleration using (i) v = u + at

(ii) s = ut + ½ at2

(iii) v2 = u2 + 2as

Distance and Displacement

A: Fill in the blank with the correct answer

1. Physical quantities can be divided into 2 :

(a) Scalar quantity

(b) vector quantity

2. Distance is scalar quantity which has magnitude and no direction

3. Displacement is a vector quantity which has magnitude and direction

4. The SI unit for both physical quantities is metre / m

B: Complete the table below :

Aspect Distance Displacement

Definition Total route by a motionDistance taken with

consideration of direction

Type of quantity Scalar quantity Vector quantity

SI unit Metre / m

Speed and Velocity

C: Complete the table below

Aspect Speed Velocity

Definition Rate of change of distanceRate of change of displacement

Type of physical quantity

Scalar Vector

FormulaSpeed = Total distance Time

Velocity = total Displacement

time

SI unit m s-1

m s-1

Symbol u, v u, v

u denotes initial speedv denotes final speedAverage speed = Total distance Time

u denotes initial velocityv denotes final velocityWe usually consider the forward motion ( to the right ) as positive and the backward ( to the left) as negative )

Acceleration

D: Fill in the blank with the correct answer.

1. Acceleration is the rate of change of Velocity

2. Acceleration , a = Final velocity - Initial velocity Time taken

3. The SI unit of acceleration is ms-2

4. Acceleration is a vector quantity

5. Acceleration occurs when an object moves with increasing velocity.

6. Deceleration occurs when an object moves with decreasing velocity

E: Solve the problem.

1. A car starts from points from point O and moves to U, 50 m to the north in 60 s. The

car then moves to B, 120 m to the west in 40 s. Finally, it stops.

Calculate the :

(a) total distance moved by the car

total distance

= 120 m + 50 m

= 170 m

(b) displacement of the car

Displacement = Distance OB

= (1202 + 502)1/2

= 130 m

(c) speed of the car when it is moves to the north

Speed of the car when it moves to the north = 50 m = 5 m s-1

60 s 6(d) velocity of the car

velocity = Displacementtime

a = v – u t

5m

120 m

50m

B U

O

= 130 m = 1.3 m s-1

(40 + 60) s

(e) average speed of the car

= total distance Time

= 170 m__

(40 + 60) s

= 1.7 m s-1

2. A bus stops at a station to pick passengers up. It then moves and attains a velocity of 15 m s-1 in 8 s. What is the acceleration of the bus ?

acceleration , a = v – ut

= 15 – 08

= 1.9 m s-2

F. Complete the table below with the correct answer

Increasing short increasing far same

Speed decreasing Low high decreasing

Pattern Explanation(a) Direction of motion

Initial Final

The distance between the dots is the same It shows that the object is moving with constant speed

(b) Direction of motion

Initial Final

The distance between the dots is short. It shows that the speed of the object is low

(c) Direction of motion

Initial Final

The distance between the dots is far. It shows that the object is moving at a high speed

(d) Direction of motion

Initial Final

The distance between the dots is Increasing. It shows that the speed of the object is increasing.

(e) Direction of motion

Initial Final

The distance between the dots is decreasing It shows that the speed of the object is decreasing.

G: Determine the acceleration of a trolley from the ticker tape

1. The ticker tape is divided into 5 parts. Every part has 2 ticks as shown in figure below.

Find the acceleration.

. . . . . . .

.

. . . . . . .

.

. . . . . . .

.

. . . . . .

.

. . . . . . . . . . . . . . . .

.

1 cm 5 cm

A B C D E F

Solution :

Step Solution

1. Time taken of one part One part = 2 Ticks = 2 x 0.02 s =0.04 s

2. initial velocity, u = s initial

t initial

u = 1.0 cm 0.04 s

= 25 cm s-1

3. Final velocity, v = s final

t final

v = 5.0 cm 0.04 s

= 125 cm s-1

4. Determine the total time Total time = ( Total parts – 1) x time of one part

Total time, t = ( 5-1 ) x 0.04 s = 4 x 0.04 s = 0.16 s

5. Acceleration, a = v – u t

Acceleration , a = v – u t = 125 – 25 / 0.16

= 750 cm s-2

The equations of motion

1. Complete the table below

Physical Quantity Symbol

Displacement s

Final Velocity v

Initial velocity u

acceleration a

2. List the equations of linear motion.

(i)

ii)

iii)

iv) 1. Solve the following problems using the equations of linear motion

(a) A car moves from rest to a velocity of 10 ms-1 in 5 s.Calculate the acceleration of the car

Solution :

Given : u = 0 , v = 10 ms-1 , t = 5 s . a = ?

a = v – u t= 10 – 0 = 2.0 m s-2

5

(b) 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.

Solution :

(c) A van that is traveling with velocity 16 m s-1 decelerates until it comes to a stop.If the

distance traveled is 8 m, calculate the deceleration of the van.

Solution :

2.2 ANALYSING MOTION GRAPH

A student is able to :2.2.1 Plot and interpret displacement-time and velocity-time graphs2.2.2 Deduce from the shape of a displacement-time graph when a body is:

i) at rest

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

s = 800 m

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

ii) moving with uniform velocityiii) moving with non-uniform velocity

2.2.3 Determine distance, displacement and velocity from a displacement-time graph2.2.4 Deduce from the shape of a velocity-time graph when a body is:

i) at restii) moving with uniform velocityiii) moving with uniform acceleration

2.2.5 Determine distance, displacement and velocity and acceleration from a velocity-time graph2.2.6 Solve problems on linear motion with uniform acceleration

A: Describe the motion of an object as shown in the following motion graphs.

(a) The Displacement-Time Graph

Graph of s against t Explanation

s/m

0 t/s

The displacement of the object from a fixed point is

constant Therefore, the velocity of the object is zero

s/m

0 t/s

The gradient of the graph = Velocity of the object.

The gradient of the graph is constant , therefore the

velocity of the object is constant

s/m

0 t/s

The gradient of the graph increases with time.

Therefore the velocity of the object increases with

time

(b) The velocity – time graph

Graph of v against t Explanation

v/m s-1

Velocity of object is zero The object is stationary

t/s

v/m s-1

t/s

The object is moving with uniform velocity

v/m s-1

t/s

The gradient of the graph = acceleration

The gradient of the graph is constant.

Therefore, the acceleration of the object is constant

v/m s-1

t/s

Area under the graph = displacement

B. Describe the motion of a runner who is running in a straight line

s/m

O t/s 5 10 12 15

Solution :

Motion of the runnerO - A Running with a uniform velocity of 3 m s-1

A-B At rest / velocity is zeroB - C Uniform velocity of 5.0 m s-1

20

25

15

10

5

A B

D

C

v/m s-1

t/sA

B C

D

A 5 10 15 20 25

C - D Running with a velocity of -8.3 m s-1 and return to starting point

C: Study the velocity-time graph.

Calculate:- (i) the acceleration, a, for sections AB, BC and CD

(ii) Displacement

(i) Acceleration , a for AB = 10 – 0 = 1.0 m s-2

10 Acceleration, a for BC = 0

Acceleration , a for CD = 0 – 10 25 - 20 = -10 m s-2

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(ii) Displacement = Area under the graph = ( ½ x 10 x 10) + (10 x 10) + ( ½ x 5 x 10 ) = ( 50 ) + ( 100 ) + ( 25 ) = 175 m

2.3 UNDERSTANDING INERTIA

A student is able to :2.3.1 Explain what inertia is2.3.2 Relate mass to inertia2.3.3 Give examples to reduce the negative effects of inertia

10

5

A. Fill and underline the correct answer

1.

When the bus stops suddenly our feet are brought to rest but due to inertia, our body

tends to continue its (forward/backward) motion. This causes our body to jerk

forward. (thrown forward)

2.

When the bus moves suddenly from rest our feet are carried (forward/backward) but

due to inertia of our body tends to keep us (rest/moving) . This causes our body to

fall backwards.

3. 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

4. The mass of an object is the amount or quantity of matter contained in the object.

5. The SI unit of mass is kg

6. The mass of an object is constant wherever it is measured. It is different from

weight which is affected by the force of gravity.

7. An object with a larger mass has a larger Inertia.

Effects of inertia

B. Explain the positive effect by using the clues given.Clue : Pour out tomato sauceThe chili sauce inside the

bottle move together with the bottle. When the bottle stops suddenly, the chili sauce flow

out

Clue : Drying a wet umbrella

The umbrella is rotated and stopped abruptly to shake off the droplet of water

C . Match the correct explanation to each of the characteristics.

Characteristic Explanation

(a) The tank which carries liquid in a lorry should be divided into smaller tanks

Hold the passengers to their seat during collision

(b) the part between the driver’s seat and load should have strong steel structure

Prevent the driver from hitting the steering in an accident

(c) Safety belts Prevent the load from being thrown to the front

(d) Airbag To reduce the effects of inertia when stopped suddenly.

2.4 ANALYSNG MOMENTUM

A student is able to :2.4.1 Define the momentum of an object2.4.2 Define momentum(p) as the product of mass (m) and velocity(v) ie p=mv2.4.3 State the principle of conservation of momentum2.4.4 Describe the applications of conservation of momentum2.4.5 Solve problems involving momentum

Positive effectOf inertia

Clue: Running zig-zig when chased by a bull.

This is because the inertia of the bull is large due to its big mass. This makes the bull difficult to change directions suddenly.

Clue : Tightening a hammer headThe hammer head can be tightened by hitting the wooden handle on the floor

A. Fill in the blank with the correct answer

1. Momentum is defined as the product of mass and velocity

2. The formula of momentum is given by :

Momentum = mass x velocity

p = mv

3. Momentum is a vector quantity

4. The SI unit of momentum is kg m s -1

5. Momentum increases/decreases , when the speed increases/decreases

6. The principle of conservation of momentum states that the total momentum in a

closed system of object is constant

7. The total momentum before the collision is equal to the total momentum after the

collision if no external force acts on the system

8. The principle of conservation of momentum can be applied in

(a) elastic collision

(b) inelastic collision

(c)explosion

B. Fill in the blank

1. Collision I : Both bodies separates after collision. Momentum is conserved.

Before collision after collision

Momentum : m1u1 + m2u2 = m1v1 + m2v2

2. Collision II : Both objects move together after collision. …………. Is conserved.

m1

m2 m1m2

u2 v2

u1

m1 m2 m1 + m2

u2 = 0

u1

v

Before collision after collision

Momentum : m1 u1 + m2 u2 = (m1 + m2) v

3. explosion : Before the explosion, the objects are at rest

Two or more object in contact will be separated after the explosion

Before explosion after explosion

Momentum : (m1 + m2)u = m1 vv + m2 v2

Therefore m1 vv + m2 v2 = 0

B. Complete the table below

Elastic collision Inelastic collision Explosion

(a) Both object will separated after collision

Both object will join together or separated after collision

Two or more objects in contact will be separated after the collision

(b) Total momentum is Conserved

The total momentum is conserved

The total momentum is conserved

(c) The total kinetic energy is conserved

Total kinetic energy is not conserved

Total kinetic energy is not conserved

(c) Total energy is Conserved

Total energy is conserved Total energy is conserved

C. Solve the problem

1. 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 the same direction. Car A and B move separately after collision. If Car A moves at

25 m s-1 after collision, determine the velocity of Car B after collision.

Solution : Given : mA = 100 kg , uA = 30 m s-1, vA = 25 m s-1, mB = 90 kg, = 20 m s-1 , vB = ?

mAuA + mBuB = mAvA + mBvB (100)(30) + (90)(20) = (100)(25) + (90)(vB)

vB = 25.56 m s-1

2 : A trolley of mass 4 kg moves at 3 m s-1 and collide with a trolley of mass 2 kg which is

moving in the opposite direction at 1 m s-1. After the collision, both trolleys move together

with the same velocity. What is their common velocity ?

Solution : m1u1 + m2u2 = (m1 + m2)v

(m1 + m2), u = 0 v1

m2

v2

Please note that separation of objects after a collision does NOT mean an elastic collision, it can be inelastic as well. Only when kinetic energy is conserved, then the collision is elastic.

4)(3) + (2) (-1) = (4 + 2 ) v

v = 5/3 m s-1

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, 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