Physics Module Form 4
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Physical Quantity
Distance, s
Displacement, s
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Definition, Quantity, Symbol and unit Distance is the total path length travelled from one location to another. Quantity: scalar SI unit: meter (m) (a) The distance in a specified direction. (b) the distance between two locations measured along the shortest path connecting them in a specific direction. (c) The distance of its final position from its initial position in a specified direction. Quantity: vector SI unit: meter (m) Speed is the rate of change of distance
Speed,v Speed =
Dis tan ce time SI unit: m s -1
Quantity: scalar
Velocity is the rate of change of displacement. Velocity, v
Displacement time Velocity = Direction of velocity is the direction of displacement Quantity : Vector SI unit: m s -1
Average speed
Average velocity
v=
TotalDis tan t TotalTime
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Example: A car moves at an average speed / velocity of 20 ms -1 On average, the car moves a distance/ displacement of 20 m in 1 second for the whole journey.
Displacement TotalTime
Uniform speed
Speed that remains the same in magnitude without considering its direction
Uniform velocity
Velocity that remains the same in magnitude and direction
An object has a nonuniform velocity if
(a) The direction of motion changes or the motion is not linear.
Acceleration, a
When the velocity of an object increases, the object is said to be accelerating.
(b) The magnitude of its velocity changes.
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Acceleration is defined as the rate of change of velocity
Change in velocity Time taken Final velocity,v - Initial velocity,u = Time taken,t
Acceleration= Unit: ms-2
Acceleration is positive
The velocity of an object increases from an initial velocity, u, to a higher final velocity, v
Deceleration
acceleration is negative.
The rate of decrease in speed in a specified direction. The velocity of an object decreases from an initial velocity, u, to a lower final velocity, v.
Zero acceleration
An object moving at a constants velocity, that is, the magnitude and direction of its velocity remain unchanged – is not accelerating
Constant acceleration
Velocity increases at a uniform rate. When a car moves at a constant or uniform acceleration of 5 ms -2, its velocity increases by 5 ms -1 for every second that the car is in motion. 1. Constant = uniform 2. increasing velocity = acceleration 3. decreasing velocity = deceleration 4. zero velocity = object at stationary / at rest 5. negative velocity = object moves in opposite direction 6. zero acceleration = constant velocity 7. negative acceleration = deceleration
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Comparisons between distance and displacement Distance Total path length travelled from one location to another
Displacement The distance between two locations measured along the shortest path connecting them in specific direction
Scalar quantity
Vector quantity
It has magnitude but no direction
It has both magnitude and direction
SI unit meter
SI unit : meter
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Comparisons between speed and velocity
Speed The rate of change of distance Scalar quantity
Velocity The rate of change of displacement Vector quantity
It has magnitude but no direction
It has both magnitude and direction
SI unit : m s
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SI unit : m s
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Fill in the blanks: 1. A steady speed of 10 ms -1 = A distance of 10 m is travelled every second. 2. A steady velocity of -10 ms -1 = A displacement of 10 m is travelled every 1 second to the left. -1
3. A steady acceleration of 4 ms -2 = Speed goes up by 4 ms every 1 second. 4. A steady deceleration of 4 ms -2 = speed goes down by 4 ms-1 every 1 second 5. A steady velocity of 10 ms -1 = A displacement of 10 m is travelled every 1 second to the right.
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Physics Module Form 4
Chapter 2 – Forces & Motion
Example 1 Every day Rahim walks from his house to the junction which is 1.5km from his house. Then he turns back and stops at warung Pak Din which is 0.5 km from his house.
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Example 2 Every morning Amirul walks to Ahmad’s house which is situated 80 m to the east of Amirul’s house. They then walk towards their school which is 60 m to the south of Ahmad’s house. (a)What is the distance travelled by Amirul and his displacement from his house? Distance = (80 +60 ) m = 140 m Displacement = 100 m tan θ =
80 =1.333 60
θ = 53.1º
(b)If the total time taken by Amirul to travel from his house to Ahmad’s house and then to school is 15 minutes, what is his speed and velocity? (a)What is Rahim’s displacement from his house • when he reaches the junction. 1.5 km to the right • When he is at warung Pak Din. 0.5 km to the left.
Speed =
140m =0.156 in ms-1 15 60 s
Velocity =
100m = 0.111 ms-1 15 60 s
(b)After breakfast, Rahim walks back to his house. w hen he reaches home, (i) what is the total distance travelled by Rahim? (1.5 + 1.5 + 0.5+0.5 ) km = 4.0 km (ii) what is Rahim’s total displacement from his house? 1.5 +( -1.5) +(- 0.5 )+0.5 km = 0 km
Example 3 Salim running in a race covers 60 m in 12 s. (a) What is his speed in ms-1 Speed =
Example 4 An aeroplane flies towards the north with a velocity 300 km hr -1 in one hour. Then, the plane moves to the east with the velocity 400 km hr -1 in one hour.
60m = 5 ms-1 12 s
(b) If he takes 40 s to complete the race, what is his distance covered?
(a)What is the average speed of the plane? Average speed = (300 km hr -1 + 4 00 km hr -1 ) / 2 = 350 km hr -1 (b)What is the average velocity of the plane?
distance covered = 40 s × 5 ms-1 = 200 m
Average velocity = 250 km hr -1
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Tan θ =
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400 = 1.333 θ = 300
(c)What is the difference between average speed and average velocity of the plane? Average speed is a scalar quantity. Average velocity is a vector quantity Example 5 The speedometer reading for a car travelling due north shows 80 km hr -1. Another car travelling at 80 km hr -1 towards south. Is the speed of both cars same? Is the velocity of both cars same? The speed of both cars are the same but the velocity of both cars are different with opposite direction
A ticker timer Use: 12 V a.c. power supply 1 tick = time interval between two dots. The time taken to make 50 ticks on the ticker tape is 1 second. Hence, the time interval between 2 consecutive dots is 1/50 = 0.02 s. 1 tick = 0.02 s
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Relating displacement, velocity, acceleration and time using ticker tape. VELOCITY
FORMULA Time, t = 10 dicks x 0.02 s = 0.2 s displacement, s = x cm velocity =
ACCELERATION Initial velocity, u = final velocity, v = acceleration, a =
Elapsed time, t = (5 – 1) x 0.2 s = 0.8 s or t = (50 – 10) ticks x 0.02 s = 0.8 s
TICKER TAPE AND CHARTS
TYPE OF MOTION
Constant velocity – slow moving Constant velocity – fast moving
Distance between the dots increases uniformly the velocity is of the object is increasing uniformly The object is moving at a uniform / constant acceleration.
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- Distance between the dots decrease uniformly - The velocity of the object is decreasing Uniformly - The object is experiencing uniform / constant decceleration
Example 6 The diagram above shows a ticker tape chart for a moving trolley. The frequency of the ticker-timer used is 50 Hz. Each section has 10 dots-spacing. (a) What is the time between two dots. Time = 1/50 s = 0.02 s (b) What is the time for one strips. 0.02 s × 10 = 0.2 s (c) What is the initial velocity 2 cm / 0.2 s = 10 ms-1 (d) What is the final velocity. 12 cm / 0.2 s = 60 ms-1 (e) What is the time interval to change from initial velocity to final velocity? ( 11 - 1) × 0.2 s = 2 s (f) What is the acceleration of the object. a=
vu 60 10 -2 = ms = 25 ms-2 t 2
THE EQUATIONS OF MOTION
v u at 1 s ut at 2 2 2 2 v u 2as
u = initial velocity v = final velocity t = time taken s = displacement a = constant acceleration
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Velocity is obtained from the gradient of the graph. A – B : gradient of the graph is positive and constant velocity is constant. B – C : gradient of the graph = 0 the velocity = 0, object is at rest. C – D : gradient of the graph negative and constant. The velocity is negative and object moves in the opposite direction.
VELOCITY-TIME GRAPH
GRAPH
s versus t
Area below graph
Distance / displacement
Positive gradient
Constant Acceleration (A – B)
Negative gradient
Constant Deceleration (C – D)
Zero gradient
Constant velocity / zero acceleration (B – C)
v versus t
Zero velocity
Negative constant velocity
Positive Constant velocity
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a versus t
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GRAPH
s versus t
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v versus t
Constant acceleration
Constant deceleration
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a versus t
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Example 1 Contoh 11
Example 2
Based on the s – t graph above: (a) Calculate the velocity at (i) AB (ii) BC (i) 5 ms-1 (ii) 0 ms-1
(a) Calculate the acceleration at: (i) JK (ii) KL (iii) LM (i) 2 ms-2 (ii) -1 ms-2 (iii) (iii) (iii)
CD - 10 ms-1
(b) Describe the motion of the object at: (i) AB (ii) BC (iii) CD -1 (i) constant velocity 5 ms (ii) at rest / 0 ms-1 (iii) constant velocity of 10 ms-1in opposite direction
(b) Describe the motion of the object at: (i) JK (ii) KL (iii) LM
Calculate the total displacement. Displacement = area under the graph = 100 m + 150 m + 100 m + 25 m = 375 m
(ii) total displacement 50 m + (- 50 m) = 0
(c) Calculate the average velocity. Average velocity = 375 m / 40 s = 9.375 ms-1
(d) Calculate the average speed
0 ms-1
(i) constant acceleration of 2 ms-2-2 (ii) constant deceleration of 1ms (iii) (iii) zero acceleration or constant velocity
(c)Find: (i) total distance 50 m + 50 m = 100 m
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100m = 2.86 ms-1 35s
(ii) the average velocity of the moving particle. 0
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Inertia
The inertia of an object is the tendency of the object to remain at rest or, if moving, to continue its motion.
Newton’s first law
Every object continues in its state of rest or of uniform motion unless it is acted upon by an external force.
Relation between inertia and mass
The larger the mass, the larger the inertia
SITUATIONS INVOLVING INERTIA SITUATION
EXPLANATION EEEEEEEEJNVJLKN When the cardboard is pulled away quickly, the coin drops straight into DNFLJKVNDFLKJNB the glass. VJKL;DFN BLK;XC The inertia of the coin maintains its NB[F state at rest. The coin falls into the glassNDPnDSFJ[POJDE]Odue to gravity. JBD]AOP[FKBOP[DF LMB NOPGFMB LKFGNKLB FGNMNKL’ MCVL Chilli sauce in the bottle can be BNM’CXLB easily poured out if the bottle is moved down fast with a suddenNFGNKEPLANATION stop. The sauce inside the bottle moves together with the bottle. When the bottle stops suddenly, the sauce continues in its state of motion due to the effect of its inertia.
Body moves forward when the car stops suddenly The passengers were in a state of motion when the car was moving. When the car stopped suddenly, the inertia in the passengers made them maintain their state of motion. Thus when the car stop, the passengers moved forward.
A boy runs away from a cow in a zig- zag motion. The cow has a large inertia making it difficult to change direction.
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The head of hammer is secured tightly to its handle by knocking one end of the handle, held vertically, on a hard surface. This causes the hammer head to continue on its downward motion. When the handle has been stopped, so that the top end of the handle is slotted deeper into the hammer head.
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The drop of water on a wet umbrella will fall when the boy rotates the umbrella.
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This is because the drop of water on the surface of the umbrella moves simultaneously as the umbrella is rotated. When the umbrella stops rotating, the inertia of the drop of water will continue to maintain its motion.
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Ways to reduce the negative effects of inertia
1. Safety in a car: (a)Safety belt secure the driver to their seats. When the car stops suddenly, the seat belt provides the external force that prevents the driver from being thrown forward. (b)Headrest to prevent injuries to the neck during rearend collisions. The inertia of the head tends to keep in its state of rest when the body is moved suddenly. (c)An air bag is fitted inside the steering wheel. It provides a cushion to prevent the driver from hitting the steering wheel or dashboard during a collision.
2. Furniture carried by a lorry normally are tied up together by string. When the lorry starts to move suddenly, the furniture are more difficult to fall off due to their inertia because their combined mass has increased.
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Definition
Momentum = Mass x velocity = mv SI unit: kg ms-1
Principle of Conservation of Momentum
In the absence of an external force, the total momentum of a system remains unchanged.
Elastic Collision
ƒ ƒ ƒ
Inelastic collision
ƒ The two objects combine and move together with a common velocity after the collision. ƒ Momentum is conserved. ƒ Kinetic energy is not conserved. ƒ Total energy is conserved.
Both objects move independently at their respective velocities after the collision. Momentum is conserved. Kinetic energy is conserved. Total energy is conserved.
Total Momentum Before = total momentum after m1u1 + m2u2 = m1 v1 + m2 v2
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Total Momentum Before = Total Momentum After m1 u1 + m2 u2 = ( m1 + m2 ) v
Explosion Before explosion both object stick together and at rest. After collision, both object move at opposite direction. Total Momentum before collision is zero
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Total Momentum after collision : m1v1 + m2v2
Physics Module Form 4
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From the law of conservation of momentum: Total Momentum = Total Momentum Before collision after collision 0 = m1v1 + m2v2 m1v1 = - m2v2 Negative sign means opposite direction
EXAMPLES OF EXPLOSION (Principle Of Conservation Of Momentum) When a rifle is fired, the bullet of mass m, moves with a high velocity, v. This creates a momentum in the forward direction. From the principle of conservation of momentum, an equal but opposite momentum is produced to recoil the riffle backward.
Application in the jet engine: A high-speed hot gases are ejected from the back with high momentum. This produces an equal and opposite momentum to propel the jet plane forward.
The launching of rocket Mixture of hydrogen and oxygen fuels burn explosively in the combustion chamber. Jets of hot gases are expelled at very high speed through the exhaust. These high speed hot gases produce a large amount of momentum downward. By conservation of momentum, an equal but opposite momentum is produced and acted on the rocket, propelling the rocket upwards.
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In a swamp area, a fan boat is used. The fan produces a high speed movement of air backward. This produces a large momentum backward. By conservation of momentum, an equal but opposite momentum is produced and acted on the boat. So the boat will move forward. A squid propels by expelling water at high velocity. Water enters through a large opening and exits through a small tube. The water is forced out at a high speed backward. Total Mom. before= Total Mom. after 0 =Mom water + Mom squid 0 = mwvw + msvs - mwvw = msvs The magnitude of the momentum of water and squid are equal but opposite direction. This causes the quid to jet forward.
Example Car A of mass 1000 kg moving at 20 ms -1 collides with a car B of mass 1200 kg moving at 10 m s -1 in same direction. If the car B is shunted forwards at 15 m s -1 by the impact, what is the velocity, v, of the car A immediately after the crash? 1000 kg x 20 ms -1 + 1200 kg x 10 ms -1 = 1000 kg x v + 1200 kg x 15 ms -1 v= 14 ms -1
Example Before collision MA = 4 kg MB = 2 kg UA = 10 ms -1 r i g h t UB = 8 ms -1 l e f t
After collision
VB 4 ms-1 right
Calculate the value of VA . [4 x 10 + 2 x (-8)]kgms -1 =[ 4 x v + 2 x 4 ] kgms -1 VA = 4 ms -1 right
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Example
Example A truck of mass 1200 kg moving at 30 ms-1 collides with a car of mass 1000 kg which is travelling in the opposite direction at 20 ms-1. After the collision, the two vehicles move together. What is the velocity of both vehicles immediately after collision? 1200 kg x 30 ms
-1
+ 1000 kg x (-20 ms -1) = ( 1200 kg + 1000kg) v v = 7.27 ms -1 to the right
A man fires a pistol which has a mass of 1.5 kg. If the mass of the bullet is 10 g and it reaches a velocity of 300 ms -1 after shooting, what is the recoil velocity of the pistol? 0 = 1.5 kg x v + 0.01 kg x 300 ms -1 v = -2 ms -1 Or it recoiled with 2 ms -1 to the left
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Chapter 2 – Forces & Motion
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Balanced Force When the forces acting on an object are balanced, they cancel each other out. The net force is zero.
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Example:
Effect : the object is at rest [velocity = 0] or moves at constant velocity [ a = 0]
Unbalanced Force/ Resultant Force
When the forces acting on an object are not balanced, there must be a net force acting on it. The net force is known as the unbalanced force or the resultant force. Weight, W = Lift, U Thrust, F = drag, G
Effect : Can cause a body to - change it state at rest (an object will accelerate - change it state of motion (a moving object will decelerate or change its direction)
Newton’s Second Law of Motion
The acceleration produced by a 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. The direction of the acceleration is the same as that of the net force. Force = Mass x Acceleration F = ma
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Experiment to Find The Relationship between Force, Mass & Acceleration a& F
Relationship between
a& m
Situation
Both men are pushing the same mass but man A puts greater effort. So he moves faster.
Both men exerted the same strength. But man B moves faster than man A.
Inference
The acceleration produced by an object depends on the net force applied to it.
The acceleration produced by an object depends on the mass
Hypothesis
The acceleration of the object increases when the force applied increases
The acceleration of the object decreases when the mass of the object increases
Force
Mass
Acceleration
Acceleration
Mass
Force
Variables: Manipulated : Responding : Constant : Apparatus and Material
Ticker tape, elastic cords, ticker timer, trolleys, power supply, friction compensated runway and meter ruler.
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Procedure : - Controlling manipulated variables.
An elastic cord is hooked over the trolley. The elastic cord is stretched until the end of the trolley. The trolley is pulled down the runway with the elastic cord being kept stretched by the same amount of force
An elastic cord is hooked over a trolley. The elastic cord is stretched until the end of the trolley. The trolley is pulled down the runway with the elastic cord being kept stretched by the same amount of force
-Controlling responding variables.
Determine the acceleration by analyzing the ticker tape. Acceleration
Determine the acceleration by analyzing the ticker tape.
Acceleration -Repeating experiment.
Tabulation of data
v u a t
Repeat the experiment by using two , three, four and five elastic cords
Force, F/No of elastic cord
1 2 3 4 5
Acceleration a
Repeat the experiment by using two, three, four and five trolleys.
Acceleration, a/ ms-2 Mass, m/ no of trolleys
1 2 3 4 5
Analysing Result
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v u t
Mass, m/g
1/m, g-1
Acceleration/ ms-2
Physics Module Form 4
Chapter 2 – Forces & Motion
1. What force is required to move a 2 kg object with an acceleration of 3 m s-2, if (a) the object is on a smooth surface? (b) The object is on a surface where the average force of friction acting on the object is 2 N?
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2. Ali applies a force of 50 N to move a 10 kg table at a constant velocity. What is the frictional force acting on the table? Answer: 50 N
(a) force = 6 N (b) net force = (6 – 2) N = 4N
3. A car of mass 1200 kg travelling at 20 ms -1 is brought to rest over a distance of 30 m. Find (a) the average deceleration, (b) the average braking force. (a) u = 20 ms -1 v = 0 s = 30 m a = - 6.67 ms-2 (b) force = 1200 x 6.67 N = 8000 N
a=?
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4. Which of the following systems will produce maximum acceleration? D
Physics Module Form 4
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Chapter 2 – Forces & Motion
IMPULSE
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IMPULSIVE
Impulse
The change of momentum mv - mu Unit : kgms-1 or Ns
Impulsive Force
The rate of change of momentum in a collision or explosion Impulsive force =
FORCE
m = mass u = initial velocity v = final velocity t = time
change of momentum mv mu time t Effect of time
Impulsive force Unit = N is inversely proportional to time of contact
Longer period of time →Impulsive force decrease Shorter period of time →Impulsive force increase
Situations for Reducing Impulsive Force in Sports Situations
Explanation Thick mattress with soft surfaces are used in events such as high jump so that the time interval of impact on landing is extended, thus reducing the impulsive force. This can prevent injuries to the participants.
Goal keepers will wear gloves to increase the collision time. This will reduce the impulsive force.
A high jumper will bend his legs upon landing. This is to increase the time of impact in order to reduce the impulsive force acting on his legs. This will reduce the chance of getting serious injury.
A baseball player must catch the ball in the direction of the motion of the ball. Moving his hand backwards when catching the ball prolongs the time for the momentum to change so as to reduce the impulsive force.
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Situation of Increasing Impulsive Force Situations
Explanation A karate expert can break a thick wooden slab with his bare hand that moves at a very fast speed. The short impact time results in a large impulsive force on the wooden slab.
A massive hammer head moving at a fast speed is brought to rest upon hitting the nail within a short time interval. The large change in momentum within a short time interval produces a large impulsive force which drives the nail into the wood. A football must have enough air pressure in it so the contact time is short. The impulsive force acted on the ball will be bigger and the ball will move faster and further.
Pestle and mortar are made of stone. When a pestle is used to pound chillies, the hard surfaces of both the pestle and mortar cause the pestle to be stopped in a very short time. A large impulsive force is resulted and thus causes these spices to be crushed easily.
Example 1 A 60 kg resident jumps from the first floor of a burning house. His velocity just before landing on the ground is 6 ms-1. (a) Calculate the impulse when his legs hit the ground. (b) What is the impulsive force on the resident’s legs if he bends upon landing and takes 0.5 s to stop? (c) What is the impulsive force on the resident’s legs if he does not bend and stops in 0.05 s? (d) What is the advantage of bending his legs upon landing? Example 2 Rooney kicks a ball with a force of 1500 N. The time of contact of his boot with the ball is 0.01 s. What is the impulse delivered to the ball? If the mass of the ball is 0.5 kg, what is the velocity of the ball?
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Answer: (a) Impulse = 60 kg x ( 6 ms-1 - 0 ) = 360 Ns
360 Ns (b) Impulsive force = 0.5s =7200 N (c) He experienced a greater Impulsive force of 7200 N and he might injured his legs (d) Increase the reaction time so as to reduce impulsive force
(a) Impulse = 1500N x 0.01 s = 15 Ns (b) velocity =
15 Ns = 30 ms-1 0.5kg
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Safety features in vehicles Head rest
Crash resistant door pillars
Windscreen
Crumple zones Anti-lock brake system (ABS)
Traction control
Front bumper
Air bags
Component
Function
Headrest Air bag
To reduce the inertia effect of the driver’s head. Absorbing impact by increasing the amount of time the driver’s head to come to the steering. So that the impulsive force can be reduce
Windscreen
To protect the driver (shattered proof)
Crumple zone
Can be compressed during accident. So it can increase the amount of time the car takes to come to a complete stop. So it can reduce the impulsive force.
Front bumper ABS
Absorb the shock from the accident. Made from steel, aluminium, plastic or rubber.
Side impact bar
Prevents the collapse of the front and back of the car into the passenger compartment. Also gives good protection from a side impact
Seat belt
To reduce the effect of inertia by avoiding the driver from thrown forward.
Enables drivers to quickly stop the car without causing the brakes to lock.
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Gravitational Force
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Objects fall because they are pulled towards the Earth by the force of gravity. This force is known as the pull of gravity or the earth’s gravitational force. The earth’s gravitational force tends to pull everything towards its centre.
Free fall
An object is falling freely when it is falling under the force of gravity only. A piece of paper does not fall freely because its fall is affected by air resistance. An object falls freely only in vacuum. The absence of air means there is no air resistance to oppose the motion of the object. In vacuum, both light and heavy objects fall freely. They fall with the same acceleration i.e. The acceleration due to gravity, g. Objects dropped under the influence of the pull of gravity with constant acceleration. This acceleration is known as the gravitational acceleration, g.
Acceleration due to gravity, g
-2 The standard value of the gravitational acceleration, g is 9.81 m s . -2 The value of g is often taken to be 10 m s for simplicity. The magnitude of the acceleration due to gravity depends on the strength of the gravitational field.
Gravitational field
The gravitational field is the 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. The gravitational field strength is defined as the gravitational force which acts on a mass of 1 kilogram. g=
F m
-1 Its unit is N kg .
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-1 Gravitational field strength, g = 10 N kg -2 Acceleration due to gravity, g = 10 m s The approximate value of g can therefore be written either as -1 or as 10 N kg . Weight
Comparison between weight & mass
10 m s
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The gravitational force acting on the object. Weight = mass x gravitational acceleration W = mg SI unit : Newton, N and it is a vector quantity Mass The mass of an object is the amount of matter in the object
Weight The weight of an object is the force of gravity acting on the object.
Constant everywhere
Varies with the magnitude of gravitational field strength, g of the location
A scalar quantity A base quantity
A vector quantity A derived quantity
SI unit: kg
SI unit : Newton, N
The difference between a fall in air and a free fall in a vacuum of a coin and a feather. Both the coin and the feather are released simultaneously from the same height. At vacuum state: There is no air resistance. The coin and the feather will fall freely. Only gravitational force acted on the objects. Both will fall at the same time.
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At normal state: Both coin and feather will fall because of gravitational force. Air resistance effected by the surface area of a fallen object. The feather that has large area will have more air resistance. The coin will fall at first.
Physics Module Form 4
Chapter 2 – Forces & Motion
(a) The two spheres are falling with an acceleration. The distance between two successive images of the sphere increases showing that the two spheres are falling with increasing velocity; falling with an acceleration.
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The two spheres are falling down with the same acceleration The two spheres are at the same level at all times. Thus, a heavy object and a light object fall with the same gravitational acceleration Gravitational acceleration is independent of mass
Two steel spheres are falling under gravity. The two spheres are dropped at the same time from the same height. Motion graph for free fall object Free fall object
Object thrown upward
Example 1 A coconut takes 2.0 s to fall to the ground. What is (a) its speed when it strikes the ground (b) ) the height of the coconut tree
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Object thrown upward and fall
(a) t = 2 s u = 0 g = 10 v=? v = u + gt = 0 + 10 x 2 = 20 ms-1 (b) s = ut + ½ at2 = 0 + ½ (10) 22 = 20 m
Physics Module Form 4
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Chapter 2 – Forces & Motion
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E Q U I L I B R I U M
Forces in Equilibrium
When an object is in equilibrium, the resultant force acting on it is zero. The object will either be 1. at rest 2. move with constant velocity.
rd Newton’s 3 Law
Action is equal to reaction
Examples( Label the forces acted on the objects) Paste more picture
Paste more picture
Resultant Force
A single force that represents the combined effect of two of more forces in magnitude and direction.
Addition of Forces
Resultant force, F = F1
+
Resultant force, F = F1
+ - F2
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Physics Module Form 4
Chapter 2 – Forces & Motion
GCKL 2011
Two forces acting at a point at an angle [Parallelogram method]
STEP 1 : Using ruler and protractor, draw the two forces F1 and F2 from a point.
STEP 3 Draw the diagonal of the parallelogram. The diagonal represent the resultant force, F in magnitude and direction.
STEP 2 Complete the parallelogram scale: 1 cm = ……
Resolution of Forces
A force F can be resolved into components which are perpendicular to each other: (a) horizontal component , FX (b) vertical component, FY Inclined Plane
Fx = F cos θ Fy = F sin θ
Component of weight parallel to the plane = mg sin θ Component of weight normal to the plane = mg cos θ
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Physics Module Form 4
Chapter 2 – Forces & Motion
GCKL 2011
Find the resultant force
17 N
5N (d)
(e)
7N
FR
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Physics Module Form 4
Chapter 2 – Forces & Motion
GCKL 2011
Lift Stationary Lift
Lift accelerate upward
Lift accelerate downward
Resultant Force =
Resultant Force =
Resultant Force =
The reading of weighing scale =
The reading of weighing scale =
The reading of weighing scale =
Pulley
1. Find the resultant force, F
40 -30 = 10 N
30-2 = 28 N
2. Find the moving mass, m
4 + 3 = 7 kg
3+ 4 = 4 kg
3. Find the acceleration, a
40 -30 = (3+4)a 10 =7a a =10/ 7 ms-2
30 -2 = (4+3 )a 28 = 7a a = 4 ms-2
4. Find string tension, T
T- 3 (10) = 3 a T = 30 + 3 (10/7) =240 /7 N
30 – T = 3 (a) T =30- 12 = 18 N
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Physics Module Form 4
2.10
Chapter 2 – Forces & Motion
GCKL 2011
W O R K , E N E R G Y , P O W E R & E F FI C I E N CY
Work
Work done is the product of an applied force and the displacement of an object in the direction of the applied force W = Fs
W = work, F = force
s = displacement
The SI unit of work is the joule, J 1 joule of work is done when a force of 1 N moves an object 1 m in the direction of the force
The displacement, s of the object is in the direction of the force, F
The displacement , s of the object is not in the direction of the force, F
W = Fs s
F
W= F s Example 1 A boy pushing his bicycle with a force of 25 N through a distance of 3 m.
Calculate the work done by the boy. 75 Nm
Example 2 A girl is lifting up a 3 kg flower pot steadily to a height of 0.4 m.
Example 3 A man is pulling a crate of fish along theW floor a force of = (Fwith cos θ) s 40 N through a distance of 6 m.
What is the work done by the girl? 12 Nm
What is the work done in pulling the crate? 40 N cos 50º x 6 Nm
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Physics Module Form 4
Concept Power
Energy
Chapter 2 – Forces & Motion
D efwork is done, The rate at which or the amount ofinwork done per iti second. on
Potential Energy
GCKL 2011
Formula & Unit P=
W
t p = power, W = work / energy t = time
Energy is the capacity to do work. An object that can do work has energy Work is done because a force is applied and the objects move. This is accompanied by the transfer of energy from one object to another object. Therefore, when work is done, energy is transferred from one object to another. The work done is equal to the amount of energy transferred.
Gravitational potential energy is the energy of an object due to its higher position in the gravitational field.
m = mass h = height g = gravitational acceleration E = mgh
Kinetic Energy
Kinetic energy is the energy of an object due to its motion.
m = mass v = velocity E = ½ mv
No work is done when: The object is stationary. A student carrying his bag while waiting at the bus stop
The direction of motion of the object is perpendicular to that of the applied force.
A waiter2is- 32 carrying a tray of food and walking
2
No force is applied on the object in the direction of displacement (the object moves because of its own inertia) A satellite orbiting in space. There is no friction in space. No force is acting in the direction of movement of the satellite.
Physics Module Form 4
Chapter 2 – Forces & Motion
Principle of Conservation of Energy
GCKL 2011
Energy can be changed from one form to another, but it cannot be created or destroyed. The energy can be transformed from one form to another, total energy in a system is constant. Total energy before = total energy after Example 4 A worker is pulling a wooden block of weight W, with a force of P along a frictionless plank at height of h. The distance travelled by the block is x. Calculate the work done by the worker to pull the block. [Px = Wh]
Example 5 A student of mass m is climbing up a flight of stairs which has the height of h. He takes t seconds. What is the power of the student? [
mgh t
Example 6
Example 7
A stone is thrown upward with -1 initial velocity of 20 ms . What is the maximum height which can be reached by the stone? [ 10m ]
A ball is released from point A of height 0.8 m so that it can roll along a curve frictionless track. What is the velocity of the ball when it reaches point B? [4 ms-1]
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Physics Module Form 4
Chapter 2 – Forces & Motion
GCKL 2011
Example 8 A trolley is released from rest at point X along a frictionless track. What is the velocity of the trolley at point Y? [ v2 = 30( ms-1)2] [ v = 5.48 ms-1]
Example 9 A ball moves upwards along a frictionless track of height 1.5 m -1 with a velocity of 6 ms . What is its velocity at point B? [v2 = 30( ms-1)2 v = 5.48 ms-1]
Example 10
A boy of mass 20 kg sits at the top of a concrete slide of height 2.5 m. When he slides down the slope, he does work to overcome friction of 140 J. What is his velocity at the end of the slope? [6 ms-1]
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