chapter 2 force and motion

Chapter 2 [Force and Motion]

2 Physics Department SMK Sultan Ismail, Johor Bahru.

2.1 Analysing Linear Motion 1. Linear motion is motion in a straight line. Distance and Displacement

1. Distance is the total length of the path traveled by an object. The diagram above

shows the distance traveled from town X to town Y is 2000 m. 2. Distance is a scalar quantity. It has magnitude but no direction. 3. Displacement is the distance of its final position from its initial position in specified

direction. The distance taken from town X to town Y is 1200 m to the east. When you stated displacement make sure you write the direction.

4. Displacement is a vector quantity. It involves both magnitude and direction. 5. Both distance and displacement have the same SI unit which is meter, m. 6. Now you know the difference between distance and displacement. Example 1 To test your understanding of this distinction, consider the motion depicted (dipaparkan) in the diagram below. A physics teacher walks 4 meters East, 2 meters South, 4 meters West, and finally 2 meters North. (i) what is the distance (ii) what is the displacement

Solution (i) distance = 12m (ii) displacement = 0 m

Even though the physics teacher has walked a total distance of 12 meters, her displacement is 0 meters. During the course of her motion, she has "covered 12 meters of ground" (with distance = 12 m). Yet, when she is finished walking, she is not "out of place" – i.e., there is no displacement for her motion (displacement = 0 m). Displacement, being a vector quantity, must give attention to direction. Displacement is a vector quantity. It involves both magnitude and direction. For example, if the displacement in forward direction is written as +4 m, then the displacement in opposite direction can be written as -4 m. Therefore 4 meters east is canceled by the 4 meters west; and the 2 meters south is canceled by the 2 meters north.



Example 2 Ahmad walks to Chong’s house which is situated 80 m to the east of Ahmad’s house. They then walk towards their school which is 60 m to the south of Chong’s house. What is the distance traveled by Ahmad and his displacement from his house. Solution It’s easy to solve this problem by using a diagram.

Distance traveled by Ahmad = total length of the path traveled = PQ + QR = 80 m + 60 m = 140 m We know that displacement is the distance of its final position from its initial position in specified direction so distance of Ahmad’s final position, R from his initial position, P

obtained by using Pythagoras’ theorem PR = 22QRPQ +

= 22 6080 + = 100 m Displacement, being (adalah) a vector quantity, must give attention to direction so the direction of the displacement obtained by using, tan θ = 60/80 = 0.75 θ = tan-1 0.75 θ = 36.90 The displacement can be written as 100 m at 36.90 south-east or S53.10E

Ahmad’s

house, P

Chong’s

house, Q

School, R

80 m

60 m



Speed and Velocity Just as distance and displacement have distinctly different meanings (despite their similarities), so do speed and velocity. 1. Speed is the rate of change of distance. Speed during the course of a motion is often

computed using the following equation: So the equation can be written as,

Speed = takentime

distanceofchange

2. Velocity is the rate of change of displacement. Velocity is often computed using the

equation:

Velocity = takentime

ntdisplacemeofchange

3. Speed is a scalar quantity whereas velocity is vector quantity. 4. Both speed and velocity have the same SI unit. They are measured in meter per

second or m s-1. Other unit may be in cm s-1 or km h-1. 5. The average speed during the course of a motion is often computed using the

following equation:

Average speed = takentimeTotal

travelleddistanceTotal

6. Meanwhile, the average velocity is often computed using the equation:

Average velocity = takenime

ntdisplacemeTotal

T



Example 1 A man running in a race covers 60 m in 12 s. (a) What is his speed in,

(i) m s-1 (ii) km h-1

(b) If he takes 40 s to complete the race, what is his distance covered? (c) Another man runs with a speed of 7.5 m s-1, how long did he take to complete the race? Solution

(a) (i) speed = takentime

distanceofchange

= 12

60

= 5 m s-1

(ii) speed = 5 x [

60x60

11000

1

]

= 18 km h-1 (b) Distance = speed x time = 5 x 40 = 200 m

(c) Time = speed

distance

= 7.5

200

= 26.7 s



Example 2 The physics teacher walks 4 meters East, 2 meters South, 4 meters West, and finally 2 meters North. The entire motion lasted for 24 seconds. Determine the average speed and the average velocity.

Solution

(i) Average speed= 24

12

= 0.5 m s-1

(ii) Average velocity = takentime

ntdisplacemetotal

= 24

0

= 0 m s-1 The physics teacher walked a distance of 12 meters in 24 seconds; thus, her average speed was 0.50 m/s. However, since her displacement is 0 meters, her average velocity is 0 m/s. Remember that displacement refers to the change in position and that velocity is based upon this position change. In this case of the teacher's motion, there is a position change of 0 meters and thus an average velocity of 0 m/s.



Acceleration and Deceleration 1. Acceleration is defined as the rate of change of velocity. 2. It can be written as;

Acceleration, a = takentime

velocityinchange

= takentime

velocity) initial - velocity(final

a = t

uv -

3. Acceleration is a vector quantity. The SI unit for acceleration is meter per square

second or m s-2. 4. The acceleration is positive if the velocity increases with time. The acceleration is

negative if the velocity decreases with time. It is also called deceleration. 5. Figure 2.1, shows that the car experiences acceleration in region 1, in region 2 the car

has a constant velocity and then decelerate in region 3.

Figure 2.1

Example 1 A vehicle accelerates uniformly from rest to a speed of 25 m s-1 in 100 s along a straight road. It then decelerates uniformly at 0.2 m s-2 for 60 s. Find (a) the initial acceleration (b) the final speed Solution

(a) Initial acceleration, a = t

uv

= 100

025

= 0.25 m s-2



(b) final speed, v = at + u = (-0.2x60) + 25 = 13 m s-1 Analysing of motion 1. In the laboratory, a ticker-timer as shown in figure 2.2, with a trolley and ticker tape

are used to study the motion of an object for a short time interval. (Refer practical book; Hands-on Activity 2.3 & 2.4)

2. The time taken to make 50 dots on the ticker tape is 1 s. Hence, the time interval

between 2 consecutive dots is 50

1 = 0.02 s

3. To determine the time interval of motion of the object:

Time interval = Number of tick x 0.02 s



4. The following shows the different types of motion recorded on the ticker tape and tape chart.

Ticker tape and chart Characteristics Inference

- The separation between dots stays the same.

- The length of the strips of the tape chart is equal.

The object is moving at a constant velocity.

- The distance between the dots increases uniformly.

- The length of the strips of the tape increase uniformly.

The velocity of the object is increasing uniformly, i.e the object is moving at a constant acceleration.

- The distance between the dots and the length of strips of the tape decreases uniformly.

The velocity of the object is decreasing uniformly, i.e the object is moving at a decelerating uniformly.

5. Change in distance between dots indicates a changing velocity and thus, acceleration.

A constant distance between dots represents a constant velocity and no acceleration.



Example 1 Figure 2.3 shows a ticker tape chart obtained in an experiment to study the motion of trolley on an inclined plane. Calculate the acceleration of the trolley.

Solution Time for each strip = 5 x 0.02 s = 0.1 s

Time interval, t between u and v, = (5-1) strips x 0.1 s = 4 x 0.1 s = 0.4 s

Initial velocity, u

= s0.1

cm2.0

= 20 cm s-1 Final velocity, v

= cm0.1

cm10.0

= 100 cm s-1

Acceleration, a = t

uv −

= 0.4

20)(100 −

= 0.4

80

= 200 cm s-2 / 2 m s-2



Example 2 Figure 2.4 shows a strip of ticker tape pulled through a ticker-timer by a freely falling metal sphere. The ticker-timer vibrates at a frequency of 50 Hz. Determine the acceleration of the sphere.

Solution

Initial velocity, u = 0.02

0.8

= 40 cm s-1

Time interval, t between u and v, = (6-1) x 0.02 = 5 x 0.02 s = 0.1 s

Final velocity, v = 0.02

2.4

= 120 cm s-1

Acceleration, a = t

uv −

= 0.1

40)(120

= 800 cm s-2 / 8 m s-2

Equations of Liner Motion with Uniform Acceleration For an object in linear motion with uniform acceleration, problems involving the displacement, velocity, acceleration and time of motion can be solve by using the equation of motion. 1. v = u + at

2. s = 2

1(u + v)t

3. s = ut + 2

1at2

4. v2 = u2 + 2as

s = displacement u = initial velocity v = final velocity a = constant acceleration t = time interval



Example 1 Ali is driving a car at velocity of 30 m s-1. On seeing a student crossing the road, Ali steps on his brakes to stop the car. The speed of the car decreases uniformly and stops after traveling 150 m. (a) What is the deceleration of the car when the brakes are applied? (b) What is the time interval before the car stops? Solution u = 30 m s-1, v = 0 m s-1, s = 150 m, a = ?, t = ? (a) v2 = u2 + 2as

0 = 302 + (2 x a x 150)

a = - 300

900

= - 3 m s-2

(b) v = u + at 0 = 30 +(-3)t t = 10 s



2.2 Analysing Motion Graphs Motion Graphs 1. There are two main types of linear motion graphs:

(a) displacement-time (b) velocity-time

Displacement-Time Graph 1. We can analyze the velocity of an object by plotting a graph of displacement

against time. 2. A student walks at a constant velocity from position A to reach position B in 200 s.

He rests for 100 s at position B and then walks back to position A using the same straight path. He reaches position A after 200 s.

Figure 2.1

3. The graph below shows the change in the distance and direction with time of the

student.

Figure 2.2



(a) On a displacement-time graph, the gradient of the graph is equal to the velocity of the object.

(b) Gradient of the graph in section I,

= x

y

∆

∆

= s200

m100

= 0.5 m s-1

(c) On a displacement-time graph, a horizontal line (gradient = 0) shows the object is not moving.

(d) In section II of the graph, the student remains at position B from 200 s to 300 s.

(e) Gradient of the graph in section III,

= - x

y

∆

∆

= - s200

m100

= - 0.5 m s-1

(f) The negative sign shows that the direction of motion is opposite to the original direction. (velocity is a vector quantity)

(g) At t = 500 s, the graphs intersects the t-axis. The displacement at this moment is zero that is the student has returned to the original position.



4. The various displacement-time graphs are shown as follows;

(a) Object at rest s

t 0

- There is no change in the displacement over time.

- The straight line graph is parallel to the time axis.

- velocity = gradient of graph = 0 m s-1

(b) Object moving with constant speed s

t 0

- The rate of change of displacement is constant.

- The straight line graph has a constant gradient.

(c) Object moving with acceleration s

t 0

- The rate of change of displacement is increasing.

- The gradient of the curve is increasing showing that the velocity is increasing.

- The object is moving with constant acceleration.

(d) Object moving with deceleration s

t 0

- The gradient is decreasing with time.

- The velocity is decreasing with time.

- The object is moving with constant deceleration.



Velocity-Time Graph 1. A car starts from rest and accelerates for 20 s until it reaches a velocity of 60 m s-1.

The driver maintains this velocity for 20 seconds. The velocity of the car is then gradually reduced until it stops at t = 60 s.

Figure 2.3

2. The graph below shows how the velocity of the car changes over a certain period of

time.

Figure 2.4

(a) On a velocity-time graph, the gradient of the graph is equal to the

acceleration of the object. (b) In section I, the acceleration of the car;

a = x

y

∆

∆

= s20

sm0)(60 1−

= 3 m s-2

(c) On a velocity-time graph, a horizontal line represents a constant velocity. The gradient of the graph is equal to 0.

(d) In section II, the car travels at a constant velocity of 60 m s-1 from t = 20 s to t = 40 s.



(e) In section III, gradient of the graph;

= - x

y

∆

∆

= s20

sm60)(0 1−

−

= - 3 m s-2

(f) The negative sign indicates a deceleration. (g) On the velocity-time graph, the area under the graph is equal to the distance

traveled.

(h) In section I, the area under the graph (the shaded triangle), = ½ x base x height = ½ x 20 x 60 = 600 m.

(i) In section II, area under the graph(the shaded rectangle), = 60 x 20 = 1200 m

(j) In section III, area under the graph, = ½ x 20 x 60 = 600 m



3. The various velocity-time graphs are shown below. (a) Object at rest

v

t 0

- The gradient of the graph represents acceleration.

- The gradient of the graph is zero and hence the acceleration of the object is always zero.

- The area under the graph represents displacement traveled.

- The area under the graph is zero and hence the displacement traveled is zero.

(b) Object moving with constant speed

v

t 0

- The velocity stays the same. The object is moving with constant velocity.

- Since the gradient is equal to zero, the acceleration is zero.

(c) Object moving with constant acceleration

v

t 0

- The velocity is increasing uniformly with time.

- The gradient of the graph is constant and hence the object is moving with constant acceleration.

(d) Object moving with increasing acceleration

v

t 0

- The gradient of the graph is increasing.

- The acceleration of the object is increasing.

(e) Object moving with decreasing acceleration v

t 0

u

v

t 0

- The gradient is decreasing with time. Therefore the velocity is decreasing.

- The object is moving with constant deceleration.



Example 1 The velocity-time graph of a object starting from rest and traveling towards the east is as shown in figure 2.5. (a) how long does the object travel

towards the east? (b) how long does the object travel

towards the west? (c) find the average speed and the

average velocity.

Solution (a) 10 s [from t = 0 s to t = 10 s,

velocity being positive] (b) 4 s [from t = 16 s to t = 20 s,

velocity being negative] (c) Distance traveled while moving

towards the west, S1 = Area of triangle ABC = ½ x 10 x 20 = 100 m

Distance traveled while moving towards the west, S2 = Area of triangle PQR = ½ x 4 x 10 = 20 m

Average speed = takenTime

distanceTotal

= 20

20100 +

= 6 m s-1 Average velocity

= takenTime

ntdisplacemeFinal

= 20

20100 −

= 4 m s-1 to the east

Example 2 Figure 2.6 shows the motion of a motorcycle. Describe the motion of the motorcycle. Solution In section OA, the motorcycle is moving at a

constant velocity, v = 2

4 = 2 m s-1

The motorcycle then travels in the opposite direction at a velocity of – 4 m s-1 and reaches its original position at time, t = 3 s, and continues to travel in this direction at the same speed of -4 m s-1.

Figure 2.6



2.3 Understanding Inertia

1. The diagrams below show the situations involving inertia. Situation 1

Figure 1

When a driver inside a car applies brake suddenly (figure 1 a), the driver and the passengers are move forwards. When the car suddenly accelerates (figure 1 b), the driver and passengers will move backwards. Situation 2

Figure 2

Passengers in a bus will fall backwards when a bus stats suddenly from rest.

Situation 3

Figure 3

Passengers in a moving bus fall forward when the bus stops suddenly.



Situation 4

Figure 4

When the table cloth is suddenly pulled horizontally, the dishes on the table top still remain on the table. 2. The situations above show that our body has an inbuilt resistance to any change in its

state of rest or motion. This reluctance to change is called inertia. 3. If an object is at rest, it tends to stay in that position unless some force puts that

object into motion. If an object is moving, inertia makes the moving object continue to move at a constant speed in the same direction unless some external force changes the object’s motion.

4. 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.

5. The concept of inertia was explained by Sir Isaac Newton in the fist law of motion. 6. Newton’s first law of motion states that every object will continue in its state if

uniform velocity or at rest unless it is acted by an external force.



Relationship between Mass and inertia

Figure 5

1. A child and an adult each sit on similar swings as shown in figure 5. If they are given

a push, which one of them will be more difficult to be moved? When both swings are set in motion, which one of them will be more difficult to stop?

2. The adult who has more mass will show more reluctance to change her state of rest or motion. This property of the mass of a body which resists change from its state of rest or motion is known as inertia.

3. The larger the mass, the larger its inertia.

Mini Quiz Text book/page 31 Carry out hands-on activity 2.6 on page 20.

Effect of Inertia 1. The effect of inertia is often seen in our daily activities. We make use of the positive

effects of inertia to solve some of our daily problems. The negative effects of inertia that can endanger our lives need to find ways to reduce them.

2. Positive effect of inertia.

The head of a hammer can be tightened onto the wooden handle by applying a knock on the handle. The head of the

In order to pour out the chili source, the bottle is moved down fast with a sudden stop. The sauce inside the bottle moves



hammer has big mass and will remain in its state of motion, thus fitting it tighter on the handle.

together with the bottle. When the bottle stops suddenly, the sauces continue in its state of motion due to the effect of its inertia.

The inertia of the ice skater keeps her gliding on the surface in a straight line.

Droplets of water on a wet umbrella are spun off when the umbrella is rotated and stopped suddenly. The droplets of water initially move with the rotating umbrella. The inertia of the droplets of water causes them to continue moving even when the umbrella has stopped spinning.

3. Negative effect of inertia

Furniture carried by a lorry normally is tied up together by string. When the lorry starts to move suddenly, the furniture is more difficult to fall off due to their inertia because their combined mass has increased.

Strong iron structure between the driver’s cabin and the load to stop the inertial movement of the heavy load towards the driver when the lorry is brought to a halt suddenly.



If the car crashes while travelling, the inertia of the passengers causes them to continue in motion. This is a dangerous situation. Upon impact the passengers will crash into parts of the car immediately in front of them and suffer injuries.

4. Ways to reduce the negative effects of inertia.

(a) Wearing safety belts when driving. (b) An air begs is fitted inside the steering wheel. It provides a cushion to prevent

the driver from hitting the steering wheel or dashboard during a collision. (c) The oil tank of an oil tanker lorry is usually divided into a few smaller

compartments so that the effects of inertia can be reduced.

2.4 Analysing Momentum

You can catch a fast moving ping-pong ball easily with your bare hand.

A softball keeper must wear a glove to catch a hard and fast moving softball. Why is a slow moving softball much easier to catch?

1. When an object is moving with a speed in a straight line, we say that it has linear

momentum. 2. The amount of linear momentum of the object depends on its mass and velocity.



3. We define linear momentum as the product of mass and velocity. Momentum = Mass x velocity

p = m x v

4. SI unit for momentum is given as kg m s-1. It can also be written as N s (newton second) 5. Momentum is a vector quantity. The direction of the momentum is the same as the

direction of the velocity. Example:

A billiard ball A of mass 0.5 kg is moving from left to right with a velocity of 2 m s-1 while another billiard ball B of equal mass is moving from right to left with the same speed. Solution Momentum of ball A = mAvA = 0.5 x 2 = 1 kg m s-1

Momentum of ball B = mBvB = 0.5 x (-2) = -1 kg m s-1

* Negative sign shows the object is moving in the opposite direction.

Principle of Conservation of Momentum

1. The of conservation of momentum states that the total momentum in a closed system is constant, if no external force acts on the system, that is, the momentum of all objects before a collision equals the momentum of all objects after a collision.

2. The principle of conservation of momentum is true for the following:

(a) Collision of two objects

- elastic collision

- inelastic collision (b) Explosion



3. Elastic collision

- When objects collide and bounce perfectly, the collision is said to be an elastic collision (figure 2.1)

Figure 2.1

- Figure 2.2 shows the diagram of experiments to investigate the principle of conservation of momentum in elastic collisions.

Figure 2.2

(i) The runway is adjusted to compensate the friction. (ii) Trolley A with a spring-loaded piston is placed at the higher end of the

runway and trolley B is placed halfway down the runway and stayed at rest. (iii) Two ticker tapes are passed through the ticker timer, one attached to trolley

A and another attached to trolley B. (iv) The ticker-timer is switched on and trolley A is given a slight push so that it

moves down the runway at a uniform velocity and collides with trolley B which is stationery.

(v) After collision, the two trolleys move separately.

- igure 2.3 shows the ticker tapes obtained from the experiment.

Figure 2.3



4. Inelastic collision - When objects collide and attached to one another or couple together after

collision, the collision is called an inelastic collision (figure 2.4).

Figure 2.4

- Figure 2.5 shows the diagram of experiments to investigate the principle of conservation of momentum in elastic collisions.

Figure 2.5

(i) The runway is adjusted to compensate the friction. (ii) The spring loaded piston of trolley A is removed and some plasticine is

pasted onto trolley A and B. (iii) A ticker tape is attached to trolley A only. (iv) The ticker-timer is switched on and trolley A is given a slight push so that it

moves down the runway at a uniform velocity and collides with trolley B which is stationery.

(v) After collision, the two trolleys are move together.

- Figure 2.6 shows the ticker tapes obtained from the experiment.

Figure 2.6



5. Explosion

- Means the separation of objects which are initially at rest.

Figure 2.7

6. Application of Conservation of Momentum

1. Launching of rockets.

2. Jet engine

.



Example 1 Block A of mass 5 kg is moving with velocity 2 m s-1 and collides with another stationery block B of unknown mass. After the collision, block A moves with velocity 0.5 m s-1. Given that the collision is elastic. Find the momentum of block B after the collision. Solution mAuA + mBuB = mAvA + mBvB (5)(2) + 0 = (5)(0.5) + mBvB mBvB = 7.5 kg m s-1 Example 2 A truck travels at a velocity of 15 m s-1 collides head-on with a car that travels at 30 m s-1. The mass of the truck and the car are 6000 kg and 1500 kg respectively. What is the final velocity of the two vehicles after the collision if they stick together? Solution m1u1 + m2u2 = (m1+ m2)v (6000 x 15) + (1500 x -30) = (6000 + 1500)v v = 6 m s-1 Example 3

An instructor fires a pistol which has a mass of 1.5 kg. If the bullet weighs 10 g and it reaches a velocity of 300 m s-1 after shooting, what is the recoil velocity of the pistol? Solution m1u1 + m2u2 = m1v1 + m2v2 0 = 1.5(-v1) + 0.01 x 300 1.5v = 3 V = 2 m s-1



2.5 Understanding The Effect Of A Force

Figure 2.1 Effect of forces

1. We use force in our daily activities 2. Forces have magnitude and direction, that is, force is a vector quantity. 3. The SI unit of force is Newton, N. 4. Force can make things;

(a) move a stationery object. (b) speed up (accelerate) or slow down (decelerate) a moving object. (c) change the direction of a moving object. (d) stop a moving object. (e) change the shape of an object.

Balanced Force 1. An object may have several forces acting on it. But if the forces are in balance, they

cancel each other out. Then, the object behaves as if no force is applied to it. 2. When the forces are balanced, an object is either at rest, or moving at a constant

velocity. 3. Example of balanced forces;

Books on the table remain at rest because normal reaction force, R is cancel with weight of the books. The net force acting on the books is zero.



An airplane can flaying horizontally with constant speed because forward thrust, T provides by the engine is cancel by a drag, F provides by the wind and air resistance. An airplane can flying at a constant height because a lift force is balanced with its downward weight, W. When these four forces are balanced, the net force acting on the plane is zero.

Unbalanced Forces on an Object 1. When two or more forces acting on an object are not balanced, the object will

accelerate. This net force is known as the unbalanced force or the resultant force. 2. Example;

(a) When two opposite forces acting on a box, the box will move.

Relationship between Force, Mass and Acceleration (F = ma)

Figure 2.2 Figure 2.3 Figure 2.4

Figure 2.2 (a) shows a force, F acting on a mass, m. The mass moves with a constant acceleration, a.

If the force acting on the same mass is doubled, its acceleration is also doubled.

If the same force, F is acting on a mass of 2m, the acceleration, a, is reduced by half.

1. The diagrams above show that the acceleration of an object is directly proportional

to the unbalanced force.

Acceleration, a α Net force, F

5 N 10 N



2. The acceleration of an object is inversely proportional to the mass.

Acceleration, a α mMass,

1

3. Combining the relationship above, we get;

F = ma 4. The relationship between F, m and a, is known as Newton’s second law of motion. 5. Newton’s second law of motion states that the acceleration produced by a net force

on an object is directly proportional to the magnitude of the net force applied and is inversely proportional to the mass of the object.

6. In the formula, F = ma,

- F is the net force and a, is the acceleration in the same direction as the force.

- Net force, F is the force applied minus all other opposing forces, such as friction. 7. Examples;

(i) Azhari applies a force of 50 N to move a 12 kg carton at a constant velocity. What is the frictional force acting on the carton?

Figure 2.5

Solution Frictional force is always acting in an opposite direction or opposing to the motion. Constant velocity → acceleration, a = 0 m s-2 Net force in the direction of velocity, F = F1 – Frictional force = 50 – f From F = ma 50 – f = 12 x 0 F = 50 N



Figure 2.6

(ii) Figure 2.5 shows a trolley of mass 2 kg placed on a rough horizontal table and

being pulled by a force of 2.0 N. The trolley moves at constant velocity. (a) What is the frictional force between the trolley and the table? (b) The force, F is then increased to 12 N. what is the acceleration of the

trolley. Solution (a) Constant velocity → acceleration, a = 0 m s-2

Net force, F = ma 0.8 – f = 2 x 0 f = 0.8 N

(b) Net force, F = ma

1.2 – 0.8 = 2 x a

a = 2

4.0

a = 0.2 m s-2

Work example in text book. page 41

2 kg

f

F = 0.8 N



2.6 Analysing Impulse And Impulsive Force

1. An impulsive force is a force that acts over a short period of time during collisions or explosions.

2. Consider an impulsive force, F and the following equations; F = ma ………………………….. (1) and

a = t

uv − ………………… (2)

Substitute (2) into (1)

F = m

−

t

uv

F = t

mumv − ………………. (3)

Hence, Impulsive force also defined as the rate of change of momentum. The SI unit for impulsive force is the Kgms-2 ot N.

3. Impulse is the product between the impulsive force, F with the time of impact, t, that is,

Rearrange (3)……. F = t

mumv −

Ft = mv – mu Hence, Impulse, Ft also defined as the change of momentum. The SI unit for impulse is Kgms-1 or Ns. Example 1 Figure 1 shows a baseball approaching a bat with an initial velocity of -30 m s-1. A force is applied by the bat to hit the ball and sends it in the opposite direction with a velocity of 50 m s-1.

Figure 1

If the mass of the baseball is 150 g and the time of contact between it and the bat is 1.6 x 10-3 s, calculate, (i) the impulse applied to the ball. (ii) the impulsive force exerted on the ball by the bat Solution



Example 2

A 45 g golf ball is hit with a force of 5300 N. The time interval of interaction between the golf ball an d the club is 0.6 ms. Calculate the velocity of the golf ball immediately after it was hit. Solution Effect of Impulsive Forces in Daily Life

1. From the formula F = t

mumv −, it can be seen the impulsive force , F is large if the

time interval, t is small,

F α t

1

2. This explains why in a car accident, the car experience a large impulsive force, which causes serious damage to the car and also injuries to the passengers in the car.

Reducing Impulsive Forces 1. From the formula of impulsive force, we know that impulsive force is inversely

proportional to the time interval. 2. An effective way to reduce the impulsive force in a collision is to extend the time

interval of the interaction. 3. A large impulsive force resulting from a car accident can cause serious injuries to its

passengers. To prevent this, cars are constructed with safety features to reduce the impact.

4. Motorcyclists should wear safety helmets and elbow and knee cushions/caps to prevent themselves from falling down and knocking onto the hard surface.

5. In sports the effects of impulsive force are reduced to prevent injuries to the participants in the games.



(a) Thick mattresses with a soft surface used in events such as the high jump, so that

the time interval of impact on landing is increased.

(b) Athletes who jump down from a height must bend their knees upon landing.

(c) When a goalkeeper catches a fast moving ball, he ends his hands to increase

the time of contact when stooping the ball. With longer time of impact, a smaller impulsive force is acting on his hand.

6. Polystyrene, cardboard and rubber foam are usually used in packing fragile goods.

The soft surface of these materials reduces the impulsive force during accidental dropping by lengthening the time of impact.



Benefits Effects of Impulsive Forces

When a hammer is used to hit a nail into wood, a large change in momentum occurs in a short time interval. It produces a large impulsive force which drives the nail into the wood.

When foodstuff such as chillies are pounded using a mortar and pestle (which are both made from stone), the action produce large impulsive force which crush the food. This occurs when the pestle is brought down at a high velocity onto the mortar with short contact times.

2.7 SAFETY FEATURES IN VEHICLES

There are several safety features installed to a vehicle in order to reduce serious injury to the driver and passengers. (a) Dashboard – covered with soft materials to reduce injury due to passengers knocking

on it.



(b) Air bags – will expand during collision so as to prevent the driver from smashing directly into the car steering. The collision time will be lengthen and the impulsive force can be reduce and prevent

(c) Safety belts – to lengthen the collision time and prevent the passengers from being thrown due to their inertial when the car is forced to slow down suddenly.

(d) Rubber bumper - absorbs impact in minor accidents, thus prevents damage to the car also lengthen the collision time and to reduce the impulsive force.

(e) Front crumple zone – design to crumple upon impact. It increases the impact time and reduce the impulsive force on the car.

(f) Anti-lock braking system (ABS) – will not immediately stop the car once the brakes are applied. The car be momentarily brought to rest so that the force of impulsive is smaller.



2.8 Gravity

1. The concept of gravitational force was introduced by Sir Isaac Newton who, on seeing an apple falls on his head.

2. According to Newton, objects fall because they are pulled towards the Earth by the force of gravity.

The pull of gravitational force; (a) keeps things on the earth. (b) brings things down to earth when they are thrown upwards (c) holds the moon in its orbit round the Earth (d) captures returning space capsules and pulls them into orbit.

3. The pull of gravity causes objects to fall with acceleration. This means that objects

that fall are moving with increasing velocity. 4. The magnitude of the acceleration due to gravity depends on the strength of the

gravitational field.



Free Fall 1. An object is falling freely when it is falling under the force of gravity only.

Figure 1

2. Figure 1 show a free fall in a vacuum cylinder (where air resistance does not exist) of

a coin and a peace of paper. The both objects reach the bottom of the cylinder at the same time.

3. A piece of tissue paper (fall in atmosphere) does not fall freely because its fall is affected by air resistance.

4. A heavier golf ball can be considered to be falling freely because the air resistance is small compared to the pull of gravity and therefore is negligible.

Tips

- when an object falls; g = 9.8 m s-2

- when an object is thrown upwards; g = -9.8 m s-2

- At the highest point, v = 0 m s-1

5. Objects dropped under the influence of the pull of gravity accelerate at a constant

rate. 6. This acceleration is known as the gravitational acceleration, g. 7. The value of g is 9.8 m s-2. In calculation, the value of g is often taken to be 10 m s-2

for simplicity. 8. The acceleration due to gravity does not depend on the mass and shape of the falling

object.



9. All objects falling freely with the same acceleration. (Hands-on Activity 2.9: To determine the value of the gravitational acceleration, g)

- Apparatus/material: Ticker timer, ticker tape, 12 V a.c electrical power supply, retort stand, weights (50 g- 250 g), G-clamp, cellophane tape.

- Procedure:

switch on the ticker timer and release the slotted weight from a height of about 1 m. calculate the acceleration from the dots on the ticker tape. repeat the steps for slotted weights of 100g, 150g, 200g and 250 g.

- Results:

Mass of slotted weight, m/g

Acceleration due to gravity, g/ ms-2

50 100 150 200 250

- gravitational acceleration, g is calculated by substituting u, v and t into the formula

g = t

uv −



Weight, W and Gravitational Field, g 1. Gravitational field is a region around the earth

in which an object experiences a force towards the centre of the earth. This force is the gravitational attraction between the object and the earth.

2. The gravitational field strength, g can be calculated as follows:

g = m

F

Where F = Gravitational force m = mass of body

3. The gravitational field strength at the surface of the earth is 9.8 N kg-1/kg ms-2. 4. The weight of an object is defined as the gravitational force acting on the object and

can be calculated as follows;

W = mg Where, m = mass g = gravitational acceleration

5. The unit of weight is Newton, N.

Example 1 A iron ball is dropped from the top of a building and takes 2 s to reach the surface of the earth. What is the height of the building? (g = 9.8 m s-2)

Solution Using the formula, s = ut + ½ gt2 = 0 + ½ x 9.8 x 22 = 19.6 m



Example 2 A rock has a mass of 20.0 kg and weight of 90.0 N on the surface of a planet. (a) What is the gravitational field strength on the surface of the planet? (b) What are the mass and the weight of the rock on the surface of the Earth where its gravitational field strength is 9.8 N kg-1?

Solution

Example 2 Have you ever seen an astronaut walking on the Moon? It is known that the acceleration due to gravity near the surface of Moon is just 1/6 of that on the surface of earth. (a) Find the weight of a 50 kg man on

the surface of Moon. (b) If the 50 kg man can jump to a height

of 50 cm on the Earth, find the maximum height reached by him on the surface of moon. (Assume that his initial speed is the same on the earth and on the moon)

Solution



2.9 Forces in Equilibrium 1. When forces act upon an object and it remains stationery or moves at a constant

velocity, the object is said to be in a state of equilibrium. 2. When equilibrium is reached, the resultant force acting on the object is zero i.e there

is no net force acting upon it. 3. Newton’s Third Law of Motion states that if a force acts upon an object, then there

will be an equal and opposite reaction acting upon the same object.

Figure 1

Based on the Figure 1 above, the aeroplane will move at a constant velocity if , Lift Force = Gravity Force Thrust Force = Drag Force

Figure 2

4. Figure 2 shows a stationery block of wood resting on a table. The forces acting on

the block of wood are: (a) The weight, W, which is acting downwards. (b) The normal reaction, R, which is acting upwards.

5. The weight, W, is balanced by the normal reaction, R. 6. Hence, the block of wood is in a state of equilibrium.



Figure 3

7. Figure 3 shows a weight hanger attached to a string and which is in a stationery

state. The forces acting on the weight hanger are: (a) The weight, W, which acts downwards (b) The tension, T, which acts upwards

8. The weight, W, is balanced by the tension T of the string. 9. Hence, the weight hanger is in a state of equilibrium.

Figure 4

10. The cat resting on an inclined plane as shown in figure 4 is also in equilibrium. The

three forces acting on the cat cancel out each other so that the resultant force is zero.



Addition of Forces 1. A resultant force is a single force that represents the combined effect of two or more

forces in magnitude and direction. The directions of the forces have to be taken into considerations when forces are added. (a) Two forces that act along the same direction.

Figure 4

The resultant force is found by adding the magnitude of the forces. Two forces acting in the same direction with magnitudes 8 N and 3 N respectively will have a resultant force of 11 N, as shown in figure 4. Resultant force, F = F1 + F2

(b) Two forces that act in opposite directions

Figure 5

The two forces with magnitude 8 N and 3 N acting respectively in opposite directions will have a resultant of 5 N, as shown in figure 5. Resultant force, F = F1 – F2

(c) Two forces acting at a point at an angle to each other, the resultant force can be determined by using the parallelogram of forces. Figure 6 shows two forces F1 and F2 acting at an angle, θ with each other. The following steps show how to determine the resultant force, F.

Figure 6

(i) Draw the forces F1 and F2 from a point with an angle of θ with each other.



(ii) Draw another two lines to complete the parallelogram. (iii) Draw the diagonal of the parallelogram. The diagonal represents the

resultant force, F and its direction, α, can be determined by measuring the angle between the diagonal with either one side of the parallelogram.

(d) If the two forces are perpendicular to each other, the resultant force can be

determined by using Pythagoras theorem and trigonometry to solve the problems. Example;

A carton is acted on by two forces of 30 N and 40 N which are at right angles to each other as shown in figure 8. Determine the resultant force acting on the carton. Solution By using the triangle method;

The resultant force, F is the magnitude of the hypotenuse of the triangle

F = 22 4030 +

= 50 N

40 N

30 N F

40 N

30 N



tan θ = 30

40

θ = 53.10 The resultant force is 50 N with bearing 53.10

Resolution Force

Figure 7

1. A single force can be resolved into two

components, that is the, (a) horizontal component, Fx (b) vertical component, Fy

2. Figure 7 shows the resolution of forces, F. 3. The magnitude of the horizontal component can be written as l Fx l = l F l cos θ. 4. The magnitude of the vertical component can be written as l Fy l = l F l sinθ. 5. The direction of the resultant force can be written as;

tan θ = x

y

F

F

θ = tan-1 (x

y

F

F)



Example 1 A tourist is pulling a bag with a force 12.0 N at an angle of 300 to the horizontal floor. What is the horizontal and vertical component of the force? Solution Fx and Fy are the vertical and horizontal components of the forces. Magnitude of Fx = 12 cos 300 = 10.4 N Magnitude of Fy = 12 sin 300 = 6.0 N

Example 2

The diagram above shows a block of wood (on a smooth surface) being pulled by a force of 12 N. What is the horizontal component of the force? Solution The horizontal component of the force = 12 x cos 600 = 6 N [Only a force of 6 N needs to act on the block of wood in order for it to move horizontal to the right]

Example 3 The figure shows a box is being pulled by a man.

What is the magnitude of the resultant force that causes the box to move horizontally? (Study worked example in the text book page 56)



Example 4

A workman pushes a carton of mass 50 kg up an inclined plane into a lorry. The inclined plane makes an angle of 450 with the horizontal floor and the frictional force between the inclined plane and the carton is 135 N. If the workman pushes the cartoon with a force of 500 N (a) Can the carton move up the inclined plane? (b) What is the acceleration of the carton? Solution (a) Applied force on the carton towards the top

of the inclined plane, F = 500 N.

- Opposing force along the entire length of the inclined plane = component of weight down the plane + frictional force

= 500 sin 450 + 135 N = 488.6 N

- The carton is able to move upwards because the applied force, F = 500 N is larger than the opposing downward force.

(b) The resultant force acting on an inclined plane,

= 500 N – 488.6 N = 11.4 N

From formula; F = ma

a = m

F

= 50

4.11

= 0.23 m s-2



Three Forces in Equilibrium Figure shows a wooden block supported by two strings. The tensions of the strings are T1 and T2 respectively. Sine the wooden block is in equilibrium, the resultant forces is zero. Hence, by taking the horizontal components of forces, T1 sin α = T2 sin β By taking the vertical components W = T1 cos α + T2 cos β

Example Figure 9 shows an aeroplane model with a mass of 450 g is hung from a ceiling with two strings.

Figure 9

Calculate the tension on each string. Assume that the system is in equilibrium. [g = 9.8 ms-2]

Solution Weight of aeroplane, W = 0.5 x 9.8

= 4.9 N By resolve the forces vertically, T sin 300 + T sin 300 = 4.9 2 T sin 300 = 4.9 2 T (0.5) = 4.9 T = 4.9 N for each string.

α β

T1

T2

W

T1 cos α + T2 cos β

T1 sin α T2 sin β



2.10 Understanding Work, Energy And Efficiency

Are you doing work when you are solving a physics problem? Are you doing work when you pick up a pen that has fallen to the floor?

Photograph below shows that work is done.

A women is pushing the

stroller A fisherman is pushing the boat towards the beach.

The rocket engine produces an upward thrust.

Work

Figure 1

1. Work, W is defined as the product of the force F and the distance travelled, s, in

the direction of the force (figure 1). Work, W = Force (F) x Distance (s) (in the direction of the force)

W = F x s

The SI unit for work is Nm or Joule (J).

2. Work is a scalar quantity, that is, a quantity which has magnitude only but not direction.

3. If no distance is traversed or if there is no motion, then no work has been done.



Figure 2

4. When the force and the direction of motion of an object are perpendicular as shown

in figure 2, the work done is equal to zero.

Figure 3

5. In such a case, the work can be determined by resolving the force along the direction

of motion as shown in figure 3.

Work, W = F cos θ x s Example 1

A nurse is pushing a patient in a wheelchair with a force of 200 N over a distance of 1.5 m. How much work is done by nurse?

Solution

Example 2

Diagram shows a workman pulling up a load onto a lorry using a smooth inclined plane. If the tension in the rope is 600 N and the inclined plane is 3 m in length, how much work is being done by the man?

Solution



6. If a graph of force versus (against) distance is drawn, then the work done is equal to the area under the graph.

Energy 1. When work is done, energy is consumed. 2. Energy is the ability to do work. 3. Mechanical energy is divided into potential energy and kinetic energy. 4. Energy is the product of the force, F and the distance, d.

Energy = Work done = Force x Distance

E = F x s 5. The SI unit for energy is the Nm or Joule (J). 6. Energy is a scalar quantity Kinetic Energy 1. Kinetic energy is the energy possessed by a moving object. Only moving objects

possess kinetic energy. 2. Assume a force, F is acting on a stationery trolley of mass, m kg moving on a smooth

surface. The force acting over a distance of s causes the trolley to achieve a velocity of v m s-1.

3. The kinetic energy, Ek of an object of mass, m kg travelling at a velocity of v ms-1 is

given as;

Kinetic energy, Ek = ½ mv2

4. The factors that affect kinetic energy are: (a) mass of the object, m (b) velocity of the object, v



Example 1 A 0.6 kg trolley moves across the floor at a velocity of 0.5 m s-1. What is the kinetic energy of the trolley? Solution

Example 2 A car of mass 950 kg accelerates from a velocity of 20 m s-1 to a velocity of 35 m s-1. What is the work done for the car to accelerate? Solution

Potential Energy 1. The potential energy of an object is the energy stored in the object because of its

position or state. 2. There are two main types of potential

energy. (a) Gravitational potential energy (b) Elastic potential energy

3. The gravitational potential energy of any object is the energy stored in the object because of its height above the earth’s surface.

4. When an object of mass, m kg is raised to a height, h meters above the earth’s surface, the object possesses gravitational potential.

5. The gravitational potential energy is equal to the work done to rise an object to a particular height. Work done, W = Fs = mg x h = mgh Gravitional potential energy, Ep = W

Ep = mgh



6. Elastic Potential Energy of an object is the energy stored in the object as result of stretching or compressing it.

Example

The diagram above shows a mass of 6 kg being pulled by a force F up a smooth inclined plane. After the body has been pulled 5 m up the ramp/platform, it is found that the vertical height of the object is 2 m. Calculate the, (a) gravitational potential energy at 5 m. (b) work done to lift the body/mass up the smooth inclined ramp. (c) value of F

Solution

Principle of Conservation of Energy 1. The principle of conservation of

energy states that the total energy in a system is constant.

2. Energy cannot be destroyed but can be converted from one form to another.

3. When a coconut of mass, m kg falls from a height of h meters to the ground, it loses its gravitational potential energy which is changed into kinetic energy of motion.



Example

Lim rides his bicycle down the slope of a hill 3 m high at an initial velocity of 2 m s-1, without pedalling. At the foot of the hill, the velocity is 6 m s-1. Given that the mass of Lim with his bicycle is 75 kg. Find (a) the initial kinetic energy of the

bicycle. (b) the initial potential energy of the bicycle (c) the work done against friction along the slope. Solution

2.11 Power

1. Power, P is the rate at which work is done or energy is changed or transferred. 2. Power is expressed as:

Power, P = takentike

doneworkd

P = t

W

3. The SI unit of power is J s-1 or watt (W).

Example

A car is moving at a constant velocity of 30 m s-1 . If the car has to overcome a frictional force of 700 N, what is the power of its engine?

Solution



Efficiency 1. Work can be done by a machine when input energy is supplied. 2. If a machine does least work from the supplied energy, it is said to be non-efficient. 3. Efficiency of a machine is expressed as:

Efficiency = inputEnergy

outputenergyUseful x 100%

4. If a machine has the efficiency of 100% it is an excellent machine in hich output work

= input work. 5. In general, the efficiency for all machines is less than 100% because of the work done

against friction when operating a machine Example A 60 N load is lifted up to 0.30 m height with a machine which needs 25N of force and the load moves 0.90 m. Calculate (a) the work which is done by the force (b) the work which is done by the machine (c) the efficiency of the machine

Solution



2.12 Elasticity 1. Some common devices like eraser and ruler will change its shape when an external

force acting on it. When the external force is removed, the objects return to its original shape and dimensions.

A rubber band can be stretched

due to its elasticity. A sponge can be compressed easily due to its elasticity

2. The property of an object that enables it to return to its original shape and dimensions

after an applied external force is removed is called elasticity. 3. The elasticity of solids is due to the strong intermolecular forces between the molecules

of the solid. Stretching a solid causes the molecules to be slightly displaced away from one another. A strong attractive intermolecular force acts between the molecules to oppose the stretching as shown in figure 2.64.

4. When the external stretching force is removed, this strong attractive intermolecular force brings the molecules back to their original positions. Therefore, the solid return to its original shape and size.

5. Compressing a solid causes the molecules in the solid to be closer to one another. 6. When a solid is compressed, the strong force of repulsion will push the atoms or

molecules of the solid back to their original position as is shown in figure 2.65.



Relationship between force and extension of a spring (Carry out Experiment 2.4 on page 41 of the practical book) Title: To investigate the relationship between force and the extension of a spring. Hypothesis: The tension of a spring is directly proportional to the applied force. Variables: Manipulated variable: Force Responding variable: Extension of the spring Materials and apparatus: Steel spring, five 50 g slotted weight, half meter rule and retort stand with clamp.

Method: 1. All the weights are removed from the spring. The original length of the spring is

measured using a half meter rule. 2. A 50 g weight is hung from the end of the spring. The length of the spring is measured

again and the extension is calculated. 3. Step 2 is repeated with 100 g, 150 g, 200 g, 250 g, and 300 g weights. 4. The graph of force against extension is plotted. Result:

Mass of slotted weight, m/g

Force on the spring, (F=mg)/N

Length of spring, L/cm

Extension, x = (L-L0)/cm

50 100 150 200 250



300

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

applied force provided that the elastic limit is not exceeded.

Figure 1

Figure 2

2. The elastic limit of a spring is defined as the maximum force that can be applied to a

spring such that the spring will be able to be restored to its original length when the force is removed.

3. If the elastic limit is exceeded, the length of the spring is longer than the original length even though the force no longer acts on it / spring will not return to its original position.

4. The results from experiment show that the graph is straight line passing through origin (figure 1) meaning that the extension of a spring is directly proportional to the applied force if the elastic limit is not exceeded.

5. The mathematical expression for Hooke’s Law is,

F αααα x Therefore, F = kx ; k is constant of the spring/force constant

k = x

F with units N m-1

6. Spring constant is a measurement of the stiffness of the spring. A spring with a spring

constant of 12 N m-1 requires a force of 12 N to produce an extension of 1 cm. 7. A spring with a large spring constant is harder to extend and is said to be stiffer. 8. A spring with a smaller force constant is easier to extend and is said to be less stiff or

softer.

Force,F/N

Extension/cm



Figure 3

- spring constant k = gradient of graph

- a larger value of k indicates a stiffer spring

- a steeper graph indicates a stiffer spring

Example The length of a spring is increased from 23.0 cm to 28.0 cm when a mass of 4 kg was hung from the end of a spring. (a) What the load on the spring in newtons? (b) What is the extension of the spring? (c) Calculate the force constant of the spring (Assume g = 10 Nkg-1) Solution

System of Spring 1. Two springs can be connected in series or in parallel.

2. When two springs are connected in

series, the applied force acts on each spring. The force due to the 30 N load acts along the system such that each spring experience a force or tension of 30 N.

3. When two springs are connected in parallel, the applied force is shared equally among the springs. The force due to the 120 N load is shared by the springs. Therefore, the tension in each spring is 60 N.



Example

Figure above shows a spring extends by 1 cm when an 8 N force is applied on it. Similar springs used to set up three systems. Calculate the total extension in each system. Solution Example

Figure 3 shows identical springs. What is the value of Y?

Solution



Elastic Potential Energy 1. Elastic potential energy is the energy stored in a spring when it is extended or

compressed. 2. When a force extends a spring, work is done. The work done on the spring is the

energy transferred to the spring and stored as elastic potential energy. 3. Consider a force, F that produces an extension, x in a spring. The work done on the

spring, W = average force x extension

= x

2

F

= xF2

1

From Hooke’s law, F = kx

Therefore, W = 2

1(kx) x

= ½ kx2 Hence, the elastic potential energy stored in a stretched spring is given by,

Ep = ½ kx2

Area under the graph represents elastic potential energy

Example 4 A 2 kg load is hung form the end of a spring with a force constant of 160 Nm-1. (a) What is the tension in the spring? (b) What is the extension of the spring? (c) Calculate the elastic potential energy stored in the spring. [Assume g = 10 N kg-1] Solution



Factors which influence the elasticity of a spring 1. Factors which influence the elasticity of a spring are as follows;

(a) type of spring material

- a spring made from a hard material requires a larger force to stretch it. Hence, the spring constant, k is greater. For example; steel spring is harder than the copper spring.

(b) diameter of the coil of spring

- a spring made of a larger diameter coil is ‘softer’. (c) diameter of the wire of the spring

- a spring coil of spring made from thicker wire is more difficult to stretch than a coil of spring made from wire that is thinner.

(d) arrangement of the spring

- a longer spring is easier to stretch compared to a shorter spring.

- springs arranged in series are easier to stretch when compared to springs arranged in parallel.

chapter 2 force and motion

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